ChemWiki - User contributions [en]

MRD:ds5517

2019-05-24T16:58:48Z

Ds5517: /* F + H2 → H + HF */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: ∂V(ri)/∂ri=0. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that it has a symmetrical configuration at which rAB = rBC, shown on Figure 1. Since the TS is a saddle point, the second derivative of the reaction path potential will be negative (it is a maximum). However, the second derivative of the potential orthogonal to the reaction path potential at the TS will be positive (it is at minimum).<ref name="CKD" />

<div>
[[File:Dori paintedit.png|thumb|none|500px|'''Fig. 1.''' Contour plot for the saddler point]]
</div>

<div>
[[File:Dori q2 pes.PNG|thumb|none|500px|'''Fig. 2.''' Potential Energy surface plot at the TS]]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.'''Contour Plot Trajectory using MEP]]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:
 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants (kcal/mol)!! E of TS (kcal/mol)!! Activation Energy (kcal/mol) !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -103.959|| -103.752|| 0.207|| [[File:Dori q8 1.png]]
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042|| [[File:Dori q8 2.PNG]]
|}

===Reaction dynamics===

====Polanyi's Empirical Rules<ref name="CKD" /> ====

Hammond's postulate states that an exothermic reaction will proceed with an early TS, which will resemble the the reactants. In contrast, in an endothermic reaction the TS will resemble the products in a late TS. Polanyi's Empirical Rules relate to Hammond's Postulate. Polanyi studied different potential energy surface trajectorys of A + BC → AB + C type reactions (like F + H2 → H + HF, as discussed above) in order to determine the dyanamics and energy changes of such reactions. Polanyi found that a reaction with an early TS (eg. F + H2 → H + HF) will favour high translational energy of the reactants which will result in high vibrational energy of the products. In contrast, a reaction with late TS (eg. H + HF → F + H2), will favour high vibrational energy of the reactants and will end up having high product translational energy.

====Release of Energy====

===== F + H2 → H + HF =====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

As discussed above, F + H2 → H + HF is exothermic, the products are lower in energy than the reactants, thus energy is released from to system to the surroundings in the form of heat. The fact that energy is conserved can be confirmed and monitored by bomb calorimetry.

Figure 12 shows the energy against time plot for a successful collision. We know that the total energy of the system is constant (energy is conserved). Therefore, we see a horizontal line for the total energy. Figure 13 shows the momentum against time plot. The kinetic energy is the sum of vibrational and translational energies. A system with an early TS such as this one will favour low vibrational and high translational energies of the reactants (Polanyi's Rules) This is shown on Fig 13.

[[File:Dori q9 energyvstime.png|thumb|none|500px|'''Fig. 12.''' energy against time plot]]
[[File:Dori momentumagainsttime.png|thumb|none|500px|'''Fig. 13.''' momentum against time plot]]

{| class="wikitable" border=1
! rFH !! rHH !! pFH !! pHH !! Reactive? !! Illustration of the trajectory !! Comments
|-
| 2.0|| 0.74 || -0.5|| -3 || No || [[File:Dori q9 p-3.png|400 px]] || Because the reaction has an early TS , it will favour low vibrational energy of the products to form the products (Polanyi's Rules). Since the system has high vibrational energy at the beginning, it crosses the TS, but then goes back and the H-H bond reforms.
|-
| 2.0|| 0.74 || -0.5|| -2 || Yes || [[File:Dori q9 p-2.png|400 px]] || As we lowered the vibrational energy of the reactants, the reaction became successful. These conditions illustrate Polanyi's Rules, as the products in an exothermic reaction with an early TS have high vibrational energy.
|-
| 2.0|| 0.74 || -0.8|| 0.1 || Yes || [[File:Dori q9 p01.png|400 px]] || We now increased the vibrational energy of the products, which is favoured for an early TS reaction, thus products (H + HF) are formed. Just like in the first case, the system crosses the activation energy barrier, then reform the reactans. However, in this case it crosses the TS once again and the products are formed.
|}

===== H + HF → F + H2=====

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

Polanyi's Rules describe the relationship between the success of a reaction, its TS (early/late) and how the translational/vibrational energies effect the system. These rules are described above.

{| class="wikitable" border=1
! rFH !! rHH !! pFH !! pHH !! Reactive? !! Illustration of the trajectory !! Comments
|-
| 0.9|| 2 || -13|| -1 || Yes || [[File:Dori q10 yes.png|400 px]] || The reactants have high enough vibrational energy for the collision to be overall successful. As shown on the trajectory, the products have lower vibrational energy than the reactants.
|-
| 0.9|| 2 || -6|| -1 || No || [[File:Dori q10 nope.png|400 px]] || In this case, the vibrational energy of the reactants is not sufficient enough and the products do not form.
|}

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

MRD:ds5517

2019-05-24T16:57:36Z

Ds5517: /* Release of Energy */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: ∂V(ri)/∂ri=0. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that it has a symmetrical configuration at which rAB = rBC, shown on Figure 1. Since the TS is a saddle point, the second derivative of the reaction path potential will be negative (it is a maximum). However, the second derivative of the potential orthogonal to the reaction path potential at the TS will be positive (it is at minimum).<ref name="CKD" />

<div>
[[File:Dori paintedit.png|thumb|none|500px|'''Fig. 1.''' Contour plot for the saddler point]]
</div>

<div>
[[File:Dori q2 pes.PNG|thumb|none|500px|'''Fig. 2.''' Potential Energy surface plot at the TS]]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.'''Contour Plot Trajectory using MEP]]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:
 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants (kcal/mol)!! E of TS (kcal/mol)!! Activation Energy (kcal/mol) !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -103.959|| -103.752|| 0.207|| [[File:Dori q8 1.png]]
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042|| [[File:Dori q8 2.PNG]]
|}

===Reaction dynamics===

====Polanyi's Empirical Rules<ref name="CKD" /> ====

Hammond's postulate states that an exothermic reaction will proceed with an early TS, which will resemble the the reactants. In contrast, in an endothermic reaction the TS will resemble the products in a late TS. Polanyi's Empirical Rules relate to Hammond's Postulate. Polanyi studied different potential energy surface trajectorys of A + BC → AB + C type reactions (like F + H2 → H + HF, as discussed above) in order to determine the dyanamics and energy changes of such reactions. Polanyi found that a reaction with an early TS (eg. F + H2 → H + HF) will favour high translational energy of the reactants which will result in high vibrational energy of the products. In contrast, a reaction with late TS (eg. H + HF → F + H2), will favour high vibrational energy of the reactants and will end up having high product translational energy.

====Release of Energy====

===== F + H2 → H + HF =====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

As discussed above, F + H2 → H + HF is exothermic, the products are lower in energy than the reactants, thus energy is released from to system to the surroundings in the form of heat. The fact that energy is conserved can be confirmed and monitored by bomb calorimetry.

Figure 12 shows the energy against time plot for a successful collision. We know that the total energy of the system is constant (energy is conserved). Therefore, we see a horizontal line for the total energy. Figure 13 shows the momentum against time plot. The kinetic energy is the sum of vibrational and translational energies. A system with an early TS such as this one will favour low vibrational and high translational energies of the reactants (Polanyi's Rules) This is shown on Fig 13.

[[File:Dori momentumagainsttime.png]]
[[File:Dori q9 energyvstime.png]]

{| class="wikitable" border=1
! rFH !! rHH !! pFH !! pHH !! Reactive? !! Illustration of the trajectory !! Comments
|-
| 2.0|| 0.74 || -0.5|| -3 || No || [[File:Dori q9 p-3.png|400 px]] || Because the reaction has an early TS , it will favour low vibrational energy of the products to form the products (Polanyi's Rules). Since the system has high vibrational energy at the beginning, it crosses the TS, but then goes back and the H-H bond reforms.
|-
| 2.0|| 0.74 || -0.5|| -2 || Yes || [[File:Dori q9 p-2.png|400 px]] || As we lowered the vibrational energy of the reactants, the reaction became successful. These conditions illustrate Polanyi's Rules, as the products in an exothermic reaction with an early TS have high vibrational energy.
|-
| 2.0|| 0.74 || -0.8|| 0.1 || Yes || [[File:Dori q9 p01.png|400 px]] || We now increased the vibrational energy of the products, which is favoured for an early TS reaction, thus products (H + HF) are formed. Just like in the first case, the system crosses the activation energy barrier, then reform the reactans. However, in this case it crosses the TS once again and the products are formed.
|}

===== H + HF → F + H2=====

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

Polanyi's Rules describe the relationship between the success of a reaction, its TS (early/late) and how the translational/vibrational energies effect the system. These rules are described above.

{| class="wikitable" border=1
! rFH !! rHH !! pFH !! pHH !! Reactive? !! Illustration of the trajectory !! Comments
|-
| 0.9|| 2 || -13|| -1 || Yes || [[File:Dori q10 yes.png|400 px]] || The reactants have high enough vibrational energy for the collision to be overall successful. As shown on the trajectory, the products have lower vibrational energy than the reactants.
|-
| 0.9|| 2 || -6|| -1 || No || [[File:Dori q10 nope.png|400 px]] || In this case, the vibrational energy of the reactants is not sufficient enough and the products do not form.
|}

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

File:Dori q9 energyvstime.png

2019-05-24T16:57:24Z

Ds5517:

File:Dori momentumagainsttime.png

2019-05-24T16:56:44Z

Ds5517:

MRD:ds5517

2019-05-24T16:48:36Z

Ds5517: /* Dynamics from the transition state region */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: ∂V(ri)/∂ri=0. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that it has a symmetrical configuration at which rAB = rBC, shown on Figure 1. Since the TS is a saddle point, the second derivative of the reaction path potential will be negative (it is a maximum). However, the second derivative of the potential orthogonal to the reaction path potential at the TS will be positive (it is at minimum).<ref name="CKD" />

<div>
[[File:Dori paintedit.png|thumb|none|500px|'''Fig. 1.''' Contour plot for the saddler point]]
</div>

<div>
[[File:Dori q2 pes.PNG|thumb|none|500px|'''Fig. 2.''' Potential Energy surface plot at the TS]]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.'''Contour Plot Trajectory using MEP]]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:
 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants (kcal/mol)!! E of TS (kcal/mol)!! Activation Energy (kcal/mol) !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -103.959|| -103.752|| 0.207|| [[File:Dori q8 1.png]]
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042|| [[File:Dori q8 2.PNG]]
|}

===Reaction dynamics===

====Polanyi's Empirical Rules<ref name="CKD" /> ====

Hammond's postulate states that an exothermic reaction will proceed with an early TS, which will resemble the the reactants. In contrast, in an endothermic reaction the TS will resemble the products in a late TS. Polanyi's Empirical Rules relate to Hammond's Postulate. Polanyi studied different potential energy surface trajectorys of A + BC → AB + C type reactions (like F + H2 → H + HF, as discussed above) in order to determine the dyanamics and energy changes of such reactions. Polanyi found that a reaction with an early TS (eg. F + H2 → H + HF) will favour high translational energy of the reactants which will result in high vibrational energy of the products. In contrast, a reaction with late TS (eg. H + HF → F + H2), will favour high vibrational energy of the reactants and will end up having high product translational energy.

====Release of Energy====

===== F + H2 → H + HF =====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

As discussed above, F + H2 → H + HF is exothermic, the products are lower in energy than the reactants, thus energy is released from to system to the surroundings in the form of heat. The fact that energy is conserved can be confirmed and monitored by bomb calorimetry.

Figure 12 shows the energy against time plot for a successful collision. We know that the total energy of the system is constant (energy is conserved). Therefore, we see a horizontal line for the total energy. Figure 13 shows the momentum against time plot. The kinetic energy is the sum of vibrational and translational energies. A system with an early TS such as this one will favour low vibrational and high translational energies of the reactants (Polanyi's Rules) This is shown on Fig 13.

{| class="wikitable" border=1
! rFH !! rHH !! pFH !! pHH !! Reactive? !! Illustration of the trajectory !! Comments
|-
| 2.0|| 0.74 || -0.5|| -3 || No || [[File:Dori q9 p-3.png|400 px]] || Because the reaction has an early TS , it will favour low vibrational energy of the products to form the products (Polanyi's Rules). Since the system has high vibrational energy at the beginning, it crosses the TS, but then goes back and the H-H bond reforms.
|-
| 2.0|| 0.74 || -0.5|| -2 || Yes || [[File:Dori q9 p-2.png|400 px]] || As we lowered the vibrational energy of the reactants, the reaction became successful. These conditions illustrate Polanyi's Rules, as the products in an exothermic reaction with an early TS have high vibrational energy.
|-
| 2.0|| 0.74 || -0.8|| 0.1 || Yes || [[File:Dori q9 p01.png|400 px]] || We now increased the vibrational energy of the products, which is favoured for an early TS reaction, thus products (H + HF) are formed. Just like in the first case, the system crosses the activation energy barrier, then reform the reactans. However, in this case it crosses the TS once again and the products are formed.
|}

===== H + HF → F + H2=====

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

Polanyi's Rules describe the relationship between the success of a reaction, its TS (early/late) and how the translational/vibrational energies effect the system. These rules are described above.

{| class="wikitable" border=1
! rFH !! rHH !! pFH !! pHH !! Reactive? !! Illustration of the trajectory !! Comments
|-
| 0.9|| 2 || -13|| -1 || Yes || [[File:Dori q10 yes.png|400 px]] || The reactants have high enough vibrational energy for the collision to be overall successful. As shown on the trajectory, the products have lower vibrational energy than the reactants.
|-
| 0.9|| 2 || -6|| -1 || No || [[File:Dori q10 nope.png|400 px]] || In this case, the vibrational energy of the reactants is not sufficient enough and the products do not form.
|}

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

MRD:ds5517

2019-05-24T16:46:22Z

Ds5517: /* Calculating the reaction path */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: ∂V(ri)/∂ri=0. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that it has a symmetrical configuration at which rAB = rBC, shown on Figure 1. Since the TS is a saddle point, the second derivative of the reaction path potential will be negative (it is a maximum). However, the second derivative of the potential orthogonal to the reaction path potential at the TS will be positive (it is at minimum)

the second derivatives in orthogonal directions are going to differ. It is negative for one (maximum), but positive for the other (minimum).

<div>
[[File:Dori paintedit.png|thumb|none|500px|'''Fig. 1.''' Contour plot for the saddler point]]
</div>

<div>
[[File:Dori q2 pes.PNG|thumb|none|500px|'''Fig. 2.''' Potential Energy surface plot at the TS]]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.'''Contour Plot Trajectory using MEP]]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:
 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants (kcal/mol)!! E of TS (kcal/mol)!! Activation Energy (kcal/mol) !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -103.959|| -103.752|| 0.207|| [[File:Dori q8 1.png]]
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042|| [[File:Dori q8 2.PNG]]
|}

===Reaction dynamics===

====Polanyi's Empirical Rules<ref name="CKD" /> ====

Hammond's postulate states that an exothermic reaction will proceed with an early TS, which will resemble the the reactants. In contrast, in an endothermic reaction the TS will resemble the products in a late TS. Polanyi's Empirical Rules relate to Hammond's Postulate. Polanyi studied different potential energy surface trajectorys of A + BC → AB + C type reactions (like F + H2 → H + HF, as discussed above) in order to determine the dyanamics and energy changes of such reactions. Polanyi found that a reaction with an early TS (eg. F + H2 → H + HF) will favour high translational energy of the reactants which will result in high vibrational energy of the products. In contrast, a reaction with late TS (eg. H + HF → F + H2), will favour high vibrational energy of the reactants and will end up having high product translational energy.

====Release of Energy====

===== F + H2 → H + HF =====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

As discussed above, F + H2 → H + HF is exothermic, the products are lower in energy than the reactants, thus energy is released from to system to the surroundings in the form of heat. The fact that energy is conserved can be confirmed and monitored by bomb calorimetry.

Figure 12 shows the energy against time plot for a successful collision. We know that the total energy of the system is constant (energy is conserved). Therefore, we see a horizontal line for the total energy. Figure 13 shows the momentum against time plot. The kinetic energy is the sum of vibrational and translational energies. A system with an early TS such as this one will favour low vibrational and high translational energies of the reactants (Polanyi's Rules) This is shown on Fig 13.

{| class="wikitable" border=1
! rFH !! rHH !! pFH !! pHH !! Reactive? !! Illustration of the trajectory !! Comments
|-
| 2.0|| 0.74 || -0.5|| -3 || No || [[File:Dori q9 p-3.png|400 px]] || Because the reaction has an early TS , it will favour low vibrational energy of the products to form the products (Polanyi's Rules). Since the system has high vibrational energy at the beginning, it crosses the TS, but then goes back and the H-H bond reforms.
|-
| 2.0|| 0.74 || -0.5|| -2 || Yes || [[File:Dori q9 p-2.png|400 px]] || As we lowered the vibrational energy of the reactants, the reaction became successful. These conditions illustrate Polanyi's Rules, as the products in an exothermic reaction with an early TS have high vibrational energy.
|-
| 2.0|| 0.74 || -0.8|| 0.1 || Yes || [[File:Dori q9 p01.png|400 px]] || We now increased the vibrational energy of the products, which is favoured for an early TS reaction, thus products (H + HF) are formed. Just like in the first case, the system crosses the activation energy barrier, then reform the reactans. However, in this case it crosses the TS once again and the products are formed.
|}

===== H + HF → F + H2=====

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

Polanyi's Rules describe the relationship between the success of a reaction, its TS (early/late) and how the translational/vibrational energies effect the system. These rules are described above.

{| class="wikitable" border=1
! rFH !! rHH !! pFH !! pHH !! Reactive? !! Illustration of the trajectory !! Comments
|-
| 0.9|| 2 || -13|| -1 || Yes || [[File:Dori q10 yes.png|400 px]] || The reactants have high enough vibrational energy for the collision to be overall successful. As shown on the trajectory, the products have lower vibrational energy than the reactants.
|-
| 0.9|| 2 || -6|| -1 || No || [[File:Dori q10 nope.png|400 px]] || In this case, the vibrational energy of the reactants is not sufficient enough and the products do not form.
|}

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

MRD:ds5517

2019-05-24T16:41:01Z

Ds5517: /* Transition State Theory */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: ∂V(ri)/∂ri=0. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that it has a symmetrical configuration at which rAB = rBC, shown on Figure 1. Since the TS is a saddle point, the second derivative of the reaction path potential will be negative (it is a maximum). However, the second derivative of the potential orthogonal to the reaction path potential at the TS will be positive (it is at minimum)

the second derivatives in orthogonal directions are going to differ. It is negative for one (maximum), but positive for the other (minimum).

<div>
[[File:Dori paintedit.png|thumb|none|500px|'''Fig. 1.''' Contour plot for the saddler point]]
</div>

<div>
[[File:Dori q2 pes.PNG|thumb|none|500px|'''Fig. 2.''' Potential Energy surface plot at the TS]]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.''']]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:
 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants (kcal/mol)!! E of TS (kcal/mol)!! Activation Energy (kcal/mol) !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -103.959|| -103.752|| 0.207|| [[File:Dori q8 1.png]]
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042|| [[File:Dori q8 2.PNG]]
|}

===Reaction dynamics===

====Polanyi's Empirical Rules<ref name="CKD" /> ====

Hammond's postulate states that an exothermic reaction will proceed with an early TS, which will resemble the the reactants. In contrast, in an endothermic reaction the TS will resemble the products in a late TS. Polanyi's Empirical Rules relate to Hammond's Postulate. Polanyi studied different potential energy surface trajectorys of A + BC → AB + C type reactions (like F + H2 → H + HF, as discussed above) in order to determine the dyanamics and energy changes of such reactions. Polanyi found that a reaction with an early TS (eg. F + H2 → H + HF) will favour high translational energy of the reactants which will result in high vibrational energy of the products. In contrast, a reaction with late TS (eg. H + HF → F + H2), will favour high vibrational energy of the reactants and will end up having high product translational energy.

====Release of Energy====

===== F + H2 → H + HF =====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

As discussed above, F + H2 → H + HF is exothermic, the products are lower in energy than the reactants, thus energy is released from to system to the surroundings in the form of heat. The fact that energy is conserved can be confirmed and monitored by bomb calorimetry.

Figure 12 shows the energy against time plot for a successful collision. We know that the total energy of the system is constant (energy is conserved). Therefore, we see a horizontal line for the total energy. Figure 13 shows the momentum against time plot. The kinetic energy is the sum of vibrational and translational energies. A system with an early TS such as this one will favour low vibrational and high translational energies of the reactants (Polanyi's Rules) This is shown on Fig 13.

{| class="wikitable" border=1
! rFH !! rHH !! pFH !! pHH !! Reactive? !! Illustration of the trajectory !! Comments
|-
| 2.0|| 0.74 || -0.5|| -3 || No || [[File:Dori q9 p-3.png|400 px]] || Because the reaction has an early TS , it will favour low vibrational energy of the products to form the products (Polanyi's Rules). Since the system has high vibrational energy at the beginning, it crosses the TS, but then goes back and the H-H bond reforms.
|-
| 2.0|| 0.74 || -0.5|| -2 || Yes || [[File:Dori q9 p-2.png|400 px]] || As we lowered the vibrational energy of the reactants, the reaction became successful. These conditions illustrate Polanyi's Rules, as the products in an exothermic reaction with an early TS have high vibrational energy.
|-
| 2.0|| 0.74 || -0.8|| 0.1 || Yes || [[File:Dori q9 p01.png|400 px]] || We now increased the vibrational energy of the products, which is favoured for an early TS reaction, thus products (H + HF) are formed. Just like in the first case, the system crosses the activation energy barrier, then reform the reactans. However, in this case it crosses the TS once again and the products are formed.
|}

===== H + HF → F + H2=====

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

Polanyi's Rules describe the relationship between the success of a reaction, its TS (early/late) and how the translational/vibrational energies effect the system. These rules are described above.

{| class="wikitable" border=1
! rFH !! rHH !! pFH !! pHH !! Reactive? !! Illustration of the trajectory !! Comments
|-
| 0.9|| 2 || -13|| -1 || Yes || [[File:Dori q10 yes.png|400 px]] || The reactants have high enough vibrational energy for the collision to be overall successful. As shown on the trajectory, the products have lower vibrational energy than the reactants.
|-
| 0.9|| 2 || -6|| -1 || No || [[File:Dori q10 nope.png|400 px]] || In this case, the vibrational energy of the reactants is not sufficient enough and the products do not form.
|}

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

MRD:ds5517

2019-05-24T16:32:50Z

Ds5517: /* Activation Energies */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: ∂V(ri)/∂ri=0. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that it has a symmetrical configuration at which rAB = rBC, shown on Figure 1. Since the TS is a saddle point, the second derivative of the reaction path potential will be negative (it is a maximum). However, the second derivative of the potential orthogonal to the reaction path potential at the TS will be positive (it is at minimum)

the second derivatives in orthogonal directions are going to differ. It is negative for one (maximum), but positive for the other (minimum).

<div>
[[File:Dori paintedit.png|thumb|none|500px|'''Fig. 1.''' Contour plot for the saddler point]]
</div>

<div>
[[File:Dori q2 pes.PNG|thumb|none|500px|'''Fig. 2.''' Potential Energy surface plot at the TS]]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.''']]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:

 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants (kcal/mol)!! E of TS (kcal/mol)!! Activation Energy (kcal/mol) !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -103.959|| -103.752|| 0.207|| [[File:Dori q8 1.png]]
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042|| [[File:Dori q8 2.PNG]]
|}

===Reaction dynamics===

====Polanyi's Empirical Rules<ref name="CKD" /> ====

Hammond's postulate states that an exothermic reaction will proceed with an early TS, which will resemble the the reactants. In contrast, in an endothermic reaction the TS will resemble the products in a late TS. Polanyi's Empirical Rules relate to Hammond's Postulate. Polanyi studied different potential energy surface trajectorys of A + BC → AB + C type reactions (like F + H2 → H + HF, as discussed above) in order to determine the dyanamics and energy changes of such reactions. Polanyi found that a reaction with an early TS (eg. F + H2 → H + HF) will favour high translational energy of the reactants which will result in high vibrational energy of the products. In contrast, a reaction with late TS (eg. H + HF → F + H2), will favour high vibrational energy of the reactants and will end up having high product translational energy.

====Release of Energy====

===== F + H2 → H + HF =====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

As discussed above, F + H2 → H + HF is exothermic, the products are lower in energy than the reactants, thus energy is released from to system to the surroundings in the form of heat. The fact that energy is conserved can be confirmed and monitored by bomb calorimetry.

Figure 12 shows the energy against time plot for a successful collision. We know that the total energy of the system is constant (energy is conserved). Therefore, we see a horizontal line for the total energy. Figure 13 shows the momentum against time plot. The kinetic energy is the sum of vibrational and translational energies. A system with an early TS such as this one will favour low vibrational and high translational energies of the reactants (Polanyi's Rules) This is shown on Fig 13.

{| class="wikitable" border=1
! rFH !! rHH !! pFH !! pHH !! Reactive? !! Illustration of the trajectory !! Comments
|-
| 2.0|| 0.74 || -0.5|| -3 || No || [[File:Dori q9 p-3.png|400 px]] || Because the reaction has an early TS , it will favour low vibrational energy of the products to form the products (Polanyi's Rules). Since the system has high vibrational energy at the beginning, it crosses the TS, but then goes back and the H-H bond reforms.
|-
| 2.0|| 0.74 || -0.5|| -2 || Yes || [[File:Dori q9 p-2.png|400 px]] || As we lowered the vibrational energy of the reactants, the reaction became successful. These conditions illustrate Polanyi's Rules, as the products in an exothermic reaction with an early TS have high vibrational energy.
|-
| 2.0|| 0.74 || -0.8|| 0.1 || Yes || [[File:Dori q9 p01.png|400 px]] || We now increased the vibrational energy of the products, which is favoured for an early TS reaction, thus products (H + HF) are formed. Just like in the first case, the system crosses the activation energy barrier, then reform the reactans. However, in this case it crosses the TS once again and the products are formed.
|}

===== H + HF → F + H2=====

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

Polanyi's Rules describe the relationship between the success of a reaction, its TS (early/late) and how the translational/vibrational energies effect the system. These rules are described above.

{| class="wikitable" border=1
! rFH !! rHH !! pFH !! pHH !! Reactive? !! Illustration of the trajectory !! Comments
|-
| 0.9|| 2 || -13|| -1 || Yes || [[File:Dori q10 yes.png|400 px]] || The reactants have high enough vibrational energy for the collision to be overall successful. As shown on the trajectory, the products have lower vibrational energy than the reactants.
|-
| 0.9|| 2 || -6|| -1 || No || [[File:Dori q10 nope.png|400 px]] || In this case, the vibrational energy of the reactants is not sufficient enough and the products do not form.
|}

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

MRD:ds5517

2019-05-24T16:30:32Z

Ds5517: /* H + HF → F + H2 */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: ∂V(ri)/∂ri=0. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that it has a symmetrical configuration at which rAB = rBC, shown on Figure 1. Since the TS is a saddle point, the second derivative of the reaction path potential will be negative (it is a maximum). However, the second derivative of the potential orthogonal to the reaction path potential at the TS will be positive (it is at minimum)

the second derivatives in orthogonal directions are going to differ. It is negative for one (maximum), but positive for the other (minimum).

<div>
[[File:Dori paintedit.png|thumb|none|500px|'''Fig. 1.''' Contour plot for the saddler point]]
</div>

<div>
[[File:Dori q2 pes.PNG|thumb|none|500px|'''Fig. 2.''' Potential Energy surface plot at the TS]]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.''']]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:

 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants !! E of TS !! Activation Energy !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -103.959 || -103.752|| 0.207 kcal/mol|| [[File:Dori q8 1.png]]
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042 kcal/mol|| [[File:Dori q8 2.PNG]]
|}

===Reaction dynamics===

====Polanyi's Empirical Rules<ref name="CKD" /> ====

Hammond's postulate states that an exothermic reaction will proceed with an early TS, which will resemble the the reactants. In contrast, in an endothermic reaction the TS will resemble the products in a late TS. Polanyi's Empirical Rules relate to Hammond's Postulate. Polanyi studied different potential energy surface trajectorys of A + BC → AB + C type reactions (like F + H2 → H + HF, as discussed above) in order to determine the dyanamics and energy changes of such reactions. Polanyi found that a reaction with an early TS (eg. F + H2 → H + HF) will favour high translational energy of the reactants which will result in high vibrational energy of the products. In contrast, a reaction with late TS (eg. H + HF → F + H2), will favour high vibrational energy of the reactants and will end up having high product translational energy.

====Release of Energy====

===== F + H2 → H + HF =====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

As discussed above, F + H2 → H + HF is exothermic, the products are lower in energy than the reactants, thus energy is released from to system to the surroundings in the form of heat. The fact that energy is conserved can be confirmed and monitored by bomb calorimetry.

Figure 12 shows the energy against time plot for a successful collision. We know that the total energy of the system is constant (energy is conserved). Therefore, we see a horizontal line for the total energy. Figure 13 shows the momentum against time plot. The kinetic energy is the sum of vibrational and translational energies. A system with an early TS such as this one will favour low vibrational and high translational energies of the reactants (Polanyi's Rules) This is shown on Fig 13.

{| class="wikitable" border=1
! rFH !! rHH !! pFH !! pHH !! Reactive? !! Illustration of the trajectory !! Comments
|-
| 2.0|| 0.74 || -0.5|| -3 || No || [[File:Dori q9 p-3.png|400 px]] || Because the reaction has an early TS , it will favour low vibrational energy of the products to form the products (Polanyi's Rules). Since the system has high vibrational energy at the beginning, it crosses the TS, but then goes back and the H-H bond reforms.
|-
| 2.0|| 0.74 || -0.5|| -2 || Yes || [[File:Dori q9 p-2.png|400 px]] || As we lowered the vibrational energy of the reactants, the reaction became successful. These conditions illustrate Polanyi's Rules, as the products in an exothermic reaction with an early TS have high vibrational energy.
|-
| 2.0|| 0.74 || -0.8|| 0.1 || Yes || [[File:Dori q9 p01.png|400 px]] || We now increased the vibrational energy of the products, which is favoured for an early TS reaction, thus products (H + HF) are formed. Just like in the first case, the system crosses the activation energy barrier, then reform the reactans. However, in this case it crosses the TS once again and the products are formed.
|}

===== H + HF → F + H2=====

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

Polanyi's Rules describe the relationship between the success of a reaction, its TS (early/late) and how the translational/vibrational energies effect the system. These rules are described above.

{| class="wikitable" border=1
! rFH !! rHH !! pFH !! pHH !! Reactive? !! Illustration of the trajectory !! Comments
|-
| 0.9|| 2 || -13|| -1 || Yes || [[File:Dori q10 yes.png|400 px]] || The reactants have high enough vibrational energy for the collision to be overall successful. As shown on the trajectory, the products have lower vibrational energy than the reactants.
|-
| 0.9|| 2 || -6|| -1 || No || [[File:Dori q10 nope.png|400 px]] || In this case, the vibrational energy of the reactants is not sufficient enough and the products do not form.
|}

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

File:Dori q10 nope.png

2019-05-24T16:27:21Z

Ds5517:

File:Dori q10 yes.png

2019-05-24T16:25:46Z

Ds5517:

MRD:ds5517

2019-05-24T16:21:37Z

Ds5517: /* Release of Energy */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: ∂V(ri)/∂ri=0. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that it has a symmetrical configuration at which rAB = rBC, shown on Figure 1. Since the TS is a saddle point, the second derivative of the reaction path potential will be negative (it is a maximum). However, the second derivative of the potential orthogonal to the reaction path potential at the TS will be positive (it is at minimum)

the second derivatives in orthogonal directions are going to differ. It is negative for one (maximum), but positive for the other (minimum).

<div>
[[File:Dori paintedit.png|thumb|none|500px|'''Fig. 1.''' Contour plot for the saddler point]]
</div>

<div>
[[File:Dori q2 pes.PNG|thumb|none|500px|'''Fig. 2.''' Potential Energy surface plot at the TS]]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.''']]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:

 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants !! E of TS !! Activation Energy !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -103.959 || -103.752|| 0.207 kcal/mol|| [[File:Dori q8 1.png]]
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042 kcal/mol|| [[File:Dori q8 2.PNG]]
|}

===Reaction dynamics===

====Polanyi's Empirical Rules<ref name="CKD" /> ====

Hammond's postulate states that an exothermic reaction will proceed with an early TS, which will resemble the the reactants. In contrast, in an endothermic reaction the TS will resemble the products in a late TS. Polanyi's Empirical Rules relate to Hammond's Postulate. Polanyi studied different potential energy surface trajectorys of A + BC → AB + C type reactions (like F + H2 → H + HF, as discussed above) in order to determine the dyanamics and energy changes of such reactions. Polanyi found that a reaction with an early TS (eg. F + H2 → H + HF) will favour high translational energy of the reactants which will result in high vibrational energy of the products. In contrast, a reaction with late TS (eg. H + HF → F + H2), will favour high vibrational energy of the reactants and will end up having high product translational energy.

====Release of Energy====

===== F + H2 → H + HF =====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

As discussed above, F + H2 → H + HF is exothermic, the products are lower in energy than the reactants, thus energy is released from to system to the surroundings in the form of heat. The fact that energy is conserved can be confirmed and monitored by bomb calorimetry.

Figure 12 shows the energy against time plot for a successful collision. We know that the total energy of the system is constant (energy is conserved). Therefore, we see a horizontal line for the total energy. Figure 13 shows the momentum against time plot. The kinetic energy is the sum of vibrational and translational energies. A system with an early TS such as this one will favour low vibrational and high translational energies of the reactants (Polanyi's Rules) This is shown on Fig 13.

{| class="wikitable" border=1
! rFH !! rHH !! pFH !! pHH !! Reactive? !! Illustration of the trajectory !! Comments
|-
| 2.0|| 0.74 || -0.5|| -3 || No || [[File:Dori q9 p-3.png|400 px]] || Because the reaction has an early TS , it will favour low vibrational energy of the products to form the products (Polanyi's Rules). Since the system has high vibrational energy at the beginning, it crosses the TS, but then goes back and the H-H bond reforms.
|-
| 2.0|| 0.74 || -0.5|| -2 || Yes || [[File:Dori q9 p-2.png|400 px]] || As we lowered the vibrational energy of the reactants, the reaction became successful. These conditions illustrate Polanyi's Rules, as the products in an exothermic reaction with an early TS have high vibrational energy.
|-
| 2.0|| 0.74 || -0.8|| 0.1 || Yes || [[File:Dori q9 p01.png|400 px]] || We now increased the vibrational energy of the products, which is favoured for an early TS reaction, thus products (H + HF) are formed. Just like in the first case, the system crosses the activation energy barrier, then reform the reactans. However, in this case it crosses the TS once again and the products are formed.
|}

===== H + HF → F + H2=====

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

Polanyi's Rules describe the relationship between the success of a reaction, its TS (early/late) and how the translational/vibrational energies effect the system.

{| class="wikitable" border=1
! rFH !! rHH !! pFH !! pHH !! Reactive? !! Illustration of the trajectory !! Comments
|-
| 2.0|| 0.74 || -0.5|| -3 || No || [[File:Dori q9 p-3.png|400 px]] || Because the reaction has an early TS , it will favour low vibrational energy of the products to form the products (Polanyi's Rules).
|-
| 2.0|| 0.74 || -0.5|| -2 || Yes || [[File:Dori q9 p-2.png|400 px]] || As we lowered the vibrational energy of the reactants, the reaction became successful. These conditions illustrate Polanyi's Rules, as the products in an exothermic reaction with an early TS have high vibrational energy.

|}

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

MRD:ds5517

2019-05-24T15:44:46Z

Ds5517: /* Reaction dynamics */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: ∂V(ri)/∂ri=0. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that it has a symmetrical configuration at which rAB = rBC, shown on Figure 1. Since the TS is a saddle point, the second derivative of the reaction path potential will be negative (it is a maximum). However, the second derivative of the potential orthogonal to the reaction path potential at the TS will be positive (it is at minimum)

the second derivatives in orthogonal directions are going to differ. It is negative for one (maximum), but positive for the other (minimum).

<div>
[[File:Dori paintedit.png|thumb|none|500px|'''Fig. 1.''' Contour plot for the saddler point]]
</div>

<div>
[[File:Dori q2 pes.PNG|thumb|none|500px|'''Fig. 2.''' Potential Energy surface plot at the TS]]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.''']]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:

 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants !! E of TS !! Activation Energy !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -103.959 || -103.752|| 0.207 kcal/mol|| [[File:Dori q8 1.png]]
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042 kcal/mol|| [[File:Dori q8 2.PNG]]
|}

===Reaction dynamics===

====Polanyi's Empirical Rules<ref name="CKD" /> ====

Hammond's postulate states that an exothermic reaction will proceed with an early TS, which will resemble the the reactants. In contrast, in an endothermic reaction the TS will resemble the products in a late TS. Polanyi's Empirical Rules relate to Hammond's Postulate. Polanyi studied different potential energy surface trajectorys of A + BC → AB + C type reactions (like F + H2 → H + HF, as discussed above) in order to determine the dyanamics and energy changes of such reactions. Polanyi found that a reaction with an early TS (eg. F + H2 → H + HF) will favour high translational energy of the reactants which will result in high vibrational energy of the products. In contrast, a reaction with late TS (eg. H + HF → F + H2), will favour high vibrational energy of the reactants and will end up having high product translational energy.

====Release of Energy====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

As discussed above, F + H2 → H + HF is exothermic, the products are lower in energy than the reactants, thus energy is released from to system to the surroundings in the form of heat. The fact that energy is conserved can be confirmed and monitored by bomb calorimetry.

Figure 12 shows the energy against time plot for a successful collision. We know that the total energy of the system is constant (energy is conserved). Therefore, we see a horizontal line for the total energy. Figure 13 shows the momentum against time plot. The kinetic energy is the sum of vibrational and translational energies. A system with an early TS such as this one will favour low vibrational and high translational energies of the reactants (Polanyi's Rules) This is shown on Fig 13.

{| class="wikitable" border=1
! rFH !! rHH !! pFH !! pHH !! Reactive? !! Illustration of the trajectory !! Comments
|-
| 2.0|| 0.74 || -0.5|| -3 || No || [[File:Dori q9 p-3.png|400 px]] || Because the reaction has an early TS , it will favour low vibrational energy of the products to form the products (Polanyi's Rules). Since the system has high vibrational energy at the beginning, the F-H bond is formed but than the H-H bond reforms.
|-
| 2.0|| 0.74 || -0.5|| -2 || Yes || [[File:Dori q9 p-2.png|400 px]] || As we lowered the vibrational energy of the reactants, the reaction became successful. These conditions illustrate Polanyi's Rules, as the products in an exothermic reaction with an early TS have high vibrational energy.
|-
| 2.0|| 0.74 || -0.8|| 0.1 || Yes || [[File:Dori q9 p01.png|400 px]] || We now increased the vibrational energy of the products, which is favoured for an early TS reaction, thus products (H + HF) are formed
|}

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

File:Dori q9 p01.png

2019-05-24T15:28:35Z

Ds5517:

MRD:ds5517

2019-05-24T15:26:13Z

Ds5517: /* Polanyi's Empirical Rules */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: ∂V(ri)/∂ri=0. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that it has a symmetrical configuration at which rAB = rBC, shown on Figure 1. Since the TS is a saddle point, the second derivative of the reaction path potential will be negative (it is a maximum). However, the second derivative of the potential orthogonal to the reaction path potential at the TS will be positive (it is at minimum)

the second derivatives in orthogonal directions are going to differ. It is negative for one (maximum), but positive for the other (minimum).

<div>
[[File:Dori paintedit.png|thumb|none|500px|'''Fig. 1.''' Contour plot for the saddler point]]
</div>

<div>
[[File:Dori q2 pes.PNG|thumb|none|500px|'''Fig. 2.''' Potential Energy surface plot at the TS]]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.''']]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:

 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants !! E of TS !! Activation Energy !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -103.959 || -103.752|| 0.207 kcal/mol|| [[File:Dori q8 1.png]]
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042 kcal/mol|| [[File:Dori q8 2.PNG]]
|}

===Reaction dynamics===

====Polanyi's Empirical Rules<ref name="CKD" /> ====

Hammond's postulate states that an exothermic reaction will proceed with an early TS, which will resemble the the reactants. In contrast, in an endothermic reaction the TS will resemble the products in a late TS. Polanyi's Empirical Rules relate to Hammond's Postulate. Polanyi studied different potential energy surface trajectorys of A + BC → AB + C type reactions (like F + H2 → H + HF, as discussed above) in order to determine the dyanamics and energy changes of such reactions. Polanyi found that a reaction with an early TS (eg. F + H2 → H + HF) will favour high translational energy of the reactants which will result in high vibrational energy of the products. In contrast, a reaction with late TS (eg. H + HF → F + H2), will favour high vibrational energy of the reactants and will end up having high product translational energy.

====Release of Energy====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

As discussed above, F + H2 → H + HF is exothermic, the products are lower in energy than the reactants, thus energy is released from to system to the surroundings in the form of heat. The fact that energy is conserved can be confirmed and monitored by bomb calorimetry.

Figure 12 shows the energy against time plot for a successful collision. We know that the total energy of the system is constant (energy is conserved). Therefore, we see a horizontal line for the total energy. Figure 13 shows the momentum against time plot. The kinetic energy is the sum of vibrational and translational energies. A system with an early TS such as this one will favour low vibrational and high translational energies of the reactants (Polanyi's Rules) This is shown on Fig 13.

{| class="wikitable" border=1
! rFH !! rHH !! pFH !! pHH !! Reactive? !! Illustration of the trajectory !! Comments
|-
| 2.0|| 0.74 || -0.5|| -3 || No || [[File:Dori q9 p-3.png|400 px]] || Because the reaction has an early TS , it will favour low vibrational energy of the products to form the products (Polanyi's Rules). Since the system has high vibrational energy at the beginning, the F-H bond is formed but than the H-H bond reforms.
|-
| 2.0|| 0.74 || -0.5|| -2 || Yes || [[File:Dori q9 p-2.png|400 px]] || As we lowered the vibrational energy of the reactants, the reaction became successful. These conditions illustrate Polanyi's Rules, as the products in an exothermic reaction with an early TS have high vibrational energy.
|}

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

MRD:ds5517

2019-05-24T15:21:47Z

Ds5517: /* Release of Energy */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: ∂V(ri)/∂ri=0. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that it has a symmetrical configuration at which rAB = rBC, shown on Figure 1. Since the TS is a saddle point, the second derivative of the reaction path potential will be negative (it is a maximum). However, the second derivative of the potential orthogonal to the reaction path potential at the TS will be positive (it is at minimum)

the second derivatives in orthogonal directions are going to differ. It is negative for one (maximum), but positive for the other (minimum).

<div>
[[File:Dori paintedit.png|thumb|none|500px|'''Fig. 1.''' Contour plot for the saddler point]]
</div>

<div>
[[File:Dori q2 pes.PNG|thumb|none|500px|'''Fig. 2.''' Potential Energy surface plot at the TS]]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.''']]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:

 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants !! E of TS !! Activation Energy !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -103.959 || -103.752|| 0.207 kcal/mol|| [[File:Dori q8 1.png]]
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042 kcal/mol|| [[File:Dori q8 2.PNG]]
|}

===Reaction dynamics===

====Polanyi's Empirical Rules<ref name="CKD" /> ====

Hammond's postulate states that an exothermic reaction will proceed with an early TS, which will resemble the the reactants. In contrast, in an endothermic reaction the TS will resemble the products in a late TS. Polanyi's Empirical Rules relate to Hammond's Postulate. Polanyi studied different potential energy surface trajectorys of A + BC → AB + C type reactions (like F + H2 → H + HF, as discussed above) in order to determine the dyanamics and energy changes of such reactions. Polanyi found that a reaction with an early TS (eg. F + H2 → H + HF) will favour high translational energy of the reactants. In contrast, a reaction with late TS (eg. H + HF → F + H2), will favour high vibrational energy of the reactants.

====Release of Energy====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

As discussed above, F + H2 → H + HF is exothermic, the products are lower in energy than the reactants, thus energy is released from to system to the surroundings in the form of heat. The fact that energy is conserved can be confirmed and monitored by bomb calorimetry.

Figure 12 shows the energy against time plot for a successful collision. We know that the total energy of the system is constant (energy is conserved). Therefore, we see a horizontal line for the total energy. Figure 13 shows the momentum against time plot. The kinetic energy is the sum of vibrational and translational energies. A system with an early TS such as this one will favour low vibrational and high translational energies of the reactants (Polanyi's Rules) This is shown on Fig 13.

{| class="wikitable" border=1
! rFH !! rHH !! pFH !! pHH !! Reactive? !! Illustration of the trajectory !! Comments
|-
| 2.0|| 0.74 || -0.5|| -3 || No || [[File:Dori q9 p-3.png|400 px]] || Because the reaction has an early TS , it will favour low vibrational energy of the products to form the products (Polanyi's Rules). Since the system has high vibrational energy at the beginning, the F-H bond is formed but than the H-H bond reforms.
|-
| 2.0|| 0.74 || -0.5|| -2 || Yes || [[File:Dori q9 p-2.png|400 px]] || As we lowered the vibrational energy of the reactants, the reaction became successful. These conditions illustrate Polanyi's Rules, as the products in an exothermic reaction with an early TS have high vibrational energy.
|}

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

MRD:ds5517

2019-05-24T15:18:02Z

Ds5517:

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: ∂V(ri)/∂ri=0. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that it has a symmetrical configuration at which rAB = rBC, shown on Figure 1. Since the TS is a saddle point, the second derivative of the reaction path potential will be negative (it is a maximum). However, the second derivative of the potential orthogonal to the reaction path potential at the TS will be positive (it is at minimum)

the second derivatives in orthogonal directions are going to differ. It is negative for one (maximum), but positive for the other (minimum).

<div>
[[File:Dori paintedit.png|thumb|none|500px|'''Fig. 1.''' Contour plot for the saddler point]]
</div>

<div>
[[File:Dori q2 pes.PNG|thumb|none|500px|'''Fig. 2.''' Potential Energy surface plot at the TS]]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.''']]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:

 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants !! E of TS !! Activation Energy !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -103.959 || -103.752|| 0.207 kcal/mol|| [[File:Dori q8 1.png]]
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042 kcal/mol|| [[File:Dori q8 2.PNG]]
|}

===Reaction dynamics===

====Polanyi's Empirical Rules<ref name="CKD" /> ====

Hammond's postulate states that an exothermic reaction will proceed with an early TS, which will resemble the the reactants. In contrast, in an endothermic reaction the TS will resemble the products in a late TS. Polanyi's Empirical Rules relate to Hammond's Postulate. Polanyi studied different potential energy surface trajectorys of A + BC → AB + C type reactions (like F + H2 → H + HF, as discussed above) in order to determine the dyanamics and energy changes of such reactions. Polanyi found that a reaction with an early TS (eg. F + H2 → H + HF) will favour high translational energy of the reactants. In contrast, a reaction with late TS (eg. H + HF → F + H2), will favour high vibrational energy of the reactants.

====Release of Energy====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

As discussed above, F + H2 → H + HF is exothermic, the products are lower in energy than the reactants, thus energy is released from to system to the surroundings in the form of heat. The fact that energy is conserved can be confirmed and monitored by bomb calorimetry.

Figure 12 shows the energy against time plot for a successful collision. We know that the total energy of the system is constant (energy is conserved). Therefore, we see a horizontal line for the total energy. Figure 13 shows the momentum against time plot. The kinetic energy is the sum of vibrational and translational energies. A system with an early TS such as this one will favour low vibrational and high translational energies of the reactants (Polanyi's Rules) This is shown on Fig 13.

{| class="wikitable" border=1
! rFH !! rHH !! pFH !! pHH !! Reactive? !! Illustration of the trajectory !! Comments
|-
| 2.0|| 0.74 || -0.5|| -3 || No || [[File:Dori q9 p-3.png|400 px]] || Because the reaction has an early TS , it will favour low vibrational energy of the products to form the products. Since the system has high vibrational energy at the beginning, the F-H bond is formed but than the H-H bond reforms.
|-
| 2.0|| 0.74 || -0.5|| -2 || Yes || [[File:Dori q9 p-2.png|400 px]] || As we lowered the vibrational energy of the reactants, the reaction became successful.
|}

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

File:Dori q9 p-2.png

2019-05-24T15:15:39Z

Ds5517:

File:Dori q9 p-3.png

2019-05-24T14:58:53Z

Ds5517:

MRD:ds5517

2019-05-24T14:53:11Z

Ds5517: /* Reaction dynamics */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: ∂V(ri)/∂ri=0. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that it has a symmetrical configuration at which rAB = rBC, shown on Figure 1. Since the TS is a saddle point, the second derivative of the reaction path potential will be negative (it is a maximum). However, the second derivative of the potential orthogonal to the reaction path potential at the TS will be positive (it is at minimum)

the second derivatives in orthogonal directions are going to differ. It is negative for one (maximum), but positive for the other (minimum).

<div>
[[File:Dori paintedit.png|thumb|none|500px|'''Fig. 1.''' Contour plot for the saddler point]]
</div>

<div>
[[File:Dori q2 pes.PNG|thumb|none|500px|'''Fig. 2.''' Potential Energy surface plot at the TS]]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.''']]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:

 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants !! E of TS !! Activation Energy !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -103.959 || -103.752|| 0.207 kcal/mol|| [[File:Dori q8 1.png]]
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042 kcal/mol|| [[File:Dori q8 2.PNG]]
|}

===Reaction dynamics===

====Polanyi's Empirical Rules<ref name="CKD" /> ====

Hammond's postulate states that an exothermic reaction will proceed with an early TS, which will resemble the the reactants. In contrast, in an endothermic reaction the TS will resemble the products in a late TS. Polanyi's Empirical Rules relate to Hammond's Postulate. Polanyi studied different potential energy surface trajectorys of A + BC → AB + C type reactions (like F + H2 → H + HF, as discussed above) in order to determine the dyanamics and energy changes of such reactions. Polanyi found that a reaction with an early TS (eg. F + H2 → H + HF) will favour high translational energy of the reactants. In contrast, a reaction with late TS (eg. H + HF → F + H2), will favour high vibrational energy of the reactants.

====Release of Energy====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

As discussed above, F + H2 → H + HF is exothermic, the products are lower in energy than the reactants, thus energy is released from to system to the surroundings in the form of heat. The fact that energy is conserved can be confirmed and monitored by bomb calorimetry.

Figure 12 shows the energy against time plot for a successful collision. We know that the total energy of the system is constant (energy is conserved). Therefore, we see a horizontal line for the total energy. Figure 13 shows the momentum against time plot. The kinetic energy is the sum of vibrational and translational energies. A system with an early TS such as this one will favour low vibrational and high translational energies of the reactants (Polanyi's Rules) This is shown on Fig 13.

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

MRD:ds5517

2019-05-24T14:32:44Z

Ds5517: /* Reaction dynamics */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: ∂V(ri)/∂ri=0. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that it has a symmetrical configuration at which rAB = rBC, shown on Figure 1. Since the TS is a saddle point, the second derivative of the reaction path potential will be negative (it is a maximum). However, the second derivative of the potential orthogonal to the reaction path potential at the TS will be positive (it is at minimum)

the second derivatives in orthogonal directions are going to differ. It is negative for one (maximum), but positive for the other (minimum).

<div>
[[File:Dori paintedit.png|thumb|none|500px|'''Fig. 1.''' Contour plot for the saddler point]]
</div>

<div>
[[File:Dori q2 pes.PNG|thumb|none|500px|'''Fig. 2.''' Potential Energy surface plot at the TS]]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.''']]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:

 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants !! E of TS !! Activation Energy !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -103.959 || -103.752|| 0.207 kcal/mol|| [[File:Dori q8 1.png]]
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042 kcal/mol|| [[File:Dori q8 2.PNG]]
|}

===Reaction dynamics===

====Polanyi's Empirical Rules<ref name="CKD" /> ====

Hammond's postulate states that an exothermic reaction will proceed with an early TS, which will resemble the the reactants. In contrast, in an endothermic reaction the TS will resemble the products in a late TS. Polanyi's Empirical Rules relate to Hammond's Postulate. Polanyi studied different potential energy surface trajectorys of A + BC → AB + C type reactions (like F + H2 → H + HF, as discussed above) in order to determine the dyanamics and energy changes of such reactions. Polanyi found that a reaction with an early TS (eg. F + H2 → H + HF) will favour high translational energy of the reactants. In contrast, a reaction with late TS (eg. H + HF → F + H2), will favour high vibrational energy of the reactants.

====Release of Energy====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

As discussed above, F + H2 → H + HF is exothermic, the products are lower in energy than the reactants, thus energy is released from to system to the surroundings in the form of heat. The fact that energy is conserved can be confirmed and monitored by bomb calorimetry.

Figure 12 shows the energy against time plot for a successful collision. We know that the total energy of the system is constant (energy is conserved). Therefore, we see a horizontal line for the total energy. Figure 13 shows the momentum against time plot. The kinetic energy is the sum of vibrational and translational energies.

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

MRD:ds5517

2019-05-24T14:14:45Z

Ds5517: /* Release of Energy */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: ∂V(ri)/∂ri=0. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that it has a symmetrical configuration at which rAB = rBC, shown on Figure 1. Since the TS is a saddle point, the second derivative of the reaction path potential will be negative (it is a maximum). However, the second derivative of the potential orthogonal to the reaction path potential at the TS will be positive (it is at minimum)

the second derivatives in orthogonal directions are going to differ. It is negative for one (maximum), but positive for the other (minimum).

<div>
[[File:Dori paintedit.png|thumb|none|500px|'''Fig. 1.''' Contour plot for the saddler point]]
</div>

<div>
[[File:Dori q2 pes.PNG|thumb|none|500px|'''Fig. 2.''' Potential Energy surface plot at the TS]]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.''']]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:

 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants !! E of TS !! Activation Energy !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -103.959 || -103.752|| 0.207 kcal/mol|| [[File:Dori q8 1.png]]
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042 kcal/mol|| [[File:Dori q8 2.PNG]]
|}

===Reaction dynamics===

====Polanyi's Empirical Rules<ref name="CKD" /> ====

Hammond's postulate states that an exothermic reaction will proceed with an early TS, which will resemble the the reactants. In contrast, in an endothermic reaction the TS will resemble the products in a late TS. Polanyi's Empirical Rules relate to Hammond's Postulate. Polanyi studied different potential energy surface trajectorys of A + BC → AB + C type reactions (like F + H2 → H + HF, as discussed above) in order to determine the dyanamics and energy changes of such reactions. Polanyi found that a reaction with an early TS (eg. F + H2 → H + HF) will favour high translational energy of the reactants. In contrast, a reaction with late TS (eg. H + HF → F + H2), will favour high vibrational energy of the reactants.

====Release of Energy====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

As discussed above,

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

MRD:ds5517

2019-05-24T14:14:12Z

Ds5517: /* Reaction dynamics */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: ∂V(ri)/∂ri=0. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that it has a symmetrical configuration at which rAB = rBC, shown on Figure 1. Since the TS is a saddle point, the second derivative of the reaction path potential will be negative (it is a maximum). However, the second derivative of the potential orthogonal to the reaction path potential at the TS will be positive (it is at minimum)

the second derivatives in orthogonal directions are going to differ. It is negative for one (maximum), but positive for the other (minimum).

<div>
[[File:Dori paintedit.png|thumb|none|500px|'''Fig. 1.''' Contour plot for the saddler point]]
</div>

<div>
[[File:Dori q2 pes.PNG|thumb|none|500px|'''Fig. 2.''' Potential Energy surface plot at the TS]]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.''']]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:

 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants !! E of TS !! Activation Energy !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -103.959 || -103.752|| 0.207 kcal/mol|| [[File:Dori q8 1.png]]
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042 kcal/mol|| [[File:Dori q8 2.PNG]]
|}

===Reaction dynamics===

====Polanyi's Empirical Rules<ref name="CKD" /> ====

Hammond's postulate states that an exothermic reaction will proceed with an early TS, which will resemble the the reactants. In contrast, in an endothermic reaction the TS will resemble the products in a late TS. Polanyi's Empirical Rules relate to Hammond's Postulate. Polanyi studied different potential energy surface trajectorys of A + BC → AB + C type reactions (like F + H2 → H + HF, as discussed above) in order to determine the dyanamics and energy changes of such reactions. Polanyi found that a reaction with an early TS (eg. F + H2 → H + HF) will favour high translational energy of the reactants. In contrast, a reaction with late TS (eg. H + HF → F + H2), will favour high vibrational energy of the reactants.

====Release of Energy====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

MRD:ds5517

2019-05-24T13:28:15Z

Ds5517: /* Trajectories from r1 = r2: locating the transition states */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: ∂V(ri)/∂ri=0. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that it has a symmetrical configuration at which rAB = rBC, shown on Figure 1. Since the TS is a saddle point, the second derivative of the reaction path potential will be negative (it is a maximum). However, the second derivative of the potential orthogonal to the reaction path potential at the TS will be positive (it is at minimum)

the second derivatives in orthogonal directions are going to differ. It is negative for one (maximum), but positive for the other (minimum).

<div>
[[File:Dori paintedit.png|thumb|none|500px|'''Fig. 1.''' Contour plot for the saddler point]]
</div>

<div>
[[File:Dori q2 pes.PNG|thumb|none|500px|'''Fig. 2.''' Potential Energy surface plot at the TS]]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.''']]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:

 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants !! E of TS !! Activation Energy !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -103.959 || -103.752|| 0.207 kcal/mol|| [[File:Dori q8 1.png]]
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042 kcal/mol|| [[File:Dori q8 2.PNG]]
|}

===Reaction dynamics===

====Polanyi's Empirical Rules<ref name="CKD" /> ====

Hammond's postulate states that an exothermic reaction will proceed with an early TS, which will resemble the the reactants. In contrast, in an endothermic reaction the TS will resemble the products in a late TS. Polanyi's Empirical Rules relate to Hammond's Postulate. Polanyi studied different potential energy surface trajectorys of A + BC → AB + C type reactions (like F + H2 → H + HF, as discussed above) in order to determine the dyanamics and energy changes of such reactions. Polanyi found that an exothermic reaction (eg. F + H2 → H + HF) will lead to high product vibrational excitation whereas an endothermic reaction (eg. H + HF → F + H2) will favour low vibrational energy, but proceeds with high translational energy.

====Release of Energy====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

MRD:ds5517

2019-05-24T13:16:12Z

Ds5517: /* Activation Energies */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: ∂V(ri)/∂ri=0. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that it has a symmetrical configuration at which rAB = rBC, shown on Figure 1. Since the TS is a saddle point, the second derivative of the reaction path potential will be negative (it is a maximum). However, the second derivative of the potential orthogonal to the reaction path potential at the TS will be positive (it is at minimum)

the second derivatives in orthogonal directions are going to differ. It is negative for one (maximum), but positive for the other (minimum).

<div>
[[File:Dori paintedit.png|thumb|none|500px|'''Fig. 1.''' Contour plot for the saddler point]]
</div>

<div>
[[File:Dori q2 pes.PNG|thumb|none|500px|'''Fig. 2.''' Potential Energy surface plot at the TS]]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.''']]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:

 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants !! E of TS !! Activation Energy !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -103.959 || -103.752|| 0.207 kcal/mol|| [[File:Dori q8 1.png]]
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042 kcal/mol|| [[File:Dori q8 2.PNG]]
|}

===Reaction dynamics===

====Polanyi's Empirical Rules<ref name="CKD" /> ====

Hammond's postulate states that an exothermic reaction will proceed with an early TS, which will resemble the the reactants. In contrast, in an endothermic reaction the TS will resemble the products in a late TS. Polanyi's Empirical Rules relate to Hammond's Postulate. Polanyi studied different potential energy surface trajectorys of A + BC → AB + C type reactions (like F + H2 → H + HF, as discussed above) in order to determine the dyanamics and energy changes of such reactions. Polanyi found that an exothermic reaction (eg. F + H2 → H + HF) will lead to high product vibrational excitation whereas an endothermic reaction (eg. H + HF → F + H2) will favour low vibrational energy, but proceeds with high translational energy.

====Release of Energy====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

File:Dori q8 2.PNG

2019-05-24T13:16:00Z

Ds5517:

File:Dori q8 1.png

2019-05-24T13:15:38Z

Ds5517:

MRD:ds5517

2019-05-24T11:30:33Z

Ds5517: /* Dynamics from the transition state region */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: ∂V(ri)/∂ri=0. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that it has a symmetrical configuration at which rAB = rBC, shown on Figure 1. Since the TS is a saddle point, the second derivative of the reaction path potential will be negative (it is a maximum). However, the second derivative of the potential orthogonal to the reaction path potential at the TS will be positive (it is at minimum)

the second derivatives in orthogonal directions are going to differ. It is negative for one (maximum), but positive for the other (minimum).

<div>
[[File:Dori paintedit.png|thumb|none|500px|'''Fig. 1.''' Contour plot for the saddler point]]
</div>

<div>
[[File:Dori q2 pes.PNG|thumb|none|500px|'''Fig. 2.''' Potential Energy surface plot at the TS]]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.''']]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:

 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants !! E of TS !! Activation Energy !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -104.010 || -103.752|| 0.258 kcal/mol||
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042 kcal/mol||
|}

===Reaction dynamics===

====Polanyi's Empirical Rules<ref name="CKD" /> ====

Hammond's postulate states that an exothermic reaction will proceed with an early TS, which will resemble the the reactants. In contrast, in an endothermic reaction the TS will resemble the products in a late TS. Polanyi's Empirical Rules relate to Hammond's Postulate. Polanyi studied different potential energy surface trajectorys of A + BC → AB + C type reactions (like F + H2 → H + HF, as discussed above) in order to determine the dyanamics and energy changes of such reactions. Polanyi found that an exothermic reaction (eg. F + H2 → H + HF) will lead to high product vibrational excitation whereas an endothermic reaction (eg. H + HF → F + H2) will favour low vibrational energy, but proceeds with high translational energy.

====Release of Energy====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

File:Dori q2 pes.PNG

2019-05-24T11:29:04Z

Ds5517:

MRD:ds5517

2019-05-24T10:37:12Z

Ds5517: /* Dynamics from the transition state region */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: ∂V(ri)/∂ri=0. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that it has a symmetrical configuration at which rAB = rBC, shown on Figure 1. Since the TS is a saddle point, the second derivative of the reaction path potential will be negative (it is a maximum). However, the second derivative of the potential orthogonal to the reaction path potential at the TS will be positive (it is at minimum)

the second derivatives in orthogonal directions are going to differ. It is negative for one (maximum), but positive for the other (minimum).

<div>
[[File:Dori paintedit.png|thumb|none|500px|'''Fig. 1.''']]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.''']]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:

 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants !! E of TS !! Activation Energy !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -104.010 || -103.752|| 0.258 kcal/mol||
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042 kcal/mol||
|}

===Reaction dynamics===

====Polanyi's Empirical Rules<ref name="CKD" /> ====

Hammond's postulate states that an exothermic reaction will proceed with an early TS, which will resemble the the reactants. In contrast, in an endothermic reaction the TS will resemble the products in a late TS. Polanyi's Empirical Rules relate to Hammond's Postulate. Polanyi studied different potential energy surface trajectorys of A + BC → AB + C type reactions (like F + H2 → H + HF, as discussed above) in order to determine the dyanamics and energy changes of such reactions. Polanyi found that an exothermic reaction (eg. F + H2 → H + HF) will lead to high product vibrational excitation whereas an endothermic reaction (eg. H + HF → F + H2) will favour low vibrational energy, but proceeds with high translational energy.

====Release of Energy====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

File:Dori paintedit.png

2019-05-24T10:36:57Z

Ds5517:

MRD:ds5517

2019-05-24T10:04:43Z

Ds5517: /* Polanyi's Empirical Rules */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: EQUATIONNNNN. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that the second derivatives in orthogonal directions are going to differ. It is negative for one (maximum), but positive for the other (minimum).

<div>
[[File:Dori minimax.PNG|thumb|none|500px|'''Fig. 1.''']]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.''']]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:

 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants !! E of TS !! Activation Energy !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -104.010 || -103.752|| 0.258 kcal/mol||
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042 kcal/mol||
|}

===Reaction dynamics===

====Polanyi's Empirical Rules<ref name="CKD" /> ====

Hammond's postulate states that an exothermic reaction will proceed with an early TS, which will resemble the the reactants. In contrast, in an endothermic reaction the TS will resemble the products in a late TS. Polanyi's Empirical Rules relate to Hammond's Postulate. Polanyi studied different potential energy surface trajectorys of A + BC → AB + C type reactions (like F + H2 → H + HF, as discussed above) in order to determine the dyanamics and energy changes of such reactions. Polanyi found that an exothermic reaction (eg. F + H2 → H + HF) will lead to high product vibrational excitation whereas an endothermic reaction (eg. H + HF → F + H2) will favour low vibrational energy, but proceeds with high translational energy.

====Release of Energy====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

MRD:ds5517

2019-05-24T09:46:41Z

Ds5517: /* Reaction dynamics */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: EQUATIONNNNN. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that the second derivatives in orthogonal directions are going to differ. It is negative for one (maximum), but positive for the other (minimum).

<div>
[[File:Dori minimax.PNG|thumb|none|500px|'''Fig. 1.''']]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.''']]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:

 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants !! E of TS !! Activation Energy !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -104.010 || -103.752|| 0.258 kcal/mol||
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042 kcal/mol||
|}

===Reaction dynamics===

====Polanyi's Empirical Rules<ref name="CKD" /> ====

====Release of Energy====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

MRD:ds5517

2019-05-24T09:43:42Z

Ds5517: /* Reaction dynamics */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: EQUATIONNNNN. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that the second derivatives in orthogonal directions are going to differ. It is negative for one (maximum), but positive for the other (minimum).

<div>
[[File:Dori minimax.PNG|thumb|none|500px|'''Fig. 1.''']]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.''']]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:

 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants !! E of TS !! Activation Energy !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -104.010 || -103.752|| 0.258 kcal/mol||
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042 kcal/mol||
|}

===Reaction dynamics===

====Polanyi's Empirical Rules ==== <ref name="CKD" />

====Release of Energy====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

MRD:ds5517

2019-05-24T09:43:10Z

Ds5517: /* Reaction dynamics */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: EQUATIONNNNN. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that the second derivatives in orthogonal directions are going to differ. It is negative for one (maximum), but positive for the other (minimum).

<div>
[[File:Dori minimax.PNG|thumb|none|500px|'''Fig. 1.''']]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.''']]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:

 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants !! E of TS !! Activation Energy !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -104.010 || -103.752|| 0.258 kcal/mol||
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042 kcal/mol||
|}

===Reaction dynamics===

'''Polanyi's Empirical Rules'''<ref name="CKD" />

====Release of Energy====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

MRD:ds5517

2019-05-24T09:31:23Z

Ds5517: /* Transition State Theory */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: EQUATIONNNNN. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that the second derivatives in orthogonal directions are going to differ. It is negative for one (maximum), but positive for the other (minimum).

<div>
[[File:Dori minimax.PNG|thumb|none|500px|'''Fig. 1.''']]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.''']]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:

 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants !! E of TS !! Activation Energy !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -104.010 || -103.752|| 0.258 kcal/mol||
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042 kcal/mol||
|}

===Reaction dynamics===

====Release of Energy====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

MRD:ds5517

2019-05-24T09:30:39Z

Ds5517:

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: EQUATIONNNNN. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that the second derivatives in orthogonal directions are going to differ. It is negative for one (maximum), but positive for the other (minimum).

<div>
[[File:Dori minimax.PNG|thumb|none|500px|'''Fig. 1.''']]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.''']]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory <ref name="CKD" />===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:

 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants !! E of TS !! Activation Energy !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -104.010 || -103.752|| 0.258 kcal/mol||
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042 kcal/mol||
|}

===Reaction dynamics===

====Release of Energy====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

==References==

<references>
<ref name="CKD">Steinfeld, J. I., Francisco, J. S. and Hase, W. L. (1989) Chemical Kinetics and Dyamics.</ref>
</references>

MRD:ds5517

2019-05-24T09:04:02Z

Ds5517: /* Transition State Theory */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: EQUATIONNNNN. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that the second derivatives in orthogonal directions are going to differ. It is negative for one (maximum), but positive for the other (minimum).

<div>
[[File:Dori minimax.PNG|thumb|none|500px|'''Fig. 1.''']]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.''']]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:

 1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
 2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
 3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
 4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
 5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants !! E of TS !! Activation Energy !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -104.010 || -103.752|| 0.258 kcal/mol||
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042 kcal/mol||
|}

===Reaction dynamics===

====Release of Energy====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

MRD:ds5517

2019-05-24T09:01:55Z

Ds5517: /* Exercise 1: H + H2 system */

==Exercise 1: H + H2 system==

===Dynamics from the transition state region===

'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state (TS) on a potential energy surface diagram is represented by the saddle point. It is a stationary point at which the gradient of the potential is zero. This means that the TS can be found by looking at the first derivative of the potentian energy surface: EQUATIONNNNN. In order to distinguish the TS from a local minimum, we have to look at the second derivative as well. As mentioned above, the TS is a saddle point, which means that the second derivatives in orthogonal directions are going to differ. It is negative for one (maximum), but positive for the other (minimum).

<div>
[[File:Dori minimax.PNG|thumb|none|500px|'''Fig. 1.''']]
</div>

===Trajectories from r1 = r2: locating the transition states ===
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

At the transition state (TS) the distances between A-B and B-C are equal. As mentioned above, the TS is stationary which results in the forces along AB and BC and the kinetic energy being zero. The value of the transition state position (rts) was found to be 0.907743 Å. This value can be rounded up to 0.908 Å, since using 7 decimal places is unneccessary.

The internuclear distances against time plot is horizontal. This confirms that the TS was found, since rBC=rAB (stationary) and the system vibrates symmetrically.

<div>
[[File:Dori TS distvstime.PNG|thumb|none|500px|'''Fig. 3.''' Internuclear distance against Time plot when rBC=rAB=0.907743 Å and pBC=pAB=0]]
</div>

===Calculating the reaction path===

'''Question 3: Comment on how the MEP and the trajectory you just calculated differ.'''

In the following section two different calculation types, Dynamic and Minimum Energy Path (MEP), are analyzed using the previously calculated value of rts=0.907743 Å. The system was slightly offset from the TS by setting rBC= rts + 0.01 = 0.917743 Å and zero initial momenta.

MEP is the lowest energy pathway from reactants to products. This method corresponds to infinitely slow motion where the mass of the atoms are not taken into account. Figures 5 and 7 show a relatively straight line for the reaction path. This indicates that the kinetic energy is zero and the atoms are not vibrating. Since the momenta/ velocities are set back to zero in each step, energy is not conserved from step to step which also results in a straight line. However, when we compare this with the Dynamic method shown on Figures 4 and 6, we can see a wavy line for reaction pathway. This indicates that the kinetic energy of the system is no longer zero, the atoms are oscillating. After considering both methods, it is safe to say that MEP is less realistic and useful.

<div>
[[File:Dori q3 dyn disttime.PNG|thumb|none|500px|'''Fig. 4.''' Internuclear distance against Time plot using Dynamics Calculation Type]]
[[File:Dori q3 MEP disttime.PNG|thumb|none|500px|'''Fig. 5.''' Internuclear distance against Time plot using MEP Calculation Type]]

[[File:Dori q3 dyn1.PNG|thumb|none|500px|'''Fig. 6.''' Contour Plot Trajectory using Dynamic ]]
[[File:Dori q3 mep.PNG|thumb|none|500px|'''Fig. 7.''']]

</div>

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || HA + HBHC → HAHB + HC. The reaction is successful. Originally, HBHC did not oscillate, but after the collision, oscillation of HAHB is observable. || [[File:Dori q4 1.PNG| 400px ]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || The activation energy is higher than the kinetic energy of the system, thus the reaction is unsuccessful, the HAHB bond is not formed. || [[File:Dori q4 2.PNG| 400px ]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || HA + HBHC → HAHB + HC. Comparing these conditions to the first one, we can see that here HBHC oscillates at the start of the reaction.|| [[File:Dori q4 3.PNG| 400px ]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || The trajectory shows that under these conditions, the HAHB product did form, however the system's vibrational energy was too high which resulted in the breakage of the HA-HB bond and the reformation of the HB-HC bond.|| [[File:Dori q4 4.PNG| 400px ]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Overall, these conditions result in the success of the reaction: HA + HBHC → HAHB + HC. As shown on the trajectory, first the reactants have enough energy to overcome the TS, but shortly after the product dissociates, followed by the reformation of the products. This is due to the high vibrational energy of the system. || [[File:Dori q4 5.PNG| 400px ]]
|}

'''Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

In conclusion, the success of a reaction does not depend only on the system having enough energy to overcome the activation energy barrier, but on other factors as well. As shown in the above table, the 4th example is higher in energy than the 1st or the 3rd, but is still unreactive.

===Transition State Theory===

'''Question 5: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The aim of the transition-state theory (TST) is to study a reaction in which the reactants are in thermal equilibrium with the TS and to determine the rate constant of such reactions. This theory is based on the statistical properties of the system. The main assumptions of the transition-state theory are the following:

1. The reactants are in thermal equilibrium (quasi-equilibrium) with the TS species.
2. The electronic and nuclear motions are assumed to be separate from each other. This is equivalent to the Born-Oppenheimer Approximation in quantum mechanics.
3. The distribution of of reactants molecules and transition states among their states is directed by the Maxwell-Bolztmann laws.
4. Once the reactants have sufficient energy to cross the activation energy barrier at the TS, the products are formed in an irreversible reaction.
5. The reaction is treated classically.

The TST is not accurate in all cases. As shown in the table above, a system can have sufficient activation energy to reach the TS, but reform the products afterwards (example 4). This is in disagreement with assumption 4 of the TST. Small systems, such as the H-H-H, cannot only be treated classically, quantum mechanics must also be taken into account to obtain correct results. This disagrees with assumption 5.

==Exercise 2: F - H - H system==

===PES inspection===

'''Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

====F + H2 → HF + H====

[[File:Dori FplusH2.PNG|thumb|none|500px|'''Fig. 8.''' Potential Energy Surface plot for reaction F + H2]]

The reaction F + H2 → HF + H is is examined in the following section. As shown on Figure 8., the reactants are higher in energy than the products, therefore the reaction is classified to be exothermic. The bond being broken is the H-H single bond (432 kJ/mol) and an H-F single bond is formed (565 kJ/mol). Since F is more electronegative than H, the new bond is more covalent and stronger, due to which the overall reaction is exothermic.

====H + HF → F + H2 ====
[[File:Dori HplusHF.PNG|thumb|none|500px|'''Fig. 9.''' Potential Energy Surface plot for reaction H + HF]]

As discussed above, the H-F single bond is stronger than the H-H single bond. The reaction H + HF → F + H2 is therefore results in a breakage of a stronger bond in order to form a weaker. Thus the reaction is endothermic. As shown on Figure 9., the products are higher in energy than the reactants.

===Position of Transition State===

'''Question 7: Locate the approximate position of the transition state.'''

According to Hammond's Postulate, The structure of the transition state will resemble either the structure of the reactants or the products, depending which is closer in energy. The lifetime of the TS is essentially zero, thus this postulate is useful in determining the structure at the energy maxima (TS). Hammond suggested that in an exothermic reaction the TS will resemble the reactants (early TS) whereas in an endothermic reaction reaction it resembles the products.

As the reaction F + H2 → HF + H is found to be exothermic, the TS will resemble the structure of H2. The bond length of H-H = 0.74 Å was used to determine the TS distances. Further decimal places were added to both the H-H and the H-F length until the forces along AB and BC equaled 0. The distances were found to be the following: H-F = 1.810790 Å and H-H = 0.744878 Å. Since the TS is stationary, the Internuclear Distance vs Time plot shows horizontal lines (there is no change in the distance between atoms).

[[File:Dori forcesTS.PNG|thumb|none|500px|'''Fig. 10.''' Forces in TS]]

[[File:Doir ts disttime.PNG|thumb|none|500px|'''Fig. 11.''' Internuclear Distance vs Time plot in TS]]

===Activation Energies===

'''Question 8: Report the activation energy for both reactions.'''

Activation energies:

The activation energies of both reactions were calculated by the difference in energy between the reactants and the transition state. Energy of reactants was found by slightly offsetting on of the r values, using the MEP calculation and the Energy vs Time graphs. The values can be found in the table below.

{| class="wikitable" border=1
! Reaction !! E of reactants !! E of TS !! Activation Energy !! Illustration of the trajectory
|-
| F + H2 → H + HF|| -104.010 || -103.752|| 0.258 kcal/mol||
|-
| H + HF → F + H2|| -133.794 || -103.752|| 30.042 kcal/mol||
|}

===Reaction dynamics===

====Release of Energy====

'''Question 9: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

'''Question 10: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

MRD:ds5517

2019-05-24T08:50:00Z

Ds5517: /* Transition State Theory */

MRD:ds5517

2019-05-23T21:53:09Z

Ds5517: /* Position of Transition State */

MRD:ds5517

2019-05-23T21:51:36Z

Ds5517: /* Position of Transition State */

MRD:ds5517

2019-05-23T21:50:26Z

Ds5517: /* Position of Transition State */

MRD:ds5517

2019-05-23T21:49:32Z

Ds5517: /* Position of Transition states */

MRD:ds5517

2019-05-23T21:46:01Z

Ds5517: /* Calculating the reaction path */

MRD:ds5517

2019-05-23T21:38:07Z

Ds5517: /* Activation Energies */

MRD:ds5517

2019-05-23T21:37:44Z

Ds5517: /* Activation Energies */

MRD:ds5517

2019-05-23T21:37:07Z

Ds5517: /* Activation Energies */

MRD:ds5517

2019-05-23T21:26:38Z

Ds5517: /* Reaction dynamics */

MRD:ds5517

2019-05-23T21:26:00Z

Ds5517: /* Activation Energies */