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MRD:M4H1M4

2020-05-22T22:54:46Z

Mb7718: /* Trajectories from r1 = r2: locating the transition state */

== [[EXERCISE 1: H + H2 system]] ==

=== Dynamics from the transition state region ===
[[File:Hdiagram.png|200px|thumb|left|H + H2]]
''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 is defined as the first order saddle point on the potential energy surface (PES) diagram. The derivative of potential energy is equal to the force acting on the atoms/molecules. At this point the gradient is equal to 0 and therefore no force is acting on the particles. This allows us to use a constrained one dimensional optimisation to calculate transition state. In order to verify this stationary point as the saddle transition state the Hessian eigenvectors can be used. Only one of the second derivatives must be negative and the other positive so that (fxxfyy - f2xy) < 0. Therefore the determinent of the Hessian matrix less than 0.

This verifies the point as a saddle rather than a local minimum of the potential energy surface.

==== Trajectories from r1 = r2: locating the transition state ====
''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.''

The the H + H2 ---> H2 + H surface is symmetrical, neither H2 or H atom are favoured.Therefore r1 is equal to r2 in the transition state. This restriction allows us to calculate the transition state position through a process of trial and error.

[[File:R1=R2.png|600px|thumb|left|Figure 1 - PES of transition state of reaction H + H2]]

[[File:Hessian.png|600px|thumb|Figure 2 - Hessian matrix and determinant for H + H2 reaction]]

[[File:Dvst01499529.png|600px|thumb|left|Figure 3 - Displacement vs Time for reaction at transition state]]Transition state position r1= r2 = 90.8 pm

As shown in figure 1, at these positions the forces action along AB and BC are approximately 0 suggesting a stationary point. Figure 3 shows two horizontal lines with gradient 0 which therefore means 0 acceleration and 0 forces acting on the atoms.

Looking at the Hessian matrix we can verify this as a saddle point. To further confirm this, trajectories were calculated with either r1 or r2 being slightly displaced from equilibrium. Figure 4 shows that increasing r2 by 1 pm pushing the reaction to the reactant side figure 5 shows increasing r2 by 1 pm favours the formation of the products.

[[File:Reactantchannel.01499529.png|200px|thumb|Figure 4 - Trajectory in reactant channel at '''r2''' = '''rts'''+1, '''r1''' = '''rts p1''' = '''p2''' = 0 g.mol-1.pm.fs-1]]

[[File:Productchannel.01499529.png|200px|thumb|Figure 5 - Trajectory in product channel at '''r1''' = '''rts'''+1 pm, '''r2''' = '''rts, p1''' = '''p2''' = 0 g.mol-1.pm.fs-1]]

==== Trajectories from r1 = rts+δ, r2 = rts ====
''Comment on how the mep and the trajectory you just calculated differ.''

Figure 6 shows a straight line suggesting the velocity and momentum do not change with time. Figure 7, the dynamic trajectory, accounts for the oscillation of the bonds.

Final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t):

r1=74.0 pm r2= 352.6 pm p1= 3.2 g.mol-1.pm.fs-1 p2= 5.1 g.mol-1.pm.fs-1
* ''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.''
''What do you observe?''

The reaction does not occur atom C reaches a constant velocity and atom B and A vibrate with constant acceleration.
[[File:MEPMT.png|200px|thumb|middle|Figure 6 - Momentum vs Time MEP]]
[[File:DYNMT.png|200px|thumb|middle|Figure 7 - Momentum vs Time dynamic]]

=== Reactive and unreactive trajectories ===
{| class="wikitable"
!
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|1
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|Yes
|Initially r1 < r2, and therefore atoms A and B are bonded to each other and atom C is the approaching atom. R1 remains almost constant as r2 begins to decrease. r1 begins to increase until both r1 and r2 are equal at the transition state. From this point r2 remains almost constant whilst r1 increases. Atom A and B have disassociated whilst a new B-C bond has been formed. This bond has kinetic energy hence the vibrations.
|[[File:1.yes.png|200px|thumb|left|Figure 8 - Trajectory set 1]]
|-
|2
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.1</nowiki>
|No
|From the initial positions, r1 oscillates around 74 pm. Distance r2, intially decreases as C approaches B. The trajectory does not go through the TS instead r2 begins to increase again. In this trajectory A and B remain bonded and atom C is repelled from atom B. This is because the momentum p2 is not large enough for a reaction to occur.
|[[File:2.no.png|200px|thumb|left|Figure 9 - Trajectory set 2]]
|-
|3
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.0</nowiki>
|Yes
|From the initial position, r2 decreases with little change in r1. r1 beings to increase until the atoms are in the transition state. r1 increases suggesting the breaking of the A-B bond. r2 continues to oscilate due to internal vibration of the B-C bond.
|[[File:3.yes.png|200px|thumb|left|Figure 10 - Trajectory set 3]]
|-
|4
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.3</nowiki>
|No
|This trajectory does not go through the transition state.Atom C approaches with too much momentum. The A-B bond breaks and the B-C bond forms. B-C vibrates with a large amount of momentum and kinetic energy. B-C vibrates with increasing r2 until passing back through to the reactant channel with the formation of A-B and the disassociation of B-C This is a case of barrier recrossing.Therefore there is no overall reaction.
|[[File:4.no.png|200px|thumb|left|Figure 11 - Trajectory set 4]]
|-
|5
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.5</nowiki>
|Yes
|Atom B and C approach each other with a large amount of momentum causing the A-B bond to break and the formation of the B-C bond.r2 begins to increase agin while the A-B bond reforms. r1 increaes again while r2 decreases with the trajectory returning to the product channel leading to an overall reaction.
|[[File:5.yes.png|200px|thumb|left|Figure 12 - Trajectory set 5]]
|}
''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?''

From this table we can conclude, trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are always reactive. However trajectories higher momenta and therefore higher kinetic energy does not necessarily mean a successful reaction.This is evident by trajectory set 4 having a kinetic energy of 76.5 kJ/mol and resulting in no reaction and set trajectory 3 having only 19.8 kJ/mol and being successful. Increasing the momentum can lead to barrier crossing which if there isn't sufficient energy to cross back to the product channel can lead to overall no reaction.

==== Transition State Theory ====
''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''

Transition state theory assumes quasi-equilibrium between the reactants and the activated transition state complex and only requires details bout the PES. The Born-Oppenheimer approximation is used and quantum tunnelling affects are assumed to be negligible. The energies of the reactants also follow the Boltzman distribution curve.<ref>T. Bligaard, J.K. Nørskov, in Chemical Bonding at Surfaces and Interfaces, 2008</ref> From the table above we can see that quantum tunneling and barrier recrossing can occur, As TST does not account for that the value of rate calculated would be an overestimate compared to the experimental value. However error due to this is only marginal.<ref>PetersBaron , in Reaction Rate Theory and Rare Events Simulations, 2017</ref>

== EXERCISE 2: F - H - H system ==

=== PES inspection ===
[[File:Fdiagram.png|200px|thumb|left|H2 + F]]
F + H2 ----> HF + H (1)

Initial conditions: p1= p2= 0 g.mol-1.pm.fs-1, r1=74.1 pm, r2= 400 pm

HF + H -----> H2 + F (2)

<nowiki> </nowiki>Initial conditions: p1=p2=0 g.mol-1.pm.fs-1 r1=400 pm, r2= 91.7 pm
[[File:FH2.png|200px|thumb|left|Figure 13 - PES of H2 + F]]
[[File:HfH.png|200px|thumb|left|Figure 14 - PES of HF + H]]
From figure we can conclude this reaction is exothermic as the formation of the H-F bond releases energy greater than the energy required to break the H-H bond. In the PES diagram as p2 decreases p1 increases the trajectory falls into the product channel at a lower energy than the products. This is reflective of the bonds being broken and formed. Reaction 2 is endothermic as the H-H bond being formed releases less energy than required to break the H-F bond. From this we can conclude the H-F bond is stronger that the H-H bond.

H-H bond energy = 435.7799 kJ/mol<ref name=":0">Luo, Yu-Ran. ''Comprehensive handbook of chemical bond energies''. CRC press, 2007.</ref>

H-F bond energy = 569.680 kJ/mol<ref name=":0" />

''Locate the approximate position of the transition state.''
[[File:TScontuor.png|200px|thumb|Figure 15 - Transition state of H2 + F]]

Fluorine is highly electronegative and so attacks the hydrogen early on in the reaction this is fitting of the exothermic characteristic of the reaction. Using the Hammond postulate, this early transition state therefore means that the structure of the transition state resembles that of the reactants. Using the liturature value of H-H bond length, r1 was set to 74.1 pm as an initial estimate.<ref>Huber, K.P.; Herzberg, G., Molecular Spectra and Molecular Structure. IV. Constants of Diatomic Molecules,, Van Nostrand Reinhold Co., 1979</ref> Trial and error was performed in order to calculate the r1 and r2 values that give the closest to 0 kJ/mol.pm of force along the axis.

Transition state position - r1= 74.5 pm r2= 181.1 pm

''Report the activation energy for both reactions.''

To calculate the activation energy of both reactions the potential energy at transition state was recorded. The energy of the reactants of the forwards reaction was measured by setting r1 as the bond length of H2, 74.1 pm and r2 being set as 500 pm. This allows us to view the energy of the reactants when they are not reacting with each other. The same was repeated for the backward reaction using the bond length of HF, 91.7 pm as r2.<ref>NIST Diatomic Spectral Database (www.physics.nist.gov/PhysRefData/MolSpec/Diatomic/index.html)</ref>

<nowiki> </nowiki>Transition state energy = -434.0 kJ/mol

Energy of Reactants H2 + F = -435.1 kJ/mol Ea1 = 1.1 kJ/mol

Energy of Reactants HF + H = -560.7 kJ/mol Ea2 = 125.6 kJ/mol

=== Reaction dynamics ===
''Set of initial conditions that results in a reactive trajectory for the F + H2'': r1= 74.5 pm, r2= 130 pm, p1= -1g.mol-1.pm.fs-1 p2= -3 g.mol-1.pm.fs-1 

''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.''

Translational energy is required in order for the reactants to collide with one another with enough energy to react. However large amounts of excess translational energy can lead to the reacts rebounding off of each other.

MRD:M4H1M4

MRD:M4H1M4

2020-05-22T22:01:58Z

Mb7718:

== [[EXERCISE 1: H + H2 system]] ==

=== Dynamics from the transition state region ===
''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 is defined as the first order saddle point on the potential energy surface (PES) diagram. The derivative of potential energy is equal to the force acting on the atoms/molecules. At this point the gradient is equal to 0 and therefore no force is acting on the particles. This allows us to use a constrained one dimensional optimisation to calculate transition state. In order to verify this stationary point as the saddle transition state the Hessian eigenvectors can be used. Only one of the second derivatives must be negative and the other positive so that (fxxfyy - f2xy) < 0. Therefore the determinent of the Hessian matrix less than 0.

This verifies the point as a saddle rather than a local minimum of the potential energy surface.

==== Trajectories from r1 = r2: locating the transition state ====
''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.''

The the H + H2 ---> H2 + H surface is symmetrical, neither H2 or H atom are favoured.Therefore r1 is equal to r2 in the transition state. This restriction allows us to calculate the transition state position through a process of trial and error.

[[File:R1=R2.png|600px|thumb|left|Figure 1]]

[[File:Hessian.png|600px|thumb|Figure 2]]

[[File:Dvst01499529.png|600px|thumb|left|Figure 3]]Transition state position r1= r2 = 90.8 pm

As shown in figure 1, at these positions the forces action along AB and BC are approximately 0 suggesting a stationary point. Figure 3 shows two horizontal lines with gradient 0 which therefore means 0 acceleration and 0 forces acting on the atoms.

Looking at the Hessian matrix we can verify this as a saddle point. To further confirm this, trajectories were calculated with either r1 or r2 being slightly displaced from equilibrium. Figure 4 shows that increasing r2 by 1 pm pushing the reaction to the reactant side figure 5 shows increasing r2 by 1 pm favours the formation of the products.

[[File:Reactantchannel.01499529.png|200px|thumb|Figure 4]]

[[File:Productchannel.01499529.png|200px|thumb|Figure 5]]

==== Trajectories from r1 = rts+δ, r2 = rts ====
''Comment on how the mep and the trajectory you just calculated differ.''

Figure 6 shows a straight line suggesting the velocity and momentum do not change with time. Figure 7, the dynamic trajectory, accounts for the oscillation of the bonds.
[[File:MEPMT.png|200px|thumb|middle|alt text]]
=== Reactive and unreactive trajectories ===
{| class="wikitable"
!
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|1
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|Yes
|Initially r1 < r2, and therefore atoms A and B are bonded to each other and atom C is the approaching atom. R1 remains almost constant as r2 begins to decrease. r1 begins to increase until both r1 and r2 are equal at the transition state. From this point r2 remains almost constant whilst r1 increases. Atom A and B have disassociated whilst a new B-C bond has been formed. This bond has kinetic energy hence the vibrations.
|[[File:1.yes.png|200px|thumb|left|Figure 6]]
|-
|2
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.1</nowiki>
|No
|From the initial positions, r1 oscillates around 74 pm. Distance r2, intially decreases as C approaches B. The trajectory does not go through the TS instead r2 begins to increase again. In this trajectory A and B remain bonded and atom C is repelled from atom B. This is because the momentum p2 is not large enough for a reaction to occur.
|[[File:2.no.png|200px|thumb|left|Figure 7]]
|-
|3
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.0</nowiki>
|Yes
|From the initial position, r2 decreases with little change in r1. r1 beings to increase until the atoms are in the transition state. r1 increases suggesting the breaking of the A-B bond. r2 continues to oscilate due to internal vibration of the B-C bond.
|[[File:3.yes.png|200px|thumb|left|Figure 8]]
|-
|4
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.3</nowiki>
|No
|This trajectory does not go through the transition state.Atom C approaches with too much momentum. The A-B bond breaks and the B-C bond forms. B-C vibrates with a large amount of momentum and kinetic energy. B-C vibrates with increasing r2 until passing back through to the reactant channel with the formation of A-B and the disassociation of B-C This is a case of barrier recrossing.Therefore there is no overall reaction.
|[[File:4.no.png|200px|thumb|left|Figure 9]]
|-
|5
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.5</nowiki>
|Yes
|Atom B and C approach each other with a large amount of momentum causing the A-B bond to break and the formation of the B-C bond.r2 begins to increase agin while the A-B bond reforms. r1 increaes again while r2 decreases with the trajectory returning to the product channel leading to an overall reaction.
|[[File:5.yes.png|200px|thumb|left|Figure 10]]
|}
''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?''

From this table we can conclude, trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are always reactive. However trajectories higher momenta and therefore higher kinetic energy does not necessarily mean a successful reaction.This is evident by trajectory set 4 having a kinetic energy of 76.5 kJ/mol and resulting in no reaction and set trajectory 3 having only 19.8 kJ/mol and being successful. Increasing the momentum can lead to barrier crossing which if there isn't sufficient energy to cross back to the product channel can lead to overall no reaction.

==== Transition State Theory ====
Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

== EXERCISE 2: F - H - H system ==

=== PES inspection ===
F + H2 ----> HF + H (1)

Initial conditions: p1= p2= 0 g.mol-1.pm.fs-1, r1=74.1 pm, r2= 400 pm

HF + H -----> H2 + F (2)

<nowiki> </nowiki>Initial conditions: p1=p2=0 g.mol-1.pm.fs-1 r1=400 pm, r2= 91.7 pm
[[File:FH2.png|200px|thumb|left|Figure 11]]
[[File:HfH.png|200px|thumb|left|Figure 12]]
From figure we can conclude this reaction is exothermic as the formation of the H-F bond releases energy greater than the energy required to break the H-H bond. In the PES diagram as p2 decreases p1 increases the trajectory falls into the product channel at a lower energy than the products. This is reflective of the bonds being broken and formed. Reaction 2 is endothermic as the H-H bond being formed releases less energy than required to break the H-F bond. From this we can conclude the H-F bond is stronger that the H-H bond.

H-H bond energy = 435.7799 kJ/mol<ref name=":0">Luo, Yu-Ran. ''Comprehensive handbook of chemical bond energies''. CRC press, 2007.</ref>

H-F bond energy = 569.680 kJ/mol<ref name=":0" />

''Locate the approximate position of the transition state.''
[[File:TScontuor.png|200px|thumb|Figure 13]]

Fluorine is highly electronegative and so attacks the hydrogen early on in the reaction this is fitting of the exothermic characteristic of the reaction. Using the Hammond postulate, this early transition state therefore means that the structure of the transition state resembles that of the reactants. Using the liturature value of H-H bond length, r1 was set to 74.1 pm as an initial estimate.<ref>Huber, K.P.; Herzberg, G., Molecular Spectra and Molecular Structure. IV. Constants of Diatomic Molecules,, Van Nostrand Reinhold Co., 1979</ref> Trial and error was performed in order to calculate the r1 and r2 values that give the closest to 0 kJ/mol.pm of force along the axis.

Transition state position - r1= 74.5 pm r2= 181.1 pm

''Report the activation energy for both reactions.''

To calculate the activation energy of both reactions the potential energy at transition state was recorded. The energy of the reactants of the forwards reaction was measured by setting r1 as the bond length of H2, 74.1 pm and r2 being set as 500 pm. This allows us to view the energy of the reactants when they are not reacting with each other. The same was repeated for the backward reaction using the bond length of HF, 91.7 pm as r2.<ref>NIST Diatomic Spectral Database (www.physics.nist.gov/PhysRefData/MolSpec/Diatomic/index.html)</ref>

<nowiki> </nowiki>Transition state energy = -434.0 kJ/mol

Energy of Reactants H2 + F = -435.1 kJ/mol Ea1 = 1.1 kJ/mol

Energy of Reactants HF + H = -560.7 kJ/mol Ea2 = 125.6 kJ/mol

=== Reaction dynamics ===
''Set of initial conditions that results in a reactive trajectory for the F + H2'': r1= 74.5 pm, r2= 130 pm, p1= -1g.mol-1.pm.fs-1 p2= -3 g.mol-1.pm.fs-1 

''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.''

Calorimetry can be used to confirm this. For this reaction the total energy is -438.6 kJ/mol.

File:DYNMT.png

2020-05-22T22:01:43Z

Mb7718:

File:MEPMT.png

2020-05-22T22:00:14Z

Mb7718:

MRD:M4H1M4

2020-05-22T21:59:42Z

Mb7718: /* Trajectories from r1 = r2: locating the transition state */

== [[EXERCISE 1: H + H2 system]] ==

=== Dynamics from the transition state region ===
''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 is defined as the first order saddle point on the potential energy surface (PES) diagram. The derivative of potential energy is equal to the force acting on the atoms/molecules. At this point the gradient is equal to 0 and therefore no force is acting on the particles. This allows us to use a constrained one dimensional optimisation to calculate transition state. In order to verify this stationary point as the saddle transition state the Hessian eigenvectors can be used. Only one of the second derivatives must be negative and the other positive so that (fxxfyy - f2xy) < 0. Therefore the determinent of the Hessian matrix less than 0.

This verifies the point as a saddle rather than a local minimum of the potential energy surface.

==== Trajectories from r1 = r2: locating the transition state ====
''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.''

The the H + H2 ---> H2 + H surface is symmetrical, neither H2 or H atom are favoured.Therefore r1 is equal to r2 in the transition state. This restriction allows us to calculate the transition state position through a process of trial and error.

[[File:R1=R2.png|600px|thumb|left|Figure 1]]

[[File:Hessian.png|600px|thumb|Figure 2]]

[[File:Dvst01499529.png|600px|thumb|left|Figure 3]]Transition state position r1= r2 = 90.8 pm

As shown in figure 1, at these positions the forces action along AB and BC are approximately 0 suggesting a stationary point. Figure 3 shows two horizontal lines with gradient 0 which therefore means 0 acceleration and 0 forces acting on the atoms.

Looking at the Hessian matrix we can verify this as a saddle point. To further confirm this, trajectories were calculated with either r1 or r2 being slightly displaced from equilibrium. Figure 4 shows that increasing r2 by 1 pm pushing the reaction to the reactant side figure 5 shows increasing r2 by 1 pm favours the formation of the products.

[[File:Reactantchannel.01499529.png|200px|thumb|Figure 4]]

[[File:Productchannel.01499529.png|200px|thumb|Figure 5]]

==== Trajectories from r1 = rts+δ, r2 = rts ====
''Comment on how the mep and the trajectory you just calculated differ.''

Figure 6 shows a straight line suggesting the velocity and momentum do not change with time. Figure 7, the dynamic trajectory, accounts for the oscillation of the bonds.

=== Reactive and unreactive trajectories ===
{| class="wikitable"
!
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|1
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|Yes
|Initially r1 < r2, and therefore atoms A and B are bonded to each other and atom C is the approaching atom. R1 remains almost constant as r2 begins to decrease. r1 begins to increase until both r1 and r2 are equal at the transition state. From this point r2 remains almost constant whilst r1 increases. Atom A and B have disassociated whilst a new B-C bond has been formed. This bond has kinetic energy hence the vibrations.
|[[File:1.yes.png|200px|thumb|left|Figure 6]]
|-
|2
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.1</nowiki>
|No
|From the initial positions, r1 oscillates around 74 pm. Distance r2, intially decreases as C approaches B. The trajectory does not go through the TS instead r2 begins to increase again. In this trajectory A and B remain bonded and atom C is repelled from atom B. This is because the momentum p2 is not large enough for a reaction to occur.
|[[File:2.no.png|200px|thumb|left|Figure 7]]
|-
|3
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.0</nowiki>
|Yes
|From the initial position, r2 decreases with little change in r1. r1 beings to increase until the atoms are in the transition state. r1 increases suggesting the breaking of the A-B bond. r2 continues to oscilate due to internal vibration of the B-C bond.
|[[File:3.yes.png|200px|thumb|left|Figure 8]]
|-
|4
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.3</nowiki>
|No
|This trajectory does not go through the transition state.Atom C approaches with too much momentum. The A-B bond breaks and the B-C bond forms. B-C vibrates with a large amount of momentum and kinetic energy. B-C vibrates with increasing r2 until passing back through to the reactant channel with the formation of A-B and the disassociation of B-C This is a case of barrier recrossing.Therefore there is no overall reaction.
|[[File:4.no.png|200px|thumb|left|Figure 9]]
|-
|5
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.5</nowiki>
|Yes
|Atom B and C approach each other with a large amount of momentum causing the A-B bond to break and the formation of the B-C bond.r2 begins to increase agin while the A-B bond reforms. r1 increaes again while r2 decreases with the trajectory returning to the product channel leading to an overall reaction.
|[[File:5.yes.png|200px|thumb|left|Figure 10]]
|}
''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?''

From this table we can conclude, trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are always reactive. However trajectories higher momenta and therefore higher kinetic energy does not necessarily mean a successful reaction.This is evident by trajectory set 4 having a kinetic energy of 76.5 kJ/mol and resulting in no reaction and set trajectory 3 having only 19.8 kJ/mol and being successful. Increasing the momentum can lead to barrier crossing which if there isn't sufficient energy to cross back to the product channel can lead to overall no reaction.

==== Transition State Theory ====
Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

== EXERCISE 2: F - H - H system ==

=== PES inspection ===
F + H2 ----> HF + H (1)

Initial conditions: p1= p2= 0 g.mol-1.pm.fs-1, r1=74.1 pm, r2= 400 pm

HF + H -----> H2 + F (2)

<nowiki> </nowiki>Initial conditions: p1=p2=0 g.mol-1.pm.fs-1 r1=400 pm, r2= 91.7 pm
[[File:FH2.png|200px|thumb|left|Figure 11]]
[[File:HfH.png|200px|thumb|left|Figure 12]]
From figure we can conclude this reaction is exothermic as the formation of the H-F bond releases energy greater than the energy required to break the H-H bond. In the PES diagram as p2 decreases p1 increases the trajectory falls into the product channel at a lower energy than the products. This is reflective of the bonds being broken and formed. Reaction 2 is endothermic as the H-H bond being formed releases less energy than required to break the H-F bond. From this we can conclude the H-F bond is stronger that the H-H bond.

H-H bond energy = 435.7799 kJ/mol<ref name=":0">Luo, Yu-Ran. ''Comprehensive handbook of chemical bond energies''. CRC press, 2007.</ref>

H-F bond energy = 569.680 kJ/mol<ref name=":0" />

''Locate the approximate position of the transition state.''
[[File:TScontuor.png|200px|thumb|Figure 13]]

Fluorine is highly electronegative and so attacks the hydrogen early on in the reaction this is fitting of the exothermic characteristic of the reaction. Using the Hammond postulate, this early transition state therefore means that the structure of the transition state resembles that of the reactants. Using the liturature value of H-H bond length, r1 was set to 74.1 pm as an initial estimate.<ref>Huber, K.P.; Herzberg, G., Molecular Spectra and Molecular Structure. IV. Constants of Diatomic Molecules,, Van Nostrand Reinhold Co., 1979</ref> Trial and error was performed in order to calculate the r1 and r2 values that give the closest to 0 kJ/mol.pm of force along the axis.

Transition state position - r1= 74.5 pm r2= 181.1 pm

''Report the activation energy for both reactions.''

To calculate the activation energy of both reactions the potential energy at transition state was recorded. The energy of the reactants of the forwards reaction was measured by setting r1 as the bond length of H2, 74.1 pm and r2 being set as 500 pm. This allows us to view the energy of the reactants when they are not reacting with each other. The same was repeated for the backward reaction using the bond length of HF, 91.7 pm as r2.<ref>NIST Diatomic Spectral Database (www.physics.nist.gov/PhysRefData/MolSpec/Diatomic/index.html)</ref>

<nowiki> </nowiki>Transition state energy = -434.0 kJ/mol

Energy of Reactants H2 + F = -435.1 kJ/mol Ea1 = 1.1 kJ/mol

Energy of Reactants HF + H = -560.7 kJ/mol Ea2 = 125.6 kJ/mol

=== Reaction dynamics ===
''Set of initial conditions that results in a reactive trajectory for the F + H2'': r1= 74.5 pm, r2= 130 pm, p1= -1g.mol-1.pm.fs-1 p2= -3 g.mol-1.pm.fs-1 

''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.''

Calorimetry can be used to confirm this. For this reaction the total energy is -438.6 kJ/mol.

MRD:M4H1M4

2020-05-22T21:56:33Z

Mb7718: /* PES inspection */

== [[EXERCISE 1: H + H2 system]] ==

=== Dynamics from the transition state region ===
''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 is defined as the first order saddle point on the potential energy surface (PES) diagram. The derivative of potential energy is equal to the force acting on the atoms/molecules. At this point the gradient is equal to 0 and therefore no force is acting on the particles. This allows us to use a constrained one dimensional optimisation to calculate transition state. In order to verify this stationary point as the saddle transition state the Hessian eigenvectors can be used. Only one of the second derivatives must be negative and the other positive so that (fxxfyy - f2xy) < 0. Therefore the determinent of the Hessian matrix less than 0.

This verifies the point as a saddle rather than a local minimum of the potential energy surface.

==== Trajectories from r1 = r2: locating the transition state ====
''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.''

The the H + H2 ---> H2 + H surface is symmetrical, neither H2 or H atom are favoured.Therefore r1 is equal to r2 in the transition state. This restriction allows us to calculate the transition state position through a process of trial and error.

[[File:R1=R2.png|600px|thumb|left|Figure 1]]

[[File:Hessian.png|600px|thumb|Figure 2]]

[[File:Dvst01499529.png|600px|thumb|left|Figure 3]]Transition state position r1= r2 = 90.8 pm

As shown in figure 1, at these positions the forces action along AB and BC are approximately 0 suggesting a stationary point. Figure 3 shows two horizontal lines with gradient 0 which therefore means 0 acceleration and 0 forces acting on the atoms.

Looking at the Hessian matrix we can verify this as a saddle point. To further confirm this, trajectories were calculated with either r1 or r2 being slightly displaced from equilibrium. Figure 4 shows that increasing r2 by 1 pm pushing the reaction to the reactant side figure 5 shows increasing r2 by 1 pm favours the formation of the products.

[[File:Reactantchannel.01499529.png|200px|thumb|Figure 4]]

[[File:Productchannel.01499529.png|200px|thumb|Figure 5]]

=== Reactive and unreactive trajectories ===
{| class="wikitable"
!
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|1
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|Yes
|Initially r1 < r2, and therefore atoms A and B are bonded to each other and atom C is the approaching atom. R1 remains almost constant as r2 begins to decrease. r1 begins to increase until both r1 and r2 are equal at the transition state. From this point r2 remains almost constant whilst r1 increases. Atom A and B have disassociated whilst a new B-C bond has been formed. This bond has kinetic energy hence the vibrations.
|[[File:1.yes.png|200px|thumb|left|Figure 6]]
|-
|2
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.1</nowiki>
|No
|From the initial positions, r1 oscillates around 74 pm. Distance r2, intially decreases as C approaches B. The trajectory does not go through the TS instead r2 begins to increase again. In this trajectory A and B remain bonded and atom C is repelled from atom B. This is because the momentum p2 is not large enough for a reaction to occur.
|[[File:2.no.png|200px|thumb|left|Figure 7]]
|-
|3
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.0</nowiki>
|Yes
|From the initial position, r2 decreases with little change in r1. r1 beings to increase until the atoms are in the transition state. r1 increases suggesting the breaking of the A-B bond. r2 continues to oscilate due to internal vibration of the B-C bond.
|[[File:3.yes.png|200px|thumb|left|Figure 8]]
|-
|4
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.3</nowiki>
|No
|This trajectory does not go through the transition state.Atom C approaches with too much momentum. The A-B bond breaks and the B-C bond forms. B-C vibrates with a large amount of momentum and kinetic energy. B-C vibrates with increasing r2 until passing back through to the reactant channel with the formation of A-B and the disassociation of B-C This is a case of barrier recrossing.Therefore there is no overall reaction.
|[[File:4.no.png|200px|thumb|left|Figure 9]]
|-
|5
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.5</nowiki>
|Yes
|Atom B and C approach each other with a large amount of momentum causing the A-B bond to break and the formation of the B-C bond.r2 begins to increase agin while the A-B bond reforms. r1 increaes again while r2 decreases with the trajectory returning to the product channel leading to an overall reaction.
|[[File:5.yes.png|200px|thumb|left|Figure 10]]
|}
''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?''

From this table we can conclude, trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are always reactive. However trajectories higher momenta and therefore higher kinetic energy does not necessarily mean a successful reaction.This is evident by trajectory set 4 having a kinetic energy of 76.5 kJ/mol and resulting in no reaction and set trajectory 3 having only 19.8 kJ/mol and being successful. Increasing the momentum can lead to barrier crossing which if there isn't sufficient energy to cross back to the product channel can lead to overall no reaction.

==== Transition State Theory ====
Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

== EXERCISE 2: F - H - H system ==

=== PES inspection ===
F + H2 ----> HF + H (1)

Initial conditions: p1= p2= 0 g.mol-1.pm.fs-1, r1=74.1 pm, r2= 400 pm

HF + H -----> H2 + F (2)

<nowiki> </nowiki>Initial conditions: p1=p2=0 g.mol-1.pm.fs-1 r1=400 pm, r2= 91.7 pm
[[File:FH2.png|200px|thumb|left|Figure 11]]
[[File:HfH.png|200px|thumb|left|Figure 12]]
From figure we can conclude this reaction is exothermic as the formation of the H-F bond releases energy greater than the energy required to break the H-H bond. In the PES diagram as p2 decreases p1 increases the trajectory falls into the product channel at a lower energy than the products. This is reflective of the bonds being broken and formed. Reaction 2 is endothermic as the H-H bond being formed releases less energy than required to break the H-F bond. From this we can conclude the H-F bond is stronger that the H-H bond.

H-H bond energy = 435.7799 kJ/mol<ref name=":0">Luo, Yu-Ran. ''Comprehensive handbook of chemical bond energies''. CRC press, 2007.</ref>

H-F bond energy = 569.680 kJ/mol<ref name=":0" />

''Locate the approximate position of the transition state.''
[[File:TScontuor.png|200px|thumb|Figure 13]]

Fluorine is highly electronegative and so attacks the hydrogen early on in the reaction this is fitting of the exothermic characteristic of the reaction. Using the Hammond postulate, this early transition state therefore means that the structure of the transition state resembles that of the reactants. Using the liturature value of H-H bond length, r1 was set to 74.1 pm as an initial estimate.<ref>Huber, K.P.; Herzberg, G., Molecular Spectra and Molecular Structure. IV. Constants of Diatomic Molecules,, Van Nostrand Reinhold Co., 1979</ref> Trial and error was performed in order to calculate the r1 and r2 values that give the closest to 0 kJ/mol.pm of force along the axis.

Transition state position - r1= 74.5 pm r2= 181.1 pm

''Report the activation energy for both reactions.''

To calculate the activation energy of both reactions the potential energy at transition state was recorded. The energy of the reactants of the forwards reaction was measured by setting r1 as the bond length of H2, 74.1 pm and r2 being set as 500 pm. This allows us to view the energy of the reactants when they are not reacting with each other. The same was repeated for the backward reaction using the bond length of HF, 91.7 pm as r2.<ref>NIST Diatomic Spectral Database (www.physics.nist.gov/PhysRefData/MolSpec/Diatomic/index.html)</ref>

<nowiki> </nowiki>Transition state energy = -434.0 kJ/mol

Energy of Reactants H2 + F = -435.1 kJ/mol Ea1 = 1.1 kJ/mol

Energy of Reactants HF + H = -560.7 kJ/mol Ea2 = 125.6 kJ/mol

=== Reaction dynamics ===
''Set of initial conditions that results in a reactive trajectory for the F + H2'': r1= 74.5 pm, r2= 130 pm, p1= -1g.mol-1.pm.fs-1 p2= -3 g.mol-1.pm.fs-1 

''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.''

Calorimetry can be used to confirm this. For this reaction the total energy is -438.6 kJ/mol.

MRD:M4H1M4

2020-05-22T21:38:38Z

Mb7718: /* EXERCISE 2: F - H - H system */

== [[EXERCISE 1: H + H2 system]] ==

=== Dynamics from the transition state region ===
''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 is defined as the first order saddle point on the potential energy surface (PES) diagram. The derivative of potential energy is equal to the force acting on the atoms/molecules. At this point the gradient is equal to 0 and therefore no force is acting on the particles. This allows us to use a constrained one dimensional optimisation to calculate transition state. In order to verify this stationary point as the saddle transition state the Hessian eigenvectors can be used. Only one of the second derivatives must be negative and the other positive so that (fxxfyy - f2xy) < 0. Therefore the determinent of the Hessian matrix less than 0.

This verifies the point as a saddle rather than a local minimum of the potential energy surface.

==== Trajectories from r1 = r2: locating the transition state ====
''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.''

The the H + H2 ---> H2 + H surface is symmetrical, neither H2 or H atom are favoured.Therefore r1 is equal to r2 in the transition state. This restriction allows us to calculate the transition state position through a process of trial and error.

[[File:R1=R2.png|600px|thumb|left|Figure 1]]

[[File:Hessian.png|600px|thumb|Figure 2]]

[[File:Dvst01499529.png|600px|thumb|left|Figure 3]]Transition state position r1= r2 = 90.8 pm

As shown in figure 1, at these positions the forces action along AB and BC are approximately 0 suggesting a stationary point. Figure 3 shows two horizontal lines with gradient 0 which therefore means 0 acceleration and 0 forces acting on the atoms.

Looking at the Hessian matrix we can verify this as a saddle point. To further confirm this, trajectories were calculated with either r1 or r2 being slightly displaced from equilibrium.

[[File:Reactantchannel.01499529.png|200px|thumb|Figure 4]]

[[File:Productchannel.01499529.png|200px|thumb|Figure 5]]

=== Reactive and unreactive trajectories ===
{| class="wikitable"
!
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|1
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|Yes
|Initially r1 < r2, and therefore atoms A and B are bonded to each other and atom C is the approaching atom. R1 remains almost constant as r2 begins to decrease. r1 begins to increase until both r1 and r2 are equal at the transition state. From this point r2 remains almost constant whilst r1 increases. Atom A and B have disassociated whilst a new B-C bond has been formed. This bond has kinetic energy hence the vibrations.
|[[File:1.yes.png|200px|thumb|left|Figure 6]]
|-
|2
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.1</nowiki>
|No
|From the initial positions, r1 oscillates around 74 pm. Distance r2, intially decreases as C approaches B. The trajectory does not go through the TS instead r2 begins to increase again. In this trajectory A and B remain bonded and atom C is repelled from atom B. This is because the momentum p2 is not large enough for a reaction to occur.
|[[File:2.no.png|200px|thumb|left|Figure 7]]
|-
|3
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.0</nowiki>
|Yes
|From the initial position, r2 decreases with little change in r1. r1 beings to increase until the atoms are in the transition state. r1 increases suggesting the breaking of the A-B bond. r2 continues to oscilate due to internal vibration of the B-C bond.
|[[File:3.yes.png|200px|thumb|left|Figure 8]]
|-
|4
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.3</nowiki>
|No
|This trajectory does not go through the transition state.Atom C approaches with too much momentum. The A-B bond breaks and the B-C bond forms. B-C vibrates with a large amount of momentum and kinetic energy. B-C vibrates with increasing r2 until passing back through to the reactant channel with the formation of A-B and the disassociation of B-C This is a case of barrier recrossing.Therefore there is no overall reaction.
|[[File:4.no.png|200px|thumb|left|Figure 9]]
|-
|5
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.5</nowiki>
|Yes
|Atom B and C approach each other with a large amount of momentum causing the A-B bond to break and the formation of the B-C bond.r2 begins to increase agin while the A-B bond reforms. r1 increaes again while r2 decreases with the trajectory returning to the product channel leading to an overall reaction.
|[[File:5.yes.png|200px|thumb|left|Figure 10]]
|}
''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?''

From this table we can conclude, trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are always reactive. However trajectories higher momenta and therefore higher kinetic energy does not necessarily mean a successful reaction.This is evident by trajectory set 4 having a kinetic energy of 76.5 kJ/mol and resulting in no reaction and set trajectory 3 having only 19.8 kJ/mol and being successful. Increasing the momentum can lead to barrier crossing which if there isn't sufficient energy to cross back to the product channel can lead to overall no reaction.

==== Transition State Theory ====
Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

== EXERCISE 2: F - H - H system ==

=== PES inspection ===
F + H2 ----> HF + H (1)

Initial conditions: p1= p2= 0 g.mol-1.pm.fs-1, r1=74.1 pm, r2= 400 pm

HF + H -----> H2 + F (2)

<nowiki> </nowiki>Initial conditions: p1=p2=0 g.mol-1.pm.fs-1 r1=400 pm, r2= 91.7 pm
[[File:FH2.png|200px|thumb|left|Figure 11]]
[[File:HfH.png|200px|thumb|left|Figure 12]]
From figure we can conclude this reaction is exothermic as the formation of the H-F bond releases energy greater than the energy required to break the H-H bond. In the PES diagram as p2 decreases p1 increases the trajectory falls into the product channel at a lower energy than the products. This is reflective of the bonds being broken and formed. Reaction 2 is endothermic as the H-H bond being formed releases less energy than required to break the H-F bond. From this we can conclude the H-F bond is stronger that the H-H bond.

H-H bond energy = 435.7799 kJ/mol<ref name=":0">Luo, Yu-Ran. ''Comprehensive handbook of chemical bond energies''. CRC press, 2007.</ref>

H-F bond energy = 569.680 kJ/mol<ref name=":0" />

''Locate the approximate position of the transition state.''
[[File:TScontuor.png|200px|thumb|Figure 13]]

Fluorine is highly electronegative and so attacks the hydrogen early on in the reaction this is fitting of the exothermic characteristic of the reaction. Using the Hammond postulate, this early transition state therefore means that the structure of the transition state resembles that of the reactants. Using the liturature value of H-H bond length, r1 was set to 74.1 pm as an initial estimate.<ref>Huber, K.P.; Herzberg, G., Molecular Spectra and Molecular Structure. IV. Constants of Diatomic Molecules,, Van Nostrand Reinhold Co., 1979</ref> Trial and error was performed in order to calculate the r1 and r2 values that give the closest to 0 kJ/mol.pm of force along the axis.

Transition state position - r1= 74.5 pm r2= 181.1 pm

''Report the activation energy for both reactions.''

To calculate the activation energy of both reactions the potential energy at transition state was recorded. The energy of the reactants of the forwards reaction was measured by setting r1 as the bond length of H2, 74.1 pm and r2 being set as 500 pm. This allows us to view the energy of the reactants when they are not reacting with each other. The same was repeated for the backward reaction using the bond length of HF, 91.7 pm as r2.<ref>NIST Diatomic Spectral Database (www.physics.nist.gov/PhysRefData/MolSpec/Diatomic/index.html)</ref>

<nowiki> </nowiki>Transition state energy = -434.0 kJ/mol

Energy of Reactants H2 + F = -435.1 kJ/mol Ea1 = 1.1 kJ/mol

Energy of Reactants HF + H = -560.7 kJ/mol Ea2 = 125.6 kJ/mol

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.

MRD:M4H1M4

2020-05-22T21:36:57Z

Mb7718: /* EXERCISE 2: F - H - H system */

== [[EXERCISE 1: H + H2 system]] ==

=== Dynamics from the transition state region ===
''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 is defined as the first order saddle point on the potential energy surface (PES) diagram. The derivative of potential energy is equal to the force acting on the atoms/molecules. At this point the gradient is equal to 0 and therefore no force is acting on the particles. This allows us to use a constrained one dimensional optimisation to calculate transition state. In order to verify this stationary point as the saddle transition state the Hessian eigenvectors can be used. Only one of the second derivatives must be negative and the other positive so that (fxxfyy - f2xy) < 0. Therefore the determinent of the Hessian matrix less than 0.

This verifies the point as a saddle rather than a local minimum of the potential energy surface.

==== Trajectories from r1 = r2: locating the transition state ====
''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.''

The the H + H2 ---> H2 + H surface is symmetrical, neither H2 or H atom are favoured.Therefore r1 is equal to r2 in the transition state. This restriction allows us to calculate the transition state position through a process of trial and error.

[[File:R1=R2.png|600px|thumb|left|Figure 1]]

[[File:Hessian.png|600px|thumb|Figure 2]]

[[File:Dvst01499529.png|600px|thumb|left|Figure 3]]Transition state position r1= r2 = 90.8 pm

As shown in figure 1, at these positions the forces action along AB and BC are approximately 0 suggesting a stationary point. Figure 3 shows two horizontal lines with gradient 0 which therefore means 0 acceleration and 0 forces acting on the atoms.

Looking at the Hessian matrix we can verify this as a saddle point. To further confirm this, trajectories were calculated with either r1 or r2 being slightly displaced from equilibrium.

[[File:Reactantchannel.01499529.png|200px|thumb|Figure 4]]

[[File:Productchannel.01499529.png|200px|thumb|Figure 5]]

=== Reactive and unreactive trajectories ===
{| class="wikitable"
!
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|1
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|Yes
|Initially r1 < r2, and therefore atoms A and B are bonded to each other and atom C is the approaching atom. R1 remains almost constant as r2 begins to decrease. r1 begins to increase until both r1 and r2 are equal at the transition state. From this point r2 remains almost constant whilst r1 increases. Atom A and B have disassociated whilst a new B-C bond has been formed. This bond has kinetic energy hence the vibrations.
|[[File:1.yes.png|200px|thumb|left|Figure 6]]
|-
|2
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.1</nowiki>
|No
|From the initial positions, r1 oscillates around 74 pm. Distance r2, intially decreases as C approaches B. The trajectory does not go through the TS instead r2 begins to increase again. In this trajectory A and B remain bonded and atom C is repelled from atom B. This is because the momentum p2 is not large enough for a reaction to occur.
|[[File:2.no.png|200px|thumb|left|Figure 7]]
|-
|3
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.0</nowiki>
|Yes
|From the initial position, r2 decreases with little change in r1. r1 beings to increase until the atoms are in the transition state. r1 increases suggesting the breaking of the A-B bond. r2 continues to oscilate due to internal vibration of the B-C bond.
|[[File:3.yes.png|200px|thumb|left|Figure 8]]
|-
|4
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.3</nowiki>
|No
|This trajectory does not go through the transition state.Atom C approaches with too much momentum. The A-B bond breaks and the B-C bond forms. B-C vibrates with a large amount of momentum and kinetic energy. B-C vibrates with increasing r2 until passing back through to the reactant channel with the formation of A-B and the disassociation of B-C This is a case of barrier recrossing.Therefore there is no overall reaction.
|[[File:4.no.png|200px|thumb|left|Figure 9]]
|-
|5
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.5</nowiki>
|Yes
|Atom B and C approach each other with a large amount of momentum causing the A-B bond to break and the formation of the B-C bond.r2 begins to increase agin while the A-B bond reforms. r1 increaes again while r2 decreases with the trajectory returning to the product channel leading to an overall reaction.
|[[File:5.yes.png|200px|thumb|left|Figure 10]]
|}
''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?''

From this table we can conclude, trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are always reactive. However trajectories higher momenta and therefore higher kinetic energy does not necessarily mean a successful reaction.This is evident by trajectory set 4 having a kinetic energy of 76.5 kJ/mol and resulting in no reaction and set trajectory 3 having only 19.8 kJ/mol and being successful. Increasing the momentum can lead to barrier crossing which if there isn't sufficient energy to cross back to the product channel can lead to overall no reaction.

==== Transition State Theory ====
Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

== EXERCISE 2: F - H - H system ==

=== PES inspection ===
F + H2 ----> HF + H (1)

Initial conditions: p1= p2= 0 g.mol-1.pm.fs-1, r1=74.1 pm, r2= 400 pm

HF + H -----> H2 + F (2)

<nowiki> </nowiki>Initial conditions: p1=p2=0 g.mol-1.pm.fs-1 r1=400 pm, r2= 91.7 pm
[[File:FH2.png|200px|thumb|left|Figure 11]]
[[File:HfH.png|200px|thumb|left|Figure 12]]
From figure we can conclude this reaction is exothermic as the formation of the H-F bond releases energy greater than the energy required to break the H-H bond. In the PES diagram as p2 decreases p1 increases the trajectory falls into the product channel at a lower energy than the products. This is reflective of the bonds being broken and formed. Reaction 2 is endothermic as the H-H bond being formed releases less energy than required to break the H-F bond. From this we can conclude the H-F bond is stronger that the H-H bond.

H-H bond energy = 435.7799 kJ/mol<ref name=":0">Luo, Yu-Ran. ''Comprehensive handbook of chemical bond energies''. CRC press, 2007.</ref>

H-F bond energy = 569.680 kJ/mol<ref name=":0" />

''Locate the approximate position of the transition state.''
[[File:TScontuor.png|200px|thumb|right|alt text]]

Fluorine is highly electronegative and so attacks the hydrogen early on in the reaction this is fitting of the exothermic characteristic of the reaction. Using the Hammond postulate, this early transition state therefore means that the structure of the transition state resembles that of the reactants. Using the liturature value of H-H bond length, r1 was set to 74.1 pm as an initial estimate.<ref>Huber, K.P.; Herzberg, G., Molecular Spectra and Molecular Structure. IV. Constants of Diatomic Molecules,, Van Nostrand Reinhold Co., 1979</ref> Trial and error was performed in order to calculate the r1 and r2 values that give the closest to 0 kJ/mol.pm of force along the axis.

Transition state position - r1= 74.5 pm r2= 181.1 pm

''Report the activation energy for both reactions.''

To calculate the activation energy of both reactions the potential energy at transition state was recorded. The energy of the reactants of the forwards reaction was measured by setting r1 as the bond length of H2, 74.1 pm and r2 being set as 500 pm. This allows us to view the energy of the reactants when they are not reacting with each other. The same was repeated for the backward reaction using the bond length of HF, 91.7 pm as r2.<ref>NIST Diatomic Spectral Database (www.physics.nist.gov/PhysRefData/MolSpec/Diatomic/index.html)</ref>

<nowiki> </nowiki>Transition state energy = -434.0 kJ/mol

Energy of Reactants H2 + F = -435.1 kJ/mol Ea1 = 1.1 kJ/mol

Energy of Reactants HF + H = -560.7 kJ/mol Ea2 = 125.6 kJ/mol

File:TScontuor.png

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