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MRD:lhl17

2020-05-22T21:08:38Z

Lhl17: /* Question 10 */

'''<nowiki/>'''

== Molecular Reaction Dynamics: Applications to Triatomic Systems ==
Louise Li

Group C, deadline: 22/05/2020

== Introduction ==
The program used for this computational lab calculates potential energy and individual trajectories using the London-Eyring-Polanyi-Sato (LEPS) potential. By analysis of these reaction paths and comparing with the prediction of Transition State Theory, we can learn more about reactions and their molecular dynamics.

== Exercise 1: H + H2 system ==

=== 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?''

On a potential energy surface (PES) diagram, the transition state is mathematically defined as the point at which the following is true:

<math> { \partial V(r)\over \partial r} = 0 </math>

While a one-dimensional potential energy curve (PEC) could simply be approximated by quadratic curves, the PES is multi-dimensional function (with the energy on the z axis and bond distances on the x/y axes). This means the TS, the maximum point on the minimum reactive trajectory, can be visually identified as the "saddle point" - a maximum point in one direction and minimum in the other.

To distinguish the TS from other local minima, both the second derivative and the determinant of the Hessian Matrix at the TS should be smaller than 0. In addition, the eigenvalues of the Hessian Matrix should have opposite positive/negative signs.<ref>T. Fueno, ''Transition State: A Theoretical Approach'', CRC Press, Japan, 1999, p.31.</ref>

=== Question 2 ===
''Report your best estimate of the transition state position (<nowiki>'''</nowiki>rts<nowiki>'''</nowiki>) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.''

The TS position for the H + H2 system was found at rAB = rBC = 90.77 pm with pAB = pBC = 0 g.mol-1.pm.fm-1.
{| class="wikitable"
!Internuclear Distances vs Time
!Contour Plot
!Force analysis
|-
|[[File:Lhl17 TS1.png|none|thumb|170x170px]]With the initial conditions set at the TS with no momentum, the atoms will remain in place and not move at all. Therefore, intermolecular distances remain constant which results in horizontal lines in the graph above.
|[[File:Lhl17 TScontour.png|none|thumb|227x227px]]The red cross indicates the initial conditions input, and it is clear that no bond movement is shown here. The transition state is totally symmetric (neither endo nor exothermic, and neither early nor late according to Hammond's postulate).
|[[File:Lhl17 forces.png|none|thumb|170x170px]]The exact point was confirmed by force analysis (where forces on the molecules are minimised).
|}

=== Question 3 ===
''Comment on how the mep and the trajectory you just calculated differ.''

The minimum energy path (''mep'') is a special trajectory that corresponds to infinitely slow motion (i.e. the momenta/velocities are always reset to zero in each time step)<ref><nowiki>https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD (accessed May 2020).</nowiki></ref>. The MEP is useful for finding an equilibrium geometry which can then be used to calculate energies, although the dynamics plot is more realistic and is a good model for gas phase collision.

When comparing reaction pathways with initial conditions rAB = 90.77 pm, rBC = 91.77 pm and pAB = pBC = 0 g.mol-1.pm.fm-1, we obtain the following:
{| class="wikitable"
!Dynamic Calculation
!MEP Calculation (5000 steps)
!

Comments
|-
|[[File:Lhl17 Ex1 Dynamic displaced.png|none|thumb|Contour plot of reaction path]]
|[[File:Lhl17 Ex1 MEP displaced.png|none|thumb|Contour plot of reaction path]]
|There is no vibrational energy for the MEP plot (unlike in the dynamic plot) as the velocity and momentum are reset to 0 at each step iteration, so that kinetic energy always remains at 0. Thus, in this "perfect" theoretical pathway, there is no vibrational energy in the system so no oscillations are visualised.
|-
|[[File:Lhl17 Ex1 Dynamic displaced p.png|thumb|Momentum vs Time plot of reaction path]]
|[[File:Lhl17 Ex1 MEP displaced p.png|thumb|Momentum vs Time plot of reaction path]]
|In the dynamics calculation, we can see that the oscillation of the atoms has been included in the calculation, and increases over time.
|-
|[[File:Lhl17 Ex1 dynamics displaced d.png|thumb|Distance vs Time plot of reaction path]]
|[[File:Lhl17 Ex1 MEP displaced d.png|thumb|Distance vs Time plot of reaction path]]
|In both cases, the atoms are moving away from each other. Note that in the dynamics velocities at which they are moving remain constant.
|}
Reversing the initial/final positions and momenta reverses the direction of the reaction pathway observed. By slightly displacing the initial conditions from the TS in either direction, we can encourage product or reactant formation and observe how the trajectory evolves just away from the saddle. Like a ball placed onto a sloped surface, the system will gain momentum as soon as the simulation starts as it is not at a stable point on the PES and will thus fall to an energy minimum (Ep to Ek).

=== Question 4 ===
''Complete the table below 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?''

{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|[[File:Lhl17_Table1a.png|thumb]]Reaction path passes through the TS (at around 21 fs) and forms products. The vibrational oscillations in the product channel confirm formation of the the product.
|[[File:Lhl17 Table1.png|none|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|[[File:Lhl17 Table2a.png|thumb]]Reaction path unable to get over maximum at TS and form products. There is no successful collision and instead, the system rolls back down the potential to reactants.
|[[File:Lhl17 Table2.png|none|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|[[File:Lhl17 Table3a.png|thumb]]Momentum of incoming H atom increased compared to first example, and as before, the reaction path passes over the TS (at around 21 fs) to the product channel. There is more energy in the system overall and so stronger oscillations are observed.
|[[File:Lhl17 Table3.png|none|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|[[File:Lhl17 Table4a.png|thumb]]Initial momenta large enough for reaction path to reach the TS (at around 8 fs), but excess energy causes the product to cross the activation barrier twice. Even though the product bond does form, the system eventually returns to reactants (with significantly higher vibronic energy). This is a case of barrier recrossing.
|[[File:Lhl17 Table4.png|none|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|[[File:Lhl17 Table5a.png|thumb]]Similar to the previous example, but with the momentum of the incoming H atom increased. The system fluctuates twice between reactant and product bond formation (with TS at 9, 15 and 26 fs).
|[[File:Lhl17 Table5.png|none|thumb]]
|}
The higher kinetic energy, hence momenta does not necessarily guarantee the reaction would proceed. Higher kinetic energy can mean that the reactants can go over the activation energy barrier, but with energy being too high, newly formed products may overcome the TS barrier again in the reversed direction and reform reactants, in which case it is unreactive overall.

=== Question 5 ===
''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''

The Transition State Theory assumes that only the reactants are in constant equilibrium with the transition state structure (quasi-equilibrium), and that the energy of the particles follow a Boltzmann distribution in a thermally equibrilated reactant pool.

In addition, it is assumed that once the reactants have trajectories with a kinetic energy along the reaction coordinate greater than the activation energy, that the products will definitely form, ignoring the possibility of barrier recrossing caused by excess vibrational energy. As seen in row 4, the starting materials may be reformed instead leading to an overall unreactive path.

Furthermore, the theory fails for some reactions at low temperatures due to quantum tunneling, where the product can still be formed even if the TS is not reached (although the effect of this is negligible compared to barrier recrossing). The nuclei are also treated as classical point masses and do not have wavefunctions, which is another flaw of TST.

In the table above, only 5 possible individual trajectories of the H + H2 reaction were investigated. The true reaction rate can be thought of as a "weighted average" of all such individual paths and so, with TST predicting unreactive trajectories to in fact be reactive, it can be concluded that the theory overestimates the rate of reaction when compared with experimental values.

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

=== 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?''[[File:Lhl17 Ex2 TS surface.png|thumb|PES of F + H2, showing early transition state]]

The enthalpy change of a reaction can be calculated using the following equation:

ΔHrxn=ΣΔHbond breaking'''-'''ΣΔHbond forming

Using literature values for bond dissociation energies of H-H (436 kJ mol-1) and H-F (569 kJ mol-1)<ref>https://labs.chem.ucsb.edu/zakarian/armen/11---bonddissociationenergy.pdf (accessed May 2020)</ref> shows that the enthalpy released from breaking the H-F bond is greater than the energy required to break H-H bonds. Therefore, the reaction F + H2 → HF + H is energetically favorable and exothermic overall. This is consistent with the very small atomic radius of the F atom, and therefore stronger attraction and a stronger bond overall. The PES shows that the reactants are higher in energy than the products (where AB distance, representing the H-F bond), indicating an exothermic reaction.

The reverse reaction, H + HF is therefore endothermic, as extra external energy is required in order to break the H-F bond.

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

According to Hammond's postulate the structure of the transition state will resemble the structure of the nearest stable species (either products or reactants). Taking the first reaction F + H2 → HF + H, the transition state will resemble the reactants since it is an exothermic reaction. For HF + H → F + H2, the opposite is true: the transition state will resemble the products for the endothermic reaction.

After testing several initial distances by trial and error, and setting the initial momentum of both reactants to be 0, it was found that the transition state has the following internuclear distances: rFH = 180.5 pm and rHH = 74.5 pm (Etotal = -433.981 kJ mol-1).
{| class="wikitable"
!Internuclear distances
!Contour
!Forces
|-
|[[File:Lhl17 Ex2 TS internuclear.png|none|thumb]]To find the TS, both momenta were set to zero and the distances were optimised until minimal oscillation was reached in the internuclear distances vs time graph
|[[File:Lhl17 Ex2 TS contour.png|none|thumb]]Unlike exercise 1, the system is clearly asymmetric.
|[[File:Lhl17 forces1.png|none|thumb|272x272px]]The forces at the transition state are very close to 0 which means that the system will not move.
|}
Note that for these triatomics there are two bond lengths we can vary and thus at a saddle point there is one direction that decreases most in energy, and another that increases the most. For the transition structure of a polyatomic molecule with 3N-6 vibrational modes, there would exist one direction down and the remaining 3N-7 vibrational modes acting as sinks for the excess energy.<ref>R. D. Levine, ''Molecular Reaction Dynamics'', Cambridge University Press, Cambridge, 2005, p. 216.</ref>

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

The Ea was found by very slightly altering the TS distance for each of AB, then BC, and taking the energy difference between the TS and the system with maximum displacement between the reactants.
{| class="wikitable"
!F + H2
!HF + H
|-
|[[File:Lhl17 Ex2 Ea1.png|none|thumb]]Ea: 1.26 kJ.mol-1.
|[[File:Lhl17 Ex2 reactants 560.png|none|thumb]]Ea: 127 kJ.mol-1.
|}
Knowing that the F + H2 reaction was exothermic and had an early TS provided useful guidance when locating the exact point, since the activation energy was so small.

=== Question 9 ===
''Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Momenta vs Time”.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.''.

To investigate a reactive trajectory for the reaction F + H2 → HF + H, the following initial conditions were chosen:
{| class="wikitable"

! rFH / pm
! rHH / pm
! pFH/gmol-1pmfs-1
! pHH/gmol-1pmfs-1

|-
| 200
| 74
| -1.0
| 0
|}
This yielded the follow plots:
{| class="wikitable"
!Contour plot
!Momenta vs time
!Surface Plot
!Energy plot
|-
|[[File:Lhl17 Ex2 energyrelease contour.png|none|thumb|200x200px]]The contour plot shows strong oscillations after the system passes through the transition state, which represents the system passes its excess energy into the vibration of HF.
|[[File:Lhl17 Ex2 energyrelease p.png|none|thumb|267x267px]]
|[[File:Lhl17 Ex2 energyrelease s.png|none|thumb|200x200px]]
|[[File:Lhl17 Ex2 release e.png|none|thumb|267x267px]]
|}
The conservation of energy suggests that total energy is constant through the course of reaction. From the exothermic energy vs time graph below, the loss in potential energy from the breaking of H2 bonds is mirrored by the increase in Ek, which can be either translational, rotational or vibrational.

Calorimetry can be used to measure the transfer of both Etrans and Evib (Erot is ignored here) from the products to the walls of the calorimeter in the form of heat and the heat transferred can then be measured via a temperature change. We would expect an increase in temperature for exothermic reactions and a decrease for endothermic ones, and the total energy change can be calculated using ΔQ=cΔΤ.

Meanwhile, FTIR spectroscopy can also be used to quantify Evib. Before the collision, the atoms lie in the ground state - however, during the collision, they are excited to occupy higher vibrational modes as Ep is converted to Ek. Overtone and peak wavenumbers in the IR spectrum can be used to identify the changes in Evib of the IR active product and over time, the vibrational states eventually relax back to the ground state.
=== 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 Empirical rules state that vibrational energy is more efficient at activating a late transition state (present in endothermic reactions, according to Hammond's postulate), than translational energy.

On the other hand, translational energy is more efficient in activating an early transition state in an exothermic reaction than vibrational energy as the trajectory can simply "fall" into the lower energy region. Therefore for a successful exothermic reaction, the molecules need an excess of translational (kinetic) energy compared to vibrational (oscillatory) energy.

To investigate this, a calculation starting on the side of the reactants of F + H2, at the bottom of the well with rHF = 200 pm, rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1 was set up. Several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 were explored:
{| class="wikitable"
!pHH/gmol-1pmfs-1
!Etot/ kJ.mol-1
!Reactive trajectory
!Contour plot
|-
|<nowiki>-6.1</nowiki>
|<nowiki>-402.5</nowiki>
|No
|[[File:Lhl17 -6.1.png|none|thumb]]
|-
|<nowiki>-5</nowiki>
|<nowiki>-413.6</nowiki>
|Yes
|[[File:Lhl17 -5.png|none|thumb]]
|-
|<nowiki>-3</nowiki>
|<nowiki>-427.6</nowiki>
|No
|[[File:Lhl17 -3.png|none|thumb]]
|-
|0
|<nowiki>-433.6</nowiki>
|Yes
|[[File:Lhl17 0.png|none|thumb]]
|-
|3
|<nowiki>-421.6</nowiki>
|Yes
|[[File:Lhl17 3.png|thumb]]
|-
|5
|<nowiki>-403.6</nowiki>
|No
|[[File:Lhl17 5.png|none|thumb]]
|-
|6.1
|<nowiki>-390.3</nowiki>
|No
|[[File:Lhl17 6.1.png|none|thumb]]
|}
For the same initial position, the momentum pFH (represents the translational energy) was increased slightly to -1.6 g.mol-1.pm.fs-1,and pHH (representing vibrational energy) was decreased to 0.2 g.mol-1.pm.fs-1. Although the overall energy of this simulation (-432.4 kJ mol-1) was similar to a successful trajectory, such as row 4 in the table above, the reaction did not go to completion. This confirms that the form of energy in the system is important to determine whether the reaction goes to completion.

Barrier recrossing means products are able to cross back and become reactants again, as shown by the contour plots, and results in unsuccessful reactions:
{| class="wikitable"
!Contour
!Momentum
!Energy
|-
|[[File:Lhl17 Ex2 reduceE contour.png|none|thumb]]
|[[File:Lhl17 Ex2 reduceE m.png|none|thumb]]
|[[File:Lhl17 Ex2 reduceE E.png|none|thumb]]
|}
[[File:Lhl17 q10.png|none|thumb|A reactive trajectory for the reverse reaction, H + HF with initial conditions rHF = 92 pm, rHH = 200 pm, pHF = 15.3 g.mol-1.pm.fm-1 and pHH -1 g.mol-1.pm.fm-1]]
The above trajectory was found by increasing the vibrational energy of the HF bond while maintaining a low pHH.

== References ==

MRD:lhl17

2020-05-22T20:55:16Z

Lhl17: /* Question 10 */

'''<nowiki/>'''

== Molecular Reaction Dynamics: Applications to Triatomic Systems ==
Louise Li

Group C, deadline: 22/05/2020

== Introduction ==
The program used for this computational lab calculates potential energy and individual trajectories using the London-Eyring-Polanyi-Sato (LEPS) potential. By analysis of these reaction paths and comparing with the prediction of Transition State Theory, we can learn more about reactions and their molecular dynamics.

== Exercise 1: H + H2 system ==

=== 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?''

On a potential energy surface (PES) diagram, the transition state is mathematically defined as the point at which the following is true:

<math> { \partial V(r)\over \partial r} = 0 </math>

While a one-dimensional potential energy curve (PEC) could simply be approximated by quadratic curves, the PES is multi-dimensional function (with the energy on the z axis and bond distances on the x/y axes). This means the TS, the maximum point on the minimum reactive trajectory, can be visually identified as the "saddle point" - a maximum point in one direction and minimum in the other.

To distinguish the TS from other local minima, both the second derivative and the determinant of the Hessian Matrix at the TS should be smaller than 0. In addition, the eigenvalues of the Hessian Matrix should have opposite positive/negative signs.<ref>T. Fueno, ''Transition State: A Theoretical Approach'', CRC Press, Japan, 1999, p.31.</ref>

=== Question 2 ===
''Report your best estimate of the transition state position (<nowiki>'''</nowiki>rts<nowiki>'''</nowiki>) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.''

The TS position for the H + H2 system was found at rAB = rBC = 90.77 pm with pAB = pBC = 0 g.mol-1.pm.fm-1.
{| class="wikitable"
!Internuclear Distances vs Time
!Contour Plot
!Force analysis
|-
|[[File:Lhl17 TS1.png|none|thumb|170x170px]]With the initial conditions set at the TS with no momentum, the atoms will remain in place and not move at all. Therefore, intermolecular distances remain constant which results in horizontal lines in the graph above.
|[[File:Lhl17 TScontour.png|none|thumb|227x227px]]The red cross indicates the initial conditions input, and it is clear that no bond movement is shown here. The transition state is totally symmetric (neither endo nor exothermic, and neither early nor late according to Hammond's postulate).
|[[File:Lhl17 forces.png|none|thumb|170x170px]]The exact point was confirmed by force analysis (where forces on the molecules are minimised).
|}

=== Question 3 ===
''Comment on how the mep and the trajectory you just calculated differ.''

The minimum energy path (''mep'') is a special trajectory that corresponds to infinitely slow motion (i.e. the momenta/velocities are always reset to zero in each time step)<ref><nowiki>https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD (accessed May 2020).</nowiki></ref>. The MEP is useful for finding an equilibrium geometry which can then be used to calculate energies, although the dynamics plot is more realistic and is a good model for gas phase collision.

When comparing reaction pathways with initial conditions rAB = 90.77 pm, rBC = 91.77 pm and pAB = pBC = 0 g.mol-1.pm.fm-1, we obtain the following:
{| class="wikitable"
!Dynamic Calculation
!MEP Calculation (5000 steps)
!

Comments
|-
|[[File:Lhl17 Ex1 Dynamic displaced.png|none|thumb|Contour plot of reaction path]]
|[[File:Lhl17 Ex1 MEP displaced.png|none|thumb|Contour plot of reaction path]]
|There is no vibrational energy for the MEP plot (unlike in the dynamic plot) as the velocity and momentum are reset to 0 at each step iteration, so that kinetic energy always remains at 0. Thus, in this "perfect" theoretical pathway, there is no vibrational energy in the system so no oscillations are visualised.
|-
|[[File:Lhl17 Ex1 Dynamic displaced p.png|thumb|Momentum vs Time plot of reaction path]]
|[[File:Lhl17 Ex1 MEP displaced p.png|thumb|Momentum vs Time plot of reaction path]]
|In the dynamics calculation, we can see that the oscillation of the atoms has been included in the calculation, and increases over time.
|-
|[[File:Lhl17 Ex1 dynamics displaced d.png|thumb|Distance vs Time plot of reaction path]]
|[[File:Lhl17 Ex1 MEP displaced d.png|thumb|Distance vs Time plot of reaction path]]
|In both cases, the atoms are moving away from each other. Note that in the dynamics velocities at which they are moving remain constant.
|}
Reversing the initial/final positions and momenta reverses the direction of the reaction pathway observed. By slightly displacing the initial conditions from the TS in either direction, we can encourage product or reactant formation and observe how the trajectory evolves just away from the saddle. Like a ball placed onto a sloped surface, the system will gain momentum as soon as the simulation starts as it is not at a stable point on the PES and will thus fall to an energy minimum (Ep to Ek).

=== Question 4 ===
''Complete the table below 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?''

{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|[[File:Lhl17_Table1a.png|thumb]]Reaction path passes through the TS (at around 21 fs) and forms products. The vibrational oscillations in the product channel confirm formation of the the product.
|[[File:Lhl17 Table1.png|none|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|[[File:Lhl17 Table2a.png|thumb]]Reaction path unable to get over maximum at TS and form products. There is no successful collision and instead, the system rolls back down the potential to reactants.
|[[File:Lhl17 Table2.png|none|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|[[File:Lhl17 Table3a.png|thumb]]Momentum of incoming H atom increased compared to first example, and as before, the reaction path passes over the TS (at around 21 fs) to the product channel. There is more energy in the system overall and so stronger oscillations are observed.
|[[File:Lhl17 Table3.png|none|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|[[File:Lhl17 Table4a.png|thumb]]Initial momenta large enough for reaction path to reach the TS (at around 8 fs), but excess energy causes the product to cross the activation barrier twice. Even though the product bond does form, the system eventually returns to reactants (with significantly higher vibronic energy). This is a case of barrier recrossing.
|[[File:Lhl17 Table4.png|none|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|[[File:Lhl17 Table5a.png|thumb]]Similar to the previous example, but with the momentum of the incoming H atom increased. The system fluctuates twice between reactant and product bond formation (with TS at 9, 15 and 26 fs).
|[[File:Lhl17 Table5.png|none|thumb]]
|}
The higher kinetic energy, hence momenta does not necessarily guarantee the reaction would proceed. Higher kinetic energy can mean that the reactants can go over the activation energy barrier, but with energy being too high, newly formed products may overcome the TS barrier again in the reversed direction and reform reactants, in which case it is unreactive overall.

=== Question 5 ===
''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''

The Transition State Theory assumes that only the reactants are in constant equilibrium with the transition state structure (quasi-equilibrium), and that the energy of the particles follow a Boltzmann distribution in a thermally equibrilated reactant pool.

In addition, it is assumed that once the reactants have trajectories with a kinetic energy along the reaction coordinate greater than the activation energy, that the products will definitely form, ignoring the possibility of barrier recrossing caused by excess vibrational energy. As seen in row 4, the starting materials may be reformed instead leading to an overall unreactive path.

Furthermore, the theory fails for some reactions at low temperatures due to quantum tunneling, where the product can still be formed even if the TS is not reached (although the effect of this is negligible compared to barrier recrossing). The nuclei are also treated as classical point masses and do not have wavefunctions, which is another flaw of TST.

In the table above, only 5 possible individual trajectories of the H + H2 reaction were investigated. The true reaction rate can be thought of as a "weighted average" of all such individual paths and so, with TST predicting unreactive trajectories to in fact be reactive, it can be concluded that the theory overestimates the rate of reaction when compared with experimental values.

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

=== 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?''[[File:Lhl17 Ex2 TS surface.png|thumb|PES of F + H2, showing early transition state]]

The enthalpy change of a reaction can be calculated using the following equation:

ΔHrxn=ΣΔHbond breaking'''-'''ΣΔHbond forming

Using literature values for bond dissociation energies of H-H (436 kJ mol-1) and H-F (569 kJ mol-1)<ref>https://labs.chem.ucsb.edu/zakarian/armen/11---bonddissociationenergy.pdf (accessed May 2020)</ref> shows that the enthalpy released from breaking the H-F bond is greater than the energy required to break H-H bonds. Therefore, the reaction F + H2 → HF + H is energetically favorable and exothermic overall. This is consistent with the very small atomic radius of the F atom, and therefore stronger attraction and a stronger bond overall. The PES shows that the reactants are higher in energy than the products (where AB distance, representing the H-F bond), indicating an exothermic reaction.

The reverse reaction, H + HF is therefore endothermic, as extra external energy is required in order to break the H-F bond.

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

According to Hammond's postulate the structure of the transition state will resemble the structure of the nearest stable species (either products or reactants). Taking the first reaction F + H2 → HF + H, the transition state will resemble the reactants since it is an exothermic reaction. For HF + H → F + H2, the opposite is true: the transition state will resemble the products for the endothermic reaction.

After testing several initial distances by trial and error, and setting the initial momentum of both reactants to be 0, it was found that the transition state has the following internuclear distances: rFH = 180.5 pm and rHH = 74.5 pm (Etotal = -433.981 kJ mol-1).
{| class="wikitable"
!Internuclear distances
!Contour
!Forces
|-
|[[File:Lhl17 Ex2 TS internuclear.png|none|thumb]]To find the TS, both momenta were set to zero and the distances were optimised until minimal oscillation was reached in the internuclear distances vs time graph
|[[File:Lhl17 Ex2 TS contour.png|none|thumb]]Unlike exercise 1, the system is clearly asymmetric.
|[[File:Lhl17 forces1.png|none|thumb|272x272px]]The forces at the transition state are very close to 0 which means that the system will not move.
|}
Note that for these triatomics there are two bond lengths we can vary and thus at a saddle point there is one direction that decreases most in energy, and another that increases the most. For the transition structure of a polyatomic molecule with 3N-6 vibrational modes, there would exist one direction down and the remaining 3N-7 vibrational modes acting as sinks for the excess energy.<ref>R. D. Levine, ''Molecular Reaction Dynamics'', Cambridge University Press, Cambridge, 2005, p. 216.</ref>

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

The Ea was found by very slightly altering the TS distance for each of AB, then BC, and taking the energy difference between the TS and the system with maximum displacement between the reactants.
{| class="wikitable"
!F + H2
!HF + H
|-
|[[File:Lhl17 Ex2 Ea1.png|none|thumb]]Ea: 1.26 kJ.mol-1.
|[[File:Lhl17 Ex2 reactants 560.png|none|thumb]]Ea: 127 kJ.mol-1.
|}
Knowing that the F + H2 reaction was exothermic and had an early TS provided useful guidance when locating the exact point, since the activation energy was so small.

=== Question 9 ===
''Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Momenta vs Time”.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.''.

To investigate a reactive trajectory for the reaction F + H2 → HF + H, the following initial conditions were chosen:
{| class="wikitable"

! rFH / pm
! rHH / pm
! pFH/gmol-1pmfs-1
! pHH/gmol-1pmfs-1

|-
| 200
| 74
| -1.0
| 0
|}
This yielded the follow plots:
{| class="wikitable"
!Contour plot
!Momenta vs time
!Surface Plot
!Energy plot
|-
|[[File:Lhl17 Ex2 energyrelease contour.png|none|thumb|200x200px]]The contour plot shows strong oscillations after the system passes through the transition state, which represents the system passes its excess energy into the vibration of HF.
|[[File:Lhl17 Ex2 energyrelease p.png|none|thumb|267x267px]]
|[[File:Lhl17 Ex2 energyrelease s.png|none|thumb|200x200px]]
|[[File:Lhl17 Ex2 release e.png|none|thumb|267x267px]]
|}
The conservation of energy suggests that total energy is constant through the course of reaction. From the exothermic energy vs time graph below, the loss in potential energy from the breaking of H2 bonds is mirrored by the increase in Ek, which can be either translational, rotational or vibrational.

Calorimetry can be used to measure the transfer of both Etrans and Evib (Erot is ignored here) from the products to the walls of the calorimeter in the form of heat and the heat transferred can then be measured via a temperature change. We would expect an increase in temperature for exothermic reactions and a decrease for endothermic ones, and the total energy change can be calculated using ΔQ=cΔΤ.

Meanwhile, FTIR spectroscopy can also be used to quantify Evib. Before the collision, the atoms lie in the ground state - however, during the collision, they are excited to occupy higher vibrational modes as Ep is converted to Ek. Overtone and peak wavenumbers in the IR spectrum can be used to identify the changes in Evib of the IR active product and over time, the vibrational states eventually relax back to the ground state.
=== 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 Empirical rules state that vibrational energy is more efficient at activating a late transition state (present in endothermic reactions, according to Hammond's postulate), than translational energy.

On the other hand, translational energy is more efficient in activating an early transition state in an exothermic reaction than vibrational energy as the trajectory can simply "fall" into the lower energy region. Therefore for a successful exothermic reaction, the molecules need an excess of translational (kinetic) energy compared to vibrational (oscillatory) energy.

To investigate this, a calculation starting on the side of the reactants of F + H2, at the bottom of the well with rHF = 200 pm, rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1 was set up. Several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 were explored:
{| class="wikitable"
!pHH/gmol-1pmfs-1
!Etot/ kJ.mol-1
!Reactive trajectory
!Contour plot
|-
|<nowiki>-6.1</nowiki>
|<nowiki>-402.5</nowiki>
|No
|[[File:Lhl17 -6.1.png|none|thumb]]
|-
|<nowiki>-5</nowiki>
|<nowiki>-413.6</nowiki>
|Yes
|[[File:Lhl17 -5.png|none|thumb]]
|-
|<nowiki>-3</nowiki>
|<nowiki>-427.6</nowiki>
|No
|[[File:Lhl17 -3.png|none|thumb]]
|-
|0
|<nowiki>-433.6</nowiki>
|Yes
|[[File:Lhl17 0.png|none|thumb]]
|-
|3
|<nowiki>-421.6</nowiki>
|Yes
|[[File:Lhl17 3.png|thumb]]
|-
|5
|<nowiki>-403.6</nowiki>
|No
|[[File:Lhl17 5.png|none|thumb]]
|-
|6.1
|<nowiki>-390.3</nowiki>
|No
|[[File:Lhl17 6.1.png|none|thumb]]
|}
For the same initial position, the momentum pFH (represents the translational energy) was increased slightly to -1.6 g.mol-1.pm.fs-1,and pHH (representing vibrational energy decreased to 0.2 g.mol-1.pm.fs-1. Although the overall energy of this simulation (-432.4 kJ mol-1) was similar to a successful trajectory, such as row 4 in the table above, the reaction did not go to completion. This confirms that the form of energy in the system is important to determine whether the reaction goes to completion.

Barrier recrossing means products are able to cross back and become reactants again, as shown by the contour plots, and results in unsuccessful reactions:
{| class="wikitable"
!Contour
!Momentum
!Energy
|-
|[[File:Lhl17 Ex2 reduceE contour.png|none|thumb]]
|[[File:Lhl17 Ex2 reduceE m.png|none|thumb]]
|[[File:Lhl17 Ex2 reduceE E.png|none|thumb]]
|}
[[File:Lhl17 q10.png|none|thumb|A reactive trajectory for the reverse reaction, H + HF with initial conditions rHF = 92 pm, rHH = 200 pm, pHF = 15.3 g.mol-1.pm.fm-1 and pHH -1 g.mol-1.pm.fm-1]]
The above trajectory was found by increasing the vibrational energy of the HF bond while maintaining a low pHH.

== References ==

2020-05-22T11:12:13Z

Lhl17: /* Question 8 */

'''<nowiki/>'''

== Molecular Reaction Dynamics: Applications to Triatomic Systems ==

== Exercise 1: H + H2 system ==

=== 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 program calculates potential energy using It is a London-Eyring-Polanyi-Sato (LEPS) potential and calculates individual trajectories. By analysis of these reaction paths and using TST we can analyse reactions and the molecular reaction dynamics.

On a potential energy surface (PES) diagram, the transition state is mathematically defined as the point at which the following is true:

<math> { \partial V(r)\over \partial r} = 0 </math>

While a one-dimensional potential energy curve (PEC) could simply be approximated by quadratic curves, the PES is multi-dimensional function (with the energy on the z axis and bond distances on the x/y axes). This means the TS, the maximum point on the minimum reactive trajectory, can be visually identified as the "saddle point" - a maximum point in one direction and minimum in the other.

However, to distinguish the TS from other local minima, both the second derivative and the determinant of the Hessian Matrix at the TS should be smaller than 0. In addition, the eigenvalues of the Hessian Matrix should have opposite positive/negative signs.<ref><nowiki>https://books.google.co.uk/books?id=6Q1oF3dJTHYC&pg=PA31&lpg=PA31&dq=hessian+matrix+transitions+tate&source=bl&ots=ppp83307EI&sig=ACfU3U0lPGQ3nR2k0AIxgv41i8Ed2KNZ3Q&hl=en&sa=X&ved=2ahUKEwi3rdn7iMfpAhXFoXEKHcw1CFIQ6AEwAXoECAkQAQ#v=onepage&q=hessian%20matrix%20transitions%20tate&f=false</nowiki></ref>

=== Question 2 ===
''Report your best estimate of the transition state position (<nowiki>'''</nowiki>rts<nowiki>'''</nowiki>) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.''

The TS position for the H + H2 system was found at rAB = rBC = 90.77 pm with pAB = pBC = 0 g.mol-1.pm.fm-1.
{| class="wikitable"
!Internuclear Distances vs Time
!Contour Plot
!Force analysis
|-
|[[File:Lhl17 TS1.png|none|thumb|170x170px]]Starting the TS with no momentum with no forces on the atoms, they will remain in place and not move at all. Intermolecular distances remain constant which results in horizontal lines in the graph above.
|[[File:Lhl17 TScontour.png|none|thumb|227x227px]]The red cross indicates the initial conditions input, and it is clear that no bond movement is shown here. The transition state is totally symmetric (neither endo nor exothermic, and neither early nor late according to Hammond's postulate).
|[[File:Lhl17 forces.png|none|thumb|170x170px]]The exact point was confirmed by force analysis (where forces on the molecules are minimised).
|}

=== Question 3 ===
''Comment on how the mep and the trajectory you just calculated differ.''

The minimum energy path (''mep'') is a very special trajectory that corresponds to infinitely slow motion (i.e. the momenta/velocities are always reset to zero in each time step). By slightly displacing the initial conditions from the TS in either direction, it possible to encourage product or reactant formation and observe how the trajectory evolves just away from the saddle. Like a ball rolling along a hilly surface, the system will gain momentum as soon as the simulation starts. as the system is not at a stable point on the PES it will fall to an energy minimum, then potential energy is converted into kinetic. By comparing this same system in a MEP and Dynamic simulation, the produced trajectories should illustrate the differences between both approaches.

When calculating the minimum energy pathway using calculation type MEP and Dynamics, we obtain the following results (with rAB = 90.77 pm, rBC = 91.77 pm and pAB = pBC = 0 g.mol-1.pm.fm-1):
{| class="wikitable"
!Dynamic Calculation
!MEP Calculation (5000 steps)
!

Comments
|-
|[[File:Lhl17 Ex1 Dynamic displaced.png|none|thumb|Contour plot of reaction path]]
|[[File:Lhl17 Ex1 MEP displaced.png|none|thumb|Contour plot of reaction path]]
|There is no vibrational energy for the MEP plot, unlike in the dynamic plot. This is an indication of how different MEP is algorithmically set - for MEP velocity is reset to 0 at each time step, so that kinetic energy always remains at 0. Thus, there is no vibrational energy in the system so no oscillation.
|-
|[[File:Lhl17 Ex1 Dynamic displaced p.png|thumb|Momentum vs Time plot of reaction path]]
|[[File:Lhl17 Ex1 MEP displaced p.png|thumb|Momentum vs Time plot of reaction path]]
|In Dynamics, we can see that the oscillation of the atoms has been included in the calculation, and increases over time.
|-
|[[File:Lhl17 Ex1 dynamics displaced d.png|thumb|Distance vs Time plot of reaction path]]
|[[File:Lhl17 Ex1 MEP displaced d.png|thumb|Distance vs Time plot of reaction path]]
|In both cases, the atoms are moving away from each other. Note that in the dynamics velocities at which they are moving remain constant.
|}

=== Question 4 ===
''Complete the table below 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?''

{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|[[File:Lhl17_Table1a.png|thumb]]Reaction path passes through the TS (at around 21 fs) and forms products. The vibrational oscillations in the product channel confirm formation of the the product.
|[[File:Lhl17 Table1.png|none|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|[[File:Lhl17 Table2a.png|thumb]]Reaction path unable to get over maximum at TS and form products. There is no successful collision and instead, the system rolls back down the potential to reactants.
|[[File:Lhl17 Table2.png|none|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|[[File:Lhl17 Table3a.png|thumb]]Momentum of incoming H atom increased compared to first example, and as before, the reaction path passes over the TS (at around 21 fs) to the product channel. More oscillations observed in the product than reactant channel.
|[[File:Lhl17 Table3.png|none|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|[[File:Lhl17 Table4a.png|thumb]]Initial momenta large enough for reaction path to reach the TS (at around 8 fs), but excess energy causes the product to cross the activation barrier twice. Even though the product bond does form, the system eventually returns to reactants (with significantly higher vibronic energy).
|[[File:Lhl17 Table4.png|none|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|[[File:Lhl17 Table5a.png|thumb]]Similar to the previous example, but with the momentum of the incoming H atom increased. The system fluctuates twice between reactant and product bond formation (with TS at 9, 15 and 26 fs), which suggests the TS in this scenario is far from the lowest energy saddle point.
|[[File:Lhl17 Table5.png|none|thumb]]
|}
The higher kinetic energy, hence momenta does not necessarily guarantee the reaction would proceed. Higher kinetic energy can mean that the reactants can go over the activation energy barrier, but with energy being too high, products newly formed might overcome the transition state barrier again in the reversed direction and reform reactants, in which case it is not reactive overall.

=== Question 5 ===
''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''

The Transition State Theory assumes that only the reactants are in constant equilibrium with the transition state structure (quasi-equilibrium), and that the energy of the particles follow a Boltzmann distribution in a thermally equibrilated reactant pool.

In addition, it is assumed that once the reactants have trajectories with a kinetic energy along the reaction coordinate greater than the activation energy, that the products will definitely form, ignoring the possibility of barrier recrossing caused by excess vibrational energy. As seen in row 4, the starting materials may be reformed instead leading to an overall unreactive path.

Furthermore, the theory fails for some reactions at low temperatures due to quantum tunneling, where the product can still be formed even if the TS is not reached (the effect of this is negligible compared to barrier recrossing). The nuclei are also treated as classical point masses and don't have wavefunctions, which is another flaw of TST.

In the table above, only 5 possible individual trajectories of the H + H2 reaction were investigated. The true reaction rate can be thought of as a "weighted average" of all such individual paths and so, with TST predicting unreactive trajectories to in face be reactive, it can be concluded that the theory overestimates the rate of reaction when compared with experimental values.

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

=== 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?''[[File:Lhl17 Ex2 TS surface.png|thumb|PES of F + H2, showing early transition state]]The enthalpy change of a reaction can be calculated using the following equation:

ΔHrxn=ΣΔHbond breaking'''-'''ΣΔHbond forming

Using literature values for bond dissociation energies of H-H (436 kJ mol-1) and H-F (569 kJ mol-1)<ref>https://labs.chem.ucsb.edu/zakarian/armen/11---bonddissociationenergy.pdf</ref> shows that the enthalpy released from breaking the H-F bond is greater than the energy required to break H-H bonds. Therefore, the reaction F + H2 → HF + H is energetically favorable and exothermic overall. This is consistent with the very small atomic radius of the F atom, and therefore stronger attraction and a stronger bond overall. The PES shows that the reactants are higher in energy than the products (where AB distance, representing the H-F bond), indicating an exothermic reaction.

The reverse reaction, H + HF is therefore endothermic, as extra external energy is required in order to break the H-F bond.

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

According to Hammond's postulate the structure of the transition state will resemble the structure of the nearest stable species (either products or reactants). Taking the first reaction F + H2 → HF + H, the transition state will resemble the reactants since it is an exothermic reaction. For HF + H → F + H2, the opposite is true: the transition state will resemble the products for the endothermic reaction.

After testing several initial distances by trial and error, initial momentum of both reactants = 0, it was found that the transition state has the following internuclear distances: rFH = 180.5 pm and rHH = 74.5 pm.: -433.981 kJ mol-1
{| class="wikitable"
!Internuclear distances
!Contour
!Forces
|-
|[[File:Lhl17 Ex2 TS internuclear.png|none|thumb]]To find the TS, both momenta were set to zero and the distances were optimised until minimal oscillation was reached in the internuclear distances vs time graph
|[[File:Lhl17 Ex2 TS contour.png|none|thumb]]Unlike exercise 1, the system is clearly asymmetric.
|[[File:Lhl17 forces1.png|none|thumb|272x272px]]the forces at the transition state are very close to 0 which means that the system will not move if no momentum or change in position is given to it.
|}
Note that for these triatomics there are two bond lengths we can vary and thus at a saddle point there is one direction that decreases most in energy, and another that increases the most. For the transition structure of a polyatomic molecule with 3N-6 vibrational modes, there would exist one direction down and the remaining 3N-7 vibrational modes acting as sinks for the excess energy.<ref><nowiki>https://books.google.co.uk/books?id=FVqyS31OM7sC&pg=PA216&lpg=PA216&dq=polyatomics+saddle+point+3n-7&source=bl&ots=BKxrckRrr2&sig=ACfU3U1lawEEVNA3oh5R-wa3h0t8iKPKZQ&hl=en&sa=X&ved=2ahUKEwiAq5rAmMfpAhXKRxUIHY0xCM0Q6AEwAXoECAsQAQ#v=onepage&q=polyatomics%20saddle%20point%203n-7&f=false</nowiki></ref>

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

The Ea was found by very slightly altering the TS distance for each of AB, then BC, and taking the energy difference between the TS and the system with maximum displacement between the reactants.
{| class="wikitable"
!F + H2
!HF + H
|-
|[[File:Lhl17 Ex2 Ea1.png|none|thumb]]Ea: 1.26 kJ.mol-1.
|[[File:Lhl17 Ex2 reactants 560.png|none|thumb]]Ea: 127 kJ.mol-1.
|}
=== Question 9 ===
''Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Momenta vs Time”.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.''.

To investigate a reactive trajectory for the reaction H2 + F -> HF + H, the following initial conditions were chosen:
{| class="wikitable"

! rFH / pm
! rHH / pm
! pFH/gmol-1pmfs-1
! pHH/gmol-1pmfs-1

|-
| 200
| 74.5
| -1.0
| 0
|}
The conservation of energy suggests that total energy is constant through the course of reaction. From the Energy Vs Time graph of the exothermic H2 and F reaction below, the loss in potential energy from the breaking of H2 bonds is mirrored by the increase in kinetic energy. The kinetic energy can be of three forms: translational, rotational or vibrational.

Calorimetry can be used to measuring the transfer of both translational and kinetic energy from the products to the walls of the calorimeter in the form of heat. The heat transferred can then be measured in the form of change in temperature with a thermal sensor. We would expect an increase in temperature for exothermic reactions and a decrease for endothermic ones, and the energy can be calculated using ΔQ=cΔΤ.

Meanwhile, IR spectroscopy can be used to quantify the vibrational energy. Vibrational exictation and peak wavenumbers in the infrared spectrum can be used to identify the changes in Evib of the IR active HF molecule.
{| class="wikitable"
!Contour plot
!Momenta vs time
!Surface Plot
!Energy plot
|-
|[[File:Lhl17 Ex2 energyrelease contour.png|none|thumb|200x200px]]The contour plot shows strong oscillations after the system passes through the transition state, which represents the system passes its excess energy into the vibration of HF.
|[[File:Lhl17 Ex2 energyrelease p.png|none|thumb|267x267px]]
|[[File:Lhl17 Ex2 energyrelease s.png|none|thumb|200x200px]]
|[[File:Lhl17 Ex2 release e.png|none|thumb|267x267px]]
|}
=== 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 Empirical rules state that vibrational energy is more efficient at activating a late transition state than translational energy (endothermic reaction - Hammonds Postulate) and that translational energy is more efficient in activating an early transition state in an exothermic reaction than vibrational energy. Therefore for a successful exothermic reaction, the molecules need an excess of translational (kinetic) energy compared to vibrational (oscillatory) energy.

To investigate this, a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1 was set up. Several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 were explored: (explore values also close to these limits).

The table below shows the reactivity and energy of the reaction F + H2 with differing initial momenta. The initial positions are set to AB = 200 pm and BC= 74 pm where A is F, and B and C are H.
{| class="wikitable"
!pHH/gmol-1pmfs-1
!Etot/ kJ.mol-1
!Reactive trajectory
!Contour plot
|-
|<nowiki>-6.1</nowiki>
|<nowiki>-402.5</nowiki>
|No
|[[File:Lhl17 -6.1.png|none|thumb]]
|-
|<nowiki>-5</nowiki>
|<nowiki>-413.6</nowiki>
|Yes
|[[File:Lhl17 -5.png|none|thumb]]
|-
|<nowiki>-3</nowiki>
|<nowiki>-427.6</nowiki>
|No
|[[File:Lhl17 -3.png|none|thumb]]
|-
|0
|<nowiki>-433.6</nowiki>
|Yes
|[[File:Lhl17 0.png|none|thumb]]
|-
|3
|<nowiki>-421.6</nowiki>
|Yes
|[[File:Lhl17 3.png|thumb]]
|-
|5
|<nowiki>-403.6</nowiki>
|No
|[[File:Lhl17 5.png|none|thumb]]
|-
|6.1
|<nowiki>-390.3</nowiki>
|No
|[[File:Lhl17 6.1.png|none|thumb]]
|}
For the same initial position, the momentum pFH (represents the translational energy) was increased slightly to -1.6 g.mol-1.pm.fs-1,and pHH (representing vibrational energy decreased to 0.2 g.mol-1.pm.fs-1. Although the overall energy of this simulation (-432.4 kJ mol-1) was similar to a successful trajectory, such as row 4 in the table above, the reaction did not go to completion. This confirms that the form of energy in the system is important to determine whether the reaction goes to completion.

Barrier recrossing means products are able to cross back and become reactants again, as shown by the contour plots, and results in unsuccessful reactions.
{| class="wikitable"
!Contour
!Momentum
!Energy
|-
|[[File:Lhl17 Ex2 reduceE contour.png|none|thumb]]
|[[File:Lhl17 Ex2 reduceE m.png|none|thumb]]
|[[File:Lhl17 Ex2 reduceE E.png|none|thumb]]
|}

== References ==

File:Lhl17 -3.png

2020-05-22T10:53:52Z

Lhl17:

File:Lhl17 3.png

2020-05-22T10:53:33Z

Lhl17:

MRD:lhl17

2020-05-22T10:15:07Z

Lhl17: /* Question 3 */

'''<nowiki/>'''

== Molecular Reaction Dynamics: Applications to Triatomic Systems ==

== Exercise 1: H + H2 system ==

=== 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 program calculates potential energy using It is a London-Eyring-Polanyi-Sato (LEPS) potential and calculates individual trajectories. By analysis of these reaction paths and using TST we can analyse reactions and the molecular reaction dynamics.

On a potential energy surface (PES) diagram, the transition state is mathematically defined as the point at which the following is true:

<math> { \partial V(r)\over \partial r} = 0 </math>

While a one-dimensional potential energy curve (PEC) could simply be approximated by quadratic curves, the PES is multi-dimensional function (with the energy on the z axis and bond distances on the x/y axes). This means the TS, the maximum point on the minimum reactive trajectory, can be visually identified as the "saddle point" - a maximum point in one direction and minimum in the other.

However, to distinguish the TS from other local minima, both the second derivative and the determinant of the Hessian Matrix at the TS should be smaller than 0. In addition, the eigenvalues of the Hessian Matrix should have opposite positive/negative signs.<ref><nowiki>https://books.google.co.uk/books?id=6Q1oF3dJTHYC&pg=PA31&lpg=PA31&dq=hessian+matrix+transitions+tate&source=bl&ots=ppp83307EI&sig=ACfU3U0lPGQ3nR2k0AIxgv41i8Ed2KNZ3Q&hl=en&sa=X&ved=2ahUKEwi3rdn7iMfpAhXFoXEKHcw1CFIQ6AEwAXoECAkQAQ#v=onepage&q=hessian%20matrix%20transitions%20tate&f=false</nowiki></ref>

=== Question 2 ===
''Report your best estimate of the transition state position (<nowiki>'''</nowiki>rts<nowiki>'''</nowiki>) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.''

The TS position for the H + H2 system was found at rAB = rBC = 90.77 pm with pAB = pBC = 0 g.mol-1.pm.fm-1.
{| class="wikitable"
!Internuclear Distances vs Time
!Contour Plot
!Force analysis
|-
|[[File:Lhl17 TS1.png|none|thumb|170x170px]]Starting the TS with no momentum with no forces on the atoms, they will remain in place and not move at all. Intermolecular distances remain constant which results in horizontal lines in the graph above.
|[[File:Lhl17 TScontour.png|none|thumb|227x227px]]The red cross indicates the initial conditions input, and it is clear that no bond movement is shown here. The transition state is totally symmetric (neither endo nor exothermic, and neither early nor late according to Hammond's postulate).
|[[File:Lhl17 forces.png|none|thumb|170x170px]]The exact point was confirmed by force analysis (where forces on the molecules are minimised).
|}

=== Question 3 ===
''Comment on how the mep and the trajectory you just calculated differ.''

The minimum energy path (''mep'') is a very special trajectory that corresponds to infinitely slow motion (i.e. the momenta/velocities are always reset to zero in each time step). By slightly displacing the initial conditions from the TS in either direction, it possible to encourage product or reactant formation and observe how the trajectory evolves just away from the saddle. Like a ball rolling along a hilly surface, the system will gain momentum as soon as the simulation starts. as the system is not at a stable point on the PES it will fall to an energy minimum, then potential energy is converted into kinetic. By comparing this same system in a MEP and Dynamic simulation, the produced trajectories should illustrate the differences between both approaches.

When calculating the minimum energy pathway using calculation type MEP and Dynamics, we obtain the following results (with rAB = 90.77 pm, rBC = 91.77 pm and pAB = pBC = 0 g.mol-1.pm.fm-1):
{| class="wikitable"
!Dynamic Calculation
!MEP Calculation (5000 steps)
!

Comments
|-
|[[File:Lhl17 Ex1 Dynamic displaced.png|none|thumb|Contour plot of reaction path]]
|[[File:Lhl17 Ex1 MEP displaced.png|none|thumb|Contour plot of reaction path]]
|There is no vibrational energy for the MEP plot, unlike in the dynamic plot. This is an indication of how different MEP is algorithmically set - for MEP velocity is reset to 0 at each time step, so that kinetic energy always remains at 0. Thus, there is no vibrational energy in the system so no oscillation.
|-
|[[File:Lhl17 Ex1 Dynamic displaced p.png|thumb|Momentum vs Time plot of reaction path]]
|[[File:Lhl17 Ex1 MEP displaced p.png|thumb|Momentum vs Time plot of reaction path]]
|In Dynamics, we can see that the oscillation of the atoms has been included in the calculation, and increases over time.
|-
|[[File:Lhl17 Ex1 dynamics displaced d.png|thumb|Distance vs Time plot of reaction path]]
|[[File:Lhl17 Ex1 MEP displaced d.png|thumb|Distance vs Time plot of reaction path]]
|In both cases, the atoms are moving away from each other. Note that in the dynamics velocities at which they are moving remain constant.
|}

=== Question 4 ===
''Complete the table below 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?''

{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|[[File:Lhl17_Table1a.png|thumb]]Reaction path passes through the TS (at around 21 fs) and forms products. The vibrational oscillations in the product channel confirm formation of the the product.
|[[File:Lhl17 Table1.png|none|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|[[File:Lhl17 Table2a.png|thumb]]Reaction path unable to get over maximum at TS and form products. There is no successful collision and instead, the system rolls back down the potential to reactants.
|[[File:Lhl17 Table2.png|none|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|[[File:Lhl17 Table3a.png|thumb]]Momentum of incoming H atom increased compared to first example, and as before, the reaction path passes over the TS (at around 21 fs) to the product channel. More oscillations observed in the product than reactant channel.
|[[File:Lhl17 Table3.png|none|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|[[File:Lhl17 Table4a.png|thumb]]Initial momenta large enough for reaction path to reach the TS (at around 8 fs), but excess energy causes the product to cross the activation barrier twice. Even though the product bond does form, the system eventually returns to reactants (with significantly higher vibronic energy).
|[[File:Lhl17 Table4.png|none|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|[[File:Lhl17 Table5a.png|thumb]]Similar to the previous example, but with the momentum of the incoming H atom increased. The system fluctuates twice between reactant and product bond formation (with TS at 9, 15 and 26 fs), which suggests the TS in this scenario is far from the lowest energy saddle point.
|[[File:Lhl17 Table5.png|none|thumb]]
|}
The higher kinetic energy, hence momenta does not necessarily guarantee the reaction would proceed. Higher kinetic energy can mean that the reactants can go over the activation energy barrier, but with energy being too high, products newly formed might overcome the transition state barrier again in the reversed direction and reform reactants, in which case it is not reactive overall.

=== Question 5 ===
''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''

The Transition State Theory assumes that only the reactants are in constant equilibrium with the transition state structure (quasi-equilibrium), and that the energy of the particles follow a Boltzmann distribution.

In addition, it is assumed that once the reactants have trajectories with a kinetic energy along the reaction coordinate greater than the activation energy, that the products will definitely form, ignoring the possibility of barrier recrossing caused by excess vibrational energy. As seen in row 4, the starting materials may be reformed instead leading to an overall unreactive path.

Furthermore, the theory fails for some reactions at low temperatures due to quantum tunneling, where the product can still be formed even if the TS is not reached (the effect of this is negligible compared to barrier recrossing). The nuclei are also treated as classical point masses and don't have wavefunctions, which is another flaw of TST.

In the table above, only 5 possible individual trajectories of the H + H2 reaction were investigated. The true reaction rate can be thought of as a "weighted average" of all such individual paths and so, with TST predicting unreactive trajectories to in face be reactive, it can be concluded that the theory overestimates the rate of reaction when compared with experimental values.

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

=== 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?''[[File:Lhl17 Ex2 TS surface.png|thumb|PES of F + H2, showing early transition state]]The enthalpy change of a reaction can be calculated using the following equation:

ΔHrxn=ΣΔHbond breaking'''-'''ΣΔHbond forming

Using literature values for bond dissociation energies of H-H (436 kJ mol-1) and H-F (569 kJ mol-1)<ref>https://labs.chem.ucsb.edu/zakarian/armen/11---bonddissociationenergy.pdf</ref> shows that the enthalpy released from breaking the H-F bond is greater than the energy required to break H-H bonds. Therefore, the reaction F + H2 → HF + H is energetically favorable and exothermic overall. This is consistent with the very small atomic radius of the F atom, and therefore stronger attraction and a stronger bond overall. The PES shows that the reactants are higher in energy than the products (where AB distance, representing the H-F bond), indicating an exothermic reaction.

The reverse reaction, H + HF is therefore endothermic, as extra external energy is required in order to break the H-F bond.

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

According to Hammond's postulate the structure of the transition state will resemble the structure of the nearest stable species (either products or reactants). Taking the first reaction F + H2 → HF + H, the transition state will resemble the reactants since it is an exothermic reaction. For HF + H → F + H2, the opposite is true: the transition state will resemble the products for the endothermic reaction.

After testing several initial distances by trial and error, initial momentum of both reactants = 0, it was found that the transition state has the following internuclear distances: rFH = 180.5 pm and rHH = 74.5 pm.: -433.981 kJ mol-1
{| class="wikitable"
!Internuclear distances
!Contour
!Forces
|-
|[[File:Lhl17 Ex2 TS internuclear.png|none|thumb]]To find the TS, both momenta were set to zero and the distances were optimised until minimal oscillation was reached in the internuclear distances vs time graph
|[[File:Lhl17 Ex2 TS contour.png|none|thumb]]Unlike exercise 1, the system is clearly asymmetric.
|[[File:Lhl17 forces1.png|none|thumb|272x272px]]the forces at the transition state are very close to 0 which means that the system will not move if no momentum or change in position is given to it.
|}
Note that for these triatomics there are two bond lengths we can vary and thus at a saddle point there is one direction that decreases most in energy, and another that increases the most. For the transition structure of a polyatomic molecule with 3N-6 vibrational modes, there would exist one direction down and the remaining 3N-7 vibrational modes acting as sinks for the excess energy.<ref><nowiki>https://books.google.co.uk/books?id=FVqyS31OM7sC&pg=PA216&lpg=PA216&dq=polyatomics+saddle+point+3n-7&source=bl&ots=BKxrckRrr2&sig=ACfU3U1lawEEVNA3oh5R-wa3h0t8iKPKZQ&hl=en&sa=X&ved=2ahUKEwiAq5rAmMfpAhXKRxUIHY0xCM0Q6AEwAXoECAsQAQ#v=onepage&q=polyatomics%20saddle%20point%203n-7&f=false</nowiki></ref>

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

The Ea was found by altering the TS distance for each of AB, then BC by 1 pm. The energy difference between the TS system with maximum displacement between the reactants (F-H=750 pm, energy=-435.100 kJ.mol-1).
{| class="wikitable"
!F + H2
!HF + H
|-
|[[File:Lhl17 Ex2 Ea1.png|none|thumb]]Ea: 1.158 kJ.mol-1.
|[[File:Lhl17 Ex2 reactants 560.png|none|thumb]]Ea: 127 kJ.mol-1.
|}
=== Question 9 ===
''Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Momenta vs Time”.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.''.

To investigate a reactive trajectory for the reaction H2 + F -> HF + H, the following initial conditions were chosen:
{| class="wikitable"

! rFH / pm
! rHH / pm
! pFH/gmol-1pmfs-1
! pHH/gmol-1pmfs-1

|-
| 200
| 74.5
| -1.0
| 0
|}
The conservation of energy suggests that total energy is constant through the course of reaction. From the Energy Vs Time graph of the exothermic H2 and F reaction below, the loss in potential energy from the breaking of H2 bonds is mirrored by the increase in kinetic energy. The kinetic energy can be of three forms: translational, rotational or vibrational.

Calorimetry can be used to measuring the transfer of both translational and kinetic energy from the products to the walls of the calorimeter in the form of heat. The heat transferred can then be measured in the form of change in temperature with a thermal sensor. We would expect an increase in temperature for exothermic reactions and a decrease for endothermic ones, and the energy can be calculated using ΔQ=cΔΤ.

Meanwhile, IR spectroscopy can be used to quantify the vibrational energy. Vibrational exictation and peak wavenumbers in the infrared spectrum can be used to identify the changes in Evib of the IR active HF molecule.
{| class="wikitable"
!Contour plot
!Momenta vs time
!Surface Plot
!Energy plot
|-
|[[File:Lhl17 Ex2 energyrelease contour.png|none|thumb|200x200px]]The contour plot shows strong oscillations after the system passes through the transition state, which represents the system passes its excess energy into the vibration of HF.
|[[File:Lhl17 Ex2 energyrelease p.png|none|thumb|267x267px]]
|[[File:Lhl17 Ex2 energyrelease s.png|none|thumb|200x200px]]
|[[File:Lhl17 Ex2 release e.png|none|thumb|267x267px]]
|}
=== 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 Empirical rules state that vibrational energy is more efficient at activating a late transition state than translational energy (endothermic reaction - Hammonds Postulate) and that translational energy is more efficient in activating an early transition state in an exothermic reaction than vibrational energy. Therefore for a successful exothermic reaction, the molecules need an excess of translational (kinetic) energy compared to vibrational (oscillatory) energy.

To investigate thiis, a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration.

The table below shows the reactivity and energy of the reaction H2 + F with differing initial momenta. The initial positions are set to AB = 74 pm and BC= 200 pm where A and B are H and C is F.
{| class="wikitable"
!pHH/gmol-1pmfs-1
!Etot/ kJ.mol-1
!Reactive trajectory
!Contour plot
|-
|<nowiki>-6.1</nowiki>
|<nowiki>-402.5</nowiki>
|No
|[[File:Lhl17 -6.1.png|none|thumb]]
|-
|<nowiki>-5</nowiki>
|
|Yes
|[[File:Lhl17 -5.png|none|thumb]]
|-
|
|
|
|
|-
|0
|<nowiki>-433.6</nowiki>
|Yes
|[[File:Lhl17 0.png|none|thumb]]
|-
|
|
|
|
|-
|5
|
|No
|[[File:Lhl17 5.png|none|thumb]]
|-
|6
|
|No
|[[File:Lhl17 6.1.png|none|thumb]]
|}
For the same initial position, the momentum pFH ( represents the translational energy) was increased slightly to -1.6 g.mol-1.pm.fs-1,and pHH (representing vibrational energy decreased to 0.2 g.mol-1.pm.fs-1. Although the overall energy was similar to a successful trajectory (-432.4 here, ) the reaction did not go to completion. This shows that the type of energy in the system is important to determine whether the reaction is successful. In the reactions above with total energy being large but still having an unsuccessful reaction is mainly due to the reversibility where the products are able to cross back and become reactants again, as shown by the contour plots.

If you start a trajectory from close to the transition structure it will gain momentum as it moves in the reactant or product channel. At some molecular geometry in that channel, you could reverse the momentum (change the sign) and you should get a trajectory that goes back to the transition structure
{| class="wikitable"
!Contour
!Momentum
!Energy
|-
|[[File:Lhl17 Ex2 reduceE contour.png|none|thumb]]
|[[File:Lhl17 Ex2 reduceE m.png|none|thumb]]
|[[File:Lhl17 Ex2 reduceE E.png|none|thumb]]
|}

== References ==

MRD:lhl17

2020-05-22T10:11:19Z

Lhl17: /* Question 3 */

'''<nowiki/>'''

== Molecular Reaction Dynamics: Applications to Triatomic Systems ==

== Exercise 1: H + H2 system ==

=== 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 program calculates potential energy using It is a London-Eyring-Polanyi-Sato (LEPS) potential and calculates individual trajectories. By analysis of these reaction paths and using TST we can analyse reactions and the molecular reaction dynamics.

On a potential energy surface (PES) diagram, the transition state is mathematically defined as the point at which the following is true:

<math> { \partial V(r)\over \partial r} = 0 </math>

While a one-dimensional potential energy curve (PEC) could simply be approximated by quadratic curves, the PES is multi-dimensional function (with the energy on the z axis and bond distances on the x/y axes). This means the TS, the maximum point on the minimum reactive trajectory, can be visually identified as the "saddle point" - a maximum point in one direction and minimum in the other.

However, to distinguish the TS from other local minima, both the second derivative and the determinant of the Hessian Matrix at the TS should be smaller than 0. In addition, the eigenvalues of the Hessian Matrix should have opposite positive/negative signs.<ref><nowiki>https://books.google.co.uk/books?id=6Q1oF3dJTHYC&pg=PA31&lpg=PA31&dq=hessian+matrix+transitions+tate&source=bl&ots=ppp83307EI&sig=ACfU3U0lPGQ3nR2k0AIxgv41i8Ed2KNZ3Q&hl=en&sa=X&ved=2ahUKEwi3rdn7iMfpAhXFoXEKHcw1CFIQ6AEwAXoECAkQAQ#v=onepage&q=hessian%20matrix%20transitions%20tate&f=false</nowiki></ref>

=== Question 2 ===
''Report your best estimate of the transition state position (<nowiki>'''</nowiki>rts<nowiki>'''</nowiki>) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.''

The TS position for the H + H2 system was found at rAB = rBC = 90.77 pm with pAB = pBC = 0 g.mol-1.pm.fm-1.
{| class="wikitable"
!Internuclear Distances vs Time
!Contour Plot
!Force analysis
|-
|[[File:Lhl17 TS1.png|none|thumb|170x170px]]Starting the TS with no momentum with no forces on the atoms, they will remain in place and not move at all. Intermolecular distances remain constant which results in horizontal lines in the graph above.
|[[File:Lhl17 TScontour.png|none|thumb|227x227px]]The red cross indicates the initial conditions input, and it is clear that no bond movement is shown here. The transition state is totally symmetric (neither endo nor exothermic, and neither early nor late according to Hammond's postulate).
|[[File:Lhl17 forces.png|none|thumb|170x170px]]The exact point was confirmed by force analysis (where forces on the molecules are minimised).
|}

=== Question 3 ===
''Comment on how the mep and the trajectory you just calculated differ.''

The minimum energy path (''mep'') is a very special trajectory that corresponds to infinitely slow motion (i.e. the momenta/velocities are always reset to zero in each time step). By slightly displacing the initial conditions from the TS in either direction, it possible to encourage product or reactant formation and observe how the trajectory evolves just away from the saddle. Like a ball rolling along a hilly surface, the system will gain momentum as soon as the simulation starts. as the system is not at a stable point on the PES it will fall to an energy minimum, then potential energy is converted into kinetic. By comparing this same system in a MEP and Dynamic simulation, the produced trajectories should illustrate the differences between both approaches.

When calculating the minimum energy pathway using calculation type MEP and Dynamics, we obtain the following results (with rAB = 90.77 pm, rBC = 91.77 pm and pAB = pBC = 0 g.mol-1.pm.fm-1):
[[File:Lhl17 Animation 2020-05-22 11-06-21.mp4|none|thumb]]
{| class="wikitable"
!Dynamic Calculation
!MEP Calculation (5000 steps)
!

Comments
|-
|[[File:Lhl17 Ex1 Dynamic displaced.png|none|thumb|Contour plot of reaction path]]
|[[File:Lhl17 Ex1 MEP displaced.png|none|thumb|Contour plot of reaction path]]
|There is no vibrational energy for the MEP plot, unlike in the dynamic plot. This is an indication of how different MEP is algorithmically set - for MEP velocity is reset to 0 at each time step, so that kinetic energy always remains at 0. Thus, there is no vibrational energy in the system so no oscillation.
|-
|[[File:Lhl17 Ex1 Dynamic displaced p.png|thumb|Momentum vs Time plot of reaction path]]
|[[File:Lhl17 Ex1 MEP displaced p.png|thumb|Momentum vs Time plot of reaction path]]
|In Dynamics, we can see that the oscillation of the atoms has been included in the calculation, and increases over time.
|-
|[[File:Lhl17 Ex1 dynamics displaced d.png|thumb|Distance vs Time plot of reaction path]]
|[[File:Lhl17 Ex1 MEP displaced d.png|thumb|Distance vs Time plot of reaction path]]
|In both cases, the atoms are moving away from each other. Note that in the dynamics velocities at which they are moving remain constant.
|}

=== Question 4 ===
''Complete the table below 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?''

{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|[[File:Lhl17_Table1a.png|thumb]]Reaction path passes through the TS (at around 21 fs) and forms products. The vibrational oscillations in the product channel confirm formation of the the product.
|[[File:Lhl17 Table1.png|none|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|[[File:Lhl17 Table2a.png|thumb]]Reaction path unable to get over maximum at TS and form products. There is no successful collision and instead, the system rolls back down the potential to reactants.
|[[File:Lhl17 Table2.png|none|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|[[File:Lhl17 Table3a.png|thumb]]Momentum of incoming H atom increased compared to first example, and as before, the reaction path passes over the TS (at around 21 fs) to the product channel. More oscillations observed in the product than reactant channel.
|[[File:Lhl17 Table3.png|none|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|[[File:Lhl17 Table4a.png|thumb]]Initial momenta large enough for reaction path to reach the TS (at around 8 fs), but excess energy causes the product to cross the activation barrier twice. Even though the product bond does form, the system eventually returns to reactants (with significantly higher vibronic energy).
|[[File:Lhl17 Table4.png|none|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|[[File:Lhl17 Table5a.png|thumb]]Similar to the previous example, but with the momentum of the incoming H atom increased. The system fluctuates twice between reactant and product bond formation (with TS at 9, 15 and 26 fs), which suggests the TS in this scenario is far from the lowest energy saddle point.
|[[File:Lhl17 Table5.png|none|thumb]]
|}
The higher kinetic energy, hence momenta does not necessarily guarantee the reaction would proceed. Higher kinetic energy can mean that the reactants can go over the activation energy barrier, but with energy being too high, products newly formed might overcome the transition state barrier again in the reversed direction and reform reactants, in which case it is not reactive overall.

=== Question 5 ===
''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''

The Transition State Theory assumes that only the reactants are in constant equilibrium with the transition state structure (quasi-equilibrium), and that the energy of the particles follow a Boltzmann distribution.

In addition, it is assumed that once the reactants have trajectories with a kinetic energy along the reaction coordinate greater than the activation energy, that the products will definitely form, ignoring the possibility of barrier recrossing caused by excess vibrational energy. As seen in row 4, the starting materials may be reformed instead leading to an overall unreactive path.

Furthermore, the theory fails for some reactions at low temperatures due to quantum tunneling, where the product can still be formed even if the TS is not reached (the effect of this is negligible compared to barrier recrossing). The nuclei are also treated as classical point masses and don't have wavefunctions, which is another flaw of TST.

In the table above, only 5 possible individual trajectories of the H + H2 reaction were investigated. The true reaction rate can be thought of as a "weighted average" of all such individual paths and so, with TST predicting unreactive trajectories to in face be reactive, it can be concluded that the theory overestimates the rate of reaction when compared with experimental values.

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

=== 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?''[[File:Lhl17 Ex2 TS surface.png|thumb|PES of F + H2, showing early transition state]]The enthalpy change of a reaction can be calculated using the following equation:

ΔHrxn=ΣΔHbond breaking'''-'''ΣΔHbond forming

Using literature values for bond dissociation energies of H-H (436 kJ mol-1) and H-F (569 kJ mol-1)<ref>https://labs.chem.ucsb.edu/zakarian/armen/11---bonddissociationenergy.pdf</ref> shows that the enthalpy released from breaking the H-F bond is greater than the energy required to break H-H bonds. Therefore, the reaction F + H2 → HF + H is energetically favorable and exothermic overall. This is consistent with the very small atomic radius of the F atom, and therefore stronger attraction and a stronger bond overall. The PES shows that the reactants are higher in energy than the products (where AB distance, representing the H-F bond), indicating an exothermic reaction.

The reverse reaction, H + HF is therefore endothermic, as extra external energy is required in order to break the H-F bond.

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

According to Hammond's postulate the structure of the transition state will resemble the structure of the nearest stable species (either products or reactants). Taking the first reaction F + H2 → HF + H, the transition state will resemble the reactants since it is an exothermic reaction. For HF + H → F + H2, the opposite is true: the transition state will resemble the products for the endothermic reaction.

After testing several initial distances by trial and error, initial momentum of both reactants = 0, it was found that the transition state has the following internuclear distances: rFH = 180.5 pm and rHH = 74.5 pm.: -433.981
{| class="wikitable"
!Internuclear distances
!Contour
!Forces
|-
|[[File:Lhl17 Ex2 TS internuclear.png|none|thumb]]To find the TS, both momenta were set to zero and the distances were optimised until minimal oscillation was reached in the internuclear distances vs time graph
|[[File:Lhl17 Ex2 TS contour.png|none|thumb]]Unlike exercise 1, the system is clearly asymmetric.
|[[File:Lhl17 forces1.png|none|thumb|272x272px]]the forces at the transition state are very close to 0 which means that the system will not move if no momentum or change in position is given to it.
|}
Note that for these triatomics there are two bond lengths we can vary and thus at a saddle point there is one direction that decreases most in energy, and another that increases the most. For the transition structure of a polyatomic molecule with 3N-6 vibrational modes, there would exist one direction down and the remaining 3N-7 vibrational modes acting as sinks for the excess energy.<ref><nowiki>https://books.google.co.uk/books?id=FVqyS31OM7sC&pg=PA216&lpg=PA216&dq=polyatomics+saddle+point+3n-7&source=bl&ots=BKxrckRrr2&sig=ACfU3U1lawEEVNA3oh5R-wa3h0t8iKPKZQ&hl=en&sa=X&ved=2ahUKEwiAq5rAmMfpAhXKRxUIHY0xCM0Q6AEwAXoECAsQAQ#v=onepage&q=polyatomics%20saddle%20point%203n-7&f=false</nowiki></ref>

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

The Ea was found by altering the TS distance for each of AB, then BC by 1 pm. The energy difference between the TS system with maximum displacement between the reactants (F-H=750 pm, energy=-435.100 kJ.mol-1).
{| class="wikitable"
!F + H2
!HF + H
|-
|[[File:Lhl17 Ex2 Ea1.png|none|thumb]]Ea: 1.158 kJ.mol-1.
|[[File:Lhl17 Ex2 reactants 560.png|none|thumb]]Ea: 127 kJ.mol-1..
|}
=== Question 9 ===
''Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Momenta vs Time”.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.''.

To investigate a reactive trajectory for the reaction H2 + F -> HF + H, the following initial conditions were chosen:
{| class="wikitable"

! rFH / pm
! rHH / pm
! pFH/gmol-1pmfs-1
! pHH/gmol-1pmfs-1

|-
| 200
| 74.5
| -1.0
| 0
|}
The conservation of energy suggests that total energy is constant through the course of reaction. From the Energy Vs Time graph of the exothermic H2 and F reaction below, the loss in potential energy from the breaking of H2 bonds is mirrored by the increase in kinetic energy. The kinetic energy can be of three forms: translational, rotational or vibrational.

Calorimetry can be used to measuring the transfer of both translational and kinetic energy from the products to the walls of the calorimeter in the form of heat. The heat transferred can then be measured in the form of change in temperature with a thermal sensor. We would expect an increase in temperature for exothermic reactions and a decrease for endothermic ones, and the energy can be calculated using ΔQ=cΔΤ.

Meanwhile, IR spectroscopy can be used to quantify the vibrational energy. Vibrational exictation and peak wavenumbers in the infrared spectrum can be used to identify the changes in Evib of the IR active HF molecule.
{| class="wikitable"
!Contour plot
!Momenta vs time
!Surface Plot
!Energy plot
|-
|[[File:Lhl17 Ex2 energyrelease contour.png|none|thumb|200x200px]]The contour plot shows strong oscillations after the system passes through the transition state, which represents the system passes its excess energy into the vibration of HF.
|[[File:Lhl17 Ex2 energyrelease p.png|none|thumb|267x267px]]
|[[File:Lhl17 Ex2 energyrelease s.png|none|thumb|200x200px]]
|[[File:Lhl17 Ex2 release e.png|none|thumb|267x267px]]
|}
=== 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 Empirical rules state that vibrational energy is more efficient at activating a late transition state than translational energy (endothermic reaction - Hammonds Postulate) and that translational energy is more efficient in activating an early transition state in an exothermic reaction than vibrational energy. Therefore for a successful exothermic reaction, the molecules need an excess of translational (kinetic) energy compared to vibrational (oscillatory) energy.

To investigate thiis, a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration.

The table below shows the reactivity and energy of the reaction H2 + F with differing initial momenta. The initial positions are set to AB = 74 pm and BC= 200 pm where A and B are H and C is F.
{| class="wikitable"
!pHH/gmol-1pmfs-1
!Etot/ kJ.mol-1
!Reactive trajectory
!Contour plot
|-
|<nowiki>-6.1</nowiki>
|<nowiki>-402.5</nowiki>
|No
|[[File:Lhl17 -6.1.png|none|thumb]]
|-
|<nowiki>-5</nowiki>
|
|Yes
|[[File:Lhl17 -5.png|none|thumb]]
|-
|
|
|
|
|-
|0
|<nowiki>-433.6</nowiki>
|Yes
|[[File:Lhl17 0.png|none|thumb]]
|-
|
|
|
|
|-
|5
|
|No
|[[File:Lhl17 5.png|none|thumb]]
|-
|6
|
|No
|[[File:Lhl17 6.1.png|none|thumb]]
|}
For the same initial position, the momentum pFH ( represents the translational energy) was increased slightly to -1.6 g.mol-1.pm.fs-1,and pHH (representing vibrational energy decreased to 0.2 g.mol-1.pm.fs-1. Although the overall energy was similar to a successful trajectory (-432.4 here, ) the reaction did not go to completion. This shows that the type of energy in the system is important to determine whether the reaction is successful. In the reactions above with total energy being large but still having an unsuccessful reaction is mainly due to the reversibility where the products are able to cross back and become reactants again, as shown by the contour plots.

If you start a trajectory from close to the transition structure it will gain momentum as it moves in the reactant or product channel. At some molecular geometry in that channel, you could reverse the momentum (change the sign) and you should get a trajectory that goes back to the transition structure
{| class="wikitable"
!Contour
!Momentum
!Energy
|-
|[[File:Lhl17 Ex2 reduceE contour.png|none|thumb]]
|[[File:Lhl17 Ex2 reduceE m.png|none|thumb]]
|[[File:Lhl17 Ex2 reduceE E.png|none|thumb]]
|}

== References ==

2020-05-21T14:19:56Z

Lhl17:

MRD:lhl17

2020-05-21T14:04:47Z

Lhl17: /* Question 1 */

2020-05-20T17:14:03Z

Lhl17: /* Exercise 2: F - H - H system */

MRD:lhl17

2020-05-20T17:08:18Z

Lhl17:

File:Lhl17 Ex2 reactive contour.png

2020-05-20T16:57:04Z

Lhl17: