ChemWiki - User contributions [en]

MRD:01576020

2020-05-15T21:11:15Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
Modelling the dynamics of molecular reactions has been enabled by the emergence of computers and the field of computational chemistry is an ever-expanding field. Despite most approaches being purely theoretical, ''ab initio'', calculations, the insight given is invaluable for understanding the pathways of chemical reactions. It is often also not necessary to include quantum mechanics, as atoms are of sufficient mass to be reasonably accurately modelled by Newton's equations of motion. Isolated systems in the gas phase only experience interatomic interactions which can be represented by a potential energy surface (PES) which is a function only of atomic positions. The force on a given interatomic coordinate can be calculated as follows:

<math> F = - { \partial V(r_1,r_2,...)\over \partial r_i}</math>

where F = <math>{dp_i \over dt}</math>

Calculating the forces and adjusting the system in consecutive steps results in a trajectory which shows how a system evolves in time. Such calculations have almost endless application from chemical biology and medicine<ref>{{cite book|author1=Pregeljc, Domen|author2=Jug, Urska|author3=Mavri, Janez|author4=Stare, Jernej|lastauthoramp=yes|year=2018|title=Why does the Y326I mutant of monoamine oxidase B decompose an endogenous amphetamine at a slower rate than the wild type enzyme? Reaction step elucidated by multiscale molecular simulations.|journal= Phys Chem Chem Phys.|volume= 10|issue= 6|pages= 4181-4188|doi= 10.1039/c7cp07069a}}</ref> to materials science<ref>{{cite book|author1=Artritha, Nonguch|author2=Urban, Alexander|lastauthoramp=yes|year=2016|title=An implementation of artificial neural-network potentials for atomistic materials simulations: Performance for TiO2|journal= Phys Chem Chem Phys.|volume= 114|pages= 135-150|doi= 10.1016/j.commatsci.2015.11.047}}</ref>.

It is worth mentioning some main aspects about the transition state (TS) theory here as it is central to the concepts used here. It is purely classical - it ignores any quantum mechanical effects such as tunnelling, which is incredibly small anyway. It also assumes that the transition state structure is in equilibrium with the reactants, which is called a quasi-equilibrium. TS theory also assumes that the reactants have energies which follow a Boltzmann distribution, which is not a bad assumption as long as the system has had time to equilibrate. This assumption, however is not very applicable to the system studied here. Another crucial assumption of the TS theory, developed in the 1930s, is that once the TS is surpassed, the products will necessarily be formed, which was not the case in some simulations performed here. This latter assumption is also the reason why theoretically derived reaction rates using the TS state theory overestimate experimentally obtained reaction rates. This can be improved by adding a correcting factor, but to do that one must already have extensive knowledge of the system in question. The TS is defined as the maximum on the minimum energy path going from reactants to products. This is called a saddle point and the net force on the system is 0.<ref>{{cite book|author1=Laidler, Keith J.|lastauthoramp=yes|year=2016|title=Chemical Kinetics}}</ref><ref>{{cite book|author1=Atkins, Peter|author2=de Paula, Julio|lastauthoramp=yes|year=2014|title=Atkins' Physical Chemistry}}</ref>

We will deal here with triatomic linear systems, as depicted in the following Figure 1.
[[File:Y2C8.png|thumb|centre|500px| Figure 1. A triatomic system used here. The atoms will either all be Hs or in the last case one H will be replaced with a F. Figure taken from https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD]]

Terms like reactive and unreactive trajectories will be used often. A reactive trajectory is one which leads the reaction to completeness, i. e. the products are formed. In contrast an unreactive trajectory will not form products. The work here is divided into two exercises. Questions from the script are coloured blue for clarity and the answers follow below.

==EXERCISE 1==
===Dynamics from the TS region===

{{fontcolor|blue|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?}}
<div>The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its divergence being the zero vector at that point. To further characterise a point, one needs to evaluate the second derivatives at that point to determine whether it is a maximum or a minimum. The second derivative at the transition state will be negative, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the first derivative is zero and the second is positive, the point in question is a local minimum, which could be the lowest energy structure of the reactants or the products for example.</div>

{{fontcolor|blue|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.}}
<div>A simple system, consisting of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|Figure 2. By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this figure. The transition state is the minimum of this line.]]

The minimum of this curve is the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if the initial structure corresponds to the TS structure. By changing the initial internuclear distances, we can indentify the TS as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Figure 3. Internuclear distance vs. time plots are shown in this figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Figure 4. Contour plot with the initial geometry being the transition state geometry and initial momenta = 0 g.mol-1.pm.fs-1.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the TS the atoms are separated by more than 90 pm. Conversely, in case (c), the distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in close proximity to 90.8 pm.

But we can get a more accurate value now that we know where to look for the transition state. With some interval bisection, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 kJ/mol.pm along AB and -0.000 kJ/mol.pm along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

===Dynamics vs. MEP===

<div>{{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In a minimum energy path (MEP) calculation, the momentum of all particles is set to 0 in each step. Therefore there are no vibrations of the molecule at any point of the simulation. Of course this is unphysical as energy is lost, but it can be useful for some applications. In a dynamics simulation, on the other hand, vibrations of the molecule are observed as total energy of the system is constant.</div>

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Figure 5. Internuclear distance vs. time plot for r1 = 91.8 pm and r2 = 90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. Molecule BC vibrates, as can be seen from the oscillating distance between atoms B and C. AB and AC distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Figure 6. Momenta vs. time plot for r1 = 91.8 pm and r2 = 90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. The AB momentum increases and ultimately moves with constant momentum away from the molecule BC. Molecule BC vibrates, so the momentum BC oscillates.]]

Using the initial conditions r1=rts and r2=rts+1, the system would roll into the other valley on the PES with similar behaviour to what is described in the figures on the left and right.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translational and vibrational). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill so the initial point of the previous simulation will not be reached.

===Reactive and Unreactive Trajectories===

{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point, fluorine and hydrogen collided a few times, but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected. There is also no general way to determine a reactive trajectory. This is influenced by many factors. There were significant vibrations present in the product molecule, which is a result of conservation of energy and the fact that the reaction is exothermic.

{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
As mentioned in the introduction, one of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the reactants again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===

{{fontcolor|blue|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?Locate the approximate position of the transition state.}}
[[File:MRD 01576020 MEP HF+H.png|thumb|right|Figure 7. Surface plot for the MEP calculation to determine the energy of the HF + H system. A similar calculation was performed to find the energy of the H2 + F system.]]
The F + H2 --> HF + H reaction is exothermic and H + HF --> H2 + F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than the H-H bond. This makes sense as the former has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is a fluorine atom. For these initial conditions, the forces were: along AB: 0.000 kJ.mol-1.pm-1 and along BC: 0.001 kJ.mol-1.pm-1. Due to these minimum forces, this structure is very close to the TS structure. One can notice that the distance between the H atoms in the TS structure is very close to their bond distance. This phenomenon is expressed in Hammond's postulate which says that the TS structure will resemble that of the species close to it in energy. Since this reaction is very exothermic, the TS structure is close to the structure of reactants which are a H2 molecule and a F atom.

{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. These were calculated by first finding the energy of the TS structure, which was -433.981 kJ/mol. Energies of the reactants and products were then obtained by performing a MEP simulation starting slightly off the TS structure in both directions (towards the reactants and towards the products, the one for the products is shown in the figure on the right). The calculated energy of the HF + H complex was -360.662 kJ/mol and that of the H2 + F complex was -435.031 kJ/mol. These values were used to calculate the activation energies given above. The obtained values are only approximate as we cannot move the atom infinitely far from the molecule but are nevertheless in good agreement with experimental findings.

===Reaction Dynamics===
<div>A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g.mol-1.pm.fs-1.
{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.</div>

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left|Figure 8. A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms the products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|Figure 9. A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 fs of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is basically constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the BC pair which is moving apart, meaning that the total energy of the system is conserved. ]]

Let's now setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, an rFH = 238 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. Note that we are putting more energy into the system than the activation energy. The table below shows some observations and contour plots.

{| class="wikitable" border=1
! pFH/ g.mol-1.pm.fs-1 !! pHH/ g.mol-1.pm.fs-1 !! Einitial /kJ.mol-1 !! Reactive? !! Observations !! Illustration of the trajectory
|-
| -1.0 || -6.1 || -402.504 || No ||Here the system passed through the transition state structure, the hydrogen atom collided with the fluorine and then reformed reactants. || [[File:MRD 01576020 fhh table 1.png|400 px]]
|-
| -1.0 || -6.0 || -404.098 || No ||This is very similar to the case above. || [[File:MRD 01576020 fhh table 2.png|400 px]]
|-
| -1.0 || -5.0 || -414.098 || No ||This trajectory is also not reactive and is quite similar to the two above, again there was one collision with the fluorine atom. || [[File:MRD 01576020 fhh table 3.png|400 px]]
|-
| -1.0 || -4.1 || -421.388 || No ||Here, the fluorine atom did not collide with any atoms and the transition state was not surpassed. || [[File:MRD 01576020 fhh table 4.png|400 px]]
|-
| -1.0 || 0.0 || -434.098 || No ||Since even less energy was given the picture was very similar to the one above, but the hydrogen atom approached the fluorine even less. || [[File:MRD 01576020 fhh table 5.png|400 px]]
|-
| -1.0 || 2.1 || -427.588 || No ||Slightly more energy than above so the atoms were closer but the transition state was still not reached. || [[File:MRD 01576020 fhh table 6.png|400 px]]
|-
| -1.0 || 4.1 || -413.188 || No ||This trajectory was more interesting, the fluorine collided twice with one of the hydrogen atoms, but the reactants were reformed afterwards. || [[File:MRD 01576020 fhh table 7.png|400 px]]
|-
| -1.0 || 4.5 || -409.348 || Yes ||Interestingly, this trajectory was reactive despite the fact that the system has smaller energy than in some other cases. The total energy is therefore far from being the only parameter that influences the success of a given chemical reaction. || [[File:MRD 01576020 fhh table 11.png|400 px]]
|-
| -1.0 || 6.0 || -392.098 || No ||The system was very close to forming products here, fluorine and hydrogen collided 7 times, but the system rolled into the valley of reactants again. || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 6.1 || -390.788 || No ||The fluorine and hydrogen only collided once in this simulation. There is a pattern emerging in this table ... || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 7.0 || -378.098 || No ||This was very similar to the one above, there was one collision and then the reactants were reformed.|| [[File:MRD 01576020 fhh table 10.png|400 px]]
|}

From the table above it is obvious that whether the reaction will result in the formation of products is clearly not dependent only on the energy of vibration of the H2 molecule. We have also seen that these conditions where the molecule and the fluorine atom approach each other rather slowly rarely result in producing products. Maybe increasing the momentum pFH will help ...

{{fontcolor|blue|For the same initial position, increase slightly the momentum AB, and considerably reduce the overall energy of the system by reducing the momentum BC. What do you observe now?}}
There is very little vibration in the H2 molecule now and the F atom and the H2 molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.

Let's now focus on the reverse reaction, H + HF. When there is very little vibrational energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below (Polanyi's empirical rules).

An example of a reactive trajectory for this reaction would be produced by the following initial conditions: AB distance = 230 pm, BC distance = 74.2 pm, AB momentum = -1 g.mol-1.pm.fs-1 and BC momentum = 14.5 g.mol-1.pm.fs-1. The animation of this rather interesting trajectory is below.

[[File:MRD 01576020 animation rt2 GIF.gif|1000 px|thumb|left|Figure 10. A GIF showing the animation of the reactive trajectory of the HF + H reaction. The system passes the saddle point a few times but eventually forms the products.]]
[[File:MRD 01576020 skew.png|600 px|thumb|center|Figure 11. A skew plot of the same reactive trajectory as animated on the left.]]

{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions for a reactive trajectory. These rules tell us what modes need to have sufficient energy for certain types of TSs for a reaction to proceed. For early TSs, translational energy is more efficient than vibrational energy. For late TSs, on the other hand, vibrational energy will be more efficient. Why this is so is discussed in the figure below.

[[File:MRD 01576020 polanyi.png|800 px|thumb|center|Figure 12. A figure representing the main idea of Polanyi's empirical rules. Translational energy is more important for reactions with an early TS, and vibrational energy is more efficient for reactions with a late TS.]]

==Conclusions==
Two linear triatomic systems, HHH and HHF, were studied here. Their reaction trajectories were of particular interest. Transition state theory was used and the Polanyi's empirical rules were applied in finding suitable initial conditions for reactions. It was shown that even with a relatively simplistic model, without even invoking quantum mechanics, a remarkable amount of information can be gathered about a system and the insight into the dynamics of molecular reactions is eye-opening.

==References==

MRD:01576020

2020-05-15T20:59:58Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
Modelling the dynamics of molecular reactions has been enabled by the emergence of computers and the field of computational chemistry is an ever-expanding field. Despite most approaches being purely theoretical, ''ab initio'', calculations, the insight given is invaluable for understanding the pathways of chemical reactions. It is often also not necessary to include quantum mechanics, as atoms are of sufficient mass to be reasonably accurately modelled by Newton's equations of motion. Isolated systems in the gas phase only experience interatomic interactions which can be represented by a potential energy surface (PES) which is a function only of atomic positions. The force on a given interatomic coordinate can be calculated as follows:

<math> F = - { \partial V(r_1,r_2,...)\over \partial r_i}</math>

where F = <math>{dp_i \over dt}</math>

Calculating the forces and adjusting the system in consecutive steps results in a trajectory which shows how a system evolves in time. Such calculations have almost endless application from chemical biology and medicine<ref>{{cite book|author1=Pregeljc, Domen|author2=Jug, Urska|author3=Mavri, Janez|author4=Stare, Jernej|lastauthoramp=yes|year=2018|title=Why does the Y326I mutant of monoamine oxidase B decompose an endogenous amphetamine at a slower rate than the wild type enzyme? Reaction step elucidated by multiscale molecular simulations.|journal= Phys Chem Chem Phys.|volume= 10|issue= 6|pages= 4181-4188|doi= 10.1039/c7cp07069a}}</ref> to materials science<ref>{{cite book|author1=Artritha, Nonguch|author2=Urban, Alexander|lastauthoramp=yes|year=2016|title=An implementation of artificial neural-network potentials for atomistic materials simulations: Performance for TiO2|journal= Phys Chem Chem Phys.|volume= 114|pages= 135-150|doi= 10.1016/j.commatsci.2015.11.047}}</ref>.

It is worth mentioning some main aspects about the transition state (TS) theory here as it is central to the concepts used here. It is purely classical - it ignores any quantum mechanical effects such as tunnelling, which is incredibly small anyway. It also assumes that the transition state structure is in equilibrium with the reactants, which is called a quasi-equilibrium. TS theory also assumes that the reactants have energies which follow a Boltzmann distribution, which is not a bad assumption as long as the system had time to equilibrate. This assumption, however is not very applicable to the system studied here. Another crucial assumption of the TS theory, developed in the 1930s, is that once the TS is surpassed, the products will necessarily be formed, which was not the case in some simulations performed here. This latter assumption is also the reason why theoretically derived reaction rates using the TS state theory overestimate experimentally obtained reaction rates. This can be improved by adding a correcting factor, but to do that one must already have extensive knowledge of the system in question. The transition point is defined as the maximum on the minimum energy path going from reactants to products. This is called a saddle point and the net force on the system is 0.<ref>{{cite book|author1=Laidler, Keith J.|lastauthoramp=yes|year=2016|title=Chemical Kinetics}}</ref><ref>{{cite book|author1=Atkins, Peter|author2=de Paula, Julio|lastauthoramp=yes|year=2014|title=Atkins' Physical Chemistry}}</ref>

We will deal here with triatomic linear systems, as depicted in the following Figure 1.
[[File:Y2C8.png|thumb|centre|500px| Figure 1. A triatomic system used here. The atoms will either all be Hs or in the last case one H will be replaced with a F. Figure taken from https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD]]

Terms like reactive and unreactive trajectories will be used often. A reactive trajectory is one which leads the reaction to completeness, i. e. the products are formed. In contrast an unreactive trajectory will not form products. The work here is divided into two exercises. Questions from the script are coloured blue for clarity and the answers follow below.

==EXERCISE 1==
===Dynamics from the TS region===

{{fontcolor|blue|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?}}
<div>The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its divergence being the zero vector at that point. To further characterise a point, one needs to evaluate the second derivatives at that point to determine whether it is a maximum or a minimum. The second derivative at the transition state will be negative, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the first derivative is zero and the second is positive, the point in question is a local minimum, which could be the lowest energy structure of the reactants or the products for example.</div>

{{fontcolor|blue|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.}}
<div>A simple system, consisting of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|Figure 2. By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this figure. The transition state is the minimum of this line.]]

The minimum of this curve is the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if the initial structure corresponds to the TS structure. By changing the initial internuclear distances, we can indentify the TS as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Figure 3. Internuclear distance vs. time plots are shown in this figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Figure 4. Contour plot with the initial geometry being the transition state geometry and initial momenta = 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the TS the atoms are separated by more than 90 pm. Conversely, in case (c), the distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in close proximity of 90.8 pm.

But we can get a more accurate value now that we know where to look for the transition state. With some interval bisection, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

===Dynamics vs. MEP===

<div>{{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In a minimum energy path (MEP) calculation, the momentum of all particles is set to 0 in each step. Therefore there are no vibrations of the molecule at any point of the simulation. Of course this is unphysical as energy is lost, but it can be useful for some applications. In a dynamics simulation, on the other hand, vibrations of the molecule are observed as total energy of the system is constant.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Figure 5. Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. Molecule BC vibrates, as can be seen from the oscillating distance between atoms B and C. AB and AC distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Figure 6. Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. The AB momentum increases and ultimately moves with constant momentum away from the molecule BC. Molecule BC vibrates, so the momentum BC oscillates.]]</div>

Using the initial conditions r1=rts and r2=r<sub<ts+1, the system would roll into the other valley on the PES with similar behaviour to what is described in the figures on the left and right.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translational and vibrational). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill so the initial point of the previous simulation will not be reached.

===Reactive and Unreactive Trajectories===

{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point, fluorine and hydrogen collided a few times, but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected. There is also no general way to determine a reactive trajectory. This is influenced by many factors. There were significant vibrations present in the product molecule, which is a result of conservation of energy and the fact that the reaction is exothermic.

{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
As mentioned in the introduction, one of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the reactants again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===

{{fontcolor|blue|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?Locate the approximate position of the transition state.}}
[[File:MRD 01576020 MEP HF+H.png|thumb|right|Figure 7. Surface plot for the MEP calculation to determine the energy of the HF + H system. A similar calculation was performed to find the energy of the H2 + F system.]]
The F + H2 --> HF + H reaction is exothermic and H + HF --> H2 + F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than the H-H bond. This makes sense as the former has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is a fluorine atom. For these initial conditions, the forces were: along AB: 0.000 kJ.mol-1.pm-1 and along BC: 0.001 kJ.mol-1.pm-1. Due to these minimum forces, this structure is very close to the TS structure. One can notice that the distance between the H atoms in the TS structure is very close to their bond distance. This phenomenon is expressed in Hammond's postulate which says that the TS structure will resemble that of the species close to it in energy. Since this reaction is very exothermic, the TS structure is close to the structure of reactants which are a H2 molecule and a F atom.

{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. These were calculated by first finding the energy of the TS structure, which was -433.981 kJ/mol. Energies of the reactants and products were then obtained by performing a MEP simulation starting slightly off the TS structure in both directions (towards the reactants and towards the products, the one for the products is shown in the figure on the right). The calculated energy of the HF + H complex was -360.662 kJ/mol and that of the H2 + F complex was -435.031 kJ/mol. These values were used to calculate the activation energies given above. The obtained values are only approximate as we cannot move the atom infinitely far from the molecule but are nevertheless in good agreement with experimental findings.

===Reaction Dynamics===
<div>A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g.mol-1.pm.fs-1.
{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.</div>

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left|Figure 8. A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms the products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|Figure 9. A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 fs of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is basically constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the BC pair which is moving apart, meaning that the total energy of the system is conserved. ]]

Let's now setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, an rFH = 238 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. Note that we are putting more energy into the system than the activation energy. The table below shows some observations and contour plots.

{| class="wikitable" border=1
! pFH/ g.mol-1.pm.fs-1 !! pHH/ g.mol-1.pm.fs-1 !! Einitial /kJ.mol-1 !! Reactive? !! Observations !! Illustration of the trajectory
|-
| -1.0 || -6.1 || -402.504 || No ||Here the system passed through the transition state structure, the hydrogen atom collided with the fluorine and then reformed reactants. || [[File:MRD 01576020 fhh table 1.png|400 px]]
|-
| -1.0 || -6.0 || -404.098 || No ||This is very similar to the case above. || [[File:MRD 01576020 fhh table 2.png|400 px]]
|-
| -1.0 || -5.0 || -414.098 || No ||This trajectory is also not reactive and is quite similar to the two above, again there was one collision with the fluorine atom. || [[File:MRD 01576020 fhh table 3.png|400 px]]
|-
| -1.0 || -4.1 || -421.388 || No ||Here, the fluorine atom did not collide with any atoms and the transition state was not surpassed. || [[File:MRD 01576020 fhh table 4.png|400 px]]
|-
| -1.0 || 0.0 || -434.098 || No ||Since even less energy was given the picture was very similar to the one above, but the hydrogen atom approached the fluorine even less. || [[File:MRD 01576020 fhh table 5.png|400 px]]
|-
| -1.0 || 2.1 || -427.588 || No ||Slightly more energy than above so the atoms were closer but the transition state was still not reached. || [[File:MRD 01576020 fhh table 6.png|400 px]]
|-
| -1.0 || 4.1 || -413.188 || No ||This trajectory was more interesting, the fluorine collided twice with one of the hydrogen atoms, but the reactants were reformed afterwards. || [[File:MRD 01576020 fhh table 7.png|400 px]]
|-
| -1.0 || 4.5 || -409.348 || Yes ||Interestingly, this trajectory was reactive despite the fact that the system has smaller energy than in some other cases. The total energy is therefore far from being the only parameter that influences the success of a given chemical reaction. || [[File:MRD 01576020 fhh table 11.png|400 px]]
|-
| -1.0 || 6.0 || -392.098 || No ||The system was very close to forming products here, fluorine and hydrogen collided 7 times, but the system rolled into the valley of reactants again. || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 6.1 || -390.788 || No ||The fluorine and hydrogen only collided once in this simulation. There is a pattern emerging in this table ... || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 7.0 || -378.098 || No ||This was very similar to the one above, there was one collision and then the reactants were reformed.|| [[File:MRD 01576020 fhh table 10.png|400 px]]
|}

From the table above it is obvious that whether the reaction will result in the formation of products is clearly not dependent only on the energy of vibration of the H2 molecule. We have also seen that these conditions where the molecule and the fluorine atom approach each other rather slowly rarely result in producing products. Maybe increasing the momentum pFH will help ...

{{fontcolor|blue|For the same initial position, increase slightly the momentum AB, and considerably reduce the overall energy of the system by reducing the momentum BC. What do you observe now?}}
There is very little vibration in the H2 molecule now and the F atom and the H2 molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.

Let's now focus on the reverse reaction, H + HF. When there is very little vibrational energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below (Polanyi's empirical rules).

An example of a reactive trajectory for this reaction would be produced by the following initial conditions: AB distance = 230 pm, BC distance = 74.2 pm, AB momentum = -1 g.mol-1.pm.fs-1 and BC momentum = 14.5 g.mol-1.pm.fs-1. The animation of this rather interesting trajectory is below.

[[File:MRD 01576020 animation rt2 GIF.gif|1000 px|thumb|left|Figure 10. A GIF showing the animation of the reactive trajectory of the HF + H reaction. The system passes the saddle point a few times but eventually forms the products.]]
[[File:MRD 01576020 skew.png|600 px|thumb|center|Figure 11. A skew plot of the same reactive trajectory as animated on the left.]]

{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions for a reactive trajectory. These rules tell us what modes need to have sufficient energy for certain types of TSs for a reaction to proceed. For early TSs, translational energy is more efficient than vibrational energy. For late TSs, on the other hand, vibrational energy will be more efficient. Why this is so is discussed in the figure below.

[[File:MRD 01576020 polanyi.png|800 px|thumb|center|Figure 12. A figure representing the main idea of Polanyi's empirical rules. Translational energy is more important for reactions with an early TS, and vibrational energy is more efficient for reactions with a late TS.]]

==Conclusions==
Two linear triatomic systems, HHH and HHF, were studied here. Their reaction trajectories were of particular interest. Transition state theory was used and the Polanyi's empirical rules were applied in finding suitable initial conditions for reactions. It was shown that even with a relatively simplistic model, without even invoking quantum mechanics, a remarkable amount of information can be gathered about a system and the insight into the dynamics of molecular reactions is eye-opening.

==References==

MRD:01576020

2020-05-15T20:58:07Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
Modelling the dynamics of molecular reactions has been enabled by the emergence of computers and the field of computational chemistry is an ever-expanding field. Despite most approaches being purely theoretical, ''ab initio'', calculations, the insight given is invaluable for understanding the pathways of chemical reactions. It is often also not necessary to include quantum mechanics, as atoms are of sufficient mass to be reasonably accurately modelled by Newton's equations of motion. Isolated systems in the gas phase only experience interatomic interactions which can be represented by a potential energy surface (PES) which is a function only of atomic positions. The force on a given interatomic coordinate can be calculated as follows:

<math> F = - { \partial V(r_1,r_2,...)\over \partial r_i}</math>

where F = <math>{dp_i \over dt}</math>

Calculating the forces and adjusting the system in consecutive steps results in a trajectory which shows how a system evolves in time. Such calculations have almost endless application from chemical biology and medicine<ref>{{cite book|author1=Pregeljc, Domen|author2=Jug, Urska|author3=Mavri, Janez|author4=Stare, Jernej|lastauthoramp=yes|year=2018|title=Why does the Y326I mutant of monoamine oxidase B decompose an endogenous amphetamine at a slower rate than the wild type enzyme? Reaction step elucidated by multiscale molecular simulations.|journal= Phys Chem Chem Phys.|volume= 10|issue= 6|pages= 4181-4188|doi= 10.1039/c7cp07069a}}</ref> to materials science<ref>{{cite book|author1=Artritha, Nonguch|author2=Urban, Alexander|lastauthoramp=yes|year=2016|title=An implementation of artificial neural-network potentials for atomistic materials simulations: Performance for TiO2|journal= Phys Chem Chem Phys.|volume= 114|pages= 135-150|doi= 10.1016/j.commatsci.2015.11.047}}</ref>.

It is worth mentioning some main aspects about the transition state (TS) theory here as it is central to the concepts used here. It is purely classical - it ignores any quantum mechanical effects such as tunnelling, which is incredibly small anyway. It also assumes that the transition state structure is in equilibrium with the reactants, which is called a quasi-equilibrium. TS theory also assumes that the reactants have energies which follow a Boltzmann distribution, which is not a bad assumption as long as the system had time to equilibrate. This assumption, however is not very applicable to the system studied here. Another crucial assumption of the TS theory, developed in the 1930s, is that once the TS is surpassed, the products will necessarily be formed, which was not the case in some simulations performed here. This latter assumption is also the reason why theoretically derived reaction rates using the TS state theory overestimate experimentally obtained reaction rates. This can be improved by adding a correcting factor, but to do that one must already have extensive knowledge of the system in question. The transition point is defined as the maximum on the minimum energy path going from reactants to products. This is called a saddle point and the net force on the system is 0.<ref>{{cite book|author1=Laidler, Keith J.|lastauthoramp=yes|year=2016|title=Chemical Kinetics}}</ref><ref>{{cite book|author1=Atkins, Peter|author2=de Paula, Julio|lastauthoramp=yes|year=2014|title=Atkins' Physical Chemistry}}</ref>

We will deal here with triatomic linear systems, as depicted in the following Figure 1.
[[File:Y2C8.png|thumb|centre|500px| Figure 1. A triatomic system used here. The atoms will either all be Hs or in the last case one H will be replaced with a F. Figure taken from https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD]]

Terms like reactive and unreactive trajectories will be used often. A reactive trajectory is one which leads the reaction to completeness, i. e. the products are formed. In contrast an unreactive trajectory will not form products. The work here is divided into two exercises. Questions from the script are coloured blue for clarity and the answers follow below.

==EXERCISE 1==
===Dynamics from the TS region===

{{fontcolor|blue|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?}}
<div>The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its divergence being the zero vector at that point. To further characterise a point, one needs to evaluate the second derivatives at that point to determine whether it is a maximum or a minimum. The second derivative at the transition state will be negative, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the first derivative is zero and the second is positive, the point in question is a local minimum, which could be the lowest energy structure of the reactants or the products for example.</div>

{{fontcolor|blue|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.}}
<div>A simple system, consisting of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|Figure 2. By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this figure. The transition state is the minimum of this line.]]

The minimum of this curve is the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if the initial structure corresponds to the TS structure. By changing the initial internuclear distances, we can indentify the TS as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Figure 3. Internuclear distance vs. time plots are shown in this figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Figure 4. Contour plot with the initial geometry being the transition state geometry and initial momenta = 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the TS the atoms are separated by more than 90 pm. Conversely, in case (c), the distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in close proximity of 90.8 pm.

But we can get a more accurate value now that we know where to look for the transition state. With some interval bisection, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

===Dynamics vs. MEP===

<div>{{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In a minimum energy path (MEP) calculation, the momentum of all particles is set to 0 in each step. Therefore there are no vibrations of the molecule at any point of the simulation. Of course this is unphysical as energy is lost, but it can be useful for some applications. In a dynamics simulation, on the other hand, vibrations of the molecule are observed as total energy of the system is constant.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Figure 5. Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. Molecule BC vibrates, as can be seen from the oscillating distance between atoms B and C. AB and AC distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. The AB momentum increases and ultimately moves with constant momentum away from the molecule BC. Molecule BC vibrates, so the momentum BC oscillates.]]</div>

Using the initial conditions r1=rts and r2=r<sub<ts+1, the system would roll into the other valley on the PES with similar behaviour to what is described in the figures on the left and right.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translational and vibrational). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill so the initial point of the previous simulation will not be reached.

===Reactive and Unreactive Trajectories===

{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point, fluorine and hydrogen collided a few times, but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected. There is also no general way to determine a reactive trajectory. This is influenced by many factors. There were significant vibrations present in the product molecule, which is a result of conservation of energy and the fact that the reaction is exothermic.

{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
As mentioned in the introduction, one of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the reactants again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===

{{fontcolor|blue|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?Locate the approximate position of the transition state.}}
[[File:MRD 01576020 MEP HF+H.png|thumb|right|Figure 6.Surface plot for the MEP calculation to determine the energy of the HF + H system. A similar calculation was performed to find the energy of the H2 + F system.]]
The F + H2 --> HF + H reaction is exothermic and H + HF --> H2 + F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than the H-H bond. This makes sense as the former has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is a fluorine atom. For these initial conditions, the forces were: along AB: 0.000 kJ.mol-1.pm-1 and along BC: 0.001 kJ.mol-1.pm-1. Due to these minimum forces, this structure is very close to the TS structure. One can notice that the distance between the H atoms in the TS structure is very close to their bond distance. This phenomenon is expressed in Hammond's postulate which says that the TS structure will resemble that of the species close to it in energy. Since this reaction is very exothermic, the TS structure is close to the structure of reactants which are a H2 molecule and a F atom.

{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. These were calculated by first finding the energy of the TS structure, which was -433.981 kJ/mol. Energies of the reactants and products were then obtained by performing a MEP simulation starting slightly off the TS structure in both directions (towards the reactants and towards the products, the one for the products is shown in the figure on the right). The calculated energy of the HF + H complex was -360.662 kJ/mol and that of the H2 + F complex was -435.031 kJ/mol. These values were used to calculate the activation energies given above. The obtained values are only approximate as we cannot move the atom infinitely far from the molecule but are nevertheless in good agreement with experimental findings.

===Reaction Dynamics===
<div>A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g.mol-1.pm.fs-1.
{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.</div>

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left|Figure 7. A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms the products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|Figure 8. A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 fs of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is basically constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the BC pair which is moving apart, meaning that the total energy of the system is conserved. ]]

Let's now setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, an rFH = 238 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. Note that we are putting more energy into the system than the activation energy. The table below shows some observations and contour plots.

{| class="wikitable" border=1
! pFH/ g.mol-1.pm.fs-1 !! pHH/ g.mol-1.pm.fs-1 !! Einitial /kJ.mol-1 !! Reactive? !! Observations !! Illustration of the trajectory
|-
| -1.0 || -6.1 || -402.504 || No ||Here the system passed through the transition state structure, the hydrogen atom collided with the fluorine and then reformed reactants. || [[File:MRD 01576020 fhh table 1.png|400 px]]
|-
| -1.0 || -6.0 || -404.098 || No ||This is very similar to the case above. || [[File:MRD 01576020 fhh table 2.png|400 px]]
|-
| -1.0 || -5.0 || -414.098 || No ||This trajectory is also not reactive and is quite similar to the two above, again there was one collision with the fluorine atom. || [[File:MRD 01576020 fhh table 3.png|400 px]]
|-
| -1.0 || -4.1 || -421.388 || No ||Here, the fluorine atom did not collide with any atoms and the transition state was not surpassed. || [[File:MRD 01576020 fhh table 4.png|400 px]]
|-
| -1.0 || 0.0 || -434.098 || No ||Since even less energy was given the picture was very similar to the one above, but the hydrogen atom approached the fluorine even less. || [[File:MRD 01576020 fhh table 5.png|400 px]]
|-
| -1.0 || 2.1 || -427.588 || No ||Slightly more energy than above so the atoms were closer but the transition state was still not reached. || [[File:MRD 01576020 fhh table 6.png|400 px]]
|-
| -1.0 || 4.1 || -413.188 || No ||This trajectory was more interesting, the fluorine collided twice with one of the hydrogen atoms, but the reactants were reformed afterwards. || [[File:MRD 01576020 fhh table 7.png|400 px]]
|-
| -1.0 || 4.5 || -409.348 || Yes ||Interestingly, this trajectory was reactive despite the fact that the system has smaller energy than in some other cases. The total energy is therefore far from being the only parameter that influences the success of a given chemical reaction. || [[File:MRD 01576020 fhh table 11.png|400 px]]
|-
| -1.0 || 6.0 || -392.098 || No ||The system was very close to forming products here, fluorine and hydrogen collided 7 times, but the system rolled into the valley of reactants again. || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 6.1 || -390.788 || No ||The fluorine and hydrogen only collided once in this simulation. There is a pattern emerging in this table ... || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 7.0 || -378.098 || No ||This was very similar to the one above, there was one collision and then the reactants were reformed.|| [[File:MRD 01576020 fhh table 10.png|400 px]]
|}

From the table above it is obvious that whether the reaction will result in the formation of products is clearly not dependent only on the energy of vibration of the H2 molecule. We have also seen that these conditions where the molecule and the fluorine atom approach each other rather slowly rarely result in producing products. Maybe increasing the momentum pFH will help ...

{{fontcolor|blue|For the same initial position, increase slightly the momentum AB, and considerably reduce the overall energy of the system by reducing the momentum BC. What do you observe now?}}
There is very little vibration in the H2 molecule now and the F atom and the H2 molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.

Let's now focus on the reverse reaction, H + HF. When there is very little vibrational energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below (Polanyi's empirical rules).

An example of a reactive trajectory for this reaction would be produced by the following initial conditions: AB distance = 230 pm, BC distance = 74.2 pm, AB momentum = -1 g.mol-1.pm.fs-1 and BC momentum = 14.5 g.mol-1.pm.fs-1. The animation of this rather interesting trajectory is below.

[[File:MRD 01576020 animation rt2 GIF.gif|1000 px|thumb|left|Figure 9. A GIF showing the animation of the reactive trajectory of the HF + H reaction. The system passes the saddle point a few times but eventually forms the products.]]
[[File:MRD 01576020 skew.png|600 px|thumb|center|Figure 10. A skew plot of the same reactive trajectory as animated on the left.]]

{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions for a reactive trajectory. These rules tell us what modes need to have sufficient energy for certain types of TSs for a reaction to proceed. For early TSs, translational energy is more efficient than vibrational energy. For late TSs, on the other hand, vibrational energy will be more efficient. Why this is so is discussed in the figure below.

[[File:MRD 01576020 polanyi.png|800 px|thumb|center|Figure 11. A figure representing the main idea of Polanyi's empirical rules. Translational energy is more important for reactions with an early TS, and vibrational energy is more efficient for reactions with a late TS.]]

==Conclusions==
Two linear triatomic systems, HHH and HHF, were studied here. Their reaction trajectories were of particular interest. Transition state theory was used and the Polanyi's empirical rules were applied in finding suitable initial conditions for reactions. It was shown that even with a relatively simplistic model, without even invoking quantum mechanics, a remarkable amount of information can be gathered about a system and the insight into the dynamics of molecular reactions is eye-opening.

==References==

MRD:01576020

2020-05-15T20:53:08Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
Modelling the dynamics of molecular reactions has been enabled by the emergence of computers and the field of computational chemistry is an ever-expanding field. Despite most approaches being purely theoretical, ''ab initio'', calculations, the insight given is invaluable for understanding the pathways of chemical reactions. It is often also not necessary to include quantum mechanics, as atoms are of sufficient mass to be reasonably accurately modelled by Newton's equations of motion. Isolated systems in the gas phase only experience interatomic interactions which can be represented by a potential energy surface (PES) which is a function only of atomic positions. The force on a given interatomic coordinate can be calculated as follows:

<math> F = - { \partial V(r_1,r_2,...)\over \partial r_i}</math>

where F = <math>{dp_i \over dt}</math>

Calculating the forces and adjusting the system in consecutive steps results in a trajectory which shows how a system evolves in time. Such calculations have almost endless application from chemical biology and medicine<ref>{{cite book|author1=Pregeljc, Domen|author2=Jug, Urska|author3=Mavri, Janez|author4=Stare, Jernej|lastauthoramp=yes|year=2018|title=Why does the Y326I mutant of monoamine oxidase B decompose an endogenous amphetamine at a slower rate than the wild type enzyme? Reaction step elucidated by multiscale molecular simulations.|journal= Phys Chem Chem Phys.|volume= 10|issue= 6|pages= 4181-4188|doi= 10.1039/c7cp07069a}}</ref> to materials science<ref>{{cite book|author1=Artritha, Nonguch|author2=Urban, Alexander|lastauthoramp=yes|year=2016|title=An implementation of artificial neural-network potentials for atomistic materials simulations: Performance for TiO2|journal= Phys Chem Chem Phys.|volume= 114|pages= 135-150|doi= 10.1016/j.commatsci.2015.11.047}}</ref>.

It is worth mentioning some main aspects about the transition state (TS) theory here as it is central to the concepts used here. It is purely classical - it ignores any quantum mechanical effects such as tunnelling, which is incredibly small anyway. It also assumes that the transition state structure is in equilibrium with the reactants, which is called a quasi-equilibrium. TS theory also assumes that the reactants have energies which follow a Boltzmann distribution, which is not a bad assumption as long as the system had time to equilibrate. This assumption, however is not very applicable to the system studied here. Another crucial assumption of the TS theory, developed in the 1930s, is that once the TS is surpassed, the products will necessarily be formed, which was not the case in some simulations performed here. This latter assumption is also the reason why theoretically derived reaction rates using the TS state theory overestimate experimentally obtained reaction rates. This can be improved by adding a correcting factor, but to do that one must already have extensive knowledge of the system in question. The transition point is defined as the maximum on the minimum energy path going from reactants to products. This is called a saddle point and the net force on the system is 0.<ref>{{cite book|author1=Laidler, Keith J.|lastauthoramp=yes|year=2016|title=Chemical Kinetics}}</ref><ref>{{cite book|author1=Atkins, Peter|author2=de Paula, Julio|lastauthoramp=yes|year=2014|title=Atkins' Physical Chemistry}}</ref>

We will deal here with triatomic linear systems, as depicted in the following Figure 1.
[[File:Y2C8.png|thumb|centre|500px| Figure 1. A triatomic system used here. The atoms will either all be Hs or in the last case one H will be replaced with a F. Figure taken from https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD]]

Terms like reactive and unreactive trajectories will be used often. A reactive trajectory is one which leads the reaction to completeness, i. e. the products are formed. In contrast an unreactive trajectory will not form products. The work here is divided into two exercises. Questions from the script are coloured blue for clarity and the answers follow below.

==EXERCISE 1==
===Dynamics from the TS region===

{{fontcolor|blue|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?}}
<div>The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its divergence being the zero vector at that point. To further characterise a point, one needs to evaluate the second derivatives at that point to determine whether it is a maximum or a minimum. The second derivative at the transition state will be negative, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the first derivative is zero and the second is positive, the point in question is a local minimum, which could be the lowest energy structure of the reactants or the products for example.</div>

{{fontcolor|blue|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.}}
<div>A simple system, consisting of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|Figure 2. By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this figure. The transition state is the minimum of this line.]]

The minimum of this curve is the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if the initial structure corresponds to the TS structure. By changing the initial internuclear distances, we can indentify the TS as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Figure 3. Internuclear distance vs. time plots are shown in this figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Figure 4. Contour plot with the initial geometry being the transition state geometry and initial momenta = 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the TS the atoms are separated by more than 90 pm. Conversely, in case (c), the distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in close proximity of 90.8 pm.

But we can get a more accurate value now that we know where to look for the transition state. With some interval bisection, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

===Dynamics vs. MEP===

<div>{{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In a minimum energy path (MEP) calculation, the momentum of all particles is set to 0 in each step. Therefore there are no vibrations of the molecule at any point of the simulation. Of course this is unphysical as energy is lost, but it can be useful for some applications. In a dynamics simulation, on the other hand, vibrations of the molecule are observed as total energy of the system is constant.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Figure 5. Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. Molecule BC vibrates, as can be seen from the oscillating distance between atoms B and C. AB and AC distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. The AB momentum increases and ultimately moves with constant momentum away from the molecule BC. Molecule BC vibrates, so the momentum BC oscillates.]]</div>

Using the initial conditions r1=rts and r2=r<sub<ts+1, the system would roll into the other valley on the PES with similar behaviour to what is described in the figures on the left and right.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translational and vibrational). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill so the initial point of the previous simulation will not be reached.

===Reactive and Unreactive Trajectories===

{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point, fluorine and hydrogen collided a few times, but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected. There is also no general way to determine a reactive trajectory. This is influenced by many factors. There were significant vibrations present in the product molecule, which is a result of conservation of energy and the fact that the reaction is exothermic.

{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
As mentioned in the introduction, one of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the reactants again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===

{{fontcolor|blue|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?Locate the approximate position of the transition state.}}
[[File:MRD 01576020 MEP HF+H.png|thumb|right|Figure 6.Surface plot for the MEP calculation to determine the energy of the HF + H system. A similar calculation was performed to find the energy of the H2 + F system.]]
The F + H2 --> HF + H reaction is exothermic and H + HF --> H2 + F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than the H-H bond. This makes sense as the former has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is a fluorine atom. For these initial conditions, the forces were: along AB: 0.000 kJ.mol-1.pm-1 and along BC: 0.001 kJ.mol-1.pm-1. Due to these minimum forces, this structure is very close to the TS structure. One can notice that the distance between the H atoms in the TS structure is very close to their bond distance. This phenomenon is expressed in Hammond's postulate which says that the TS structure will resemble that of the species close to it in energy. Since this reaction is very exothermic, the TS structure is close to the structure of reactants which are a H2 molecule and a F atom.

{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. These were calculated by first finding the energy of the TS structure, which was -433.981 kJ/mol. Energies of the reactants and products were then obtained by performing a MEP simulation starting slightly off the TS structure in both directions (towards the reactants and towards the products, the one for the products is shown in the figure on the right). The calculated energy of the HF + H complex was -360.662 kJ/mol and that of the H2 + F complex was -435.031 kJ/mol. These values were used to calculate the activation energies given above. The obtained values are only approximate as we cannot move the atom infinitely far from the molecule but are nevertheless in good agreement with experimental findings.

===Reaction Dynamics===
<div>A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g.mol-1.pm.fs-1.
{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.</div>

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left|Figure 7. A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms the products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|Figure 8. A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 fs of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is basically constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the BC pair which is moving apart, meaning that the total energy of the system is conserved. ]]

Let's now setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, an rFH = 238 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. Note that we are putting more energy into the system than the activation energy. The table below shows some observations and contour plots.

{| class="wikitable" border=1
! pFH/ g.mol-1.pm.fs-1 !! pHH/ g.mol-1.pm.fs-1 !! Einitial /kJ.mol-1 !! Reactive? !! Observations !! Illustration of the trajectory
|-
| -1.0 || -6.1 || -402.504 || No ||Here the system passed through the transition state structure, the hydrogen atom collided with the fluorine and then reformed reactants. || [[File:MRD 01576020 fhh table 1.png|400 px]]
|-
| -1.0 || -6.0 || -404.098 || No ||This is very similar to the case above. || [[File:MRD 01576020 fhh table 2.png|400 px]]
|-
| -1.0 || -5.0 || -414.098 || No ||This trajectory is also not reactive and is quite similar to the two above, again there was one collision with the fluorine atom. || [[File:MRD 01576020 fhh table 3.png|400 px]]
|-
| -1.0 || -4.1 || -421.388 || No ||Here, the fluorine atom did not collide with any atoms and the transition state was not surpassed. || [[File:MRD 01576020 fhh table 4.png|400 px]]
|-
| -1.0 || 0.0 || -434.098 || No ||Since even less energy was given the picture was very similar to the one above, but the hydrogen atom approached the fluorine even less. || [[File:MRD 01576020 fhh table 5.png|400 px]]
|-
| -1.0 || 2.1 || -427.588 || No ||Slightly more energy than above so the atoms were closer but the transition state was still not reached. || [[File:MRD 01576020 fhh table 6.png|400 px]]
|-
| -1.0 || 4.1 || -413.188 || No ||This trajectory was more interesting, the fluorine collided twice with one of the hydrogen atoms, but the reactants were reformed afterwards. || [[File:MRD 01576020 fhh table 7.png|400 px]]
|-
| -1.0 || 4.5 || -409.348 || Yes ||Interestingly, this trajectory was reactive despite the fact that the system has smaller energy than in some other cases. The total energy is therefore far from being the only parameter that influences the success of a given chemical reaction. || [[File:MRD 01576020 fhh table 11.png|400 px]]
|-
| -1.0 || 6.0 || -392.098 || No ||The system was very close to forming products here, fluorine and hydrogen collided 7 times, but the system rolled into the valley of reactants again. || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 6.1 || -390.788 || No ||The fluorine and hydrogen only collided once in this simulation. There is a pattern emerging in this table ... || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 7.0 || -378.098 || No ||This was very similar to the one above, there was one collision and then the reactants were reformed.|| [[File:MRD 01576020 fhh table 10.png|400 px]]
|}

From the table above it is obvious that whether the reaction will result in the formation of products is clearly not dependent only on the energy of vibration of the H2 molecule. We have also seen that these conditions where the molecule and the fluorine atom approach each other rather slowly rarely result in producing products. Maybe increasing the momentum pFH will help ...

{{fontcolor|blue|For the same initial position, increase slightly the momentum AB, and considerably reduce the overall energy of the system by reducing the momentum BC. What do you observe now?}}
There is very little vibration in the H2 molecule now and the F atom and the H2 molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.

Let's now focus on the reverse reaction, H + HF. When there is very little vibrational energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below (Polanyi's empirical rules).

An example of a reactive trajectory for this reaction would be produced by the following initial conditions: AB distance = 230 pm, BC distance = 74.2 pm, AB momentum = -1 g.mol-1.pm.fs-1 and BC momentum = 14.5 g.mol-1.pm.fs-1. The animation of this rather interesting trajectory is below.

[[File:MRD 01576020 animation rt2 GIF.gif|1000 px|thumb|left|Figure 9. A GIF showing the animation of the reactive trajectory of the HF + H reaction. The system passes the saddle point a few times but eventually forms the products.]]
[[File:MRD 01576020 skew.png|600 px|thumb|center|Figure 10. A skew plot of the same reactive trajectory as animated on the left.]]

{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions for a reactive trajectory. These rules tell us what modes need to have sufficient energy for certain types of TSs for a reaction to proceed. For early TSs, translational energy is more efficient than vibrational energy. For late TSs, on the other hand, vibrational energy will be more efficient. Why this is so is discussed in the figure below.

[[File:MRD 01576020 polanyi.png|800 px|thumb|center|Figure 11. A figure representing the main idea of Polanyi's empirical rules. Translational energy is more important for reactions with an early TS, and vibrational energy is more efficient for reactions with a late TS.]]

==References==

MRD:01576020

2020-05-15T20:43:51Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
Modelling the dynamics of molecular reactions has been enabled by the emergence of computers and the field of computational chemistry is an ever-expanding field. Despite most approaches being purely theoretical, ''ab initio'', calculations, the insight given is invaluable for understanding the pathways of chemical reactions. It is often also not necessary to include quantum mechanics, as atoms are of sufficient mass to be reasonably accurately modelled by Newton's equations of motion. Isolated systems in the gas phase only experience interatomic interactions which can be represented by a potential energy surface (PES) which is a function only of atomic positions. The force on a given interatomic coordinate can be calculated as follows:

<math> F = - { \partial V(r_1,r_2,...)\over \partial r_i}</math> (1)

where F = <math>{dp_i \over dt}</math> (2)

Calculating the forces and adjusting the system in consecutive steps results in a trajectory which shows how a system evolves in time. Such calculations have almost endless application from chemical biology and medicine<ref>{{cite book|author1=Pregeljc, Domen|author2=Jug, Urska|author3=Mavri, Janez|author4=Stare, Jernej|lastauthoramp=yes|year=2018|title=Why does the Y326I mutant of monoamine oxidase B decompose an endogenous amphetamine at a slower rate than the wild type enzyme? Reaction step elucidated by multiscale molecular simulations.|journal= Phys Chem Chem Phys.|volume= 10|issue= 6|pages= 4181-4188|doi= 10.1039/c7cp07069a}}</ref> to materials science<ref>{{cite book|author1=Artritha, Nonguch|author2=Urban, Alexander|lastauthoramp=yes|year=2016|title=An implementation of artificial neural-network potentials for atomistic materials simulations: Performance for TiO2|journal= Phys Chem Chem Phys.|volume= 114|pages= 135-150|doi= 10.1016/j.commatsci.2015.11.047}}</ref>.

TRANSITION STATE THEORY
It is worth mentioning some main aspects about the transition state (TS) theory here as it is central to the concepts used here. It is purely classical - it ignores any quantum mechanical effects such as tunnelling, which is incredibly small anyway. It also assumes that the transition state structure is in equilibrium with the reactants, which is called a quasi-equilibrium. TS theory also assumes that the reactants have energies which follow a Boltzmann distribution, which is not a bad assumption as long as the system had time to equilibrate. This assumption, however is not very applicable to the system studied here. Another crucial assumption of the TS theory, developed in the 1930s, is that once the TS is surpassed, the products will necessarily be formed, which was not the case in some simulations performed here. This latter assumption is also the reason why theoretically derived reaction rates using the TS state theory overestimate experimentally obtained reaction rates. This can be improved by adding a correcting factor, but to do that one must already have extensive knowledge of the system in question. The transition point is defined as the maximum on the minimum energy path going from reactants to products. This is called a saddle point and the net force on the system is 0.

We will deal here with triatomic linear systems, as depicted in the following Figure.
[[File:Y2C8.png|thumb|centre|500px| A triatomic system used here. The atoms will either all be Hs or in the last case one H will be replaced with a F. Figure taken from https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD]]

Terms like reactive and unreactive trajectories will be used often. A reactive trajectory is one which leads the reaction to completeness, i. e. the products are formed. In contrast an unreactive trajectory will not form products. The work here is divided into two exercises. Questions from the script are coloured blue for clarity and the answers follow below.

==EXERCISE 1==
===Dynamics from the TS region===

{{fontcolor|blue|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?}}
<div>The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its divergence being the zero vector at that point. To further characterise a point, one needs to evaluate the second derivatives at that point to determine whether it is a maximum or a minimum. The second derivative at the transition state will be negative, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the first derivative is zero and the second is positive, the point in question is a local minimum, which could be the lowest energy structure of the reactants or the products for example.</div>

{{fontcolor|blue|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.}}
<div>A simple system, consisting of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this figure. The transition state is the minimum of this line.]]

The minimum of this curve is the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if the initial structure corresponds to the TS structure. By changing the initial internuclear distances, we can indentify the TS as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Internuclear distance vs. time plots are shown in this figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Contour plot with the initial geometry being the transition state geometry and initial momenta = 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the TS the atoms are separated by more than 90 pm. Conversely, in case (c), the distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in close proximity of 90.8 pm.

But we can get a more accurate value now that we know where to look for the transition state. With some interval bisection, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

===Dynamics vs. MEP===

<div>{{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In a minimum energy path (MEP) calculation, the momentum of all particles is set to 0 in each step. Therefore there are no vibrations of the molecule at any point of the simulation. Of course this is unphysical as energy is lost, but it can be useful for some applications. In a dynamics simulation, on the other hand, vibrations of the molecule are observed as total energy of the system is constant.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. Molecule BC vibrates, as can be seen from the oscillating distance between atoms B and C. AB and AC distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. The AB momentum increases and ultimately moves with constant momentum away from the molecule BC. Molecule BC vibrates, so the momentum BC oscillates.]]</div>

Using the initial conditions r1=rts and r2=r<sub<ts+1, the system would roll into the other valley on the PES with similar behaviour to what is described in the figures on the left and right.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translational and vibrational). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill so the initial point of the previous simulation will not be reached.

===Reactive and Unreactive Trajectories===

{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point, fluorine and hydrogen collided a few times, but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected. There is also no general way to determine a reactive trajectory. This is influenced by many factors. There were significant vibrations present in the product molecule, which is a result of conservation of energy and the fact that the reaction is exothermic.

{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
As mentioned in the introduction, one of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the reactants again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===

{{fontcolor|blue|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?Locate the approximate position of the transition state.}}
[[File:MRD 01576020 MEP HF+H.png|thumb|right|Surface plot for the MEP calculation to determine the energy of the HF + H system. A similar calculation was performed to find the energy of the H2 + F system.]]
The F + H2 --> HF + H reaction is exothermic and H + HF --> H2 + F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than the H-H bond. This makes sense as the former has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is a fluorine atom. For these initial conditions, the forces were: along AB: 0.000 kJ.mol-1.pm-1 and along BC: 0.001 kJ.mol-1.pm-1. Due to these minimum forces, this structure is very close to the TS structure. One can notice that the distance between the H atoms in the TS structure is very close to their bond distance. This phenomenon is expressed in Hammond's postulate which says that the TS structure will resemble that of the species close to it in energy. Since this reaction is very exothermic, the TS structure is close to the structure of reactants which are a H2 molecule and a F atom.

{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. These were calculated by first finding the energy of the TS structure, which was -433.981 kJ/mol. Energies of the reactants and products were then obtained by performing a MEP simulation starting slightly off the TS structure in both directions (towards the reactants and towards the products, the one for the products is shown in the figure on the right). The calculated energy of the HF + H complex was -360.662 kJ/mol and that of the H2 + F complex was -435.031 kJ/mol. These values were used to calculate the activation energies given above. The obtained values are only approximate as we cannot move the atom infinitely far from the molecule but are nevertheless in good agreement with experimental findings.

===Reaction Dynamics===
<div>A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g.mol-1.pm.fs-1.
{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.</div>

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left| A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms the products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 fs of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is basically constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the BC pair which is moving apart, meaning that the total energy of the system is conserved. ]]

Let's now setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, an rFH = 238 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. Note that we are putting more energy into the system than the activation energy. The table below shows some observations and contour plots.

{| class="wikitable" border=1
! pFH/ g.mol-1.pm.fs-1 !! pHH/ g.mol-1.pm.fs-1 !! Einitial /kJ.mol-1 !! Reactive? !! Observations !! Illustration of the trajectory
|-
| -1.0 || -6.1 || -402.504 || No ||Here the system passed through the transition state structure, the hydrogen atom collided with the fluorine and then reformed reactants. || [[File:MRD 01576020 fhh table 1.png|400 px]]
|-
| -1.0 || -6.0 || -404.098 || No ||This is very similar to the case above. || [[File:MRD 01576020 fhh table 2.png|400 px]]
|-
| -1.0 || -5.0 || -414.098 || No ||This trajectory is also not reactive and is quite similar to the two above, again there was one collision with the fluorine atom. || [[File:MRD 01576020 fhh table 3.png|400 px]]
|-
| -1.0 || -4.1 || -421.388 || No ||Here, the fluorine atom did not collide with any atoms and the transition state was not surpassed. || [[File:MRD 01576020 fhh table 4.png|400 px]]
|-
| -1.0 || 0.0 || -434.098 || No ||Since even less energy was given the picture was very similar to the one above, but the hydrogen atom approached the fluorine even less. || [[File:MRD 01576020 fhh table 5.png|400 px]]
|-
| -1.0 || 2.1 || -427.588 || No ||Slightly more energy than above so the atoms were closer but the transition state was still not reached. || [[File:MRD 01576020 fhh table 6.png|400 px]]
|-
| -1.0 || 4.1 || -413.188 || No ||This trajectory was more interesting, the fluorine collided twice with one of the hydrogen atoms, but the reactants were reformed afterwards. || [[File:MRD 01576020 fhh table 7.png|400 px]]
|-
| -1.0 || 4.5 || -409.348 || Yes ||Interestingly, this trajectory was reactive despite the fact that the system has smaller energy than in some other cases. The total energy is therefore far from being the only parameter that influences the success of a given chemical reaction. || [[File:MRD 01576020 fhh table 11.png|400 px]]
|-
| -1.0 || 6.0 || -392.098 || No ||The system was very close to forming products here, fluorine and hydrogen collided 7 times, but the system rolled into the valley of reactants again. || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 6.1 || -390.788 || No ||The fluorine and hydrogen only collided once in this simulation. There is a pattern emerging in this table ... || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 7.0 || -378.098 || No ||This was very similar to the one above, there was one collision and then the reactants were reformed.|| [[File:MRD 01576020 fhh table 10.png|400 px]]
|}

From the table above it is obvious that whether the reaction will result in the formation of products is clearly not dependent only on the energy of vibration of the H2 molecule. We have also seen that these conditions where the molecule and the fluorine atom approach each other rather slowly rarely result in producing products. Maybe increasing the momentum pFH will help ...

{{fontcolor|blue|For the same initial position, increase slightly the momentum AB, and considerably reduce the overall energy of the system by reducing the momentum BC. What do you observe now?}}
There is very little vibration in the H2 molecule now and the F atom and the H2 molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.

Let's now focus on the reverse reaction, H + HF. When there is very little vibrational energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below (Polanyi's empirical rules).

An example of a reactive trajectory for this reaction would be produced by the following initial conditions: AB distance = 230 pm, BC distance = 74.2 pm, AB momentum = -1 g.mol-1.pm.fs-1 and BC momentum = 14.5 g.mol-1.pm.fs-1. The animation of this rather interesting trajectory is below.

[[File:MRD 01576020 animation rt2 GIF.gif|1000 px|thumb|left| A GIF showing the animation of the reactive trajectory of the HF + H reaction. The system passes the saddle point a few times but eventually forms the products.]]
[[File:MRD 01576020 skew.png|600 px|thumb|center| A skew plot of the same reactive trajectory as animated on the left.]]

{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions for a reactive trajectory. These rules tell us what modes need to have sufficient energy for certain types of TSs for a reaction to proceed. For early TSs, translational energy is more efficient than vibrational energy. For late TSs, on the other hand, vibrational energy will be more efficient. Why this is so is discussed in the figure below.

[[File:MRD 01576020 polanyi.png|800 px|thumb|center| A figure representing the main idea of Polanyi's empirical rules. Translational energy is more important for reactions with an early TS, and vibrational energy is more efficient for reactions with a late TS.]]

==References==

File:MRD 01576020 polanyi.png

2020-05-15T20:40:45Z

Dp3618:

File:MRD 01576020 skew.png

2020-05-15T20:26:26Z

Dp3618:

MRD:01576020

2020-05-15T20:15:28Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
Modelling the dynamics of molecular reactions has been enabled by the emergence of computers and the field of computational chemistry is an ever-expanding field. Despite most approaches being purely theoretical, ''ab initio'', calculations, the insight given is invaluable for understanding the pathways of chemical reactions. It is often also not necessary to include quantum mechanics, as atoms are of sufficient mass to be reasonably accurately modelled by Newton's equations of motion. Isolated systems in the gas phase only experience interatomic interactions which can be represented by a potential energy surface (PES) which is a function only of atomic positions. The force on a given interatomic coordinate can be calculated as follows:

<math> F = - { \partial V(r_1,r_2,...)\over \partial r_i}</math> (1)

where F = <math>{dp_i \over dt}</math> (2)

Calculating the forces and adjusting the system in consecutive steps results in a trajectory which shows how a system evolves in time. Such calculations have almost endless application from chemical biology and medicine<ref>{{cite article|author1=Pregeljc, Domen|author2=Jug, Urska|author3=Mavri, Janez|author4=Stare, Jernej|lastauthoramp=yes|year=2018|title=Why does the Y326I mutant of monoamine oxidase B decompose an endogenous amphetamine at a slower rate than the wild type enzyme? Reaction step elucidated by multiscale molecular simulations.|journal= Phys Chem Chem Phys.|volume= 10|issue= 6|pages= 4181-4188|doi= 10.1039/c7cp07069a}}</ref> to materials science<ref>{{cite article|author1=Artritha, Nonguch|author2=Urban, Alexander|lastauthoramp=yes|year=2016|title=An implementation of artificial neural-network potentials for atomistic materials simulations: Performance for TiO2|journal= Phys Chem Chem Phys.|volume= 114|pages= 135-150|doi= 10.1016/j.commatsci.2015.11.047
}}</ref>.

TRANSITION STATE THEORY
It is worth mentioning some main aspects about the transition state (TS) theory here as it is central to the concepts used here. It is purely classical - it ignores any quantum mechanical effects such as tunnelling, which is incredibly small anyway. It also assumes that the transition state structure is in equilibrium with the reactants, which is called a quasi-equilibrium. TS theory also assumes that the reactants have energies which follow a Boltzmann distribution, which is not a bad assumption as long as the system had time to equilibrate. This assumption, however is not very applicable to the system studied here. Another crucial assumption of the TS theory, developed in the 1930s, is that once the TS is surpassed, the products will necessarily be formed, which was not the case in some simulations performed here. This latter assumption is also the reason why theoretically derived reaction rates using the TS state theory overestimate experimentally obtained reaction rates. This can be improved by adding a correcting factor, but to do that one must already have extensive knowledge of the system in question. The transition point is defined as the maximum on the minimum energy path going from reactants to products. This is called a saddle point and the net force on the system is 0.

We will deal here with triatomic linear systems, as depicted in the following Figure.
[[File:Y2C8.png|thumb|centre|500px| A triatomic system used here. The atoms will either all be Hs or in the last case one H will be replaced with a F. Figure taken from https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD.]]

Terms like reactive and unreactive trajectories will be used often. A reactive trajectory is one which leads the reaction to completeness, i. e. the products are formed. In contrast an unreactive trajectory will not form products. The work here is divided into two exercises. Questions from the script are coloured blue for clarity and the answers follow below.

==EXERCISE 1==
===Dynamics from the TS region===

{{fontcolor|blue|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?}}
<div>The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its divergence being the zero vector at that point. To further characterise a point, one needs to evaluate the second derivatives at that point to determine whether it is a maximum or a minimum. The second derivative at the transition state will be negative, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the first derivative is zero and the second is positive, the point in question is a local minimum, which could be the lowest energy structure of the reactants or the products for example.</div>

{{fontcolor|blue|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.}}
<div>A simple system, consisting of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this figure. The transition state is the minimum of this line.]]

The minimum of this curve is the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if the initial structure corresponds to the TS structure. By changing the initial internuclear distances, we can indentify the TS as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Internuclear distance vs. time plots are shown in this figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Contour plot with the initial geometry being the transition state geometry and initial momenta = 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the TS the atoms are separated by more than 90 pm. Conversely, in case (c), the distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in close proximity of 90.8 pm.

But we can get a more accurate value now that we know where to look for the transition state. With some interval bisection, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

===Dynamics vs. MEP===

<div>{{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In a minimum energy path (MEP) calculation, the momentum of all particles is set to 0 in each step. Therefore there are no vibrations of the molecule at any point of the simulation. Of course this is unphysical as energy is lost, but it can be useful for some applications. In a dynamics simulation, on the other hand, vibrations of the molecule are observed as total energy of the system is constant.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. Molecule BC vibrates, as can be seen from the oscillating distance between atoms B and C. AB and AC distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. The AB momentum increases and ultimately moves with constant momentum away from the molecule BC. Molecule BC vibrates, so the momentum BC oscillates.]]</div>

Using the initial conditions r1=rts and r2=r<sub<ts+1, the system would roll into the other valley on the PES with similar behaviour to what is described in the figures on the left and right.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translational and vibrational). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill so the initial point of the previous simulation will not be reached.

===Reactive and Unreactive Trajectories===

{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point, fluorine and hydrogen collided a few times, but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected. There is also no general way to determine a reactive trajectory. This is influenced by many factors. There were significant vibrations present in the product molecule, which is a result of conservation of energy and the fact that the reaction is exothermic.

{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
As mentioned in the introduction, one of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the reactants again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===

{{fontcolor|blue|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?Locate the approximate position of the transition state.}}
[[File:MRD 01576020 MEP HF+H.png|thumb|right|Surface plot for the MEP calculation to determine the energy of the HF + H system. A similar calculation was performed to find the energy of the H2 + F system.]]
The F + H2 --> HF + H reaction is exothermic and H + HF --> H2 + F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than the H-H bond. This makes sense as the former has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is a fluorine atom. For these initial conditions, the forces were: along AB: 0.000 kJ.mol-1.pm-1 and along BC: 0.001 kJ.mol-1.pm-1. Due to these minimum forces, this structure is very close to the TS structure. One can notice that the distance between the H atoms in the TS structure is very close to their bond distance. This phenomenon is expressed in Hammond's postulate which says that the TS structure will resemble that of the species close to it in energy. Since this reaction is very exothermic, the TS structure is close to the structure of reactants which are a H2 molecule and a F atom.

{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. These were calculated by first finding the energy of the TS structure, which was -433.981 kJ/mol. Energies of the reactants and products were then obtained by performing a MEP simulation starting slightly off the TS structure in both directions (towards the reactants and towards the products, the one for the products is shown in the figure on the right). The calculated energy of the HF + H complex was -360.662 kJ/mol and that of the H2 + F complex was -435.031 kJ/mol. These values were used to calculate the activation energies given above. The obtained values are only approximate as we cannot move the atom infinitely far from the molecule but are nevertheless in good agreement with experimental findings.

===Reaction Dynamics===
<div>A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g.mol-1.pm.fs-1.
{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.</div>

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left| A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms the products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 fs of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is basically constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the BC pair which is moving apart, meaning that the total energy of the system is conserved. ]]

Let's now setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, an rFH = 238 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. Note that we are putting more energy into the system than the activation energy. The table below shows some observations and contour plots.

{| class="wikitable" border=1
! pFH/ g.mol-1.pm.fs-1 !! pHH/ g.mol-1.pm.fs-1 !! Einitial /kJ.mol-1 !! Reactive? !! Observations !! Illustration of the trajectory
|-
| -1.0 || -6.1 || -402.504 || No ||Here the system passed through the transition state structure, the hydrogen atom collided with the fluorine and then reformed reactants. || [[File:MRD 01576020 fhh table 1.png|400 px]]
|-
| -1.0 || -6.0 || -404.098 || No ||This is very similar to the case above. || [[File:MRD 01576020 fhh table 2.png|400 px]]
|-
| -1.0 || -5.0 || -414.098 || No ||This trajectory is also not reactive and is quite similar to the two above, again there was one collision with the fluorine atom. || [[File:MRD 01576020 fhh table 3.png|400 px]]
|-
| -1.0 || -4.1 || -421.388 || No ||Here, the fluorine atom did not collide with any atoms and the transition state was not surpassed. || [[File:MRD 01576020 fhh table 4.png|400 px]]
|-
| -1.0 || 0.0 || -434.098 || No ||Since even less energy was given the picture was very similar to the one above, but the hydrogen atom approached the fluorine even less. || [[File:MRD 01576020 fhh table 5.png|400 px]]
|-
| -1.0 || 2.1 || -427.588 || No ||Slightly more energy than above so the atoms were closer but the transition state was still not reached. || [[File:MRD 01576020 fhh table 6.png|400 px]]
|-
| -1.0 || 4.1 || -413.188 || No ||This trajectory was more interesting, the fluorine collided twice with one of the hydrogen atoms, but the reactants were reformed afterwards. || [[File:MRD 01576020 fhh table 7.png|400 px]]
|-
| -1.0 || 4.5 || -409.348 || Yes ||Interestingly, this trajectory was reactive despite the fact that the system has smaller energy than in some other cases. The total energy is therefore far from being the only parameter that influences the success of a given chemical reaction. || [[File:MRD 01576020 fhh table 11.png|400 px]]
|-
| -1.0 || 6.0 || -392.098 || No ||The system was very close to forming products here, fluorine and hydrogen collided 7 times, but the system rolled into the valley of reactants again. || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 6.1 || -390.788 || No ||The fluorine and hydrogen only collided once in this simulation. There is a pattern emerging in this table ... || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 7.0 || -378.098 || No ||This was very similar to the one above, there was one collision and then the reactants were reformed.|| [[File:MRD 01576020 fhh table 10.png|400 px]]
|}

From the table above it is obvious that whether the reaction will result in the formation of products is clearly not dependent only on the energy of vibration of the H2 molecule. We have also seen that these conditions where the molecule and the fluorine atom approach each other rather slowly rarely result in producing products. Maybe increasing the momentum pFH will help ...

{{fontcolor|blue|For the same initial position, increase slightly the momentum AB, and considerably reduce the overall energy of the system by reducing the momentum BC. What do you observe now?}}
There is very little vibration in the H2 molecule now and the F atom and the H2 molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.

Let's now focus on the reverse reaction, H + HF. When there is very little vibrational energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below (Polanyi's empirical rules).

An example of a reactive trajectory for this reaction would be produced by the following initial conditions: AB distance = 230 pm, BC distance = 74.2 pm, AB momentum = -1 g.mol-1.pm.fs-1 and BC momentum = 14.5 g.mol-1.pm.fs-1. The animation of this rather interesting trajectory is below.

[[File:MRD 01576020 animation rt2 GIF.gif|1000 px|thumb|center| A GIF showing the animation of the reactive trajectory of the HF + h reaction. The system passes the saddle point a few times but eventually forms the products.]]

{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions for a reactive trajectory. These rules tell us what modes need to have sufficient energy for certain types of TSs for a reaction to proceed. For early TSs, translational energy is more efficient than vibrational energy. For late TSs, on the other hand, vibrational energy will be more efficient. Why this is so is discussed in the figure to the right.

==References==

MRD:01576020

2020-05-15T20:00:42Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
Modelling the dynamics of molecular reactions has been enabled by the emergence of computers and the field of computational chemistry is an ever-expanding field. Despite most approaches being purely theoretical, ''ab initio'', calculations, the insight given is invaluable for understanding the pathways of chemical reactions. It is often also not necessary to include quantum mechanics, as atoms are of sufficient mass to be reasonably accurately modelled by Newton's equations of motion. Isolated systems in the gas phase only experience interatomic interactions which can be represented by a potential energy surface (PES) which is a function only of atomic positions. The force on a given interatomic coordinate can be calculated as follows:

<math> F = - { \partial V(r_1,r_2,...)\over \partial r_i}</math> (1)

where F = <math>{dp_i \over dt}</math> (2)

Calculating the forces and adjusting the system in consecutive steps results in a trajectory which shows how a system evolves in time. Such calculations have almost endless application from chemical biology and medicine{{cite journal |last1=Pregeljc |first1=Domen |last2=Jug |first2=Urska|last3=Mavri |first3=Janez|last4=Stare |first4=Jernej |date= |title=Why does the Y326I mutant of monoamine oxidase B decompose an endogenous amphetamine at a slower rate than the wild type enzyme? Reaction step elucidated by multiscale molecular simulations. |url= |journal= Phys Chem Chem Phys.|volume= 10|issue= 6|pages= 4181-4188|doi= 10.1039/c7cp07069a|access-date=}}

TRANSITION STATE THEORY
It is worth mentioning some main aspects about the transition state (TS) theory here as it is central to the concepts used here. It is purely classical - it ignores any quantum mechanical effects such as tunnelling, which is incredibly small anyway. It also assumes that the transition state structure is in equilibrium with the reactants, which is called a quasi-equilibrium. TS theory also assumes that the reactants have energies which follow a Boltzmann distribution, which is not a bad assumption as long as the system had time to equilibrate. This assumption, however is not very applicable to the system studied here. Another crucial assumption of the TS theory, developed in the 1930s, is that once the TS is surpassed, the products will necessarily be formed, which was not the case in some simulations performed here. This latter assumption is also the reason why theoretically derived reaction rates using the TS state theory overestimate experimentally obtained reaction rates. This can be improved by adding a correcting factor, but to do that one must already have extensive knowledge of the system in question. The transition point is defined as the maximum on the minimum energy path going from reactants to products. This is called a saddle point and the net force on the system is 0.

We will deal here with triatomic linear systems, as depicted in the following Figure.
[[File:Y2C8.png|thumb|centre|500px| A triatomic system used here. The atoms will either all be Hs or in the last case one H will be replaced with a F. Figure taken from https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD.]]

Terms like reactive and unreactive trajectories will be used often. A reactive trajectory is one which leads the reaction to completeness, i. e. the products are formed. In contrast an unreactive trajectory will not form products. The work here is divided into two exercises. Questions from the script are coloured blue for clarity and the answers follow below.

==EXERCISE 1==
===Dynamics from the TS region===

{{fontcolor|blue|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?}}
<div>The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its divergence being the zero vector at that point. To further characterise a point, one needs to evaluate the second derivatives at that point to determine whether it is a maximum or a minimum. The second derivative at the transition state will be negative, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the first derivative is zero and the second is positive, the point in question is a local minimum, which could be the lowest energy structure of the reactants or the products for example.</div>

{{fontcolor|blue|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.}}
<div>A simple system, consisting of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this figure. The transition state is the minimum of this line.]]

The minimum of this curve is the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if the initial structure corresponds to the TS structure. By changing the initial internuclear distances, we can indentify the TS as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Internuclear distance vs. time plots are shown in this figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Contour plot with the initial geometry being the transition state geometry and initial momenta = 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the TS the atoms are separated by more than 90 pm. Conversely, in case (c), the distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in close proximity of 90.8 pm.

But we can get a more accurate value now that we know where to look for the transition state. With some interval bisection, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

===Dynamics vs. MEP===

<div>{{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In a minimum energy path (MEP) calculation, the momentum of all particles is set to 0 in each step. Therefore there are no vibrations of the molecule at any point of the simulation. Of course this is unphysical as energy is lost, but it can be useful for some applications. In a dynamics simulation, on the other hand, vibrations of the molecule are observed as total energy of the system is constant.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. Molecule BC vibrates, as can be seen from the oscillating distance between atoms B and C. AB and AC distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. The AB momentum increases and ultimately moves with constant momentum away from the molecule BC. Molecule BC vibrates, so the momentum BC oscillates.]]</div>

Using the initial conditions r1=rts and r2=r<sub<ts+1, the system would roll into the other valley on the PES with similar behaviour to what is described in the figures on the left and right.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translational and vibrational). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill so the initial point of the previous simulation will not be reached.

===Reactive and Unreactive Trajectories===

{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point, fluorine and hydrogen collided a few times, but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected. There is also no general way to determine a reactive trajectory. This is influenced by many factors. There were significant vibrations present in the product molecule, which is a result of conservation of energy and the fact that the reaction is exothermic.

{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
As mentioned in the introduction, one of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the reactants again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===

{{fontcolor|blue|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?Locate the approximate position of the transition state.}}
[[File:MRD 01576020 MEP HF+H.png|thumb|right|Surface plot for the MEP calculation to determine the energy of the HF + H system. A similar calculation was performed to find the energy of the H2 + F system.]]
The F + H2 --> HF + H reaction is exothermic and H + HF --> H2 + F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than the H-H bond. This makes sense as the former has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is a fluorine atom. For these initial conditions, the forces were: along AB: 0.000 kJ.mol-1.pm-1 and along BC: 0.001 kJ.mol-1.pm-1. Due to these minimum forces, this structure is very close to the TS structure. One can notice that the distance between the H atoms in the TS structure is very close to their bond distance. This phenomenon is expressed in Hammond's postulate which says that the TS structure will resemble that of the species close to it in energy. Since this reaction is very exothermic, the TS structure is close to the structure of reactants which are a H2 molecule and a F atom.

{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. These were calculated by first finding the energy of the TS structure, which was -433.981 kJ/mol. Energies of the reactants and products were then obtained by performing a MEP simulation starting slightly off the TS structure in both directions (towards the reactants and towards the products, the one for the products is shown in the figure on the right). The calculated energy of the HF + H complex was -360.662 kJ/mol and that of the H2 + F complex was -435.031 kJ/mol. These values were used to calculate the activation energies given above. The obtained values are only approximate as we cannot move the atom infinitely far from the molecule but are nevertheless in good agreement with experimental findings.

===Reaction Dynamics===
<div>A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g.mol-1.pm.fs-1.
{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.</div>

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left| A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms the products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 fs of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is basically constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the BC pair which is moving apart, meaning that the total energy of the system is conserved. ]]

Let's now setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, an rFH = 238 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. Note that we are putting more energy into the system than the activation energy. The table below shows some observations and contour plots.

{| class="wikitable" border=1
! pFH/ g.mol-1.pm.fs-1 !! pHH/ g.mol-1.pm.fs-1 !! Einitial /kJ.mol-1 !! Reactive? !! Observations !! Illustration of the trajectory
|-
| -1.0 || -6.1 || -402.504 || No ||Here the system passed through the transition state structure, the hydrogen atom collided with the fluorine and then reformed reactants. || [[File:MRD 01576020 fhh table 1.png|400 px]]
|-
| -1.0 || -6.0 || -404.098 || No ||This is very similar to the case above. || [[File:MRD 01576020 fhh table 2.png|400 px]]
|-
| -1.0 || -5.0 || -414.098 || No ||This trajectory is also not reactive and is quite similar to the two above, again there was one collision with the fluorine atom. || [[File:MRD 01576020 fhh table 3.png|400 px]]
|-
| -1.0 || -4.1 || -421.388 || No ||Here, the fluorine atom did not collide with any atoms and the transition state was not surpassed. || [[File:MRD 01576020 fhh table 4.png|400 px]]
|-
| -1.0 || 0.0 || -434.098 || No ||Since even less energy was given the picture was very similar to the one above, but the hydrogen atom approached the fluorine even less. || [[File:MRD 01576020 fhh table 5.png|400 px]]
|-
| -1.0 || 2.1 || -427.588 || No ||Slightly more energy than above so the atoms were closer but the transition state was still not reached. || [[File:MRD 01576020 fhh table 6.png|400 px]]
|-
| -1.0 || 4.1 || -413.188 || No ||This trajectory was more interesting, the fluorine collided twice with one of the hydrogen atoms, but the reactants were reformed afterwards. || [[File:MRD 01576020 fhh table 7.png|400 px]]
|-
| -1.0 || 4.5 || -409.348 || Yes ||Interestingly, this trajectory was reactive despite the fact that the system has smaller energy than in some other cases. The total energy is therefore far from being the only parameter that influences the success of a given chemical reaction. || [[File:MRD 01576020 fhh table 11.png|400 px]]
|-
| -1.0 || 6.0 || -392.098 || No ||The system was very close to forming products here, fluorine and hydrogen collided 7 times, but the system rolled into the valley of reactants again. || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 6.1 || -390.788 || No ||The fluorine and hydrogen only collided once in this simulation. There is a pattern emerging in this table ... || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 7.0 || -378.098 || No ||This was very similar to the one above, there was one collision and then the reactants were reformed.|| [[File:MRD 01576020 fhh table 10.png|400 px]]
|}

From the table above it is obvious that whether the reaction will result in the formation of products is clearly not dependent only on the energy of vibration of the H2 molecule. We have also seen that these conditions where the molecule and the fluorine atom approach each other rather slowly rarely result in producing products. Maybe increasing the momentum pFH will help ...

{{fontcolor|blue|For the same initial position, increase slightly the momentum AB, and considerably reduce the overall energy of the system by reducing the momentum BC. What do you observe now?}}
There is very little vibration in the H2 molecule now and the F atom and the H2 molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.

Let's now focus on the reverse reaction, H + HF. When there is very little vibrational energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below (Polanyi's empirical rules).

An example of a reactive trajectory for this reaction would be produced by the following initial conditions: AB distance = 230 pm, BC distance = 74.2 pm, AB momentum = -1 g.mol-1.pm.fs-1 and BC momentum = 14.5 g.mol-1.pm.fs-1. The animation of this rather interesting trajectory is below.

[[File:MRD 01576020 animation rt2 GIF.gif|1000 px|thumb|center| A GIF showing the animation of the reactive trajectory of the HF + h reaction. The system passes the saddle point a few times but eventually forms the products.]]

{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions for a reactive trajectory. These rules tell us what modes need to have sufficient energy for certain types of TSs for a reaction to proceed. For early TSs, translational energy is more efficient than vibrational energy. For late TSs, on the other hand, vibrational energy will be more efficient. Why this is so is discussed in the figure to the right.

File:MRD 01576020 animation rt2 GIF.gif

2020-05-15T19:58:46Z

Dp3618:

MRD:01576020

2020-05-15T19:48:28Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
Modelling the dynamics of molecular reactions has been enabled by the emergence of computers and the field of computational chemistry is an ever-expanding field. Despite most approaches being purely theoretical, ''ab initio'', calculations, the insight given is invaluable for understanding the pathways of chemical reactions. It is often also not necessary to include quantum mechanics, as atoms are of sufficient mass to be reasonably accurately modelled by Newton's equations of motion. Isolated systems in the gas phase only experience interatomic interactions which can be represented by a potential energy surface (PES) which is a function only of atomic positions. The force on a given interatomic coordinate can be calculated as follows:

<math> F = - { \partial V(r_1,r_2,...)\over \partial r_i}</math> (1)

where F = <math>{dp_i \over dt}</math> (2)

Calculating the forces and adjusting the system in consecutive steps results in a trajectory which shows how a system evolves in time. Such calculations have almost endless application from chemical biology and medicine{{cite journal |last1=Pregeljc |first1=Domen |last2=Jug |first2=Urska|last3=Mavri |first3=Janez|last4=Stare |first4=Jernej |date= |title=Why does the Y326I mutant of monoamine oxidase B decompose an endogenous amphetamine at a slower rate than the wild type enzyme? Reaction step elucidated by multiscale molecular simulations. |url= |journal= Phys Chem Chem Phys.|volume= 10|issue= 6|pages= 4181-4188|doi= 10.1039/c7cp07069a|access-date=}}

TRANSITION STATE THEORY
It is worth mentioning some main aspects about the transition state (TS) theory here as it is central to the concepts used here. It is purely classical - it ignores any quantum mechanical effects such as tunnelling, which is incredibly small anyway. It also assumes that the transition state structure is in equilibrium with the reactants, which is called a quasi-equilibrium. TS theory also assumes that the reactants have energies which follow a Boltzmann distribution, which is not a bad assumption as long as the system had time to equilibrate. This assumption, however is not very applicable to the system studied here. Another crucial assumption of the TS theory, developed in the 1930s, is that once the TS is surpassed, the products will necessarily be formed, which was not the case in some simulations performed here. This latter assumption is also the reason why theoretically derived reaction rates using the TS state theory overestimate experimentally obtained reaction rates. This can be improved by adding a correcting factor, but to do that one must already have extensive knowledge of the system in question. The transition point is defined as the maximum on the minimum energy path going from reactants to products. This is called a saddle point and the net force on the system is 0.

We will deal here with triatomic linear systems, as depicted in the following Figure.
[[File:Y2C8.png|thumb|centre|500px| A triatomic system used here. The atoms will either all be Hs or in the last case one H will be replaced with a F. Figure taken from https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD.]]

Terms like reactive and unreactive trajectories will be used often. A reactive trajectory is one which leads the reaction to completeness, i. e. the products are formed. In contrast an unreactive trajectory will not form products. The work here is divided into two exercises. Questions from the script are coloured blue for clarity and the answers follow below.

==EXERCISE 1==
===Dynamics from the TS region===

{{fontcolor|blue|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?}}
<div>The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its divergence being the zero vector at that point. To further characterise a point, one needs to evaluate the second derivatives at that point to determine whether it is a maximum or a minimum. The second derivative at the transition state will be negative, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the first derivative is zero and the second is positive, the point in question is a local minimum, which could be the lowest energy structure of the reactants or the products for example.</div>

{{fontcolor|blue|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.}}
<div>A simple system, consisting of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this figure. The transition state is the minimum of this line.]]

The minimum of this curve is the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if the initial structure corresponds to the TS structure. By changing the initial internuclear distances, we can indentify the TS as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Internuclear distance vs. time plots are shown in this figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Contour plot with the initial geometry being the transition state geometry and initial momenta = 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the TS the atoms are separated by more than 90 pm. Conversely, in case (c), the distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in close proximity of 90.8 pm.

But we can get a more accurate value now that we know where to look for the transition state. With some interval bisection, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

===Dynamics vs. MEP===

<div>{{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In a minimum energy path (MEP) calculation, the momentum of all particles is set to 0 in each step. Therefore there are no vibrations of the molecule at any point of the simulation. Of course this is unphysical as energy is lost, but it can be useful for some applications. In a dynamics simulation, on the other hand, vibrations of the molecule are observed as total energy of the system is constant.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. Molecule BC vibrates, as can be seen from the oscillating distance between atoms B and C. AB and AC distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. The AB momentum increases and ultimately moves with constant momentum away from the molecule BC. Molecule BC vibrates, so the momentum BC oscillates.]]</div>

Using the initial conditions r1=rts and r2=r<sub<ts+1, the system would roll into the other valley on the PES with similar behaviour to what is described in the figures on the left and right.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translational and vibrational). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill so the initial point of the previous simulation will not be reached.

===Reactive and Unreactive Trajectories===

{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point, fluorine and hydrogen collided a few times, but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected. There is also no general way to determine a reactive trajectory. This is influenced by many factors. There were significant vibrations present in the product molecule, which is a result of conservation of energy and the fact that the reaction is exothermic.

{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
As mentioned in the introduction, one of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the reactants again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===

{{fontcolor|blue|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?Locate the approximate position of the transition state.}}
[[File:MRD 01576020 MEP HF+H.png|thumb|right|Surface plot for the MEP calculation to determine the energy of the HF + H system. A similar calculation was performed to find the energy of the H2 + F system.]]
The F + H2 --> HF + H reaction is exothermic and H + HF --> H2 + F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than the H-H bond. This makes sense as the former has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is a fluorine atom. For these initial conditions, the forces were: along AB: 0.000 kJ.mol-1.pm-1 and along BC: 0.001 kJ.mol-1.pm-1. Due to these minimum forces, this structure is very close to the TS structure. One can notice that the distance between the H atoms in the TS structure is very close to their bond distance. This phenomenon is expressed in Hammond's postulate which says that the TS structure will resemble that of the species close to it in energy. Since this reaction is very exothermic, the TS structure is close to the structure of reactants which are a H2 molecule and a F atom.

{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. These were calculated by first finding the energy of the TS structure, which was -433.981 kJ/mol. Energies of the reactants and products were then obtained by performing a MEP simulation starting slightly off the TS structure in both directions (towards the reactants and towards the products, the one for the products is shown in the figure on the right). The calculated energy of the HF + H complex was -360.662 kJ/mol and that of the H2 + F complex was -435.031 kJ/mol. These values were used to calculate the activation energies given above. The obtained values are only approximate as we cannot move the atom infinitely far from the molecule but are nevertheless in good agreement with experimental findings.

===Reaction Dynamics===
<div>A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g.mol-1.pm.fs-1.
{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.</div>

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left| A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 fs of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is basically constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the BC pair which is moving apart, meaning that the total energy of the system is conserved. ]]

Let's now setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, an rFH = 238 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. Note that we are putting more energy into the system than the activation energy. The table below shows some observations and contour plots.

{| class="wikitable" border=1
! pFH/ g.mol-1.pm.fs-1 !! pHH/ g.mol-1.pm.fs-1 !! Einitial /kJ.mol-1 !! Reactive? !! Observations !! Illustration of the trajectory
|-
| -1.0 || -6.1 || -402.504 || No ||Here the system passed through the transition state structure, the hydrogen atom collided with the fluorine and then reformed reactants. || [[File:MRD 01576020 fhh table 1.png|400 px]]
|-
| -1.0 || -6.0 || -404.098 || No ||This is very similar to the case above. || [[File:MRD 01576020 fhh table 2.png|400 px]]
|-
| -1.0 || -5.0 || -414.098 || No ||This trajectory is also not reactive and is quite similar to the two above, again there was one collision with the fluorine atom. || [[File:MRD 01576020 fhh table 3.png|400 px]]
|-
| -1.0 || -4.1 || -421.388 || No ||Here, the fluorine atom did not collide with any atoms and the transition state was not surpassed. || [[File:MRD 01576020 fhh table 4.png|400 px]]
|-
| -1.0 || 0.0 || -434.098 || No ||Since even less energy was given the picture was very similar to the one above, but the hydrogen atom approached the fluorine even less. || [[File:MRD 01576020 fhh table 5.png|400 px]]
|-
| -1.0 || 2.1 || -427.588 || No ||Slightly more energy than above so the atoms were closer but the transition state was still not reached. || [[File:MRD 01576020 fhh table 6.png|400 px]]
|-
| -1.0 || 4.1 || -413.188 || No ||This trajectory was more interesting, the fluorine collided twice with one of the hydrogen atoms, but the reactants were reformed afterwards. || [[File:MRD 01576020 fhh table 7.png|400 px]]
|-
| -1.0 || 4.5 || -409.348 || Yes ||Interestingly, this trajectory was reactive despite the fact that the system has smaller energy than in some other cases. The total energy is therefore far from being the only parameter that influences the success of a given chemical reaction. || [[File:MRD 01576020 fhh table 11.png|400 px]]
|-
| -1.0 || 6.0 || -392.098 || No ||The system was very close to forming products here, fluorine and hydrogen collided 7 times, but the system rolled into the valley of reactants again. || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 6.1 || -390.788 || No ||The fluorine and hydrogen only collided once in this simulation. There is a pattern emerging in this table ... || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 7.0 || -378.098 || No ||This was very similar to the one above, there was one collision and then the reactants were reformed.|| [[File:MRD 01576020 fhh table 10.png|400 px]]
|}

From the table above it is obvious that whether the reaction will result in the formation of products is clearly not dependent only on the energy of vibration of the H2 molecule. We have also seen that these conditions where the molecule and the fluorine atom approach each other rather slowly rarely result in producing products. Maybe increasing the momentum pFH will help ...

{{fontcolor|blue|For the same initial position, increase slightly the momentum AB, and considerably reduce the overall energy of the system by reducing the momentum BC. What do you observe now?}}
There is very little vibration in the H2 molecule now and the F atom and the H2 molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.

Let's now focus on the reverse reaction, H + HF. When there is very little vibrational energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below (Polanyi's empirical rules).

An example of a reactive trajectory for this reaction would be produced by the following initial conditions: AB distance = 230 pm, BC distance = 74.2 pm, AB momentum = -1 g.mol-1.pm.fs-1 and BC momentum = 14.5 g.mol-1.pm.fs-1. The animation of this rather interesting trajectory is below.

{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions for a reactive trajectory. These rules tell us what modes need to have sufficient energy for certain types of TSs for a reaction to proceed. For early TSs, translational energy is more efficient than vibrational energy. For late TSs, on the other hand, vibrational energy will be more efficient. Why this is so is discussed in the figure to the right.

MRD:01576020

2020-05-15T19:39:30Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
Modelling the dynamics of molecular reactions has been enabled by the emergence of computers and the field of computational chemistry is an ever-expanding field. Despite most approaches being purely theoretical, ''ab initio'', calculations, the insight given is invaluable for understanding the pathways of chemical reactions. It is often also not necessary to include quantum mechanics, as atoms are of sufficient mass to be reasonably accurately modelled by Newton's equations of motion. Isolated systems in the gas phase only experience interatomic interactions which can be represented by a potential energy surface (PES) which is a function only of atomic positions. The force on a given interatomic coordinate can be calculated as follows:

<math> F = - { \partial V(r_1,r_2,...)\over \partial r_i}</math> (1)

where F = <math>{dp_i \over dt}</math> (2)

Calculating the forces and adjusting the system in consecutive steps results in a trajectory which shows how a system evolves in time. Such calculations have almost endless application from chemical biology and medicine{{cite journal |last1=Pregeljc |first1=Domen |last2=Jug |first2=Urska|last3=Mavri |first3=Janez|last4=Stare |first4=Jernej |date= |title=Why does the Y326I mutant of monoamine oxidase B decompose an endogenous amphetamine at a slower rate than the wild type enzyme? Reaction step elucidated by multiscale molecular simulations. |url= |journal= Phys Chem Chem Phys.|volume= 10|issue= 6|pages= 4181-4188|doi= 10.1039/c7cp07069a|access-date=}}

TRANSITION STATE THEORY
It is worth mentioning some main aspects about the transition state (TS) theory here as it is central to the concepts used here. It is purely classical - it ignores any quantum mechanical effects such as tunnelling, which is incredibly small anyway. It also assumes that the transition state structure is in equilibrium with the reactants, which is called a quasi-equilibrium. TS theory also assumes that the reactants have energies which follow a Boltzmann distribution, which is not a bad assumption as long as the system had time to equilibrate. This assumption, however is not very applicable to the system studied here. Another crucial assumption of the TS theory, developed in the 1930s, is that once the TS is surpassed, the products will necessarily be formed, which was not the case in some simulations performed here. This latter assumption is also the reason why theoretically derived reaction rates using the TS state theory overestimate experimentally obtained reaction rates. This can be improved by adding a correcting factor, but to do that one must already have extensive knowledge of the system in question. The transition point is defined as the maximum on the minimum energy path going from reactants to products. This is called a saddle point and the net force on the system is 0.

We will deal here with triatomic linear systems, as depicted in the following Figure.
[[File:Y2C8.png|thumb|centre|500px| A triatomic system used here. The atoms will either all be Hs or in the last case one H will be replaced with a F. Figure taken from https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD.]]

Terms like reactive and unreactive trajectories will be used often. A reactive trajectory is one which leads the reaction to completeness, i. e. the products are formed. In contrast an unreactive trajectory will not form products. The work here is divided into two exercises. Questions from the script are coloured blue for clarity and the answers follow below.

==EXERCISE 1==
===Dynamics from the TS region===

{{fontcolor|blue|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?}}
<div>The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its divergence being the zero vector at that point. To further characterise a point, one needs to evaluate the second derivatives at that point to determine whether it is a maximum or a minimum. The second derivative at the transition state will be negative, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the first derivative is zero and the second is positive, the point in question is a local minimum, which could be the lowest energy structure of the reactants or the products for example.</div>

{{fontcolor|blue|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.}}
<div>A simple system, consisting of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this figure. The transition state is the minimum of this line.]]

The minimum of this curve is the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if the initial structure corresponds to the TS structure. By changing the initial internuclear distances, we can indentify the TS as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Internuclear distance vs. time plots are shown in this figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Contour plot with the initial geometry being the transition state geometry and initial momenta = 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the TS the atoms are separated by more than 90 pm. Conversely, in case (c), the distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in close proximity of 90.8 pm.

But we can get a more accurate value now that we know where to look for the transition state. With some interval bisection, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

===Dynamics vs. MEP===

<div>{{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In a minimum energy path (MEP) calculation, the momentum of all particles is set to 0 in each step. Therefore there are no vibrations of the molecule at any point of the simulation. Of course this is unphysical as energy is lost, but it can be useful for some applications. In a dynamics simulation, on the other hand, vibrations of the molecule are observed as total energy of the system is constant.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. Molecule BC vibrates, as can be seen from the oscillating distance between atoms B and C. AB and AC distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. The AB momentum increases and ultimately moves with constant momentum away from the molecule BC. Molecule BC vibrates, so the momentum BC oscillates.]]</div>

Using the initial conditions r1=rts and r2=r<sub<ts+1, the system would roll into the other valley on the PES with similar behaviour to what is described in the figures on the left and right.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translational and vibrational). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill so the initial point of the previous simulation will not be reached.

===Reactive and Unreactive Trajectories===

{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point, fluorine and hydrogen collided a few times, but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected. There is also no general way to determine a reactive trajectory. This is influenced by many factors. There were significant vibrations present in the product molecule, which is a result of conservation of energy and the fact that the reaction is exothermic.

{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
As mentioned in the introduction, one of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the reactants again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===

{{fontcolor|blue|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?Locate the approximate position of the transition state.}}
[[File:MRD 01576020 MEP HF+H.png|thumb|right|Surface plot for the MEP calculation to determine the energy of the HF + H system. A similar calculation was performed to find the energy of the H2 + F system.]]
The F + H2 --> HF + H reaction is exothermic and H + HF --> H2 + F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than the H-H bond. This makes sense as the former has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is a fluorine atom. For these initial conditions, the forces were: along AB: 0.000 kJ.mol-1.pm-1 and along BC: 0.001 kJ.mol-1.pm-1. Due to these minimum forces, this structure is very close to the TS structure. One can notice that the distance between the H atoms in the TS structure is very close to their bond distance. This phenomenon is expressed in Hammond's postulate which says that the TS structure will resemble that of the species close to it in energy. Since this reaction is very exothermic, the TS structure is close to the structure of reactants which are a H2 molecule and a F atom.

{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. These were calculated by first finding the energy of the TS structure, which was -433.981 kJ/mol. Energies of the reactants and products were then obtained by performing a MEP simulation starting slightly off the TS structure in both directions (towards the reactants and towards the products, the one for the products is shown in the figure on the right). The calculated energy of the HF + H complex was -360.662 kJ/mol and that of the H2 + F complex was -435.031 kJ/mol. These values were used to calculate the activation energies given above. The obtained values are only approximate as we cannot move the atom infinitely far from the molecule but are nevertheless in good agreement with experimental findings.

===Reaction Dynamics===
<div>A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g.mol-1.pm.fs-1.
{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.</div>

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left| A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 fs of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is basically constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the BC pair which is moving apart, meaning that the total energy of the system is conserved. ]]

Let's now setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, an rFH = 238 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. Note that we are putting more energy into the system than the activation energy. The table below shows some observations and contour plots.

{| class="wikitable" border=1
! pFH/ g.mol-1.pm.fs-1 !! pHH/ g.mol-1.pm.fs-1 !! Einitial /kJ.mol-1 !! Reactive? !! Observations !! Illustration of the trajectory
|-
| -1.0 || -6.1 || -402.504 || No ||Here the system passed through the transition state structure, the hydrogen atom collided with the fluorine and then reformed reactants. || [[File:MRD 01576020 fhh table 1.png|400 px]]
|-
| -1.0 || -6.0 || -404.098 || No ||This is very similar to the case above. || [[File:MRD 01576020 fhh table 2.png|400 px]]
|-
| -1.0 || -5.0 || -414.098 || No ||This trajectory is also not reactive and is quite similar to the two above, again there was one collision with the fluorine atom. || [[File:MRD 01576020 fhh table 3.png|400 px]]
|-
| -1.0 || -4.1 || -421.388 || No ||Here, the fluorine atom did not collide with any atoms and the transition state was not surpassed. || [[File:MRD 01576020 fhh table 4.png|400 px]]
|-
| -1.0 || 0.0 || -434.098 || No ||Since even less energy was given the picture was very similar to the one above, but the hydrogen atom approached the fluorine even less. || [[File:MRD 01576020 fhh table 5.png|400 px]]
|-
| -1.0 || 2.1 || -427.588 || No ||Slightly more energy than above so the atoms were closer but the transition state was still not reached. || [[File:MRD 01576020 fhh table 6.png|400 px]]
|-
| -1.0 || 4.1 || -413.188 || No ||This trajectory was more interesting, the fluorine collided twice with one of the hydrogen atoms, but the reactants were reformed afterwards. || [[File:MRD 01576020 fhh table 7.png|400 px]]
|-
| -1.0 || 4.5 || -409.348 || Yes ||Interestingly, this trajectory was reactive despite the fact that the system has smaller energy than in some other cases. The total energy is therefore far from being the only parameter that influences the success of a given chemical reaction. || [[File:MRD 01576020 fhh table 11.png|400 px]]
|-
| -1.0 || 6.0 || -392.098 || No ||The system was very close to forming products here, fluorine and hydrogen collided 7 times, but the system rolled into the valley of reactants again. || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 6.1 || -390.788 || No ||The fluorine and hydrogen only collided once in this simulation. There is a pattern emerging in this table ... || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 7.0 || -378.098 || No ||This was very similar to the one above, there was one collision and then the reactants were reformed.|| [[File:MRD 01576020 fhh table 10.png|400 px]]
|}

From the table above it is obvious that whether the reaction will result in the formation of products is clearly not dependent only on the energy of vibration of the H2 molecule. We have also seen that these conditions where the molecule and the fluorine atom approach each other rather slowly rarely result in producing products. Maybe increasing the momentum pFH will help ...

{{fontcolor|blue|For the same initial position, increase slightly the momentum AB, and considerably reduce the overall energy of the system by reducing the momentum BC. What do you observe now?}}
There is very little vibration in the H2 molecule now and the F atom and the H2 molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.

Let's now focus on the reverse reaction. When there is very little vibrational energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below (Polanyi's empirical rules).

An example of a reactive trajectory for this reaction would be produced by the following initial conditions: AB distance = 212

{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions for a reactive trajectory. These rules tell us what modes need to have sufficient energy for certain types of TSs for a reaction to proceed. For early TSs, translational energy is more efficient than vibrational energy. For late TSs, on the other hand, vibrational energy will be more efficient. Why this is so is discussed in the figure to the right.

MRD:01576020

2020-05-15T19:30:50Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
Modelling the dynamics of molecular reactions has been enabled by the emergence of computers and the field of computational chemistry is an ever-expanding field. Despite most approaches being purely theoretical, ''ab initio'', calculations, the insight given is invaluable for understanding the pathways of chemical reactions. It is often also not necessary to include quantum mechanics, as atoms are of sufficient mass to be reasonably accurately modelled by Newton's equations of motion. Isolated systems in the gas phase only experience interatomic interactions which can be represented by a potential energy surface (PES) which is a function only of atomic positions. The force on a given interatomic coordinate can be calculated as follows:

<math> F = - { \partial V(r_1,r_2,...)\over \partial r_i}</math> (1)

where F = <math>{dp_i \over dt}</math> (2)

Calculating the forces and adjusting the system in consecutive steps results in a trajectory which shows how a system evolves in time. Such calculations have almost endless application from chemical biology and medicine{{cite journal |last1=Pregeljc |first1=Domen |last2=Jug |first2=Urska|last3=Mavri |first3=Janez|last4=Stare |first4=Jernej |date= |title=Why does the Y326I mutant of monoamine oxidase B decompose an endogenous amphetamine at a slower rate than the wild type enzyme? Reaction step elucidated by multiscale molecular simulations. |url= |journal= Phys Chem Chem Phys.|volume= 10|issue= 6|pages= 4181-4188|doi= 10.1039/c7cp07069a|access-date=}}

TRANSITION STATE THEORY
It is worth mentioning some main aspects about the transition state (TS) theory here as it is central to the concepts used here. It is purely classical - it ignores any quantum mechanical effects such as tunnelling, which is incredibly small anyway. It also assumes that the transition state structure is in equilibrium with the reactants, which is called a quasi-equilibrium. TS theory also assumes that the reactants have energies which follow a Boltzmann distribution, which is not a bad assumption as long as the system had time to equilibrate. This assumption, however is not very applicable to the system studied here. Another crucial assumption of the TS theory, developed in the 1930s, is that once the TS is surpassed, the products will necessarily be formed, which was not the case in some simulations performed here. This latter assumption is also the reason why theoretically derived reaction rates using the TS state theory overestimate experimentally obtained reaction rates. This can be improved by adding a correcting factor, but to do that one must already have extensive knowledge of the system in question. The transition point is defined as the maximum on the minimum energy path going from reactants to products. This is called a saddle point and the net force on the system is 0.

We will deal here with triatomic linear systems, as depicted in the following Figure.
[[File:Y2C8.png|thumb|centre|500px| A triatomic system used here. The atoms will either all be Hs or in the last case one H will be replaced with a F. Figure taken from https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD.]]

Terms like reactive and unreactive trajectories will be used often. A reactive trajectory is one which leads the reaction to completeness, i. e. the products are formed. In contrast an unreactive trajectory will not form products. The work here is divided into two exercises. Questions from the script are coloured blue for clarity and the answers follow below.

==EXERCISE 1==
===Dynamics from the TS region===

{{fontcolor|blue|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?}}
<div>The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its divergence being the zero vector at that point. To further characterise a point, one needs to evaluate the second derivatives at that point to determine whether it is a maximum or a minimum. The second derivative at the transition state will be negative, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the first derivative is zero and the second is positive, the point in question is a local minimum, which could be the lowest energy structure of the reactants or the products for example.</div>

{{fontcolor|blue|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.}}
<div>A simple system, consisting of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this figure. The transition state is the minimum of this line.]]

The minimum of this curve is the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if the initial structure corresponds to the TS structure. By changing the initial internuclear distances, we can indentify the TS as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Internuclear distance vs. time plots are shown in this figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Contour plot with the initial geometry being the transition state geometry and initial momenta = 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the TS the atoms are separated by more than 90 pm. Conversely, in case (c), the distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in close proximity of 90.8 pm.

But we can get a more accurate value now that we know where to look for the transition state. With some interval bisection, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

===Dynamics vs. MEP===

<div>{{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In a minimum energy path (MEP) calculation, the momentum of all particles is set to 0 in each step. Therefore there are no vibrations of the molecule at any point of the simulation. Of course this is unphysical as energy is lost, but it can be useful for some applications. In a dynamics simulation, on the other hand, vibrations of the molecule are observed as total energy of the system is constant.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. Molecule BC vibrates, as can be seen from the oscillating distance between atoms B and C. AB and AC distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. The AB momentum increases and ultimately moves with constant momentum away from the molecule BC. Molecule BC vibrates, so the momentum BC oscillates.]]</div>

Using the initial conditions r1=rts and r2=r<sub<ts+1, the system would roll into the other valley on the PES with similar behaviour to what is described in the figures on the left and right.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translational and vibrational). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill so the initial point of the previous simulation will not be reached.

===Reactive and Unreactive Trajectories===

{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point, fluorine and hydrogen collided a few times, but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected. There is also no general way to determine a reactive trajectory. This is influenced by many factors. There were significant vibrations present in the product molecule, which is a result of conservation of energy and the fact that the reaction is exothermic.

{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
As mentioned in the introduction, one of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the reactants again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===

{{fontcolor|blue|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?Locate the approximate position of the transition state.}}
[[File:MRD 01576020 MEP HF+H.png|thumb|right|Surface plot for the MEP calculation to determine the energy of the HF + H system. A similar calculation was performed to find the energy of the H2 + F system.]]
The F + H2 --> HF + H reaction is exothermic and H + HF --> H2 + F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than the H-H bond. This makes sense as the former has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is a fluorine atom. For these initial conditions, the forces were: along AB: 0.000 kJ.mol-1.pm-1 and along BC: 0.001 kJ.mol-1.pm-1. Due to these minimum forces, this structure is very close to the TS structure. One can notice that the distance between the H atoms in the TS structure is very close to their bond distance. This phenomenon is expressed in Hammond's postulate which says that the TS structure will resemble that of the species close to it in energy. Since this reaction is very exothermic, the TS structure is close to the structure of reactants which are a H2 molecule and a F atom.

{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. These were calculated by first finding the energy of the TS structure, which was -433.981 kJ/mol. Energies of the reactants and products were then obtained by performing a MEP simulation starting slightly off the TS structure in both directions (towards the reactants and towards the products, the one for the products is shown in the figure on the right). The calculated energy of the HF + H complex was -360.662 kJ/mol and that of the H2 + F complex was -435.031 kJ/mol. These values were used to calculate the activation energies given above. The obtained values are only approximate as we cannot move the atom infinitely far from the molecule but are nevertheless in good agreement with experimental findings.

===Reaction Dynamics===
<div>A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g.mol-1.pm.fs-1.
{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.</div>

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left| A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 fs of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is basically constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the BC pair which is moving apart, meaning that the total energy of the system is conserved. ]]

Let's now setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, an rFH = 238 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. Note that we are putting more energy into the system than the activation energy. The table below shows some observations and contour plots.

{| class="wikitable" border=1
! pFH/ g.mol-1.pm.fs-1 !! pHH/ g.mol-1.pm.fs-1 !! Einitial /kJ.mol-1 !! Reactive? !! Observations !! Illustration of the trajectory
|-
| -1.0 || -6.1 || -402.504 || No ||Here the system passed through the transition state structure, the hydrogen atom collided with the fluorine and then reformed reactants. || [[File:MRD 01576020 fhh table 1.png|400 px]]
|-
| -1.0 || -6.0 || -404.098 || No ||This is very similar to the case above. || [[File:MRD 01576020 fhh table 2.png|400 px]]
|-
| -1.0 || -5.0 || -414.098 || No ||This trajectory is also not reactive and is quite similar to the two above, again there was one collision with the fluorine atom. || [[File:MRD 01576020 fhh table 3.png|400 px]]
|-
| -1.0 || -4.1 || -421.388 || No ||Here, the fluorine atom did not collide with any atoms and the transition state was not surpassed. || [[File:MRD 01576020 fhh table 4.png|400 px]]
|-
| -1.0 || 0.0 || -434.098 || No ||Since even less energy was given the picture was very similar to the one above, but the hydrogen atom approached the fluorine even less. || [[File:MRD 01576020 fhh table 5.png|400 px]]
|-
| -1.0 || 2.1 || -427.588 || No ||Slightly more energy than above so the atoms were closer but the transition state was still not reached. || [[File:MRD 01576020 fhh table 6.png|400 px]]
|-
| -1.0 || 4.1 || -413.188 || No ||This trajectory was more interesting, the fluorine collided twice with one of the hydrogen atoms, but the reactants were reformed afterwards. || [[File:MRD 01576020 fhh table 7.png|400 px]]
|-
| -1.0 || 4.5 || -409.348 || Yes ||Interestingly, this trajectory was reactive despite the fact that the system has smaller energy than in some other cases. The total energy is therefore far from being the only parameter that influences the success of a given chemical reaction. || [[File:MRD 01576020 fhh table 11.png|400 px]]
|-
| -1.0 || 6.0 || -392.098 || No ||The system was very close to forming products here, fluorine and hydrogen collided 7 times, but the system rolled into the valley of reactants again. || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 6.1 || -390.788 || No ||The fluorine and hydrogen only collided once in this simulation. There is a pattern emerging in this table ... || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 7.0 || -378.098 || No ||This was very similar to the one above, there was one collision and then the reactants were reformed.|| [[File:MRD 01576020 fhh table 10.png|400 px]]
|}

From the table above it is obvious that whether the reaction will result in the formation of products is clearly not dependent only on the energy of vibration of the H2 molecule. We have also seen that these conditions where the molecule and the fluorine atom approach each other rather slowly rarely result in producing products. Maybe increasing the momentum pFH will help ...

{{fontcolor|blue|For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?}}

There is very little vibration in the H2 molecule now and the F atom and the H2 molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.

Let's now focus on the reverse reaction. When there is very little vibrational energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below (Polanyi's empirical rules).

An example of a reactive trajectory for this reaction would be produced by the following initial conditions: AB distance = 212

{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions for a reactive trajectory. These rules tell us what modes need to have sufficient energy for certain types of TSs for a reaction to proceed. For early TSs, translational energy is more efficient than vibrational energy. For late TSs, on the other hand, vibrational energy will be more efficient. Why this is so is discussed in the figure to the right.

MRD:01576020

2020-05-15T19:16:29Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
Modelling the dynamics of molecular reactions has been enabled by the emergence of computers and the field of computational chemistry is an ever-expanding field. Despite most approaches being purely theoretical, ''ab initio'', calculations, the insight given is invaluable for understanding the pathways of chemical reactions. It is often also not necessary to include quantum mechanics, as atoms are of sufficient mass to be reasonably accurately modelled by Newton's equations of motion. Isolated systems in the gas phase only experience interatomic interactions which can be represented by a potential energy surface (PES) which is a function only of atomic positions. The force on a given interatomic coordinate can be calculated as follows:

<math> F = - { \partial V(r_1,r_2,...)\over \partial r_i}</math> (1)

where F = <math>{dp_i \over dt}</math> (2)

Calculating the forces and adjusting the system in consecutive steps results in a trajectory which shows how a system evolves in time. Such calculations have almost endless application from chemical biology and medicine{{cite journal |last1=Pregeljc |first1=Domen |last2=Jug |first2=Urska|last3=Mavri |first3=Janez|last4=Stare |first4=Jernej |date= |title=Why does the Y326I mutant of monoamine oxidase B decompose an endogenous amphetamine at a slower rate than the wild type enzyme? Reaction step elucidated by multiscale molecular simulations. |url= |journal= Phys Chem Chem Phys.|volume= 10|issue= 6|pages= 4181-4188|doi= 10.1039/c7cp07069a|access-date=}}

TRANSITION STATE THEORY
It is worth mentioning some main aspects about the transition state (TS) theory here as it is central to the concepts used here. It is purely classical - it ignores any quantum mechanical effects such as tunnelling, which is incredibly small anyway. It also assumes that the transition state structure is in equilibrium with the reactants, which is called a quasi-equilibrium. TS theory also assumes that the reactants have energies which follow a Boltzmann distribution, which is not a bad assumption as long as the system had time to equilibrate. This assumption, however is not very applicable to the system studied here. Another crucial assumption of the TS theory, developed in the 1930s, is that once the TS is surpassed, the products will necessarily be formed, which was not the case in some simulations performed here. This latter assumption is also the reason why theoretically derived reaction rates using the TS state theory overestimate experimentally obtained reaction rates. This can be improved by adding a correcting factor, but to do that one must already have extensive knowledge of the system in question. The transition point is defined as the maximum on the minimum energy path going from reactants to products. This is called a saddle point and the net force on the system is 0.

We will deal here with triatomic linear systems, as depicted in the following Figure.
[[File:Y2C8.png|thumb|centre|500px| A triatomic system used here. The atoms will either all be Hs or in the last case one H will be replaced with a F. Figure taken from https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD.]]

Terms like reactive and unreactive trajectories will be used often. A reactive trajectory is one which leads the reaction to completeness, i. e. the products are formed. In contrast an unreactive trajectory will not form products. The work here is divided into two exercises. Questions from the script are coloured blue for clarity and the answers follow below.

==EXERCISE 1==
===Dynamics from the TS region===

{{fontcolor|blue|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?}}
<div>The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its divergence being the zero vector at that point. To further characterise a point, one needs to evaluate the second derivatives at that point to determine whether it is a maximum or a minimum. The second derivative at the transition state will be negative, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the first derivative is zero and the second is positive, the point in question is a local minimum, which could be the lowest energy structure of the reactants or the products for example.</div>

{{fontcolor|blue|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.}}
<div>A simple system, consisting of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this figure. The transition state is the minimum of this line.]]

The minimum of this curve is the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if the initial structure corresponds to the TS structure. By changing the initial internuclear distances, we can indentify the TS as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Internuclear distance vs. time plots are shown in this figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Contour plot with the initial geometry being the transition state geometry and initial momenta = 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the TS the atoms are separated by more than 90 pm. Conversely, in case (c), the distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in close proximity of 90.8 pm.

But we can get a more accurate value now that we know where to look for the transition state. With some interval bisection, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

===Dynamics vs. MEP===

<div>{{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In a minimum energy path (MEP) calculation, the momentum of all particles is set to 0 in each step. Therefore there are no vibrations of the molecule at any point of the simulation. Of course this is unphysical as energy is lost, but it can be useful for some applications. In a dynamics simulation, on the other hand, vibrations of the molecule are observed as total energy of the system is constant.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. Molecule BC vibrates, as can be seen from the oscillating distance between atoms B and C. AB and AC distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. The AB momentum increases and ultimately moves with constant momentum away from the molecule BC. Molecule BC vibrates, so the momentum BC oscillates.]]</div>

Using the initial conditions r1=rts and r2=r<sub<ts+1, the system would roll into the other valley on the PES with similar behaviour to what is described in the figures on the left and right.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translational and vibrational). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill so the initial point of the previous simulation will not be reached.

===Reactive and Unreactive Trajectories===

{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point, fluorine and hydrogen collided a few times, but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected. There is also no general way to determine a reactive trajectory. This is influenced by many factors. There were significant vibrations present in the product molecule, which is a result of conservation of energy and the fact that the reaction is exothermic.

{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
As mentioned in the introduction, one of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the reactants again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===

{{fontcolor|blue|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?Locate the approximate position of the transition state.}}
[[File:MRD 01576020 MEP HF+H.png|thumb|right|Surface plot for the MEP calculation to determine the energy of the HF + H system. A similar calculation was performed to find the energy of the H2 + F system.]]
The F + H2 --> HF + H reaction is exothermic and H + HF --> H2 + F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than the H-H bond. This makes sense as the former has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is a fluorine atom. For these initial conditions, the forces were: along AB: 0.000 kJ.mol-1.pm-1 and along BC: 0.001 kJ.mol-1.pm-1. Due to these minimum forces, this structure is very close to the TS structure. One can notice that the distance between the H atoms in the TS structure is very close to their bond distance. This phenomenon is expressed in Hammond's postulate which says that the TS structure will resemble that of the species close to it in energy. Since this reaction is very exothermic, the TS structure is close to the structure of reactants which are a H2 molecule and a F atom.

{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. These were calculated by first finding the energy of the TS structure, which was -433.981 kJ/mol. Energies of the reactants and products were then obtained by performing a MEP simulation starting slightly off the TS structure in both directions (towards the reactants and towards the products, the one for the products is shown in the figure on the right). The calculated energy of the HF + H complex was -360.662 kJ/mol and that of the H2 + F complex was -435.031 kJ/mol. These values were used to calculate the activation energies given above. The obtained values are only approximate as we cannot move the atom infinitely far from the molecule but are nevertheless in good agreement with experimental findings.

===Reaction Dynamics===
A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g mol-1 pm fs-1.
Q8{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left| A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 seconds of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the B-C pair which is moving aparth. ]]

Let's now setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, an rFH = 238 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. Note that we are putting more energy into the system than the activation energy. The table below shows some observations and contour plots.

{| class="wikitable" border=1
! pFH/ g.mol-1.pm.fs-1 !! pHH/ g.mol-1.pm.fs-1 !! Einitial /kJ.mol-1 !! Reactive? !! Observations !! Illustration of the trajectory
|-
| -1.0 || -6.1 || -402.504 || No ||Here the system passed through the transition state structure, the hydrogen atom collided with the fluorine and then reformed reactants. || [[File:MRD 01576020 fhh table 1.png|400 px]]
|-
| -1.0 || -6.0 || -404.098 || No ||This is very similar to the case above. || [[File:MRD 01576020 fhh table 2.png|400 px]]
|-
| -1.0 || -5.0 || -414.098 || No ||This trajectory is also not reactive and is quite similar to the two above, again there was one collision with the fluorine atom. || [[File:MRD 01576020 fhh table 3.png|400 px]]
|-
| -1.0 || -4.1 || -421.388 || No ||Here, the fluorine atom did not collide with any atoms and the transition state was not surpassed. || [[File:MRD 01576020 fhh table 4.png|400 px]]
|-
| -1.0 || 0.0 || -434.098 || No ||Since even less energy was given the picture was very similar to the one above, but the hydrogen atom approached the fluorine even less. || [[File:MRD 01576020 fhh table 5.png|400 px]]
|-
| -1.0 || 2.1 || -427.588 || No ||Slightly more energy than above so the atoms were closer but the transition state was still not reached. || [[File:MRD 01576020 fhh table 6.png|400 px]]
|-
| -1.0 || 4.1 || -413.188 || No ||This trajectory was more interesting, the fluorine collided twice with one of the hydrogen atoms, but the reactants were reformed afterwards. || [[File:MRD 01576020 fhh table 7.png|400 px]]
|-
| -1.0 || 4.5 || -409.348 || Yes ||Interestingly, this trajectory was reactive despite the fact that the system has smaller energy than in some other cases. The total energy is therefore far from being the only parameter that influences the success of a given chemical reaction. || [[File:MRD 01576020 fhh table 11.png|400 px]]
|-
| -1.0 || 6.0 || -392.098 || No ||The system was very close to forming products here, fluorine and hydrogen collided 7 times, but the system rolled into the valley of reactants again. || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 6.1 || -390.788 || No ||The fluorine and hydrogen only collided once in this simulation. There is a pattern emerging in this table ... || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 7.0 || -378.098 || No ||This was very similar to the one above, there was one collision and then the reactants were reformed.|| [[File:MRD 01576020 fhh table 10.png|400 px]]
|}

From the table above it is obvious that whether the reaction will result in the formation of products is clearly not dependent only on the energy of vibration of the H2 molecule. We have also seen that these conditions where the molecule and the fluorine atom approach each other rather slowly rarely result in producing products. Maybe increasing the momentum pFH will help ...

{{fontcolor|blue|For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?}}

There is very little vibration in the H2 molecule now and the F and the molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.

Let's now focus on the reverse reaction. When there is very little vibrational energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below.

An example of a reactive trajectory for this reaction would be produced by the following initial conditions: AB distance = 212

Q9{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions

MRD:01576020

2020-05-15T19:15:39Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
Modelling the dynamics of molecular reactions has been enabled by the emergence of computers and the field of computational chemistry is an ever-expanding field. Despite most approaches being purely theoretical, ''ab initio'', calculations, the insight given is invaluable for understanding the pathways of chemical reactions. It is often also not necessary to include quantum mechanics, as atoms are of sufficient mass to be reasonably accurately modelled by Newton's equations of motion. Isolated systems in the gas phase only experience interatomic interactions which can be represented by a potential energy surface (PES) which is a function only of atomic positions. The force on a given interatomic coordinate can be calculated as follows:

<math> F = - { \partial V(r_1,r_2,...)\over \partial r_i}</math> (1)

where F = <math>{dp_i \over dt}</math> (2)

Calculating the forces and adjusting the system in consecutive steps results in a trajectory which shows how a system evolves in time. Such calculations have almost endless application from chemical biology and medicine{{cite journal |last1=Pregeljc |first1=Domen |last2=Jug |first2=Urska|last3=Mavri |first3=Janez|last4=Stare |first4=Jernej |date= |title=Why does the Y326I mutant of monoamine oxidase B decompose an endogenous amphetamine at a slower rate than the wild type enzyme? Reaction step elucidated by multiscale molecular simulations. |url= |journal= Phys Chem Chem Phys.|volume= 10|issue= 6|pages= 4181-4188|doi= 10.1039/c7cp07069a|access-date=}}

TRANSITION STATE THEORY
It is worth mentioning some main aspects about the transition state (TS) theory here as it is central to the concepts used here. It is purely classical - it ignores any quantum mechanical effects such as tunnelling, which is incredibly small anyway. It also assumes that the transition state structure is in equilibrium with the reactants, which is called a quasi-equilibrium. TS theory also assumes that the reactants have energies which follow a Boltzmann distribution, which is not a bad assumption as long as the system had time to equilibrate. This assumption, however is not very applicable to the system studied here. Another crucial assumption of the TS theory, developed in the 1930s, is that once the TS is surpassed, the products will necessarily be formed, which was not the case in some simulations performed here. This latter assumption is also the reason why theoretically derived reaction rates using the TS state theory overestimate experimentally obtained reaction rates. This can be improved by adding a correcting factor, but to do that one must already have extensive knowledge of the system in question. The transition point is defined as the maximum on the minimum energy path going from reactants to products. This is called a saddle point and the net force on the system is 0.

We will deal here with triatomic linear systems, as depicted in the following Figure.
[[File:Y2C8.png|thumb|centre|500px| A triatomic system used here. The atoms will either all be Hs or in the last case one H will be replaced with a F. Figure taken from https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD.]]

Terms like reactive and unreactive trajectories will be used often. A reactive trajectory is one which leads the reaction to completeness, i. e. the products are formed. In contrast an unreactive trajectory will not form products. The work here is divided into two exercises. Questions from the script are coloured blue for clarity and the answers follow below.

==EXERCISE 1==
===Dynamics from the TS region===

{{fontcolor|blue|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?}}
<div>The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its divergence being the zero vector at that point. To further characterise a point, one needs to evaluate the second derivatives at that point to determine whether it is a maximum or a minimum. The second derivative at the transition state will be negative, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the first derivative is zero and the second is positive, the point in question is a local minimum, which could be the lowest energy structure of the reactants or the products for example.</div>

{{fontcolor|blue|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.}}
<div>A simple system, consisting of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this figure. The transition state is the minimum of this line.]]

The minimum of this curve is the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if the initial structure corresponds to the TS structure. By changing the initial internuclear distances, we can indentify the TS as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Internuclear distance vs. time plots are shown in this figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Contour plot with the initial geometry being the transition state geometry and initial momenta = 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the TS the atoms are separated by more than 90 pm. Conversely, in case (c), the distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in close proximity of 90.8 pm.

But we can get a more accurate value now that we know where to look for the transition state. With some interval bisection, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

===Dynamics vs. MEP===

<div>{{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In a minimum energy path (MEP) calculation, the momentum of all particles is set to 0 in each step. Therefore there are no vibrations of the molecule at any point of the simulation. Of course this is unphysical as energy is lost, but it can be useful for some applications. In a dynamics simulation, on the other hand, vibrations of the molecule are observed as total energy of the system is constant.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. Molecule BC vibrates, as can be seen from the oscillating distance between atoms B and C. AB and AC distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. The AB momentum increases and ultimately moves with constant momentum away from the molecule BC. Molecule BC vibrates, so the momentum BC oscillates.]]</div>

Using the initial conditions r1=rts and r2=r<sub<ts+1, the system would roll into the other valley on the PES with similar behaviour to what is described in the figures on the left and right.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translational and vibrational). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill so the initial point of the previous simulation will not be reached.

===Reactive and Unreactive Trajectories===

{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point, fluorine and hydrogen collided a few times, but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected. There is also no general way to determine a reactive trajectory. This is influenced by many factors. There were significant vibrations present in the product molecule, which is a result of conservation of energy and the fact that the reaction is exothermic.

{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
As mentioned in the introduction, one of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the reactants again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===

[[File:MRD 01576020 MEP HF+H.png|thumb|right|SUrface plot for the MEP calculation to determine the energy of the HF + H system. A similar calculation was performed to find the energy of the H2 + F system.]]
{{fontcolor|blue|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?Locate the approximate position of the transition state.}}

The F + H2 --> HF + H reaction is exothermic and H + HF --> H2 + F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than the H-H bond. This makes sense as the former has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is a fluorine atom. For these initial conditions, the forces were: along AB: 0.000 kJ.mol-1.pm-1 and along BC: 0.001 kJ.mol-1.pm-1. Due to these minimum forces, this structure is very close to the TS structure. One can notice that the distance between the H atoms in the TS structure is very close to their bond distance. This phenomenon is expressed in Hammond's postulate which says that the TS structure will resemble that of the species close to it in energy. Since this reaction is very exothermic, the TS structure is close to the structure of reactants which are a H2 molecule and a F atom.

{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. These were calculated by first finding the energy of the TS structure, which was -433.981 kJ/mol. Energies of the reactants and products were then obtained by performing a MEP simulation starting slightly off the TS structure in both directions (towards the reactants and towards the products, the one for the products is shown in the figure on the right). The calculated energy of the HF + H complex was -360.662 kJ/mol and that of the H2 + F complex was -435.031 kJ/mol. These values were used to calculate the activation energies given above. The obtained values are only approximate as we cannot move the atom infinitely far from the molecule but are nevertheless in good agreement with experimental findings.

===Reaction Dynamics===
A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g mol-1 pm fs-1.
Q8{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left| A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 seconds of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the B-C pair which is moving aparth. ]]

Let's now setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, an rFH = 238 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. Note that we are putting more energy into the system than the activation energy. The table below shows some observations and contour plots.

{| class="wikitable" border=1
! pFH/ g.mol-1.pm.fs-1 !! pHH/ g.mol-1.pm.fs-1 !! Einitial /kJ.mol-1 !! Reactive? !! Observations !! Illustration of the trajectory
|-
| -1.0 || -6.1 || -402.504 || No ||Here the system passed through the transition state structure, the hydrogen atom collided with the fluorine and then reformed reactants. || [[File:MRD 01576020 fhh table 1.png|400 px]]
|-
| -1.0 || -6.0 || -404.098 || No ||This is very similar to the case above. || [[File:MRD 01576020 fhh table 2.png|400 px]]
|-
| -1.0 || -5.0 || -414.098 || No ||This trajectory is also not reactive and is quite similar to the two above, again there was one collision with the fluorine atom. || [[File:MRD 01576020 fhh table 3.png|400 px]]
|-
| -1.0 || -4.1 || -421.388 || No ||Here, the fluorine atom did not collide with any atoms and the transition state was not surpassed. || [[File:MRD 01576020 fhh table 4.png|400 px]]
|-
| -1.0 || 0.0 || -434.098 || No ||Since even less energy was given the picture was very similar to the one above, but the hydrogen atom approached the fluorine even less. || [[File:MRD 01576020 fhh table 5.png|400 px]]
|-
| -1.0 || 2.1 || -427.588 || No ||Slightly more energy than above so the atoms were closer but the transition state was still not reached. || [[File:MRD 01576020 fhh table 6.png|400 px]]
|-
| -1.0 || 4.1 || -413.188 || No ||This trajectory was more interesting, the fluorine collided twice with one of the hydrogen atoms, but the reactants were reformed afterwards. || [[File:MRD 01576020 fhh table 7.png|400 px]]
|-
| -1.0 || 4.5 || -409.348 || Yes ||Interestingly, this trajectory was reactive despite the fact that the system has smaller energy than in some other cases. The total energy is therefore far from being the only parameter that influences the success of a given chemical reaction. || [[File:MRD 01576020 fhh table 11.png|400 px]]
|-
| -1.0 || 6.0 || -392.098 || No ||The system was very close to forming products here, fluorine and hydrogen collided 7 times, but the system rolled into the valley of reactants again. || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 6.1 || -390.788 || No ||The fluorine and hydrogen only collided once in this simulation. There is a pattern emerging in this table ... || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 7.0 || -378.098 || No ||This was very similar to the one above, there was one collision and then the reactants were reformed.|| [[File:MRD 01576020 fhh table 10.png|400 px]]
|}

From the table above it is obvious that whether the reaction will result in the formation of products is clearly not dependent only on the energy of vibration of the H2 molecule. We have also seen that these conditions where the molecule and the fluorine atom approach each other rather slowly rarely result in producing products. Maybe increasing the momentum pFH will help ...

{{fontcolor|blue|For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?}}

There is very little vibration in the H2 molecule now and the F and the molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.

Let's now focus on the reverse reaction. When there is very little vibrational energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below.

An example of a reactive trajectory for this reaction would be produced by the following initial conditions: AB distance = 212

Q9{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions

MRD:01576020

2020-05-15T19:12:34Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
Modelling the dynamics of molecular reactions has been enabled by the emergence of computers and the field of computational chemistry is an ever-expanding field. Despite most approaches being purely theoretical, ''ab initio'', calculations, the insight given is invaluable for understanding the pathways of chemical reactions. It is often also not necessary to include quantum mechanics, as atoms are of sufficient mass to be reasonably accurately modelled by Newton's equations of motion. Isolated systems in the gas phase only experience interatomic interactions which can be represented by a potential energy surface (PES) which is a function only of atomic positions. The force on a given interatomic coordinate can be calculated as follows:

<math> F = - { \partial V(r_1,r_2,...)\over \partial r_i}</math> (1)

where F = <math>{dp_i \over dt}</math> (2)

Calculating the forces and adjusting the system in consecutive steps results in a trajectory which shows how a system evolves in time. Such calculations have almost endless application from chemical biology and medicine{{cite journal |last1=Pregeljc |first1=Domen |last2=Jug |first2=Urska|last3=Mavri |first3=Janez|last4=Stare |first4=Jernej |date= |title=Why does the Y326I mutant of monoamine oxidase B decompose an endogenous amphetamine at a slower rate than the wild type enzyme? Reaction step elucidated by multiscale molecular simulations. |url= |journal= Phys Chem Chem Phys.|volume= 10|issue= 6|pages= 4181-4188|doi= 10.1039/c7cp07069a|access-date=}}

TRANSITION STATE THEORY
It is worth mentioning some main aspects about the transition state (TS) theory here as it is central to the concepts used here. It is purely classical - it ignores any quantum mechanical effects such as tunnelling, which is incredibly small anyway. It also assumes that the transition state structure is in equilibrium with the reactants, which is called a quasi-equilibrium. TS theory also assumes that the reactants have energies which follow a Boltzmann distribution, which is not a bad assumption as long as the system had time to equilibrate. This assumption, however is not very applicable to the system studied here. Another crucial assumption of the TS theory, developed in the 1930s, is that once the TS is surpassed, the products will necessarily be formed, which was not the case in some simulations performed here. This latter assumption is also the reason why theoretically derived reaction rates using the TS state theory overestimate experimentally obtained reaction rates. This can be improved by adding a correcting factor, but to do that one must already have extensive knowledge of the system in question. The transition point is defined as the maximum on the minimum energy path going from reactants to products. This is called a saddle point and the net force on the system is 0.

We will deal here with triatomic linear systems, as depicted in the following Figure.
[[File:Y2C8.png|thumb|centre|500px| A triatomic system used here. The atoms will either all be Hs or in the last case one H will be replaced with a F. Figure taken from https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD.]]

Terms like reactive and unreactive trajectories will be used often. A reactive trajectory is one which leads the reaction to completeness, i. e. the products are formed. In contrast an unreactive trajectory will not form products. The work here is divided into two exercises. Questions from the script are coloured blue for clarity and the answers follow below.

==EXERCISE 1==
===Dynamics from the TS region===

{{fontcolor|blue|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?}}
<div>The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its divergence being the zero vector at that point. To further characterise a point, one needs to evaluate the second derivatives at that point to determine whether it is a maximum or a minimum. The second derivative at the transition state will be negative, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the first derivative is zero and the second is positive, the point in question is a local minimum, which could be the lowest energy structure of the reactants or the products for example.</div>

{{fontcolor|blue|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.}}
<div>A simple system, consisting of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this figure. The transition state is the minimum of this line.]]

The minimum of this curve is the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if the initial structure corresponds to the TS structure. By changing the initial internuclear distances, we can indentify the TS as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Internuclear distance vs. time plots are shown in this figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Contour plot with the initial geometry being the transition state geometry and initial momenta = 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the TS the atoms are separated by more than 90 pm. Conversely, in case (c), the distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in close proximity of 90.8 pm.

But we can get a more accurate value now that we know where to look for the transition state. With some interval bisection, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

===Dynamics vs. MEP===

<div>{{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In a minimum energy path (MEP) calculation, the momentum of all particles is set to 0 in each step. Therefore there are no vibrations of the molecule at any point of the simulation. Of course this is unphysical as energy is lost, but it can be useful for some applications. In a dynamics simulation, on the other hand, vibrations of the molecule are observed as total energy of the system is constant.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. Molecule BC vibrates, as can be seen from the oscillating distance between atoms B and C. AB and AC distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. The AB momentum increases and ultimately moves with constant momentum away from the molecule BC. Molecule BC vibrates, so the momentum BC oscillates.]]</div>

Using the initial conditions r1=rts and r2=r<sub<ts+1, the system would roll into the other valley on the PES with similar behaviour to what is described in the figures on the left and right.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translational and vibrational). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill so the initial point of the previous simulation will not be reached.

===Reactive and Unreactive Trajectories===

{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point, fluorine and hydrogen collided a few times, but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected. There is also no general way to determine a reactive trajectory. This is influenced by many factors. There were significant vibrations present in the product molecule, which is a result of conservation of energy and the fact that the reaction is exothermic.

{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
As mentioned in the introduction, one of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the reactants again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===

{{fontcolor|blue|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?Locate the approximate position of the transition state.}}

The F+H2--> HF+H reaction is exothermic and H+HF-->H2+F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than the H-H bond. This makes sense as the former has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is a fluorine atom. For these initial conditions, the forces were: along AB: 0.000 and along BC: 0.001. Due to these minimum forces, this structure is very close to the TS structure. One can notice that the distance between the H atoms in the TS structure is very close to their bond distance. This phenomenon is expressed in Hammond's postulate which says that the TS structure will resemble that of the species close to it in energy. Since this reaction is very exothermic, the TS structure is close to the structure of reactants which are a H2 molecule and a F atom.

[[File:MRD 01576020 MEP HF+H.png|thumb|right|SUrface plot for the MEP calculation to determine the energy of the HF + H system. A similar calculation was performed to find the energy of the H2 + F system.]]

{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. These were calculated by first finding the energy of the TS structure, which was -433.981 kJ/mol. Energies of the reactants and products were then obtained by performing a MEP simulation starting slightly off the TS structure in both directions (towards the reactants and towards the products, the one for the products is shown in the figure on the right). The calculated energy of the HF + H complex was -360.662 kJ/mol and that of the H2 + F complex was -435.031 kJ/mol. These values were used to calculate the activation energies given above. The obtained values are only approximate as we cannot move the atom infinitely far from the molecule but are nevertheless in good agreement with experimental findings.

===Reaction Dynamics===
A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g mol-1 pm fs-1.
Q8{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left| A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 seconds of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the B-C pair which is moving aparth. ]]

Let's now setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, an rFH = 238 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. Note that we are putting more energy into the system than the activation energy. The table below shows some observations and contour plots.

{| class="wikitable" border=1
! pFH/ g.mol-1.pm.fs-1 !! pHH/ g.mol-1.pm.fs-1 !! Einitial /kJ.mol-1 !! Reactive? !! Observations !! Illustration of the trajectory
|-
| -1.0 || -6.1 || -402.504 || No ||Here the system passed through the transition state structure, the hydrogen atom collided with the fluorine and then reformed reactants. || [[File:MRD 01576020 fhh table 1.png|400 px]]
|-
| -1.0 || -6.0 || -404.098 || No ||This is very similar to the case above. || [[File:MRD 01576020 fhh table 2.png|400 px]]
|-
| -1.0 || -5.0 || -414.098 || No ||This trajectory is also not reactive and is quite similar to the two above, again there was one collision with the fluorine atom. || [[File:MRD 01576020 fhh table 3.png|400 px]]
|-
| -1.0 || -4.1 || -421.388 || No ||Here, the fluorine atom did not collide with any atoms and the transition state was not surpassed. || [[File:MRD 01576020 fhh table 4.png|400 px]]
|-
| -1.0 || 0.0 || -434.098 || No ||Since even less energy was given the picture was very similar to the one above, but the hydrogen atom approached the fluorine even less. || [[File:MRD 01576020 fhh table 5.png|400 px]]
|-
| -1.0 || 2.1 || -427.588 || No ||Slightly more energy than above so the atoms were closer but the transition state was still not reached. || [[File:MRD 01576020 fhh table 6.png|400 px]]
|-
| -1.0 || 4.1 || -413.188 || No ||This trajectory was more interesting, the fluorine collided twice with one of the hydrogen atoms, but the reactants were reformed afterwards. || [[File:MRD 01576020 fhh table 7.png|400 px]]
|-
| -1.0 || 4.5 || -409.348 || Yes ||Interestingly, this trajectory was reactive despite the fact that the system has smaller energy than in some other cases. The total energy is therefore far from being the only parameter that influences the success of a given chemical reaction. || [[File:MRD 01576020 fhh table 11.png|400 px]]
|-
| -1.0 || 6.0 || -392.098 || No ||The system was very close to forming products here, fluorine and hydrogen collided 7 times, but the system rolled into the valley of reactants again. || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 6.1 || -390.788 || No ||The fluorine and hydrogen only collided once in this simulation. There is a pattern emerging in this table ... || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 7.0 || -378.098 || No ||This was very similar to the one above, there was one collision and then the reactants were reformed.|| [[File:MRD 01576020 fhh table 10.png|400 px]]
|}

From the table above it is obvious that whether the reaction will result in the formation of products is clearly not dependent only on the energy of vibration of the H2 molecule. We have also seen that these conditions where the molecule and the fluorine atom approach each other rather slowly rarely result in producing products. Maybe increasing the momentum pFH will help ...

{{fontcolor|blue|For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?}}

There is very little vibration in the H2 molecule now and the F and the molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.

Let's now focus on the reverse reaction. When there is very little vibrational energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below.

An example of a reactive trajectory for this reaction would be produced by the following initial conditions: AB distance = 212

Q9{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions

File:MRD 01576020 MEP HF+H.png

2020-05-15T19:09:31Z

Dp3618:

MRD:01576020

2020-05-15T19:05:23Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
Modelling the dynamics of molecular reactions has been enabled by the emergence of computers and the field of computational chemistry is an ever-expanding field. Despite most approaches being purely theoretical, ''ab initio'', calculations, the insight given is invaluable for understanding the pathways of chemical reactions. It is often also not necessary to include quantum mechanics, as atoms are of sufficient mass to be reasonably accurately modelled by Newton's equations of motion. Isolated systems in the gas phase only experience interatomic interactions which can be represented by a potential energy surface (PES) which is a function only of atomic positions. The force on a given interatomic coordinate can be calculated as follows:

<math> F = - { \partial V(r_1,r_2,...)\over \partial r_i}</math> (1)

where F = <math>{dp_i \over dt}</math> (2)

Calculating the forces and adjusting the system in consecutive steps results in a trajectory which shows how a system evolves in time. Such calculations have almost endless application from chemical biology and medicine{{cite journal |last1=Pregeljc |first1=Domen |last2=Jug |first2=Urska|last3=Mavri |first3=Janez|last4=Stare |first4=Jernej |date= |title=Why does the Y326I mutant of monoamine oxidase B decompose an endogenous amphetamine at a slower rate than the wild type enzyme? Reaction step elucidated by multiscale molecular simulations. |url= |journal= Phys Chem Chem Phys.|volume= 10|issue= 6|pages= 4181-4188|doi= 10.1039/c7cp07069a|access-date=}}

TRANSITION STATE THEORY
It is worth mentioning some main aspects about the transition state (TS) theory here as it is central to the concepts used here. It is purely classical - it ignores any quantum mechanical effects such as tunnelling, which is incredibly small anyway. It also assumes that the transition state structure is in equilibrium with the reactants, which is called a quasi-equilibrium. TS theory also assumes that the reactants have energies which follow a Boltzmann distribution, which is not a bad assumption as long as the system had time to equilibrate. This assumption, however is not very applicable to the system studied here. Another crucial assumption of the TS theory, developed in the 1930s, is that once the TS is surpassed, the products will necessarily be formed, which was not the case in some simulations performed here. This latter assumption is also the reason why theoretically derived reaction rates using the TS state theory overestimate experimentally obtained reaction rates. This can be improved by adding a correcting factor, but to do that one must already have extensive knowledge of the system in question. The transition point is defined as the maximum on the minimum energy path going from reactants to products. This is called a saddle point and the net force on the system is 0.

We will deal here with triatomic linear systems, as depicted in the following Figure.
[[File:Y2C8.png|thumb|centre|500px| A triatomic system used here. The atoms will either all be Hs or in the last case one H will be replaced with a F. Figure taken from https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD.]]

Terms like reactive and unreactive trajectories will be used often. A reactive trajectory is one which leads the reaction to completeness, i. e. the products are formed. In contrast an unreactive trajectory will not form products. The work here is divided into two exercises. Questions from the script are coloured blue for clarity and the answers follow below.

==EXERCISE 1==
===Dynamics from the TS region===

{{fontcolor|blue|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?}}
<div>The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its divergence being the zero vector at that point. To further characterise a point, one needs to evaluate the second derivatives at that point to determine whether it is a maximum or a minimum. The second derivative at the transition state will be negative, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the first derivative is zero and the second is positive, the point in question is a local minimum, which could be the lowest energy structure of the reactants or the products for example.</div>

{{fontcolor|blue|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.}}
<div>A simple system, consisting of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this figure. The transition state is the minimum of this line.]]

The minimum of this curve is the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if the initial structure corresponds to the TS structure. By changing the initial internuclear distances, we can indentify the TS as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Internuclear distance vs. time plots are shown in this figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Contour plot with the initial geometry being the transition state geometry and initial momenta = 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the TS the atoms are separated by more than 90 pm. Conversely, in case (c), the distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in close proximity of 90.8 pm.

But we can get a more accurate value now that we know where to look for the transition state. With some interval bisection, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

===Dynamics vs. MEP===

<div>{{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In a minimum energy path (MEP) calculation, the momentum of all particles is set to 0 in each step. Therefore there are no vibrations of the molecule at any point of the simulation. Of course this is unphysical as energy is lost, but it can be useful for some applications. In a dynamics simulation, on the other hand, vibrations of the molecule are observed as total energy of the system is constant.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. Molecule BC vibrates, as can be seen from the oscillating distance between atoms B and C. AB and AC distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule BC. The AB momentum increases and ultimately moves with constant momentum away from the molecule BC. Molecule BC vibrates, so the momentum BC oscillates.]]</div>

Using the initial conditions r1=rts and r2=r<sub<ts+1, the system would roll into the other valley on the PES with similar behaviour to what is described in the figures on the left and right.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translational and vibrational). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill so the initial point of the previous simulation will not be reached.

===Reactive and Unreactive Trajectories===

{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point, fluorine and hydrogen collided a few times, but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected. There is also no general way to determine a reactive trajectory. This is influenced by many factors. There were significant vibrations present in the product molecule, which is a result of conservation of energy and the fact that the reaction is exothermic.

{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
As mentioned in the introduction, one of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the reactants again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===

{{fontcolor|blue|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?Locate the approximate position of the transition state.}}

The F+H2--> HF+H reaction is exothermic and H+HF-->H2+F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than the H-H bond. This makes sense as the former has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is a fluorine atom. For these initial conditions, the forces were: along AB: 0.000 and along BC: 0.001. Due to these minimum forces, this structure is very close to the TS structure. One can notice that the distance between the H atoms in the TS structure is very close to their bond distance. This phenomenon is expressed in Hammond's postulate which says that the TS structure will resemble that of the species close to it in energy. Since this reaction is very exothermic, the TS structure is close to the structure of reactants which are a H2 molecule and a F atom.

{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. These were calculated by first finding the energy of the TS structure, which was -433.981 kJ/mol. Energies of the reactants and products were then obtained by performing a MEP simulation starting slightly off the TS structure in both directions (towards the reactants and towards the products, the one for the products is shown in the figure on the right). The calculated energy of the HF + H complex was -360.662 kJ/mol and that of the H2 + F complex was -435.031 kJ/mol. These values were used to calculate the activation energies given above. The obtained values are only approximate as we cannot move the atom infinitely far from the molecule but are nevertheless in good agreement with experimental findings.

===Reaction Dynamics===
A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g mol-1 pm fs-1.
Q8{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left| A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 seconds of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the B-C pair which is moving aparth. ]]

Let's now setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, an rFH = 238 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. Note that we are putting more energy into the system than the activation energy. The table below shows some observations and contour plots.

{| class="wikitable" border=1
! pFH/ g.mol-1.pm.fs-1 !! pHH/ g.mol-1.pm.fs-1 !! Einitial /kJ.mol-1 !! Reactive? !! Observations !! Illustration of the trajectory
|-
| -1.0 || -6.1 || -402.504 || No ||Here the system passed through the transition state structure, the hydrogen atom collided with the fluorine and then reformed reactants. || [[File:MRD 01576020 fhh table 1.png|400 px]]
|-
| -1.0 || -6.0 || -404.098 || No ||This is very similar to the case above. || [[File:MRD 01576020 fhh table 2.png|400 px]]
|-
| -1.0 || -5.0 || -414.098 || No ||This trajectory is also not reactive and is quite similar to the two above, again there was one collision with the fluorine atom. || [[File:MRD 01576020 fhh table 3.png|400 px]]
|-
| -1.0 || -4.1 || -421.388 || No ||Here, the fluorine atom did not collide with any atoms and the transition state was not surpassed. || [[File:MRD 01576020 fhh table 4.png|400 px]]
|-
| -1.0 || 0.0 || -434.098 || No ||Since even less energy was given the picture was very similar to the one above, but the hydrogen atom approached the fluorine even less. || [[File:MRD 01576020 fhh table 5.png|400 px]]
|-
| -1.0 || 2.1 || -427.588 || No ||Slightly more energy than above so the atoms were closer but the transition state was still not reached. || [[File:MRD 01576020 fhh table 6.png|400 px]]
|-
| -1.0 || 4.1 || -413.188 || No ||This trajectory was more interesting, the fluorine collided twice with one of the hydrogen atoms, but the reactants were reformed afterwards. || [[File:MRD 01576020 fhh table 7.png|400 px]]
|-
| -1.0 || 4.5 || -409.348 || Yes ||Interestingly, this trajectory was reactive despite the fact that the system has smaller energy than in some other cases. The total energy is therefore far from being the only parameter that influences the success of a given chemical reaction. || [[File:MRD 01576020 fhh table 11.png|400 px]]
|-
| -1.0 || 6.0 || -392.098 || No ||The system was very close to forming products here, fluorine and hydrogen collided 7 times, but the system rolled into the valley of reactants again. || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 6.1 || -390.788 || No ||The fluorine and hydrogen only collided once in this simulation. There is a pattern emerging in this table ... || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 7.0 || -378.098 || No ||This was very similar to the one above, there was one collision and then the reactants were reformed.|| [[File:MRD 01576020 fhh table 10.png|400 px]]
|}

From the table above it is obvious that whether the reaction will result in the formation of products is clearly not dependent only on the energy of vibration of the H2 molecule. We have also seen that these conditions where the molecule and the fluorine atom approach each other rather slowly rarely result in producing products. Maybe increasing the momentum pFH will help ...

{{fontcolor|blue|For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?}}

There is very little vibration in the H2 molecule now and the F and the molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.

Let's now focus on the reverse reaction. When there is very little vibrational energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below.

An example of a reactive trajectory for this reaction would be produced by the following initial conditions: AB distance = 212

Q9{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions

MRD:01576020

2020-05-15T18:33:36Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
Modelling the dynamics of molecular reactions has been enabled by the emergence of computers and the field of computational chemistry is an ever-expanding field. Despite most approaches being purely theoretical, ''ab initio'', calculations, the insight given is invaluable for understanding the pathways of chemical reactions. It is often also not necessary to include quantum mechanics, as atoms are of sufficient mass to be reasonably accurately modelled by Newton's equations of motion. Isolated systems in the gas phase only experience interatomic interactions which can be represented by a potential energy surface (PES) which is a function only of atomic positions. The force on a given interatomic coordinate can be calculated as follows:

<math> F = - { \partial V(r_1,r_2,...)\over \partial r_i}</math> (1)

where F = <math>{dp_i \over dt}</math> (2)

Calculating the forces and adjusting the system in consecutive steps results in a trajectory which shows how a system evolves in time. Such calculations have almost endless application from chemical biology and medicine{{cite journal |last1=Pregeljc |first1=Domen |last2=Jug |first2=Urska|last3=Mavri |first3=Janez|last4=Stare |first4=Jernej |date= |title=Why does the Y326I mutant of monoamine oxidase B decompose an endogenous amphetamine at a slower rate than the wild type enzyme? Reaction step elucidated by multiscale molecular simulations. |url= |journal= Phys Chem Chem Phys.|volume= 10|issue= 6|pages= 4181-4188|doi= 10.1039/c7cp07069a|access-date=}}

TRANSITION STATE THEORY
It is worth mentioning some main aspects about the transition state (TS) theory here as it is central to the concepts used here. It is purely classical - it ignores any quantum mechanical effects such as tunnelling, which is incredibly small anyway. It also assumes that the transition state structure is in equilibrium with the reactants, which is called a quasi-equilibrium. TS theory also assumes that the reactants have energies which follow a Boltzmann distribution, which is not a bad assumption as long as the system had time to equilibrate. This assumption, however is not very applicable to the system studied here. Another crucial assumption of the TS theory, developed in the 1930s, is that once the TS is surpassed, the products will necessarily be formed, which was not the case in some simulations performed here. This latter assumption is also the reason why theoretically derived reaction rates using the TS state theory overestimate experimentally obtained reaction rates. This can be improved by adding a correcting factor, but to do that one must already have extensive knowledge of the system in question. The transition point is defined as the maximum on the minimum energy path going from reactants to products. This is called a saddle point and the net force on the system is 0.

We will deal here with triatomic linear systems, as depicted in the following Figure.
[[File:Y2C8.png|thumb|centre|500px| A triatomic system used here. The atoms will either all be Hs or in the last case one H will be replaced with a F. Figure taken from https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD.]]

Terms like reactive and unreactive trajectories will be used often. A reactive trajectory is one which leads the reaction to completeness, i. e. the products are formed. In contrast an unreactive trajectory will not form products. The work here is divided into two exercises. Questions from the script are coloured blue for clarity and the answers follow below.

==EXERCISE 1==
===Dynamics from the TS region===
{{fontcolor|blue|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?}}
<div>The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its divergence being the zero vector at that point. To further characterise a point, one needs to evaluate the second derivatives at that point to determine whether it is a maximum or a minimum. The second derivative at the transition state will be negative, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the first derivative is zero and the second is positive, the point in question is a local minimum, which could be the lowest energy structure of the reactants or the products for example.</div>

{{fontcolor|blue|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.}}
<div>A simple system, consisting of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this figure. The transition state is the minimum of this line.]]

The minimum of this curve is the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if the initial structure corresponds to the TS structure. By changing the initial internuclear distances, we can indentify the TS as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Internuclear distance vs. time plots are shown in this figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Contour plot with the initial geometry being the transition state geometry and initial momenta = 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the TS the atoms are separated by more than 90 pm. Conversely, in case (c), the distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in close proximity of 90.8 pm.

But we can get a more accurate value now that we know where to look for the transition state. With some interval bisection, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

===Dynamics vs. MEP===
<div>{{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In a minimum energy path (MEP) calculation, the momentum of all particles is set to 0 in each step. Therefore there are no vibrations of the molecule at any point of the simulation. Of course this is unphysical as energy is lost, but it can be useful for some applications. In a dynamics simulation, on the other hand, vibrations of the molecule are observed as total energy of the system is constant.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. Molecule B-C vibrates, as can be seen from the oscillating distance between atoms B and C. A-b and A-C distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. The A-B momentum increases and ultimately moves with constant momentum away from the molecule B-C. Molecule B-C vibrates, so the momentum B-C oscillates.]]</div>

Using the initial conditions r1=rts and r2=r<sub<ts+1, the system would roll into the other valley on the PES with similar behaviour to what is described in the figures on the left and right.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translational and vibrational). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill so the initial point of the previous simulation will not be reached.

===Reactive and Unreactive Trajectories===
<div>Reactive and Unreactive Trajectories
Q4{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected.
</div>

Q5{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
One of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the products again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===
Q6{{fontcolor|blue|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?Locate the approximate position of the transition state.}}

The F+H2--> HF+H reaction is exothermic and H+HF-->H2+F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than H-H bond. This makes sense as it has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is F. From these initial conditions, the forces were: along AB: 0.000 and along BC: 0.001. SO this structure is very close to the transition state structure.

Q7{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. This is in good agreement with experimental findings, where the H-F bond dissociation energy is reported as 566 kJ/mol and the H-H bond dissociation energy is 436 kJ/mol[ref].

===Reaction Dynamics===
A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g mol-1 pm fs-1.
Q8{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left| A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 seconds of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the B-C pair which is moving aparth. ]]

Let's now setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, an rFH = 238 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. Note that we are putting more energy into the system than the activation energy. The table below shows some observations and contour plots.

{| class="wikitable" border=1
! pFH/ g.mol-1.pm.fs-1 !! pHH/ g.mol-1.pm.fs-1 !! Einitial /kJ.mol-1 !! Reactive? !! Observations !! Illustration of the trajectory
|-
| -1.0 || -6.1 || -402.504 || No ||Here the system passed through the transition state structure, the hydrogen atom collided with the fluorine and then reformed reactants. || [[File:MRD 01576020 fhh table 1.png|400 px]]
|-
| -1.0 || -6.0 || -404.098 || No ||This is very similar to the case above. || [[File:MRD 01576020 fhh table 2.png|400 px]]
|-
| -1.0 || -5.0 || -414.098 || No ||This trajectory is also not reactive and is quite similar to the two above, again there was one collision with the fluorine atom. || [[File:MRD 01576020 fhh table 3.png|400 px]]
|-
| -1.0 || -4.1 || -421.388 || No ||Here, the fluorine atom did not collide with any atoms and the transition state was not surpassed. || [[File:MRD 01576020 fhh table 4.png|400 px]]
|-
| -1.0 || 0.0 || -434.098 || No ||Since even less energy was given the picture was very similar to the one above, but the hydrogen atom approached the fluorine even less. || [[File:MRD 01576020 fhh table 5.png|400 px]]
|-
| -1.0 || 2.1 || -427.588 || No ||Slightly more energy than above so the atoms were closer but the transition state was still not reached. || [[File:MRD 01576020 fhh table 6.png|400 px]]
|-
| -1.0 || 4.1 || -413.188 || No ||This trajectory was more interesting, the fluorine collided twice with one of the hydrogen atoms, but the reactants were reformed afterwards. || [[File:MRD 01576020 fhh table 7.png|400 px]]
|-
| -1.0 || 4.5 || -409.348 || Yes ||Interestingly, this trajectory was reactive despite the fact that the system has smaller energy than in some other cases. The total energy is therefore far from being the only parameter that influences the success of a given chemical reaction. || [[File:MRD 01576020 fhh table 11.png|400 px]]
|-
| -1.0 || 6.0 || -392.098 || No ||The system was very close to forming products here, fluorine and hydrogen collided 7 times, but the system rolled into the valley of reactants again. || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 6.1 || -390.788 || No ||The fluorine and hydrogen only collided once in this simulation. There is a pattern emerging in this table ... || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 7.0 || -378.098 || No ||This was very similar to the one above, there was one collision and then the reactants were reformed.|| [[File:MRD 01576020 fhh table 10.png|400 px]]
|}

From the table above it is obvious that whether the reaction will result in the formation of products is clearly not dependent only on the energy of vibration of the H2 molecule. We have also seen that these conditions where the molecule and the fluorine atom approach each other rather slowly rarely result in producing products. Maybe increasing the momentum pFH will help ...

{{fontcolor|blue|For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?}}

There is very little vibration in the H2 molecule now and the F and the molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.

Let's now focus on the reverse reaction. When there is very little vibrational energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below.

An example of a reactive trajectory for this reaction would be produced by the following initial conditions: AB distance = 212

Q9{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions

MRD:01576020

2020-05-15T16:56:04Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
Modelling the dynamics of molecular reactions has been enabled by the emergence of computers and the field of computational chemistry is an ever-expanding field. Despite most approaches being purely theoretical, ''ab initio'', calculations, the insight given is invaluable for understanding the pathways of chemical reactions. It is often also not necessary to include quantum mechanics, as atoms are of sufficient mass to be reasonably accurately modelled by Newton's equations of motion. Isolated systems in the gas phase only experience interatomic interactions which can be represented by a potential energy surface (PES) which is a function only of atomic positions. The force on a given interatomic coordinate can be calculated as follows:

<math> F = - { \partial V(r_1,r_2,...)\over \partial r_i}</math> (1)

where F = <math>{dp_i \over dt}</math> (2)

Calculating the forces and adjusting the system in consecutive steps results in a trajectory which shows how a system evolved in time. Such calculations have almost endless application from chemical biology and medicine{{cite journal |last1=Pregeljc |first1=Domen |last2=Jug |first2=Urska|last3=Mavri |first3=Janez|last4=Stare |first4=Jernej |date= |title=Why does the Y326I mutant of monoamine oxidase B decompose an endogenous amphetamine at a slower rate than the wild type enzyme? Reaction step elucidated by multiscale molecular simulations. |url= |journal= Phys Chem Chem Phys.|volume= 10|issue= 6|pages= 4181-4188|doi= 10.1039/c7cp07069a|access-date=}}

TRANSITION STATE THEORY
It is worth mentioning some main aspects about the transition state (TS) theory here as it is central to the concepts used here. It is purely classical - it ignores any quantum mechanical effects such as tunnelling, which is incredibly small anyway. It also assumes that the transition state structure is in equilibrium with the reactants, which is called a quasi-equilibrium. TS theory also assumes that the reactants have energies which follow a Boltzmann distribution. Another crucial assumption of the TS theory developed in the 1930s is that once the TS is surpassed, the products will necessarily be formed. This latter assumption is the reason why theoretically derived reaction rate using the TS state theory overestimate reaction rates. This can be improved by adding a correcting factor, but to do that one must already have extensive knowledge of the system in question.

We will deal here with triatomic linear systems, as depicted in the following Figure.
[[File:Y2C8.png|thumb|centre|500px| A triatomic system used here. The atoms will either all be Hs or in the last case one H will be replaced with a F. Figure taken from https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD.]]

Terms like reactive and unreactive trajectories will be often used. A reactive trajectory is one which leads the reaction to completeness, i. e. the products are formed. In contrast an unreactive trajectory will not form products. The work here is divided into two exercises.

==EXERCISE 1==
{{fontcolor|blue|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?}}
<div>Q1 The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its gradient being 0 at that point. To further characterise a point, one needs to evaluate the Laplacian at that point to determine whether it is a maximum or a minimum. The Laplacian at the transition state will be a negative value, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the gradient is zero and the Laplacian is positive, the point in question is a local minimum. [ADD SOME FIGURES AND REFERENCES]</div>

{{fontcolor|blue|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.}}
<div>Q2 A simple system, also shown in Figure X, of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following Figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this Figure. The transition state is the minimum of this line]]

The minimum of this curve in the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if we start at the transition state. By changing the initial internuclear distances, we can indentify the transition state as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Internuclear distance vs. time plots are shown in this Figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Contour plot with the initial geometry being the transition state geometry and initial momenta 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the transition state teh atoms are separated by more than 90 pm. Conversely, in case (c), The distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in the vicinity of 90.8 pm.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

But we can get a more accurate value now that we know where to look for the transition state. With some trial and error, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

===Dynamics vs. MEP===
<div>Q3 {{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In the minimum energy path (MEP) calculation, there are no vibrations of the molecule as we move away from the transition state because momenta of all particles are set to 0 in each step. In the dynamics simulation, on the other hand, vibrations of the molecule are observed. In both cases, however, the system follows the same route down the valley.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. Molecule B-C vibrates, as can be seen from the oscillating distance between atoms B and C. A-b and A-C distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. The A-B momentum increases and ultimately moves with constant momentum away from the molecule B-C. Molecule B-C vibrates, so the momentum B-C oscillates.]] </div>

Using the initial conditions r1=rts and r2=rts+1, the system would roll into the other valley on the PES with similar behaviour to what is described above.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translation and vibration). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill.

===Reactive and Unreactive Trajectories===
<div>Reactive and Unreactive Trajectories
Q4{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected.
</div>

Q5{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
One of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the products again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===
Q6{{fontcolor|blue|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?Locate the approximate position of the transition state.}}

The F+H2--> HF+H reaction is exothermic and H+HF-->H2+F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than H-H bond. This makes sense as it has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is F. From these initial conditions, the forces were: along AB: 0.000 and along BC: 0.001. SO this structure is very close to the transition state structure.

Q7{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. This is in good agreement with experimental findings, where the H-F bond dissociation energy is reported as 566 kJ/mol and the H-H bond dissociation energy is 436 kJ/mol[ref].

===Reaction Dynamics===
A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g mol-1 pm fs-1.
Q8{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left| A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 seconds of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the B-C pair which is moving aparth. ]]

Let's now setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, an rFH = 238 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. Note that we are putting more energy into the system than the activation energy. The table below shows some observations and contour plots.

{| class="wikitable" border=1
! pFH/ g.mol-1.pm.fs-1 !! pHH/ g.mol-1.pm.fs-1 !! Einitial /kJ.mol-1 !! Reactive? !! Observations !! Illustration of the trajectory
|-
| -1.0 || -6.1 || -402.504 || No ||Here the system passed through the transition state structure, the hydrogen atom collided with the fluorine and then reformed reactants. || [[File:MRD 01576020 fhh table 1.png|400 px]]
|-
| -1.0 || -6.0 || -404.098 || No ||This is very similar to the case above. || [[File:MRD 01576020 fhh table 2.png|400 px]]
|-
| -1.0 || -5.0 || -414.098 || No ||This trajectory is also not reactive and is quite similar to the two above, again there was one collision with the fluorine atom. || [[File:MRD 01576020 fhh table 3.png|400 px]]
|-
| -1.0 || -4.1 || -421.388 || No ||Here, the fluorine atom did not collide with any atoms and the transition state was not surpassed. || [[File:MRD 01576020 fhh table 4.png|400 px]]
|-
| -1.0 || 0.0 || -434.098 || No ||Since even less energy was given the picture was very similar to the one above, but the hydrogen atom approached the fluorine even less. || [[File:MRD 01576020 fhh table 5.png|400 px]]
|-
| -1.0 || 2.1 || -427.588 || No ||Slightly more energy than above so the atoms were closer but the transition state was still not reached. || [[File:MRD 01576020 fhh table 6.png|400 px]]
|-
| -1.0 || 4.1 || -413.188 || No ||This trajectory was more interesting, the fluorine collided twice with one of the hydrogen atoms, but the reactants were reformed afterwards. || [[File:MRD 01576020 fhh table 7.png|400 px]]
|-
| -1.0 || 4.5 || -409.348 || Yes ||Interestingly, this trajectory was reactive despite the fact that the system has smaller energy than in some other cases. The total energy is therefore far from being the only parameter that influences the success of a given chemical reaction. || [[File:MRD 01576020 fhh table 11.png|400 px]]
|-
| -1.0 || 6.0 || -392.098 || No ||The system was very close to forming products here, fluorine and hydrogen collided 7 times, but the system rolled into the valley of reactants again. || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 6.1 || -390.788 || No ||The fluorine and hydrogen only collided once in this simulation. There is a pattern emerging in this table ... || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 7.0 || -378.098 || No ||This was very similar to the one above, there was one collision and then the reactants were reformed.|| [[File:MRD 01576020 fhh table 10.png|400 px]]
|}

From the table above it is obvious that whether the reaction will result in the formation of products is clearly not dependent only on the energy of vibration of the H2 molecule. We have also seen that these conditions where the molecule and the fluorine atom approach each other rather slowly rarely result in producing products. Maybe increasing the momentum pFH will help ...

{{fontcolor|blue|For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?}}

There is very little vibration in the H2 molecule now and the F and the molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.

Let's now focus on the reverse reaction. When there is very little vibrational energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below.

An example of a reactive trajectory for this reaction would be produced by the following initial conditions: AB distance = 212

Q9{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions

MRD:01576020

2020-05-15T16:45:27Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
Modelling the dynamics of molecular reactions has been enabled by the emergence of computers and the field of computational chemistry is an ever-expanding field. Despite most approaches being purely theoretical, ''ab initio'', calculations, the insight given is invaluable for understanding the pathways of chemical reactions. It is often also not necessary to include quantum mechanics, as atoms are of sufficient mass to be reasonably accurately modelled by Newton's equations of motion. Isolated systems in the gas phase only experience interatomic interactions which can be represented by a potential energy surface (PES) which is a function only of atomic positions. The force on a given interatomic coordinate can be calculated as follows:

<math> F = - { \partial V(r_1,r_2,...)\over \partial r_i}</math>

where F = <math>{dp_i \over dt}</math>.

Calculating the forces and adjusting the system in consecutive steps results in a trajectory which shows how a system evolved in time. Such calculations have almost endless application from chemical biology and medicine{{cite journal |last1=Pregeljc |first1=Domen |last2=Jug |first2=Urska|last3=Mavri |first3=Janez|last4=Stare |first4=Jernej |date= |title=Why does the Y326I mutant of monoamine oxidase B decompose an endogenous amphetamine at a slower rate than the wild type enzyme? Reaction step elucidated by multiscale molecular simulations. |url= |journal= Phys Chem Chem Phys.|volume= 10|issue= 6|pages= 4181-4188|doi= 10.1039/c7cp07069a|access-date=}}

TRANSITION STATE THEORY

We will deal here with triatomic linear systems, as depicted in the following Figure.
[[File:Y2C8.png|thumb|centre|500px| A triatomic system used here. The atoms will either all be Hs or in the last case one H will be replaced with a F. Figure taken from https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD.]]

Terms like reactive and unreactive trajectories will be often used. A reactive trajectory is one which leads the reaction to completeness, i. e. the products are formed. In contrast an unreactive trajectory will not form products. The work here is divided into two exercises.

==EXERCISE 1==
{{fontcolor|blue|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?}}
<div>Q1 The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its gradient being 0 at that point. To further characterise a point, one needs to evaluate the Laplacian at that point to determine whether it is a maximum or a minimum. The Laplacian at the transition state will be a negative value, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the gradient is zero and the Laplacian is positive, the point in question is a local minimum. [ADD SOME FIGURES AND REFERENCES]</div>

{{fontcolor|blue|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.}}
<div>Q2 A simple system, also shown in Figure X, of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following Figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this Figure. The transition state is the minimum of this line]]

The minimum of this curve in the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if we start at the transition state. By changing the initial internuclear distances, we can indentify the transition state as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Internuclear distance vs. time plots are shown in this Figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Contour plot with the initial geometry being the transition state geometry and initial momenta 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the transition state teh atoms are separated by more than 90 pm. Conversely, in case (c), The distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in the vicinity of 90.8 pm.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

But we can get a more accurate value now that we know where to look for the transition state. With some trial and error, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

===Dynamics vs. MEP===
<div>Q3 {{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In the minimum energy path (MEP) calculation, there are no vibrations of the molecule as we move away from the transition state because momenta of all particles are set to 0 in each step. In the dynamics simulation, on the other hand, vibrations of the molecule are observed. In both cases, however, the system follows the same route down the valley.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. Molecule B-C vibrates, as can be seen from the oscillating distance between atoms B and C. A-b and A-C distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. The A-B momentum increases and ultimately moves with constant momentum away from the molecule B-C. Molecule B-C vibrates, so the momentum B-C oscillates.]] </div>

Using the initial conditions r1=rts and r2=rts+1, the system would roll into the other valley on the PES with similar behaviour to what is described above.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translation and vibration). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill.

===Reactive and Unreactive Trajectories===
<div>Reactive and Unreactive Trajectories
Q4{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected.
</div>

Q5{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
One of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the products again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===
Q6{{fontcolor|blue|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?Locate the approximate position of the transition state.}}

The F+H2--> HF+H reaction is exothermic and H+HF-->H2+F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than H-H bond. This makes sense as it has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is F. From these initial conditions, the forces were: along AB: 0.000 and along BC: 0.001. SO this structure is very close to the transition state structure.

Q7{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. This is in good agreement with experimental findings, where the H-F bond dissociation energy is reported as 566 kJ/mol and the H-H bond dissociation energy is 436 kJ/mol[ref].

===Reaction Dynamics===
A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g mol-1 pm fs-1.
Q8{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left| A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 seconds of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the B-C pair which is moving aparth. ]]

Let's now setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, an rFH = 238 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. Note that we are putting more energy into the system than the activation energy. The table below shows some observations and contour plots.

{| class="wikitable" border=1
! pFH/ g.mol-1.pm.fs-1 !! pHH/ g.mol-1.pm.fs-1 !! Einitial /kJ.mol-1 !! Reactive? !! Observations !! Illustration of the trajectory
|-
| -1.0 || -6.1 || -402.504 || No ||Here the system passed through the transition state structure, the hydrogen atom collided with the fluorine and then reformed reactants. || [[File:MRD 01576020 fhh table 1.png|400 px]]
|-
| -1.0 || -6.0 || -404.098 || No ||This is very similar to the case above. || [[File:MRD 01576020 fhh table 2.png|400 px]]
|-
| -1.0 || -5.0 || -414.098 || No ||This trajectory is also not reactive and is quite similar to the two above, again there was one collision with the fluorine atom. || [[File:MRD 01576020 fhh table 3.png|400 px]]
|-
| -1.0 || -4.1 || -421.388 || No ||Here, the fluorine atom did not collide with any atoms and the transition state was not surpassed. || [[File:MRD 01576020 fhh table 4.png|400 px]]
|-
| -1.0 || 0.0 || -434.098 || No ||Since even less energy was given the picture was very similar to the one above, but the hydrogen atom approached the fluorine even less. || [[File:MRD 01576020 fhh table 5.png|400 px]]
|-
| -1.0 || 2.1 || -427.588 || No ||Slightly more energy than above so the atoms were closer but the transition state was still not reached. || [[File:MRD 01576020 fhh table 6.png|400 px]]
|-
| -1.0 || 4.1 || -413.188 || No ||This trajectory was more interesting, the fluorine collided twice with one of the hydrogen atoms, but the reactants were reformed afterwards. || [[File:MRD 01576020 fhh table 7.png|400 px]]
|-
| -1.0 || 4.5 || -409.348 || Yes ||Interestingly, this trajectory was reactive despite the fact that the system has smaller energy than in some other cases. The total energy is therefore far from being the only parameter that influences the success of a given chemical reaction. || [[File:MRD 01576020 fhh table 11.png|400 px]]
|-
| -1.0 || 6.0 || -392.098 || No ||The system was very close to forming products here, fluorine and hydrogen collided 7 times, but the system rolled into the valley of reactants again. || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 6.1 || -390.788 || No ||The fluorine and hydrogen only collided once in this simulation. There is a pattern emerging in this table ... || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 7.0 || -378.098 || No ||This was very similar to the one above, there was one collision and then the reactants were reformed.|| [[File:MRD 01576020 fhh table 10.png|400 px]]
|}

From the table above it is obvious that whether the reaction will result in the formation of products is clearly not dependent only on the energy of vibration of the H2 molecule. We have also seen that these conditions where the molecule and the fluorine atom approach each other rather slowly rarely result in producing products. Maybe increasing the momentum pFH will help ...

{{fontcolor|blue|For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?}}

There is very little vibration in the H2 molecule now and the F and the molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.

Let's now focus on the reverse reaction. When there is very little vibrational energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below.

An example of a reactive trajectory for this reaction would be produced by the following initial conditions: AB distance = 212

Q9{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions

MRD:01576020

2020-05-15T16:33:02Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
Modelling the dynamics of molecular reactions has been enabled by the emergence of computers and the field of computational chemistry is an ever-expanding field. Despite most approaches being purely theoretical, ''ab initio'', calculations, the insight given is invaluable for understanding the pathways of chemical reactions. It is often also not necessary to include quantum mechanics, as atoms are of sufficient mass to be reasonably accurately modelled by Newton's equations of motion. Isolated systems in the gas phase only experience interatomic interactions which can be represented by a potential energy surface (PES) which is a function only of atomic positions. The force on a given interatomic coordinate can be calculated as follows:

<math> F = - { \partial V(r_1,r_2,...)\over \partial r_i}</math>

where F = <math>{dp_i \over dt}</math>.

Calculating the forces and adjusting the system in consecutive steps results in a trajectory which shows how a system evolved in time. Such calculations have almost endless application from chemical biology and medicine{{cite journal |vauthors= D Pregeljc, U Jug, J Mavri, J Stare|title= Why does the Y326I mutant of monoamine oxidase B decompose an endogenous amphetamine at a slower rate than the wild type enzyme? Reaction step elucidated by multiscale molecular simulations.|journal= Phys Chem Chem Phys.|volume= 20|issue= 6|pages= 4181-4188|doi= 10.1039/c7cp07069a}} [ref] to materials science[ref].
===Potential Energy Surface===
===Some nomenclature===
what momenta mean, show a diagram, what r1 r2 are ...
===Transition State Theory===

==EXERCISE 1==
{{fontcolor|blue|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?}}
<div>Q1 The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its gradient being 0 at that point. To further characterise a point, one needs to evaluate the Laplacian at that point to determine whether it is a maximum or a minimum. The Laplacian at the transition state will be a negative value, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the gradient is zero and the Laplacian is positive, the point in question is a local minimum. [ADD SOME FIGURES AND REFERENCES]</div>

{{fontcolor|blue|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.}}
<div>Q2 A simple system, also shown in Figure X, of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following Figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this Figure. The transition state is the minimum of this line]]

The minimum of this curve in the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if we start at the transition state. By changing the initial internuclear distances, we can indentify the transition state as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Internuclear distance vs. time plots are shown in this Figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Contour plot with the initial geometry being the transition state geometry and initial momenta 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the transition state teh atoms are separated by more than 90 pm. Conversely, in case (c), The distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in the vicinity of 90.8 pm.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

But we can get a more accurate value now that we know where to look for the transition state. With some trial and error, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

===Dynamics vs. MEP===
<div>Q3 {{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In the minimum energy path (MEP) calculation, there are no vibrations of the molecule as we move away from the transition state because momenta of all particles are set to 0 in each step. In the dynamics simulation, on the other hand, vibrations of the molecule are observed. In both cases, however, the system follows the same route down the valley.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. Molecule B-C vibrates, as can be seen from the oscillating distance between atoms B and C. A-b and A-C distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. The A-B momentum increases and ultimately moves with constant momentum away from the molecule B-C. Molecule B-C vibrates, so the momentum B-C oscillates.]] </div>

Using the initial conditions r1=rts and r2=rts+1, the system would roll into the other valley on the PES with similar behaviour to what is described above.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translation and vibration). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill.

===Reactive and Unreactive Trajectories===
<div>Reactive and Unreactive Trajectories
Q4{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected.
</div>

Q5{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
One of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the products again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===
Q6{{fontcolor|blue|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?Locate the approximate position of the transition state.}}

The F+H2--> HF+H reaction is exothermic and H+HF-->H2+F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than H-H bond. This makes sense as it has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is F. From these initial conditions, the forces were: along AB: 0.000 and along BC: 0.001. SO this structure is very close to the transition state structure.

Q7{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. This is in good agreement with experimental findings, where the H-F bond dissociation energy is reported as 566 kJ/mol and the H-H bond dissociation energy is 436 kJ/mol[ref].

===Reaction Dynamics===
A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g mol-1 pm fs-1.
Q8{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left| A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 seconds of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the B-C pair which is moving aparth. ]]

Let's now setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, an rFH = 238 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. Note that we are putting more energy into the system than the activation energy. The table below shows some observations and contour plots.

{| class="wikitable" border=1
! pFH/ g.mol-1.pm.fs-1 !! pHH/ g.mol-1.pm.fs-1 !! Einitial /kJ.mol-1 !! Reactive? !! Observations !! Illustration of the trajectory
|-
| -1.0 || -6.1 || -402.504 || No ||Here the system passed through the transition state structure, the hydrogen atom collided with the fluorine and then reformed reactants. || [[File:MRD 01576020 fhh table 1.png|400 px]]
|-
| -1.0 || -6.0 || -404.098 || No ||This is very similar to the case above. || [[File:MRD 01576020 fhh table 2.png|400 px]]
|-
| -1.0 || -5.0 || -414.098 || No ||This trajectory is also not reactive and is quite similar to the two above, again there was one collision with the fluorine atom. || [[File:MRD 01576020 fhh table 3.png|400 px]]
|-
| -1.0 || -4.1 || -421.388 || No ||Here, the fluorine atom did not collide with any atoms and the transition state was not surpassed. || [[File:MRD 01576020 fhh table 4.png|400 px]]
|-
| -1.0 || 0.0 || -434.098 || No ||Since even less energy was given the picture was very similar to the one above, but the hydrogen atom approached the fluorine even less. || [[File:MRD 01576020 fhh table 5.png|400 px]]
|-
| -1.0 || 2.1 || -427.588 || No ||Slightly more energy than above so the atoms were closer but the transition state was still not reached. || [[File:MRD 01576020 fhh table 6.png|400 px]]
|-
| -1.0 || 4.1 || -413.188 || No ||This trajectory was more interesting, the fluorine collided twice with one of the hydrogen atoms, but the reactants were reformed afterwards. || [[File:MRD 01576020 fhh table 7.png|400 px]]
|-
| -1.0 || 4.5 || -409.348 || Yes ||Interestingly, this trajectory was reactive despite the fact that the system has smaller energy than in some other cases. The total energy is therefore far from being the only parameter that influences the success of a given chemical reaction. || [[File:MRD 01576020 fhh table 11.png|400 px]]
|-
| -1.0 || 6.0 || -392.098 || No ||The system was very close to forming products here, fluorine and hydrogen collided 7 times, but the system rolled into the valley of reactants again. || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 6.1 || -390.788 || No ||The fluorine and hydrogen only collided once in this simulation. There is a pattern emerging in this table ... || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 7.0 || -378.098 || No ||This was very similar to the one above, there was one collision and then the reactants were reformed.|| [[File:MRD 01576020 fhh table 10.png|400 px]]
|}

From the table above it is obvious that whether the reaction will result in the formation of products is clearly not dependent only on the energy of vibration of the H2 molecule. We have also seen that these conditions where the molecule and the fluorine atom approach each other rather slowly rarely result in producing products. Maybe increasing the momentum pFH will help ...

{{fontcolor|blue|For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?}}

There is very little vibration in the H2 molecule now and the F and the molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.

Let's now focus on the reverse reaction. When there is very little vibrational energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below.

An example of a reactive trajectory for this reaction would be produced by the following initial conditions: AB distance = 212

Q9{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions

MRD:01576020

2020-05-15T16:31:32Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
Modelling the dynamics of molecular reactions has been enabled by the emergence of computers and the field of computational chemistry is an ever-expanding field. Despite most approaches being purely theoretical, ''ab initio'', calculations, the insight given is invaluable for understanding the pathways of chemical reactions. It is often also not necessary to include quantum mechanics, as atoms are of sufficient mass to be reasonably accurately modelled by Newton's equations of motion. Isolated systems in the gas phase only experience interatomic interactions which can be represented by a potential energy surface (PES) which is a function only of atomic positions. The force on a given interatomic coordinate can be calculated as follows:

<math> F = - { \partial V(r_1,r_2,...)\over \partial r_i}</math>

where F = <math>{dp_i \over dt}</math>.

Calculating the forces and adjusting the system in consecutive steps results in a trajectory which shows how a system evolved in time. Such calculations have almost endless application from chemical biology and medicine{{cite journal |vauthors= D Pregeljc, U Jug, J Mavri, J Stare|date= |title= |url= |journal= Phys Chem Chem Phys.|volume= 20|issue= 6|pages= 4181-4188|doi= 10.1039/c7cp07069a|pmc= |pmid= |access-date=}} [ref] to materials science[ref].
===Potential Energy Surface===
===Some nomenclature===
what momenta mean, show a diagram, what r1 r2 are ...
===Transition State Theory===

==EXERCISE 1==
{{fontcolor|blue|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?}}
<div>Q1 The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its gradient being 0 at that point. To further characterise a point, one needs to evaluate the Laplacian at that point to determine whether it is a maximum or a minimum. The Laplacian at the transition state will be a negative value, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the gradient is zero and the Laplacian is positive, the point in question is a local minimum. [ADD SOME FIGURES AND REFERENCES]</div>

{{fontcolor|blue|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.}}
<div>Q2 A simple system, also shown in Figure X, of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following Figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this Figure. The transition state is the minimum of this line]]

The minimum of this curve in the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if we start at the transition state. By changing the initial internuclear distances, we can indentify the transition state as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Internuclear distance vs. time plots are shown in this Figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Contour plot with the initial geometry being the transition state geometry and initial momenta 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the transition state teh atoms are separated by more than 90 pm. Conversely, in case (c), The distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in the vicinity of 90.8 pm.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

But we can get a more accurate value now that we know where to look for the transition state. With some trial and error, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

===Dynamics vs. MEP===
<div>Q3 {{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In the minimum energy path (MEP) calculation, there are no vibrations of the molecule as we move away from the transition state because momenta of all particles are set to 0 in each step. In the dynamics simulation, on the other hand, vibrations of the molecule are observed. In both cases, however, the system follows the same route down the valley.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. Molecule B-C vibrates, as can be seen from the oscillating distance between atoms B and C. A-b and A-C distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. The A-B momentum increases and ultimately moves with constant momentum away from the molecule B-C. Molecule B-C vibrates, so the momentum B-C oscillates.]] </div>

Using the initial conditions r1=rts and r2=rts+1, the system would roll into the other valley on the PES with similar behaviour to what is described above.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translation and vibration). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill.

===Reactive and Unreactive Trajectories===
<div>Reactive and Unreactive Trajectories
Q4{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected.
</div>

Q5{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
One of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the products again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===
Q6{{fontcolor|blue|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?Locate the approximate position of the transition state.}}

The F+H2--> HF+H reaction is exothermic and H+HF-->H2+F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than H-H bond. This makes sense as it has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is F. From these initial conditions, the forces were: along AB: 0.000 and along BC: 0.001. SO this structure is very close to the transition state structure.

Q7{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. This is in good agreement with experimental findings, where the H-F bond dissociation energy is reported as 566 kJ/mol and the H-H bond dissociation energy is 436 kJ/mol[ref].

===Reaction Dynamics===
A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g mol-1 pm fs-1.
Q8{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left| A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 seconds of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the B-C pair which is moving aparth. ]]

Let's now setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, an rFH = 238 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. Note that we are putting more energy into the system than the activation energy. The table below shows some observations and contour plots.

{| class="wikitable" border=1
! pFH/ g.mol-1.pm.fs-1 !! pHH/ g.mol-1.pm.fs-1 !! Einitial /kJ.mol-1 !! Reactive? !! Observations !! Illustration of the trajectory
|-
| -1.0 || -6.1 || -402.504 || No ||Here the system passed through the transition state structure, the hydrogen atom collided with the fluorine and then reformed reactants. || [[File:MRD 01576020 fhh table 1.png|400 px]]
|-
| -1.0 || -6.0 || -404.098 || No ||This is very similar to the case above. || [[File:MRD 01576020 fhh table 2.png|400 px]]
|-
| -1.0 || -5.0 || -414.098 || No ||This trajectory is also not reactive and is quite similar to the two above, again there was one collision with the fluorine atom. || [[File:MRD 01576020 fhh table 3.png|400 px]]
|-
| -1.0 || -4.1 || -421.388 || No ||Here, the fluorine atom did not collide with any atoms and the transition state was not surpassed. || [[File:MRD 01576020 fhh table 4.png|400 px]]
|-
| -1.0 || 0.0 || -434.098 || No ||Since even less energy was given the picture was very similar to the one above, but the hydrogen atom approached the fluorine even less. || [[File:MRD 01576020 fhh table 5.png|400 px]]
|-
| -1.0 || 2.1 || -427.588 || No ||Slightly more energy than above so the atoms were closer but the transition state was still not reached. || [[File:MRD 01576020 fhh table 6.png|400 px]]
|-
| -1.0 || 4.1 || -413.188 || No ||This trajectory was more interesting, the fluorine collided twice with one of the hydrogen atoms, but the reactants were reformed afterwards. || [[File:MRD 01576020 fhh table 7.png|400 px]]
|-
| -1.0 || 4.5 || -409.348 || Yes ||Interestingly, this trajectory was reactive despite the fact that the system has smaller energy than in some other cases. The total energy is therefore far from being the only parameter that influences the success of a given chemical reaction. || [[File:MRD 01576020 fhh table 11.png|400 px]]
|-
| -1.0 || 6.0 || -392.098 || No ||The system was very close to forming products here, fluorine and hydrogen collided 7 times, but the system rolled into the valley of reactants again. || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 6.1 || -390.788 || No ||The fluorine and hydrogen only collided once in this simulation. There is a pattern emerging in this table ... || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 7.0 || -378.098 || No ||This was very similar to the one above, there was one collision and then the reactants were reformed.|| [[File:MRD 01576020 fhh table 10.png|400 px]]
|}

From the table above it is obvious that whether the reaction will result in the formation of products is clearly not dependent only on the energy of vibration of the H2 molecule. We have also seen that these conditions where the molecule and the fluorine atom approach each other rather slowly rarely result in producing products. Maybe increasing the momentum pFH will help ...

{{fontcolor|blue|For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?}}

There is very little vibration in the H2 molecule now and the F and the molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.

Let's now focus on the reverse reaction. When there is very little vibrational energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below.

An example of a reactive trajectory for this reaction would be produced by the following initial conditions: AB distance = 212

Q9{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions

MRD:01576020

2020-05-15T16:25:25Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
Modelling the dynamics of molecular reactions has been enabled by the emergence of computers and the field of computational chemistry is an ever-expanding field. Despite most approaches being purely theoretical, ''ab initio'', calculations, the insight given is invaluable for understanding the pathways of chemical reactions. It is often also not necessary to include quantum mechanics, as atoms are of sufficient mass to be reasonably accurately modelled by Newton's equations of motion. Isolated systems in the gas phase only experience interatomic interactions which can be represented by a potential energy surface (PES) which is a function only of atomic positions. The force on a given interatomic coordinate can be calculated as follows:
<math> F = - { \partial V(r_1,r_2,...)\over \partial r_i}</math>

where F = <math<{dp_i \over dt}</math>
===Potential Energy Surface===
===Some nomenclature===
what momenta mean, show a diagram, what r1 r2 are ...
===Transition State Theory===

==EXERCISE 1==
{{fontcolor|blue|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?}}
<div>Q1 The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its gradient being 0 at that point. To further characterise a point, one needs to evaluate the Laplacian at that point to determine whether it is a maximum or a minimum. The Laplacian at the transition state will be a negative value, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the gradient is zero and the Laplacian is positive, the point in question is a local minimum. [ADD SOME FIGURES AND REFERENCES]</div>

{{fontcolor|blue|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.}}
<div>Q2 A simple system, also shown in Figure X, of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following Figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this Figure. The transition state is the minimum of this line]]

The minimum of this curve in the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if we start at the transition state. By changing the initial internuclear distances, we can indentify the transition state as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Internuclear distance vs. time plots are shown in this Figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Contour plot with the initial geometry being the transition state geometry and initial momenta 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the transition state teh atoms are separated by more than 90 pm. Conversely, in case (c), The distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in the vicinity of 90.8 pm.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

But we can get a more accurate value now that we know where to look for the transition state. With some trial and error, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

===Dynamics vs. MEP===
<div>Q3 {{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In the minimum energy path (MEP) calculation, there are no vibrations of the molecule as we move away from the transition state because momenta of all particles are set to 0 in each step. In the dynamics simulation, on the other hand, vibrations of the molecule are observed. In both cases, however, the system follows the same route down the valley.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. Molecule B-C vibrates, as can be seen from the oscillating distance between atoms B and C. A-b and A-C distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. The A-B momentum increases and ultimately moves with constant momentum away from the molecule B-C. Molecule B-C vibrates, so the momentum B-C oscillates.]] </div>

Using the initial conditions r1=rts and r2=rts+1, the system would roll into the other valley on the PES with similar behaviour to what is described above.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translation and vibration). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill.

===Reactive and Unreactive Trajectories===
<div>Reactive and Unreactive Trajectories
Q4{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected.
</div>

Q5{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
One of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the products again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===
Q6{{fontcolor|blue|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?Locate the approximate position of the transition state.}}

The F+H2--> HF+H reaction is exothermic and H+HF-->H2+F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than H-H bond. This makes sense as it has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is F. From these initial conditions, the forces were: along AB: 0.000 and along BC: 0.001. SO this structure is very close to the transition state structure.

Q7{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. This is in good agreement with experimental findings, where the H-F bond dissociation energy is reported as 566 kJ/mol and the H-H bond dissociation energy is 436 kJ/mol[ref].

===Reaction Dynamics===
A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g mol-1 pm fs-1.
Q8{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left| A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 seconds of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the B-C pair which is moving aparth. ]]

Let's now setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, an rFH = 238 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. Note that we are putting more energy into the system than the activation energy. The table below shows some observations and contour plots.

{| class="wikitable" border=1
! pFH/ g.mol-1.pm.fs-1 !! pHH/ g.mol-1.pm.fs-1 !! Einitial /kJ.mol-1 !! Reactive? !! Observations !! Illustration of the trajectory
|-
| -1.0 || -6.1 || -402.504 || No ||Here the system passed through the transition state structure, the hydrogen atom collided with the fluorine and then reformed reactants. || [[File:MRD 01576020 fhh table 1.png|400 px]]
|-
| -1.0 || -6.0 || -404.098 || No ||This is very similar to the case above. || [[File:MRD 01576020 fhh table 2.png|400 px]]
|-
| -1.0 || -5.0 || -414.098 || No ||This trajectory is also not reactive and is quite similar to the two above, again there was one collision with the fluorine atom. || [[File:MRD 01576020 fhh table 3.png|400 px]]
|-
| -1.0 || -4.1 || -421.388 || No ||Here, the fluorine atom did not collide with any atoms and the transition state was not surpassed. || [[File:MRD 01576020 fhh table 4.png|400 px]]
|-
| -1.0 || 0.0 || -434.098 || No ||Since even less energy was given the picture was very similar to the one above, but the hydrogen atom approached the fluorine even less. || [[File:MRD 01576020 fhh table 5.png|400 px]]
|-
| -1.0 || 2.1 || -427.588 || No ||Slightly more energy than above so the atoms were closer but the transition state was still not reached. || [[File:MRD 01576020 fhh table 6.png|400 px]]
|-
| -1.0 || 4.1 || -413.188 || No ||This trajectory was more interesting, the fluorine collided twice with one of the hydrogen atoms, but the reactants were reformed afterwards. || [[File:MRD 01576020 fhh table 7.png|400 px]]
|-
| -1.0 || 4.5 || -409.348 || Yes ||Interestingly, this trajectory was reactive despite the fact that the system has smaller energy than in some other cases. The total energy is therefore far from being the only parameter that influences the success of a given chemical reaction. || [[File:MRD 01576020 fhh table 11.png|400 px]]
|-
| -1.0 || 6.0 || -392.098 || No ||The system was very close to forming products here, fluorine and hydrogen collided 7 times, but the system rolled into the valley of reactants again. || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 6.1 || -390.788 || No ||The fluorine and hydrogen only collided once in this simulation. There is a pattern emerging in this table ... || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 7.0 || -378.098 || No ||This was very similar to the one above, there was one collision and then the reactants were reformed.|| [[File:MRD 01576020 fhh table 10.png|400 px]]
|}

From the table above it is obvious that whether the reaction will result in the formation of products is clearly not dependent only on the energy of vibration of the H2 molecule. We have also seen that these conditions where the molecule and the fluorine atom approach each other rather slowly rarely result in producing products. Maybe increasing the momentum pFH will help ...

{{fontcolor|blue|For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?}}

There is very little vibration in the H2 molecule now and the F and the molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.

Let's now focus on the reverse reaction. When there is very little vibrational energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below.

An example of a reactive trajectory for this reaction would be produced by the following initial conditions: AB distance = 212

Q9{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions

MRD:01576020

2020-05-15T16:11:23Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
===Potential Energy Surface===
===Some nomenclature===
what momenta mean, show a diagram, what r1 r2 are ...
===Transition State Theory===

==EXERCISE 1==
{{fontcolor|blue|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?}}
<div>Q1 The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its gradient being 0 at that point. To further characterise a point, one needs to evaluate the Laplacian at that point to determine whether it is a maximum or a minimum. The Laplacian at the transition state will be a negative value, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the gradient is zero and the Laplacian is positive, the point in question is a local minimum. [ADD SOME FIGURES AND REFERENCES]</div>

{{fontcolor|blue|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.}}
<div>Q2 A simple system, also shown in Figure X, of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following Figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this Figure. The transition state is the minimum of this line]]

The minimum of this curve in the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if we start at the transition state. By changing the initial internuclear distances, we can indentify the transition state as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Internuclear distance vs. time plots are shown in this Figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Contour plot with the initial geometry being the transition state geometry and initial momenta 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the transition state teh atoms are separated by more than 90 pm. Conversely, in case (c), The distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in the vicinity of 90.8 pm.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

But we can get a more accurate value now that we know where to look for the transition state. With some trial and error, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

===Dynamics vs. MEP===
<div>Q3 {{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In the minimum energy path (MEP) calculation, there are no vibrations of the molecule as we move away from the transition state because momenta of all particles are set to 0 in each step. In the dynamics simulation, on the other hand, vibrations of the molecule are observed. In both cases, however, the system follows the same route down the valley.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. Molecule B-C vibrates, as can be seen from the oscillating distance between atoms B and C. A-b and A-C distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. The A-B momentum increases and ultimately moves with constant momentum away from the molecule B-C. Molecule B-C vibrates, so the momentum B-C oscillates.]] </div>

Using the initial conditions r1=rts and r2=rts+1, the system would roll into the other valley on the PES with similar behaviour to what is described above.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translation and vibration). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill.

===Reactive and Unreactive Trajectories===
<div>Reactive and Unreactive Trajectories
Q4{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected.
</div>

Q5{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
One of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the products again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===
Q6{{fontcolor|blue|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?Locate the approximate position of the transition state.}}

The F+H2--> HF+H reaction is exothermic and H+HF-->H2+F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than H-H bond. This makes sense as it has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is F. From these initial conditions, the forces were: along AB: 0.000 and along BC: 0.001. SO this structure is very close to the transition state structure.

Q7{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. This is in good agreement with experimental findings, where the H-F bond dissociation energy is reported as 566 kJ/mol and the H-H bond dissociation energy is 436 kJ/mol[ref].

===Reaction Dynamics===
A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g mol-1 pm fs-1.
Q8{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left| A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 seconds of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the B-C pair which is moving aparth. ]]

Let's now setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, an rFH = 238 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. Note that we are putting more energy into the system than the activation energy. The table below shows some observations and contour plots.

{| class="wikitable" border=1
! pFH/ g.mol-1.pm.fs-1 !! pHH/ g.mol-1.pm.fs-1 !! Einitial /kJ.mol-1 !! Reactive? !! Observations !! Illustration of the trajectory
|-
| -1.0 || -6.1 || -402.504 || No ||Here the system passed through the transition state structure, the hydrogen atom collided with the fluorine and then reformed reactants. || [[File:MRD 01576020 fhh table 1.png|400 px]]
|-
| -1.0 || -6.0 || -404.098 || No ||This is very similar to the case above. || [[File:MRD 01576020 fhh table 2.png|400 px]]
|-
| -1.0 || -5.0 || -414.098 || No ||This trajectory is also not reactive and is quite similar to the two above, again there was one collision with the fluorine atom. || [[File:MRD 01576020 fhh table 3.png|400 px]]
|-
| -1.0 || -4.1 || -421.388 || No ||Here, the fluorine atom did not collide with any atoms and the transition state was not surpassed. || [[File:MRD 01576020 fhh table 4.png|400 px]]
|-
| -1.0 || 0.0 || -434.098 || No ||Since even less energy was given the picture was very similar to the one above, but the hydrogen atom approached the fluorine even less. || [[File:MRD 01576020 fhh table 5.png|400 px]]
|-
| -1.0 || 2.1 || -427.588 || No ||Slightly more energy than above so the atoms were closer but the transition state was still not reached. || [[File:MRD 01576020 fhh table 6.png|400 px]]
|-
| -1.0 || 4.1 || -413.188 || No ||This trajectory was more interesting, the fluorine collided twice with one of the hydrogen atoms, but the reactants were reformed afterwards. || [[File:MRD 01576020 fhh table 7.png|400 px]]
|-
| -1.0 || 4.5 || -409.348 || Yes ||Interestingly, this trajectory was reactive despite the fact that the system has smaller energy than in some other cases. The total energy is therefore far from being the only parameter that influences the success of a given chemical reaction. || [[File:MRD 01576020 fhh table 11.png|400 px]]
|-
| -1.0 || 6.0 || -392.098 || No ||The system was very close to forming products here, fluorine and hydrogen collided 7 times, but the system rolled into the valley of reactants again. || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 6.1 || -390.788 || No ||The fluorine and hydrogen only collided once in this simulation. There is a pattern emerging in this table ... || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 7.0 || -378.098 || No ||This was very similar to the one above, there was one collision and then the reactants were reformed.|| [[File:MRD 01576020 fhh table 10.png|400 px]]
|}

From the table above it is obvious that whether the reaction will result in the formation of products is clearly not dependent only on the energy of vibration of the H2 molecule. We have also seen that these conditions where the molecule and the fluorine atom approach each other rather slowly rarely result in producing products. Maybe increasing the momentum pFH will help ...

{{fontcolor|blue|For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?}}

There is very little vibration in the H2 molecule now and the F and the molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.

Let's now focus on the reverse reaction. When there is very little vibrational energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below.

An example of a reactive trajectory for this reaction would be produced by the following initial conditions: AB distance = 212

Q9{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions

File:MRD 01576020 fhh table 11.png

2020-05-15T16:10:04Z

Dp3618:

MRD:01576020

2020-05-15T14:55:25Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
===Potential Energy Surface===
===Some nomenclature===
what momenta mean, show a diagram, what r1 r2 are ...
===Transition State Theory===

==EXERCISE 1==
{{fontcolor|blue|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?}}
<div>Q1 The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its gradient being 0 at that point. To further characterise a point, one needs to evaluate the Laplacian at that point to determine whether it is a maximum or a minimum. The Laplacian at the transition state will be a negative value, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the gradient is zero and the Laplacian is positive, the point in question is a local minimum. [ADD SOME FIGURES AND REFERENCES]</div>

{{fontcolor|blue|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.}}
<div>Q2 A simple system, also shown in Figure X, of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following Figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this Figure. The transition state is the minimum of this line]]

The minimum of this curve in the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if we start at the transition state. By changing the initial internuclear distances, we can indentify the transition state as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Internuclear distance vs. time plots are shown in this Figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Contour plot with the initial geometry being the transition state geometry and initial momenta 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the transition state teh atoms are separated by more than 90 pm. Conversely, in case (c), The distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in the vicinity of 90.8 pm.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

But we can get a more accurate value now that we know where to look for the transition state. With some trial and error, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

===Dynamics vs. MEP===
<div>Q3 {{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In the minimum energy path (MEP) calculation, there are no vibrations of the molecule as we move away from the transition state because momenta of all particles are set to 0 in each step. In the dynamics simulation, on the other hand, vibrations of the molecule are observed. In both cases, however, the system follows the same route down the valley.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. Molecule B-C vibrates, as can be seen from the oscillating distance between atoms B and C. A-b and A-C distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. The A-B momentum increases and ultimately moves with constant momentum away from the molecule B-C. Molecule B-C vibrates, so the momentum B-C oscillates.]] </div>

Using the initial conditions r1=rts and r2=rts+1, the system would roll into the other valley on the PES with similar behaviour to what is described above.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translation and vibration). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill.

===Reactive and Unreactive Trajectories===
<div>Reactive and Unreactive Trajectories
Q4{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected.
</div>

Q5{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
One of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the products again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===
Q6{{fontcolor|blue|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?Locate the approximate position of the transition state.}}

The F+H2--> HF+H reaction is exothermic and H+HF-->H2+F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than H-H bond. This makes sense as it has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is F. From these initial conditions, the forces were: along AB: 0.000 and along BC: 0.001. SO this structure is very close to the transition state structure.

Q7{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. This is in good agreement with experimental findings, where the H-F bond dissociation energy is reported as 566 kJ/mol and the H-H bond dissociation energy is 436 kJ/mol[ref].

===Reaction Dynamics===
A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g mol-1 pm fs-1.
Q8{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left| A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 seconds of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the B-C pair which is moving aparth. ]]

Let's now setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, an rFH = 238 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. Note that we are putting more energy into the system than the activation energy. The table below shows some observations and contour plots.

{| class="wikitable" border=1
! pFH/ g.mol-1.pm.fs-1 !! pHH/ g.mol-1.pm.fs-1 !! Einitial /kJ.mol-1 !! Reactive? !! Observations !! Illustration of the trajectory
|-
| -1.0 || -6.1 || -402.504 || No ||Here the system passed through the transition state structure, the hydrogen atom collided with the fluorine and then reformed reactants. || [[File:MRD 01576020 fhh table 1.png|400 px]]
|-
| -1.0 || -6.0 || -404.098 || No ||This is very similar to the case above. || [[File:MRD 01576020 fhh table 2.png|400 px]]
|-
| -1.0 || -5.0 || -414.098 || No ||This trajectory is also not reactive and is quite similar to the two above, again there was one collision with the fluorine atom. || [[File:MRD 01576020 fhh table 3.png|400 px]]
|-
| -1.0 || -4.1 || -421.388 || No ||Here, the fluorine atom did not collide with any atoms and the transition state was not surpassed. || [[File:MRD 01576020 fhh table 4.png|400 px]]
|-
| -1.0 || 0.0 || -434.098 || No ||Since even less energy was given the picture was very similar to the one above, but the hydrogen atom approached the fluorine even less. || [[File:MRD 01576020 fhh table 5.png|400 px]]
|-
| -1.0 || 2.1 || -427.588 || No ||Slightly more energy than above so the atoms were closer but the transition state was still not reached. || [[File:MRD 01576020 fhh table 6.png|400 px]]
|-
| -1.0 || 4.1 || -413.188 || No ||This trajectory was more interesting, the fluorine collided twice with one of the hydrogen atoms, but the reactants were reformed afterwards. || [[File:MRD 01576020 fhh table 7.png|400 px]]
|-
| -1.0 || 6.0 || -392.098 || No ||The system was very close to forming products here, fluorine and hydrogen collided 7 times, but the system rolled into the valley of reactants again. || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 6.1 || -390.788 || No ||The fluorine and hydrogen only collided once in this simulation. There is a pattern emerging in this table ... || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 7.0 || -378.098 || No ||This was very similar to the one above, there was one collision and then the reactants were reformed.|| [[File:MRD 01576020 fhh table 10.png|400 px]]
|}

From the table above it is obvious that whether the reaction will result in the formation of products is clearly not dependent only on the energy of vibration of the H2 molecule. Maybe increasing the momentum pFH will help ...

{{fontcolor|blue|For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?}}

There is very little vibration in the H2 molecule now and the F and the molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.

Let's now focus on the reverse reaction. When there is very little vibrational energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below.

An example of a reactive trajectory for this reaction would be produced by the following initial conditions: AB distance = 212

Q9{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions

File:MRD 01576020 fhh table 10.png

2020-05-15T14:51:24Z

Dp3618:

MRD:01576020

2020-05-15T14:44:42Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
===Potential Energy Surface===
===Some nomenclature===
what momenta mean, show a diagram, what r1 r2 are ...
===Transition State Theory===

==EXERCISE 1==
{{fontcolor|blue|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?}}
<div>Q1 The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its gradient being 0 at that point. To further characterise a point, one needs to evaluate the Laplacian at that point to determine whether it is a maximum or a minimum. The Laplacian at the transition state will be a negative value, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the gradient is zero and the Laplacian is positive, the point in question is a local minimum. [ADD SOME FIGURES AND REFERENCES]</div>

{{fontcolor|blue|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.}}
<div>Q2 A simple system, also shown in Figure X, of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following Figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this Figure. The transition state is the minimum of this line]]

The minimum of this curve in the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if we start at the transition state. By changing the initial internuclear distances, we can indentify the transition state as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Internuclear distance vs. time plots are shown in this Figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Contour plot with the initial geometry being the transition state geometry and initial momenta 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the transition state teh atoms are separated by more than 90 pm. Conversely, in case (c), The distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in the vicinity of 90.8 pm.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

But we can get a more accurate value now that we know where to look for the transition state. With some trial and error, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

===Dynamics vs. MEP===
<div>Q3 {{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In the minimum energy path (MEP) calculation, there are no vibrations of the molecule as we move away from the transition state because momenta of all particles are set to 0 in each step. In the dynamics simulation, on the other hand, vibrations of the molecule are observed. In both cases, however, the system follows the same route down the valley.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. Molecule B-C vibrates, as can be seen from the oscillating distance between atoms B and C. A-b and A-C distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. The A-B momentum increases and ultimately moves with constant momentum away from the molecule B-C. Molecule B-C vibrates, so the momentum B-C oscillates.]] </div>

Using the initial conditions r1=rts and r2=rts+1, the system would roll into the other valley on the PES with similar behaviour to what is described above.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translation and vibration). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill.

===Reactive and Unreactive Trajectories===
<div>Reactive and Unreactive Trajectories
Q4{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected.
</div>

Q5{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
One of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the products again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===
Q6{{fontcolor|blue|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?Locate the approximate position of the transition state.}}

The F+H2--> HF+H reaction is exothermic and H+HF-->H2+F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than H-H bond. This makes sense as it has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is F. From these initial conditions, the forces were: along AB: 0.000 and along BC: 0.001. SO this structure is very close to the transition state structure.

Q7{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. This is in good agreement with experimental findings, where the H-F bond dissociation energy is reported as 566 kJ/mol and the H-H bond dissociation energy is 436 kJ/mol[ref].

===Reaction Dynamics===
A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g mol-1 pm fs-1.
Q8{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left| A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 seconds of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the B-C pair which is moving aparth. ]]

Let's now setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, an rFH = 238 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. Note that we are putting more energy into the system than the activation energy. The table below shows some observations and contour plots.

{| class="wikitable" border=1
! pFH/ g.mol-1.pm.fs-1 !! pHH/ g.mol-1.pm.fs-1 !! Einitial /kJ.mol-1 !! Reactive? !! Observations !! Illustration of the trajectory
|-
| -1.0 || -6.1 || -402.504 || No ||Here the system passed through the transition state structure, the hydrogen atom collided with the fluorine and then reformed reactants. || [[File:MRD 01576020 fhh table 1.png|400 px]]
|-
| -1.0 || -6.0 || -404.098 || No ||This is very similar to the case above. || [[File:MRD 01576020 fhh table 2.png|400 px]]
|-
| -1.0 || -5.0 || -414.098 || No ||This trajectory is also not reactive and is quite similar to the two above, again there was one collision with the fluorine atom. || [[File:MRD 01576020 fhh table 3.png|400 px]]
|-
| -1.0 || -4.1 || -421.388 || No ||Here, the fluorine atom did not collide with any atoms and the transition state was not surpassed. || [[File:MRD 01576020 fhh table 4.png|400 px]]
|-
| -1.0 || 0.0 || -434.098 || No ||Since even less energy was given the picture was very similar to the one above, but the hydrogen atom approached the fluorine even less. || [[File:MRD 01576020 fhh table 5.png|400 px]]
|-
| -1.0 || 2.1 || -427.588 || No ||Slightly more energy than above so the atoms were closer but the transition state was still not reached. || [[File:MRD 01576020 fhh table 6.png|400 px]]
|-
| -1.0 || 4.1 || -413.188 || No ||This trajectory was more interesting, the fluorine collided twice with one of the hydrogen atoms, but the reactants were reformed afterwards. || [[File:MRD 01576020 fhh table 7.png|400 px]]
|-
| -1.0 || 6.0 || -392.098 || No ||The system was very close to forming products here, fluorine and hydrogen collided 7 times, but the system rolled into the valley of reactants again. || [[File:MRD 01576020 fhh table 8.png|400 px]]
|-
| -1.0 || 6.1 || -x || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|-
| -1.0 || 7.0 || -x || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

{{fontcolor|blue|For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?}}

There is very little vibration in the H2 molecule now and the F and the molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.

Let's now focus on the reverse reaction. When there is very little vibrational energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below.

An example of a reactive trajectory for this reaction would be produced by the following initial conditions: AB distance = 212

Q9{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions

File:MRD 01576020 fhh table 8.png

2020-05-15T14:44:15Z

Dp3618:

File:MRD 01576020 fhh table 7.png

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File:MRD 01576020 fhh table 6.png

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File:MRD 01576020 fhh table 5.png

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File:MRD 01576020 fhh table 4.png

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File:MRD 01576020 fhh table 3.png

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File:MRD 01576020 fhh table 2.png

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File:MRD 01576020 fhh table 1.png

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MRD:01576020

2020-05-15T14:18:23Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
===Potential Energy Surface===
===Some nomenclature===
what momenta mean, show a diagram, what r1 r2 are ...
===Transition State Theory===

==EXERCISE 1==
{{fontcolor|blue|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?}}
<div>Q1 The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its gradient being 0 at that point. To further characterise a point, one needs to evaluate the Laplacian at that point to determine whether it is a maximum or a minimum. The Laplacian at the transition state will be a negative value, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the gradient is zero and the Laplacian is positive, the point in question is a local minimum. [ADD SOME FIGURES AND REFERENCES]</div>

{{fontcolor|blue|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.}}
<div>Q2 A simple system, also shown in Figure X, of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following Figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this Figure. The transition state is the minimum of this line]]

The minimum of this curve in the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if we start at the transition state. By changing the initial internuclear distances, we can indentify the transition state as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Internuclear distance vs. time plots are shown in this Figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Contour plot with the initial geometry being the transition state geometry and initial momenta 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the transition state teh atoms are separated by more than 90 pm. Conversely, in case (c), The distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in the vicinity of 90.8 pm.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

But we can get a more accurate value now that we know where to look for the transition state. With some trial and error, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

===Dynamics vs. MEP===
<div>Q3 {{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In the minimum energy path (MEP) calculation, there are no vibrations of the molecule as we move away from the transition state because momenta of all particles are set to 0 in each step. In the dynamics simulation, on the other hand, vibrations of the molecule are observed. In both cases, however, the system follows the same route down the valley.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. Molecule B-C vibrates, as can be seen from the oscillating distance between atoms B and C. A-b and A-C distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. The A-B momentum increases and ultimately moves with constant momentum away from the molecule B-C. Molecule B-C vibrates, so the momentum B-C oscillates.]] </div>

Using the initial conditions r1=rts and r2=rts+1, the system would roll into the other valley on the PES with similar behaviour to what is described above.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translation and vibration). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill.

===Reactive and Unreactive Trajectories===
<div>Reactive and Unreactive Trajectories
Q4{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected.
</div>

Q5{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
One of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the products again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===
Q6{{fontcolor|blue|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?Locate the approximate position of the transition state.}}

The F+H2--> HF+H reaction is exothermic and H+HF-->H2+F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than H-H bond. This makes sense as it has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is F. From these initial conditions, the forces were: along AB: 0.000 and along BC: 0.001. SO this structure is very close to the transition state structure.

Q7{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. This is in good agreement with experimental findings, where the H-F bond dissociation energy is reported as 566 kJ/mol and the H-H bond dissociation energy is 436 kJ/mol[ref].

===Reaction Dynamics===
A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g mol-1 pm fs-1.
Q8{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left| A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 seconds of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the B-C pair which is moving aparth. ]]

Let's now setup 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. Note that we are putting more energy into the system than the activation energy. The table below shows some observations.

{| class="wikitable" border=1
! pFH/ g.mol-1.pm.fs-1 !! pHH/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Observations !! Illustration of the trajectory
|-
| -1.0 || -6.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -1.0 || -6.0 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -1.0 || -5.0 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -1.0 || -4.1 || -357.277 || No ||In this case, the system did cross the saddle point but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -1.0 || -2.1 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|-
| -1.0 || 0.0 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|-
| -1.0 || 2.1 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|-
| -1.0 || 4.1 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|-
| -1.0 || 6.0 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|-
| -1.0 || 6.1 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|-
| -1.0 || 7.0 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

{{fontcolor|blue|For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?}}

There is very little vibration in the H2 molecule now and the F and the molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.

Let's now focus on the reverse reaction. When there is very little vibrational energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below.

An example of a reactive trajectory for this reaction would be produced by the following initial conditions: AB distance = 212

Q9{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions

MRD:01576020

2020-05-15T13:54:09Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
===Potential Energy Surface===
===Some nomenclature===
what momenta mean, show a diagram, what r1 r2 are ...
===Transition State Theory===

==EXERCISE 1==
{{fontcolor|blue|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?}}
<div>Q1 The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its gradient being 0 at that point. To further characterise a point, one needs to evaluate the Laplacian at that point to determine whether it is a maximum or a minimum. The Laplacian at the transition state will be a negative value, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the gradient is zero and the Laplacian is positive, the point in question is a local minimum. [ADD SOME FIGURES AND REFERENCES]</div>

{{fontcolor|blue|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.}}
<div>Q2 A simple system, also shown in Figure X, of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following Figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this Figure. The transition state is the minimum of this line]]

The minimum of this curve in the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if we start at the transition state. By changing the initial internuclear distances, we can indentify the transition state as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Internuclear distance vs. time plots are shown in this Figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Contour plot with the initial geometry being the transition state geometry and initial momenta 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the transition state teh atoms are separated by more than 90 pm. Conversely, in case (c), The distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in the vicinity of 90.8 pm.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

But we can get a more accurate value now that we know where to look for the transition state. With some trial and error, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

===Dynamics vs. MEP===
<div>Q3 {{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In the minimum energy path (MEP) calculation, there are no vibrations of the molecule as we move away from the transition state because momenta of all particles are set to 0 in each step. In the dynamics simulation, on the other hand, vibrations of the molecule are observed. In both cases, however, the system follows the same route down the valley.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. Molecule B-C vibrates, as can be seen from the oscillating distance between atoms B and C. A-b and A-C distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. The A-B momentum increases and ultimately moves with constant momentum away from the molecule B-C. Molecule B-C vibrates, so the momentum B-C oscillates.]] </div>

Using the initial conditions r1=rts and r2=rts+1, the system would roll into the other valley on the PES with similar behaviour to what is described above.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translation and vibration). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill.

===Reactive and Unreactive Trajectories===
<div>Reactive and Unreactive Trajectories
Q4{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected.
</div>

Q5{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
One of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the products again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===
Q6{{fontcolor|blue|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?Locate the approximate position of the transition state.}}

The F+H2--> HF+H reaction is exothermic and H+HF-->H2+F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than H-H bond. This makes sense as it has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is F. From these initial conditions, the forces were: along AB: 0.000 and along BC: 0.001. SO this structure is very close to the transition state structure.

Q7{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. This is in good agreement with experimental findings, where the H-F bond dissociation energy is reported as 566 kJ/mol and the H-H bond dissociation energy is 436 kJ/mol[ref].

===Reaction Dynamics===
A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g mol-1 pm fs-1.
Q8{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left| A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 seconds of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the B-C pair which is moving aparth. ]]

Let's now setup 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. Note that we are putting more energy into the system than the activation energy. The table below shows some observations.

{| class="wikitable" border=1
! pFH/ g.mol-1.pm.fs-1 !! pHH/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Observations !! Illustration of the trajectory
|-
| -1.0 || -6.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -1.0 || -6.0 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -1.0 || -5.0 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -1.0 || -4.1 || -357.277 || No ||In this case, the system did cross the saddle point but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -1.0 || -2.1 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|-
| -1.0 || 0.0 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|-
| -1.0 || 2.1 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|-
| -1.0 || 4.1 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|-
| -1.0 || 6.0 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|-
| -1.0 || 6.1 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|-
| -1.0 || 7.0 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

{{fontcolor|blue|For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?}}

There is very little vibration in the H2 molecule now and the F and the molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.
When there is very little energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below.

Q9{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions

MRD:01576020

2020-05-15T13:44:01Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
===Potential Energy Surface===
===Some nomenclature===
what momenta mean, show a diagram, what r1 r2 are ...
===Transition State Theory===

==EXERCISE 1==
{{fontcolor|blue|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?}}
<div>Q1 The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its gradient being 0 at that point. To further characterise a point, one needs to evaluate the Laplacian at that point to determine whether it is a maximum or a minimum. The Laplacian at the transition state will be a negative value, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the gradient is zero and the Laplacian is positive, the point in question is a local minimum. [ADD SOME FIGURES AND REFERENCES]</div>

{{fontcolor|blue|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.}}
<div>Q2 A simple system, also shown in Figure X, of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following Figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this Figure. The transition state is the minimum of this line]]

The minimum of this curve in the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if we start at the transition state. By changing the initial internuclear distances, we can indentify the transition state as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Internuclear distance vs. time plots are shown in this Figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Contour plot with the initial geometry being the transition state geometry and initial momenta 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the transition state teh atoms are separated by more than 90 pm. Conversely, in case (c), The distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in the vicinity of 90.8 pm.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

But we can get a more accurate value now that we know where to look for the transition state. With some trial and error, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

===Dynamics vs. MEP===
<div>Q3 {{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In the minimum energy path (MEP) calculation, there are no vibrations of the molecule as we move away from the transition state because momenta of all particles are set to 0 in each step. In the dynamics simulation, on the other hand, vibrations of the molecule are observed. In both cases, however, the system follows the same route down the valley.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. Molecule B-C vibrates, as can be seen from the oscillating distance between atoms B and C. A-b and A-C distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. The A-B momentum increases and ultimately moves with constant momentum away from the molecule B-C. Molecule B-C vibrates, so the momentum B-C oscillates.]] </div>

Using the initial conditions r1=rts and r2=rts+1, the system would roll into the other valley on the PES with similar behaviour to what is described above.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translation and vibration). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill.

===Reactive and Unreactive Trajectories===
<div>Reactive and Unreactive Trajectories
Q4{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected.
</div>

Q5{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
One of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the products again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===
Q6{{fontcolor|blue|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?Locate the approximate position of the transition state.}}

The F+H2--> HF+H reaction is exothermic and H+HF-->H2+F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than H-H bond. This makes sense as it has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is F. From these initial conditions, the forces were: along AB: 0.000 and along BC: 0.001. SO this structure is very close to the transition state structure.

Q7{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. This is in good agreement with experimental findings, where the H-F bond dissociation energy is reported as 566 kJ/mol and the H-H bond dissociation energy is 436 kJ/mol[ref].

===Reaction Dynamics===
A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g mol-1 pm fs-1.
Q8{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left| A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 seconds of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the B-C pair which is moving aparth. ]]

{{fontcolor|blue|Setup 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}}

{{fontcolor|blue|For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?}}

There is very little vibration in the H2 molecule now and the F and the molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.
When there is very little energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below.

Q9{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions

MRD:01576020

2020-05-15T13:41:15Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
===Potential Energy Surface===
===Some nomenclature===
what momenta mean, show a diagram, what r1 r2 are ...
===Transition State Theory===

==EXERCISE 1==
{{fontcolor|blue|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?}}
<div>Q1 The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its gradient being 0 at that point. To further characterise a point, one needs to evaluate the Laplacian at that point to determine whether it is a maximum or a minimum. The Laplacian at the transition state will be a negative value, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the gradient is zero and the Laplacian is positive, the point in question is a local minimum. [ADD SOME FIGURES AND REFERENCES]</div>

{{fontcolor|blue|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.}}
<div>Q2 A simple system, also shown in Figure X, of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following Figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this Figure. The transition state is the minimum of this line]]

The minimum of this curve in the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if we start at the transition state. By changing the initial internuclear distances, we can indentify the transition state as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Internuclear distance vs. time plots are shown in this Figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Contour plot with the initial geometry being the transition state geometry and initial momenta 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the transition state teh atoms are separated by more than 90 pm. Conversely, in case (c), The distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in the vicinity of 90.8 pm.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

But we can get a more accurate value now that we know where to look for the transition state. With some trial and error, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

===Dynamics vs. MEP===
<div>Q3 {{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In the minimum energy path (MEP) calculation, there are no vibrations of the molecule as we move away from the transition state because momenta of all particles are set to 0 in each step. In the dynamics simulation, on the other hand, vibrations of the molecule are observed. In both cases, however, the system follows the same route down the valley.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. Molecule B-C vibrates, as can be seen from the oscillating distance between atoms B and C. A-b and A-C distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. The A-B momentum increases and ultimately moves with constant momentum away from the molecule B-C. Molecule B-C vibrates, so the momentum B-C oscillates.]] </div>

Using the initial conditions r1=rts and r2=rts+1, the system would roll into the other valley on the PES with similar behaviour to what is described above.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translation and vibration). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill.

===Reactive and Unreactive Trajectories===
<div>Reactive and Unreactive Trajectories
Q4{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected.
</div>

Q5{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
One of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the products again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===
Q6{{fontcolor|blue|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?Locate the approximate position of the transition state.}}

The F+H2--> HF+H reaction is exothermic and H+HF-->H2+F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than H-H bond. This makes sense as it has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is F. From these initial conditions, the forces were: along AB: 0.000 and along BC: 0.001. SO this structure is very close to the transition state structure.

Q7{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. This is in good agreement with experimental findings, where the H-F bond dissociation energy is reported as 566 kJ/mol and the H-H bond dissociation energy is 436 kJ/mol[ref].

===Reaction Dynamics===
A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g mol-1 pm fs-1.
Q8{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This energy is released as strong vibrations of the HF molecule. Since the vibration of a H-F bond, in contrast to a H-H bond, results in a change of dipole, electromagnetic radiation from the IR part of the spectrum is emitted as the molecule vibrates. This emitted IR radiation can be experimentally observed by emission vibrational spectroscopy and the production of products can be confirmed.

[[File:MRD 01576020 animation rt GIF.gif|1000 px|thumb|left| A GIF showing the animation of the reaction. The system passes the saddle point a few times but eventually forms products.]]

[[File:MRD 01576020 rt momentavstime.png|thumb|550 px|center|A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 seconds of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the B-C pair which is moving aparth. ]]

{{fontcolor|blue|Setup 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}}

{{fontcolor|blue|For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?}}
There is very little vibration in the H2 molecule now and the F and the molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.
When there is very little energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below.

Q9{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions

MRD:01576020

2020-05-15T12:40:20Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
===Potential Energy Surface===
===Some nomenclature===
what momenta mean, show a diagram, what r1 r2 are ...
===Transition State Theory===

==EXERCISE 1==
{{fontcolor|blue|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?}}
<div>Q1 The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its gradient being 0 at that point. To further characterise a point, one needs to evaluate the Laplacian at that point to determine whether it is a maximum or a minimum. The Laplacian at the transition state will be a negative value, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the gradient is zero and the Laplacian is positive, the point in question is a local minimum. [ADD SOME FIGURES AND REFERENCES]</div>

{{fontcolor|blue|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.}}
<div>Q2 A simple system, also shown in Figure X, of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following Figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this Figure. The transition state is the minimum of this line]]

The minimum of this curve in the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if we start at the transition state. By changing the initial internuclear distances, we can indentify the transition state as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Internuclear distance vs. time plots are shown in this Figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Contour plot with the initial geometry being the transition state geometry and initial momenta 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the transition state teh atoms are separated by more than 90 pm. Conversely, in case (c), The distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in the vicinity of 90.8 pm.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

But we can get a more accurate value now that we know where to look for the transition state. With some trial and error, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

===Dynamics vs. MEP===
<div>Q3 {{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In the minimum energy path (MEP) calculation, there are no vibrations of the molecule as we move away from the transition state because momenta of all particles are set to 0 in each step. In the dynamics simulation, on the other hand, vibrations of the molecule are observed. In both cases, however, the system follows the same route down the valley.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. Molecule B-C vibrates, as can be seen from the oscillating distance between atoms B and C. A-b and A-C distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. The A-B momentum increases and ultimately moves with constant momentum away from the molecule B-C. Molecule B-C vibrates, so the momentum B-C oscillates.]] </div>

Using the initial conditions r1=rts and r2=rts+1, the system would roll into the other valley on the PES with similar behaviour to what is described above.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translation and vibration). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill.

===Reactive and Unreactive Trajectories===
<div>Reactive and Unreactive Trajectories
Q4{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected.
</div>

Q5{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
One of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the products again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===
Q6{{fontcolor|blue|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?Locate the approximate position of the transition state.}}

The F+H2--> HF+H reaction is exothermic and H+HF-->H2+F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than H-H bond. This makes sense as it has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is F. From these initial conditions, the forces were: along AB: 0.000 and along BC: 0.001. SO this structure is very close to the transition state structure.

Q7{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. This is in good agreement with experimental findings, where the H-F bond dissociation energy is reported as 566 kJ/mol and the H-H bond dissociation energy is 436 kJ/mol[ref].

===Reaction Dynamics===
A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g mol-1 pm fs-1.
Q8{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This is manifested as strong vibrations of the HF molecule.

[[File:MRD 01576020 animation rt GIF.gif|450 px|thumb]]

[[File:MRD 01576020 rt momentavstime.png|thumb|450 px|A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 seconds of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the B-C pair which is moving aparth. ]]

{{fontcolor|blue|Setup 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}}

{{fontcolor|blue|For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?}}
There is very little vibration in the H2 molecule now and the F and the molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.
When there is very little energy in the HF molecule, the trajectories are non reactive, as we increase the vibration strength more and more trajectories become reactive. Why this is so is discussed below.

Q9{{fontcolor|blue|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 are very useful when we are trying to find suitable initial conditions

MRD:01576020

2020-05-15T12:03:54Z

Dp3618:

=Molecular reaction Dynamics Lab Report=

==Introduction==
===Potential Energy Surface===
===Some nomenclature===
what momenta mean, show a diagram, what r1 r2 are ...
===Transition State Theory===

==EXERCISE 1==
{{fontcolor|blue|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?}}
<div>Q1 The transition state is defined as the local maximum on the minimum energy path the reactants take when forming products. Mathematically, an extremum of a function is identified by its gradient being 0 at that point. To further characterise a point, one needs to evaluate the Laplacian at that point to determine whether it is a maximum or a minimum. The Laplacian at the transition state will be a negative value, corresponding to a local maximum of the potential energy surface (PES). Conversely, if the gradient is zero and the Laplacian is positive, the point in question is a local minimum. [ADD SOME FIGURES AND REFERENCES]</div>

{{fontcolor|blue|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.}}
<div>Q2 A simple system, also shown in Figure X, of a H atom colliding with an H2 molecule was investigated. Firstly, the transition state position was estimated by setting the momenta (p1 and p2) to 0 and, since the system is symmetric, the positions (r1 and r2) equal. By using these initial conditions, we have constrained our system to only move along the path in black in the following Figure.</div>

[[File:MRD 01576020 p0 path.png|thumb|center|700 px|By setting both r1 and r2 equal and momenta = 0, we have constrained our system to only move along the black path shown in this Figure. The transition state is the minimum of this line]]

The minimum of this curve in the maximum of the lowest energy path. Since the gradient is zero at that point, there should be no movement of our system if we start at the transition state. By changing the initial internuclear distances, we can indentify the transition state as the distance where the internuclear distances are constant throughout the simulation.

[[File:MRD 01576020 TSs conc.png|thumb|center|1500 px|Internuclear distance vs. time plots are shown in this Figure. Due to the symmetry of the system, A-B and B-C distances overlay at all times. In graph (a), the initial distances between atoms were 90 pm. In graph (b), atoms were separated by 90.8 pm and in graph (c) the distances were 92 pm.]]

[[File:MRD 01576020 hhh TS.png|thumb|right|450 px|Contour plot with the initial geometry being the transition state geometry and initial momenta 0.]]

We can see that in cases (a) and (c) there were oscillations of the atoms present so the net force on the system was not 0, which means that these are not transition state structures. But looking at these two graphs more thoroughly, we can see that the oscillations began in opposite directions. In case (a) the initial distance was elongated at first, meaning that there was a repulsive force between the atoms and that in the transition state teh atoms are separated by more than 90 pm. Conversely, in case (c), The distance between atoms was shortened at first, denoting attraction between atoms. We know that the transition state structure will be between these two. There are no noticeable oscillations in case (b), suggesting that the distance between the atoms in the transition state structure will be in the vicinity of 90.8 pm.

The plot on the right confirms that the chosen rTS is correct, since the system did not roll into either of the valleys even in a long simulation.

But we can get a more accurate value now that we know where to look for the transition state. With some trial and error, our best estimate for the transition state is rTS = 90.775 pm. And most importantly, the forces on this initial geometry are -0.000 along AB and -0.000 along BC so this is the actual saddle point. Of course this is only true for our accuracy, there may be additional decimal points which are not 0. But force fields are not without errors either and such a system would be experimentally almost impossible to construct and study anyway so it makes no sense to improve the precision any further.

===Dynamics vs. MEP===
<div>Q3 {{fontcolor|blue|Comment on how the mep and the trajectory you just calculated differ.}}

In the minimum energy path (MEP) calculation, there are no vibrations of the molecule as we move away from the transition state because momenta of all particles are set to 0 in each step. In the dynamics simulation, on the other hand, vibrations of the molecule are observed. In both cases, however, the system follows the same route down the valley.

[[File:MRD 01576020 internucleardistance vs time.png|thumb|left|Internuclear distance vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. Molecule B-C vibrates, as can be seen from the oscillating distance between atoms B and C. A-b and A-C distances are rising linearly as the system moves towards the products.]][[File:MRD 01576020 momenta vs time.png|thumb|right|Momenta vs. time plot for r1=91.8 pm and r2=90.8 pm. As the simulation progresses, atom A moves away from the molecule B-C. The A-B momentum increases and ultimately moves with constant momentum away from the molecule B-C. Molecule B-C vibrates, so the momentum B-C oscillates.]] </div>

Using the initial conditions r1=rts and r2=rts+1, the system would roll into the other valley on the PES with similar behaviour to what is described above.

If we start another simulation, where the initial positions are the same as the final positions of the previous calculation, and we set the initial momenta to be the inverse of the final momenta, the new simulation will almost reach the transition state and then roll back through the valley. The reason that it does not reach the transition state is that we started slightly off the transition state structure in the previous calculation since otherwise our system would just stay at the transition state. Our first simulation started off with 0 momentum, and only had potential energy as it was very close to the saddle point. During the simulation the potential energy was transformed into kinetic energy (translation and vibration). We then took this kinetic energy and started transforming it into potential energy again. Since no energy was lost or gained, we reached the exact same point as we started off and then, depending on the length of simulation, the system may roll down the valley again. It will not turn back however, unless we manually revert momenta again.

If we use these initial conditions and run a MEP simulation, the first step will go back uphill, but then as the momenta are set to 0 and new momenta are calculated from the PES, the system will roll back downhill.

===Reactive and Unreactive Trajectories===
<div>Reactive and Unreactive Trajectories
Q4{{fontcolor|blue|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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot /kJ.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes ||In this simulation the molecule did not vibrate noticeably at first. The system reached the transition state and then crossed it. As it started rolling into the valley of the reactants the it accumulated some energy in the form of vibration. || [[File:MRD 01576020 table1 1.png|400 px]]
|-
| -3.1 || -4.1 || -420.077 || No ||Here, the molecule was vibrating at first and the system did not reach the saddle point but rather turned around and rolled back into the valley of reactants. || [[File:MRD 01576020 table1 2.png|400 px]]
|-
| -3.1 || -5.1 || -413.977 || Yes ||This trajectory is also reactive and is quite similar to the first one. || [[File:MRD 01576020 table1 3.png|400 px]]
|-
| -5.1 || -10.1 || -357.277 || No ||In this case, the system did cross the saddle point but then returned back and reformed reactants. There was a considerable amount of vibration present. || [[File:MRD 01576020 table1 4.png|400 px]]
|-
| -5.1 || -10.6 || -349.477 || Yes ||This trajectory is very interesting. The system crossed the transition state and had a structure similar to products but then returned back to the valley of reactants. It eventually crossed the saddle point again and ended up in the valley leading to the products with significant vibration of the product molecule. || [[File:MRD 01576020 table1 5.png|400 px]]
|}

In contrast to one of the assumptions of the transition state theory, it is obvious from the table that even if the system has sufficient energy to surpass the kinetic barrier, this is not a guarantee that the reaction will proceed as expected.
</div>

Q5{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}
One of the assumptions of transition state theory is that, once the transition state is surpassed, the system will in all cases end up in the products state. This assumption may cause some discrepancies between theoretical and experimental findings. As we have seen in some cases, even if the saddle point is reached and the system finds itself in the valley of the products, it can end up in the valley of the products again, which means that not in all cases does the system reach the products even if it had sufficient energy to overcome the activation energy. The transition state theory therefore overestimates the reaction rate.

==EXERCISE 2==
===PES inspection===
Q6{{fontcolor|blue|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?Locate the approximate position of the transition state.}}

The F+H2--> HF+H reaction is exothermic and H+HF-->H2+F is endothermic. This means that the HF molecule is more stable than the H2 molecule. This also means that the H-F bond is stronger than H-H bond. This makes sense as it has a significant ionic contribution, whereas the H-H bond is purely covalent. This nature is also reflected in the activation energies of the two reactions and the bond dissociation energies. The transition state structure is approximately: AB = 74.478 pm and BC = 181.455 pm, where C is F. From these initial conditions, the forces were: along AB: 0.000 and along BC: 0.001. SO this structure is very close to the transition state structure.

Q7{{fontcolor|blue|Report the activation energy for both reactions.}}
The activation energy for F+H2-->HF+H is approximately 1.05 kJ/mol, whereas the activation energy for the reverse reaction is 126.681 kJ/mol. This is in good agreement with experimental findings, where the H-F bond dissociation energy is reported as 566 kJ/mol and the H-H bond dissociation energy is 436 kJ/mol[ref].

===Reaction Dynamics===
A reactive set of initial conditions for the F + H2 --> HF + H reaction is the following: AB distance = 230 pm, BC distance = 74 pm, AB momentum = -5.1 g mol-1 pm fs-1.
Q8{{fontcolor|blue|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.}}
Since the reaction is exothermic and the system is isolated, there must be a way of release of the excess energy. This is manifested as strong vibrations of the HF molecule.

[[File:MRD 01576020 animation rt GIF.gif|450 px|thumb]]

[[File:MRD 01576020 rt momentavstime.png|thumb|450 px|A momenta vs. time plot of the above calculation. The momenta vary a lot around the transition state structure so there is not much useful information we can gather from the first 80 seconds of the simulation. The system was also going up and down the PES, so we cannot prove that energy was conserved. But once the system is far into the valley and the potential energy is constant, we can see that the maximum momenta in the vibrating A-B molecule is constant and so is the momentum of the B-C pair which is moving aparth. ]]

{{fontcolor|blue|Setup 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}}

{{fontocolor|blue|For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?}}
there is very little vibration in the H2 molecule now and the F and the molecule are approaching each other faster than before. The trajectory is reactive and the reaction proceeds with a considerable amount of vibrational energy in the HF molecule.

Q9{{fontcolor|blue|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.}}

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