ChemWiki - User contributions [en]

MRD:01512921

2020-05-26T09:55:41Z

Sp3418:

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surfaces (PES) of three different reactions were analysed, along with their reaction trajectories. The PES were used to identify the transition state and observe how different momenta and inter-nuclear distances affect the outcome of the reaction. It was also observed how the position of the transition state determines which type of energy, vibrational or translational, is needed to overcome the energy barrier.

Chemical reactions can be simulated by taking into consideration the relative positions, such as the distances, between the atoms involved in the reaction. In fact, the interaction that causes the atoms' motion depends on their location and is described as a PES, a function relative to the coordinates of the atoms <ref name="intro1"/>.
The region of space of the PES is separated into the reactants, where the system is before reacting, and the products regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision coordinates; for the systems analysed in this report, there are only two coordinates: r1, the distances between the atoms of the reacting molecule, and r2<ref name="intro1"/>, the distance between the atoms of the product molecule. The path of the reaction can be mapped with a reaction trajectory. A reactive trajectory will pass through a saddle point of the PES, also know as transition state(TS<ref name="intro1"/>). On the contrary, an unreactive trajectory will roll back toward the reagents upon meeting the TS<ref name="intro1"/>.

To understand the reactions and successfully predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
The only information necessary is the behaviour of the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on the following assumptions:
* It is possible to separate the motion of the collision from the other motions of the TS<ref name="tunneling" />.
* Energy distributions is in accordance with the Maxwell-Boltzmann distributionc<ref name="tunneling" />.
* it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling" />.
* the reaction is treated classically and the quantum effects are ignored<ref name="tunneling" />.

=== Exercise 1 ===

For the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

'''Q1''': On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Inter-nuclear distances vs time for the transition state.]]
'''Q2''': The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen (A or C) is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: at TS, they all approached zero, as the vibrational energy is zero due to the absence of bonds. The value was obtained by trial and error: the first distance chosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from a periodic vibration ( at 150 pm) to being stationary at 90.8 pm. This can also be observed in the “Inter-nuclear Distances vs Time” plot, where the distances between the atoms are constant in time.

'''Q3''': A MEP and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. The dynamics calculation resulted in a longer distance between atom B and C once the reaction finished; the reaction rolls toward the products.

If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This is illustrated by the following plots:
* in the Inter-nuclear distance vs time plot, the initial value of AB is equal to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
* in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increases in different ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]][[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Inter-nuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in following table was obtained. Using the final values of the reaction, a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation ended.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
[[File:forming_dist_01512921.png|thumb|centre|Plot of the Inter-nuclear distances vs time. The reaction reaches the transition state.]]

'''Q4''': for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product is not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction proceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enough to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were formed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . The last two example are cases of barrier crossing <ref name="barrier corssing"/>, which goes against one of the assumption of the conventional transition state theory.

'''Q5''':The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doesn't take into consideration the possibility of quantum tunnelling, as the conventional transition state theory is purely classical motion. <ref name="tunneling"/>
In fact, if the system tunnels through the PES, then the kinetic energy could be lower than the one needed to reach the TS, as the system can go through it: therefore, the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
'''Q1:'''

F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
* The reaction is exothermic as the energy of the reagents is higher that that of the products.

* Position of TS: AB= 74.5 pm and BC = 181pm .
* It was identified thanks to Hammond's postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds" />. In this case, the transition state is early, as the reaction is exothermic. Therefore, the TS will resemble the reagents and the separation between the hydrogens of the H2 molecule will be smaller than that of HF.
* Total energy-433.98 KJ mol-1.
* Energy of reagent:-560.592 KJ mol-1
[[File:energy_H2_01512921.png|thumb|right|Energy vs Time MEP. The transition state rolls to the reagent H2]]
* Activation energy is: -126.612
H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
* The reaction is endothermic as the energy of the reagents is lower that that of the products.
* Position of TS: HF = 95 pm, H2=250 pm
* Energy is -433.98 KJ mol-1
* Reagent energy: -434.012 KJ mol-1
* Activation energy: -0.032 KJ mol-1
[[File:energy_HF_01512921.png|thumb|left|Energy vs Time mep. The transition state rolls to the reagent HF]]

Strength of the bonds:
Strength of H2 = 436 KJ mol<ref name="strength"/>

Strength of HF = 569 KJ mol<ref name="strength"/>

When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

'''Q2,''' part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy"/>. This can be confirmed by spectroscopic methods like infrared, as it would be possible to see overtones due to the transition from the first to the second vibrational excited state.
A reactive trajectory was found at r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1. While changing value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with high vibrational energy in the products. At 3.1, the reaction has a barrier recrossing, where the product forms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reaction is not successful. Barrier recrossing is also seen at 6.1, with a successful collision. At 5.1 g.mol-1.pm.fs-1, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb exactly in the middle between Ha and F, not showing which reaction's side was preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1 and pHH=0.2 g.mol-1.pm.fs-1. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

'''Q2,''' part 2: FH + H ==> H2 +F

A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1. The skew plot of the reaction is shown.
[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

'''Q5''': The energy distribution of the products can be determined by locating the energy barrier on the reaction coordinate (The TS). For a reaction is A+ BC:
* an early barrier or TS will result in high vibrational energy in the products. In an exothermic reaction, the energy is released while AB distance is changing
* a late barrier results in low vibrational energy in the products. The energy is released after AB is formed and BC is changing, which corresponds to the formation of the products and the translational energy.

The position of the energy barrier would also help selecting what distribution of reactant energy is most likely to lead to a reaction.
* for an early TS, a molecule with high translational energy will be able to overcome the barrier, as all its motion along the reaction coordinates. On the other hand, a molecule with high vibration will not have enough energy to reach the barrier.
* for a late TS, the barrier will be overcome by vibrational energy rather than translational. In fact, a molecule with high translational energy will crash in the inner wall of the PES and bounce back<ref name="polanyi's" />.

=== Conclusions===
The investigation of the reaction dynamics, for the three reactions examined in the lab, concerned with the PES and the reaction trajectories.
Using the program, it was possible to identify the transition state of the reaction (using Hammond's postulate), the trajectories (both reactive and unreactive ones) and the activation energies (thank to the energy of the transition state and of the reactants).
A comparison was made between the rate constant from the canonical transition state theory and the experimental one. It was noted how, since the CTST is treated classicaly and hence does ot take into consideration the tunelling effect, the CTST overestimates the rate of the reaction.
By analysing the transition state, it was possible to observe how its position influences which energy, translational or vibrational, is required for the reaction to happen. An early transition state can be overcome with translational energy, while a late one will prefer high vibrational energy.

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="strength"> C. Yoder, 2020, "Common Bond Energies (D) and Bond Lengths (r)", Wired Chemist, http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html, May 2020. </ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
<ref name="polanyi's">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 9, pp 272-274. </ref>
</references>

MRD:01512921

2020-05-24T16:27:30Z

Sp3418:

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surfaces (PES) of three different reactions were analysed, along with their reaction trajectories. The PES were used to identify the transition state and observe how different momenta and inter-nuclear distances affect the outcome of the reaction. It was also observed how the position of the transition state determines which type of energy, vibrational or translational, is needed to overcome the energy barrier.

Chemical reactions can be simulated by taking into consideration the relative positions, such as the distances, between the atoms involved in the reaction. In fact, the interaction that causes the atoms' motion depends on their location and is described as a PES, a function relative to the coordinates of the atoms <ref name="intro1"/>.
The region of space of the PES is separated into the reactants, where the system is before reacting, and the products regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision coordinates; for the systems analysed in this report, there are only two coordinates: r1, the distances between the atoms of the reacting molecule, and r2<ref name="intro1"/>, the distance between the atoms of the product molecule. The path of the reaction can be mapped with a reaction trajectory. A reactive trajectory will pass through a saddle point of the PES, also know as transition state(TS<ref name="intro1"/>). On the contrary, an unreactive trajectory will roll back toward the reagents upon meeting the TS<ref name="intro1"/>.

To understand the reactions and successfully predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
The only information necessary is the behaviour of the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on the following assumptions:
* It is possible to separate the motion of the collision from the other motions of the TS<ref name="tunneling" />.
* Energy distributions is in accordance with the Maxwell-Boltzmann distributionc<ref name="tunneling" />.
* it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling" />.
* the reaction is treated classically and the quantum effects are ignored<ref name="tunneling" />.

=== Exercise 1 ===

For the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

'''Q1''': On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Inter-nuclear distances vs time for the transition state.]]
'''Q2''': The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen (A or C) is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: at TS, they all approached zero, as the vibrational energy is zero due to the absence of bonds. The value was obtained by trial and error: the first distance chosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from a periodic vibration ( at 150 pm) to being stationary at 90.8 pm. This can also be observed in the “Inter-nuclear Distances vs Time” plot, where the distances between the atoms are constant in time.

'''Q3''': A MEP and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. The dynamics calculation resulted in a longer distance between atom B and C once the reaction finished; the reaction rolls toward the products.

If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This is illustrated by the following plots:
* in the Inter-nuclear distance vs time plot, the initial value of AB is equal to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
* in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increases in different ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]][[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Inter-nuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in following table was obtained. Using the final values of the reaction, a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation ended.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
[[File:forming_dist_01512921.png|thumb|centre|Plot of the Inter-nuclear distances vs time. The reaction reaches the transition state.]]

'''Q4''': for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product is not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction proceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enough to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were formed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . The last two example are cases of barrier crossing <ref name="barrier corssing"/>, which goes against one of the assumption of the conventional transition state theory.

'''Q5''':The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doesn't take into consideration the possibility of quantum tunnelling, as the conventional transition state theory is purely classical motion. <ref name="tunneling"/>
In fact, if the system tunnels through the PES, then the kinetic energy could be lower than the one needed to reach the TS, as the system can go through it: therefore, the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
'''Q1:'''

F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
* The reaction is exothermic as the energy of the reagents is higher that that of the products.

* Position of TS: AB= 74.5 pm and BC = 181pm .
* It was identified thanks to Hammond's postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds" />. In this case, the transition state is early, as the reaction is exothermic. Therefore, the TS will resemble the reagents and the separation between the hydrogens of the H2 molecule will be smaller than that of HF.
* Total energy-433.98 KJ mol-1.
* Energy of reagent:-560.592 KJ mol-1
[[File:energy_H2_01512921.png|thumb|right|Energy vs Time MEP. The transition state rolls to the reagent H2]]
* Activation energy is: -126.612
H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
* The reaction is endothermic as the energy of the reagents is lower that that of the products.
* Position of TS: HF = 95 pm, H2=250 pm
* Energy is -433.98 KJ mol-1
* Reagent energy: -434.012 KJ mol-1
* Activation energy: -0.032 KJ mol-1
[[File:energy_HF_01512921.png|thumb|left|Energy vs Time mep. The transition state rolls to the reagent HF]]

Strength of the bonds:
Strength of H2 = 436 KJ mol<ref name="strength"/>

Strength of HF = 569 KJ mol<ref name="strength"/>

When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

'''Q2,''' part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy"/>. This can be confirmed by spectroscopic methods like infrared, as it would be possible to see overtones due to the transition from the first to the second vibrational excited state.
A reactive trajectory was found at r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1. While changing value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with high vibrational energy in the products. At 3.1, the reaction has a barrier recrossing, where the product forms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reaction is not successful. Barrier recrossing is also seen at 6.1, with a successful collision. At 5.1 g.mol-1.pm.fs-1, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb exactly in the middle between Ha and F, not showing which reaction's side was preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1 and pHH=0.2 g.mol-1.pm.fs-1. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

'''Q2,''' part 2: FH + H ==> H2 +F

A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1. The skew plot of the reaction is shown.
[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

'''Q5''': The energy distribution of the products can be determined by locating the energy barrier on the reaction coordinate (The TS). For a reaction is A+ BC:
* an early barrier or TS will result in high vibrational energy in the products. In an exothermic reaction, the energy is released while AB distance is changing
* a late barrier results in low vibrational energy in the products. The energy is released after AB is formed and BC is changing, which corresponds to the formation of the products and the translational energy.

The position of the energy barrier would also help selecting what distribution of reactant energy is most likely to lead to a reaction.
* for an early TS, a molecule with high translational energy will be able to overcome the barrier, as all its motion along the reaction coordinates. On the other hand, a molecule with high vibration will not have enough energy to reach the barrier.
* for a late TS, the barrier will be overcome by vibrational energy rather than translational. In fact, a molecule with high translational energy will crash in the inner wall of the PES and bounce back<ref name="polanyi's" />.

=== Conclusions===
The investigation of the reaction dynamics, for the three reactions examined in the lab, concerned with the PES and the reaction trajectories.
Using the program, it was possible to identify the transition state of the reaction (using Hammond's postulate), the trajectories (both reactive and unreactive ones) and the activation energies (thank to the energy of the transition state and of the reactants).
A comparison was made between the rate constant from the canonical transition state theory and the experimental one. It was noted how, since the CTST is treated classicaly and hence does ot take into consideration the tunelling effect, the CTST overestimates the rate of the reaction.
By analysing the transition state, it was possible to observe how its position influences which energy, translational or vibrational, is required for the reaction to happen. An early transition state can be overcome with translational energy, while a late one will prefer high vibrational energy.

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="strength"> C. Yoder, 2020, "Common Bond Energies (D) and Bond Lengths (r)", Wired Chemist, http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html, May 2020. </ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
<ref name="polanyi's">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 9, pp 272-274. </ref>
</references>

MRD:01512921

2020-05-22T18:09:05Z

Sp3418: /* Reaction dynamics report */

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and inter-nuclear distances affected the outcome of the reaction. It was also observed how the position of the transition state determines which type of energy, vibrational or translational, will be sufficient to overcome the energy barrier.

Chemical reactions can be simulated by taking into consideration the relative position, such as the distances, between the atom involved in the reaction. In fact, the interaction between the atoms that cause their motion depend on their location and are described as a PES, a function relative to the coordinates of the atoms <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision coordinates; for the systems analysed in this report, there are only two coordinates: r1, distances between atoms of the reacting molecule and r2<ref name="intro1"/>, distance between the atoms of the product molecule. The path of the reaction can be mapped with a reaction trajectory. A reactive trajectory will pass through a saddle point of the PES, also know as transition state(TS<ref name="intro1"/>). On the contrary, an unreactive trajectory will roll back toward the reagents upon meeting the TS<ref name="intro1"/>.

To understand the reactions and successfully predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
The only information necessary is the behaviour of the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on the following assumptions:
* It is possible to separate the motion of the collision from the other motions of the TS<ref name="tunneling" />.
* Energy distributions is in accordance with the Maxwell-Boltzmann distributionc<ref name="tunneling" />.
* it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling" />.
* the reaction is treated classically and the quantum effects are ignored<ref name="tunneling" />.

=== Exercise 1 ===

For the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

'''Q1''': On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Inter-nuclear distances vs time for the transition state.]]
'''Q2''': The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen (A or C) is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the vibrational energy is zero due to the absence of bonds. The value was obtained by trial and error: the first distance chosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from a periodic vibration ( at 150 pm) to being stationary at 90.8 pm. This can also be observed in the “Inter-nuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

'''Q3''': A MEP and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. The dynamics calculation resulted in a longer distance between atom B and C once the reaction finished: the reaction rolls toward the products.

If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This is illustrated by the following plots:
* in the Inter-nuclear distance vs time plot, the initial value of AB is equal to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
* in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in different ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]][[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Inter-nuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in following table was obtained. Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation ended.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
[[File:forming_dist_01512921.png|thumb|centre|Plot of the Inter-nuclear distances vs time. The reaction reaches the transition state.]]

'''Q4''': for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product is not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction proceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enough to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were formed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

'''Q5''':The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doesn't take into consideration the possibility of quantum tunnelling, as the conventional transition state theory is purely classical motion. <ref name="tunneling"/>
In fact, if the system tunnels through the PES, then the kinetic energy could be lower than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
'''Q1:'''

F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
* The reaction is exothermic as the energy of the reagents is higher that that of the products.

* Position of TS: AB= 74.5 pm and BC = 181pm .
* It was identified thanks to Hammond's postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds" />. In this case, the transition state is early, as the reaction is exothermic. Therefore, the TS will resemble the reagents and the separation between the hydrogen molecule will be smaller than that of HF
* Total energy-433.98 KJ mol-1.
* Energy of reagent:-560.592 KJ mol-1
[[File:energy_H2_01512921.png|thumb|right|Energy vs Time MEP. The transition state rolls to the reagent H2]]
* Activation energy is: -126.612
H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
* The reaction is endothermic as the energy of the reagents is lower that that of the products.
* Position of TS: HF = 95 pm, H2=250 pm
* Energy is -433.98 KJ mol-1
* Reagent energy: -434.012 KJ mol-1
* Activation energy: -0.032 KJ mol-1
[[File:energy_HF_01512921.png|thumb|left|Energy vs Time mep. The transition state rolls to the reagent HF]]

Strength of the bonds:
Strength of H2 = 436 KJ mol

Strength of HF = 569 KJ mol

When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

'''Q2,''' part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy" />. This can be confirmed by spectroscopic methods like infrared, as it would be possible to see overtones due to the transition from the first to the second vibrational excited state. A reactive trajectory was found at r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1. While the changing value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with high vibrational energy in the products. At 3.1, the reaction has a barrier recrossing, where the product forms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reaction is not successful. Barrier recrossing is also seen at 6.1, with a successful collision. At 5.1 g.mol-1.pm.fs-1, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb exactly in the middle between Ha and F, not showing which reaction's side is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1 and pHH=0.2 g.mol-1.pm.fs-1. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

'''Q2,''' part 2: FH + H ==> H2 +F

A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1. The skew plot of the reaction is shown.
[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

'''Q5''': The energy distribution of the products can be determined by locating the energy barrier on the reaction coordinate (The TS). For a reaction is A+ BC:
* an early barrier or TS will result in high vibrational energy in the products. In an exothermic reaction, the energy is released while AB distance is changing

* a late barrier results in low vibrational energy in the products. The energy is released after AB is formed and BC is changing, which corresponds to the formation of the products and the translational energy.
The position of the energy barrier would also help selecting what distribution of reactant energy is most likely to lead to a reaction.
* for an early TS, a molecule with high translational energy will be able to overcome the barrier, as all its motion along the reaction coordinates. On the other hand, a molecule with high vibration will not have enough energy to reach the barrier.
* for a late TS, the barrier will be overcome by vibrational energy rather than translational. In fact, a molecule with high translational energy will crash in the inner wall of the PES and bounce back<ref name="polanyi's" />.

=== Conclusions===
The investigation of the reaction dynamics for the three reactions examined in the lab concerned with the PES and the reaction trajectories.
Using the program, it was possible to identify the transition state of the reaction (using Hammond's postulate), the trajectories (both reactive and unreactive ones) and the activation energies (thank to the energy of the transition state and the of the reactants).
A comparison was made between the rate constant from the canonical transition state theory and the experimental one. It was noted how, due t the absence of tunnelling, the CTST would overestimate the rate of the reaction.
By analysing the transition state, it was possible to observe how its position influences which energy, translational or vibrational, is required for the reaction to happen. An early transition state can be overcome with translational energy, while a late one will prefer high vibrational energy.

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
<ref name="polanyi's">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 9, pp 272-274. </ref>
</references>

MRD:01512921

2020-05-22T18:08:45Z

Sp3418: /* Reaction dynamics report */

MRD:01512921

2020-05-22T18:08:15Z

Sp3418:

MRD:01512921

2020-05-22T18:06:28Z

Sp3418:

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and inter-nuclear distances affected the outcome of the reaction. It was also observed how the position of the transition state determines which type of energy, vibrational or translational, will be sufficient to overcome the energy barrier.

Chemical reactions can be simulated by taking into consideration the relative position, such as the distances, between the atom involved in the reaction. In fact, the interaction between the atoms that cause their motion depend on their location and are described as a PES, a function relative to the coordinates of the atoms <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision coordinates; for the systems analysed in this report, there are only two coordinates: r1, distances between atoms of the reacting molecule and r2<ref name="intro1"/>, distance between the atoms of the product molecule. The path of the reaction can be mapped with a reaction trajectory. A reactive trajectory will pass through a saddle point of the PES, also know as transition state(TS<ref name="intro1"/>). On the contrary, an unreactive trajectory will roll back toward the reagents upon meeting the TS<ref name="intro1"/>.

To understand the reactions and successfully predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
The only information necessary is the behaviour of the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on the following assumptions:
* It is possible to separate the motion of the collision from the other motions of the TS<ref name="tunneling" />.
* Energy distributions is in accordance with the Maxwell-Boltzmann distributionc<ref name="tunneling" />.
* it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling" />.
* the reaction is treated classically and the quantum effects are ignored<ref name="tunneling" />.

=== Exercise 1 ===

For the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

'''Q1''': On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Inter-nuclear distances vs time for the transition state.]]
'''Q2''': The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen (A or C) is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the vibrational energy is zero due to the absence of bonds. The value was obtained by trial and error: the first distance chosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from a periodic vibration ( at 150 pm) to being stationary at 90.8 pm. This can also be observed in the “Inter-nuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

'''Q3''': A MEP and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. The dynamics calculation resulted in a longer distance between atom B and C once the reaction finished: the reaction rolls toward the products.

If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This is illustrated by the following plots:
* in the Inter-nuclear distance vs time plot, the initial value of AB is equal to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
* in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in different ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]][[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Inter-nuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in following table was obtained. Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation ended.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
[[File:forming_dist_01512921.png|thumb|centre|Plot of the Inter-nuclear distances vs time. The reaction reaches the transition state.]]

'''Q4''': for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product is not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction proceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enough to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were formed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

'''Q5''':The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doesn't take into consideration the possibility of quantum tunnelling, as the conventional transition state theory is purely classical motion. <ref name="tunneling"/>
In fact, if the system tunnels through the PES, then the kinetic energy could be lower than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
'''Q1:'''

F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
* The reaction is exothermic as the energy of the reagents is higher that that of the products.

* Position of TS: AB= 74.5 pm and BC = 181pm .
* It was identified thanks to Hammond's postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds" />. In this case, the transition state is early, as the reaction is exothermic. Therefore, the TS will resemble the reagents and the separation between the hydrogen molecule will be smaller than that of HF
* Total energy-433.98 KJ mol-1.
* Energy of reagent:-560.592 KJ mol-1
[[File:energy_H2_01512921.png|thumb|right|Energy vs Time MEP. The transition state rolls to the reagent H2]]
* Activation energy is: -126.612
H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
* The reaction is endothermic as the energy of the reagents is lower that that of the products.
* Position of TS: HF = 95 pm, H2=250 pm
* Energy is -433.98 KJ mol-1
* Reagent energy: -434.012 KJ mol-1
* Activation energy: -0.032 KJ mol-1
[[File:energy_HF_01512921.png|thumb|left|Energy vs Time mep. The transition state rolls to the reagent HF]]

Strength of the bonds:
Strength of H2 = 436 KJ mol

Strength of HF = 569 KJ mol

When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

<nowiki> </nowiki>'''Q2,''' part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy" />. This can be confirmed by spectroscopic methods like infrared, as it would be possible to see overtones due to the transition from the first to the second vibrational excited state. A reactive trajectory was found at r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1. While the changing value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with high vibrational energy in the products. At 3.1, the reaction has a barrier recrossing, where the product forms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reaction is not successful. Barrier recrossing is also seen at 6.1, with a successful collision. At 5.1 g.mol-1.pm.fs-1, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb exactly in the middle between Ha and F, not showing which reaction's side is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1 and pHH=0.2 g.mol-1.pm.fs-1. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

'''Q2,''' part 2: FH + H ==> H2 +F

A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1. The skew plot of the reaction is shown.

[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

'''Q5''': The energy distribution of the products can be determined by locating the energy barrier on the reaction coordinate (The TS). For a reaction is A+ BC:
* an early barrier or TS will result in high vibrational energy in the products. In an exothermic reaction, the energy is released while AB distance is changing

* a late barrier results in low vibrational energy in the products. The energy is released after AB is formed and BC is changing, which corresponds to the formation of the products and the translational energy.
The position of the energy barrier would also help selecting what distribution of reactant energy is most likely to lead to a reaction.
* for an early TS, a molecule with high translational energy will be able to overcome the barrier, as all its motion along the reaction coordinates. On the other hand, a molecule with high vibration will not have enough energy to reach the barrier.
* for a late TS, the barrier will be overcome by vibrational energy rather than translational. In fact, a molecule with high translational energy will crash in the inner wall of the PES and bounce back<ref name="polanyi's" />.

=== Conclusions===
The investigation of the reaction dynamics for the three reactions examined in the lab concerned with the PES and the reaction trajectories.
Using the program, it was possible to identify the transition state of the reaction (using Hammond's postulate), the trajectories (both reactive and unreactive ones) and the activation energies (thank to the energy of the transition state and the of the reactants).
A comparison was made between the rate constant from the canonical transition state theory and the experimental one. It was noted how, due t the absence of tunnelling, the CTST would overestimate the rate of the reaction.
By analysing the transition state, it was possible to observe how its position influences which energy, translational or vibrational, is required for the reaction to happen. An early transition state can be overcome with translational energy, while a late one will prefer high vibrational energy.

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
<ref name="polanyi's">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 9, pp 272-274. </ref>
</references>

MRD:01512921

2020-05-22T18:05:27Z

Sp3418:

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and inter-nuclear distances affected the outcome of the reaction. It was also observed how the position of the transition state determines which type of energy, vibrational or translational, will be sufficient to overcome the energy barrier.

Chemical reactions can be simulated by taking into consideration the relative position, such as the distances, between the atom involved in the reaction. In fact, the interaction between the atoms that cause their motion depend on their location and are described as a PES, a function relative to the coordinates of the atoms <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision coordinates; for the systems analysed in this report, there are only two coordinates: r1, distances between atoms of the reacting molecule and r2<ref name="intro1"/>, distance between the atoms of the product molecule. The path of the reaction can be mapped with a reaction trajectory. A reactive trajectory will pass through a saddle point of the PES, also know as transition state(TS<ref name="intro1"/>). On the contrary, an unreactive trajectory will roll back toward the reagents upon meeting the TS<ref name="intro1"/>.

To understand the reactions and successfully predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
The only information necessary is the behaviour of the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on the following assumptions:
* It is possible to separate the motion of the collision from the other motions of the TS<ref name="tunneling" />.
* Energy distributions is in accordance with the Maxwell-Boltzmann distributionc<ref name="tunneling" />.
* it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling" />.
* the reaction is treated classically and the quantum effects are ignored<ref name="tunneling" />.

=== Exercise 1 ===

For the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

'''Q1''': On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Inter-nuclear distances vs time for the transition state.]]
'''Q2''': The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen (A or C) is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the vibrational energy is zero due to the absence of bonds. The value was obtained by trial and error: the first distance chosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from a periodic vibration ( at 150 pm) to being stationary at 90.8 pm. This can also be observed in the “Inter-nuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

'''Q3''': A MEP and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. The dynamics calculation resulted in a longer distance between atom B and C once the reaction finished: the reaction rolls toward the products.

If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This is illustrated by the following plots:
* in the Inter-nuclear distance vs time plot, the initial value of AB is equal to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
* in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in different ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]][[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Inter-nuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in following table was obtained. Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation ended.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
[[File:forming_dist_01512921.png|thumb|centre|Plot of the Inter-nuclear distances vs time. The reaction reaches the transition state.]]

'''Q4''': for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product is not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction proceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enough to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were formed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

'''Q5''':The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doesn't take into consideration the possibility of quantum tunnelling, as the conventional transition state theory is purely classical motion. <ref name="tunneling"/>
In fact, if the system tunnels through the PES, then the kinetic energy could be lower than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
'''Q1:'''

F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
* The reaction is exothermic as the energy of the reagents is higher that that of the products.

* Position of TS: AB= 74.5 pm and BC = 181pm .
* It was identified thanks to Hammond's postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds" />. In this case, the transition state is early, as the reaction is exothermic. Therefore, the TS will resemble the reagents and the separation between the hydrogen molecule will be smaller than that of HF
* Total energy-433.98 KJ mol-1.
* Energy of reagent:-560.592 KJ mol-1
[[File:energy_H2_01512921.png|thumb|right|Energy vs Time MEP. The transition state rolls to the reagent H2]]
* Activation energy is: -126.612
H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
* The reaction is endothermic as the energy of the reagents is lower that that of the products.
* Position of TS: HF = 95 pm, H2=250 pm
* Energy is -433.98 KJ mol-1
* Reagent energy: -434.012 KJ mol-1
* Activation energy: -0.032 KJ mol-1
[[File:energy_HF_01512921.png|thumb|left|Energy vs Time mep. The transition state rolls to the reagent HF]]

Strength of the bonds:
Strength of H2 = 436 KJ mol

Strength of HF = 569 KJ mol

When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

<nowiki> </nowiki>'''Q2,''' part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy" />. This can be confirmed by spectroscopic methods like infrared, as it would be possible to see overtones due to the transition from the first to the second vibrational excited state. A reactive trajectory was found at r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1. While the changing value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with high vibrational energy in the products. At 3.1, the reaction has a barrier recrossing, where the product forms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reaction is not successful. Barrier recrossing is also seen at 6.1, with a successful collision. At 5.1 g.mol-1.pm.fs-1, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb exactly in the middle between Ha and F, not showing which reaction's side is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1 and pHH=0.2 g.mol-1.pm.fs-1. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

'''Q2,''' part 2: FH + H ==> H2 +F

A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1. The skew plot of the reaction is shown.

[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

'''Q5''': The energy distribution of the products can be determined by locating the energy barrier on the reaction coordinate (The TS). For a reaction is A+ BC:
* an early barrier or TS will result in high vibrational energy in the products. In an exothermic reaction, the energy is released while AB distance is changing

* a late barrier results in low vibrational energy in the products. The energy is released after AB is formed and BC is changing, which corresponds to the formation of the products and the translational energy.
The position of the energy barrier would also help selecting what distribution of reactant energy is most likely to lead to a reaction.
* for an early TS, a molecule with high translational energy will be able to overcome the barrier, as all its motion along the reaction coordinates. On the other hand, a molecule with high vibration will not have enough energy to reach the barrier.
* for a late TS, the barrier will be overcome by vibrational energy rather than translational. In fact, a molecule with high translational energy will crash in the inner wall of the PES and bounce back<ref name="polanyi's" />.

=== Conclusions===
The investigation of the reaction dynamics for the three reactions examined in the lab concerned with the PES and the reaction trajectories.
Using the program, it was possible to identify the transition state of the reaction (using Hammond's postulate), the trajectories (both reactive and unreactive ones) and the activation energies (thank to the energy of the transition state and the of the reactants).
A comparison was made between the rate constant from the canonical transition state theory and the experimental one. It was noted how, due t the absence of tunnelling, the CTST would overestimate the rate of the reaction.
By analysing the transition state, it was possible to observe how its position influences which energy, translational or vibrational, is required for the reaction to happen. An early transition state can be overcome with translational energy, while a late one will prefer high vibrational energy.

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
<ref name="polanyi's">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 9, pp 272-274. </ref>
</references>

MRD:01512921

2020-05-22T18:05:00Z

Sp3418:

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and inter-nuclear distances affected the outcome of the reaction. It was also observed how the position of the transition state determines which type of energy, vibrational or translational, will be sufficient to overcome the energy barrier.

Chemical reactions can be simulated by taking into consideration the relative position, such as the distances, between the atom involved in the reaction. In fact, the interaction between the atoms that cause their motion depend on their location and are described as a PES, a function relative to the coordinates of the atoms <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision coordinates; for the systems analysed in this report, there are only two coordinates: r1, distances between atoms of the reacting molecule and r2<ref name="intro1"/>, distance between the atoms of the product molecule. The path of the reaction can be mapped with a reaction trajectory. A reactive trajectory will pass through a saddle point of the PES, also know as transition state(TS<ref name="intro1"/>). On the contrary, an unreactive trajectory will roll back toward the reagents upon meeting the TS<ref name="intro1"/>.

To understand the reactions and successfully predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
The only information necessary is the behaviour of the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on the following assumptions:
* It is possible to separate the motion of the collision from the other motions of the TS<ref name="tunneling" />.
* Energy distributions is in accordance with the Maxwell-Boltzmann distributionc<ref name="tunneling" />.
* it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling" />.
* the reaction is treated classically and the quantum effects are ignored<ref name="tunneling" />.

=== Exercise 1 ===

For the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

'''Q1''': On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Inter-nuclear distances vs time for the transition state.]]
'''Q2''': The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen (A or C) is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the vibrational energy is zero due to the absence of bonds. The value was obtained by trial and error: the first distance chosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from a periodic vibration ( at 150 pm) to being stationary at 90.8 pm. This can also be observed in the “Inter-nuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

'''Q3''': A MEP and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. The dynamics calculation resulted in a longer distance between atom B and C once the reaction finished: the reaction rolls toward the products.

If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This is illustrated by the following plots:
* in the Inter-nuclear distance vs time plot, the initial value of AB is equal to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
* in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in different ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]][[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Inter-nuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in following table was obtained. Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation ended.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
[[File:forming_dist_01512921.png|thumb|right|Plot of the Inter-nuclear distances vs time. The reaction reaches the transition state.]]

'''Q4''': for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product is not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction proceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enough to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were formed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

'''Q5''':The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doesn't take into consideration the possibility of quantum tunnelling, as the conventional transition state theory is purely classical motion. <ref name="tunneling"/>
In fact, if the system tunnels through the PES, then the kinetic energy could be lower than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
'''Q1:'''

F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
* The reaction is exothermic as the energy of the reagents is higher that that of the products.

* Position of TS: AB= 74.5 pm and BC = 181pm .
* It was identified thanks to Hammond's postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds" />. In this case, the transition state is early, as the reaction is exothermic. Therefore, the TS will resemble the reagents and the separation between the hydrogen molecule will be smaller than that of HF
* Total energy-433.98 KJ mol-1.
* Energy of reagent:-560.592 KJ mol-1
[[File:energy_H2_01512921.png|thumb|right|Energy vs Time MEP. The transition state rolls to the reagent H2]]
* Activation energy is: -126.612
H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
* The reaction is endothermic as the energy of the reagents is lower that that of the products.
* Position of TS: HF = 95 pm, H2=250 pm
* Energy is -433.98 KJ mol-1
* Reagent energy: -434.012 KJ mol-1
* Activation energy: -0.032 KJ mol-1
[[File:energy_HF_01512921.png|thumb|left|Energy vs Time mep. The transition state rolls to the reagent HF]]

Strength of the bonds:
Strength of H2 = 436 KJ mol

Strength of HF = 569 KJ mol

When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

<nowiki> </nowiki>'''Q2,''' part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy" />. This can be confirmed by spectroscopic methods like infrared, as it would be possible to see overtones due to the transition from the first to the second vibrational excited state. A reactive trajectory was found at r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1. While the changing value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with high vibrational energy in the products. At 3.1, the reaction has a barrier recrossing, where the product forms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reaction is not successful. Barrier recrossing is also seen at 6.1, with a successful collision. At 5.1 g.mol-1.pm.fs-1, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb exactly in the middle between Ha and F, not showing which reaction's side is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1 and pHH=0.2 g.mol-1.pm.fs-1. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

'''Q2,''' part 2: FH + H ==> H2 +F

A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1. The skew plot of the reaction is shown.

[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

'''Q5''': The energy distribution of the products can be determined by locating the energy barrier on the reaction coordinate (The TS). For a reaction is A+ BC:
* an early barrier or TS will result in high vibrational energy in the products. In an exothermic reaction, the energy is released while AB distance is changing

* a late barrier results in low vibrational energy in the products. The energy is released after AB is formed and BC is changing, which corresponds to the formation of the products and the translational energy.
The position of the energy barrier would also help selecting what distribution of reactant energy is most likely to lead to a reaction.
* for an early TS, a molecule with high translational energy will be able to overcome the barrier, as all its motion along the reaction coordinates. On the other hand, a molecule with high vibration will not have enough energy to reach the barrier.
* for a late TS, the barrier will be overcome by vibrational energy rather than translational. In fact, a molecule with high translational energy will crash in the inner wall of the PES and bounce back<ref name="polanyi's" />.

=== Conclusions===
The investigation of the reaction dynamics for the three reactions examined in the lab concerned with the PES and the reaction trajectories.
Using the program, it was possible to identify the transition state of the reaction (using Hammond's postulate), the trajectories (both reactive and unreactive ones) and the activation energies (thank to the energy of the transition state and the of the reactants).
A comparison was made between the rate constant from the canonical transition state theory and the experimental one. It was noted how, due t the absence of tunnelling, the CTST would overestimate the rate of the reaction.
By analysing the transition state, it was possible to observe how its position influences which energy, translational or vibrational, is required for the reaction to happen. An early transition state can be overcome with translational energy, while a late one will prefer high vibrational energy.

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
<ref name="polanyi's">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 9, pp 272-274. </ref>
</references>

MRD:01512921

2020-05-22T18:04:36Z

Sp3418:

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and inter-nuclear distances affected the outcome of the reaction. It was also observed how the position of the transition state determines which type of energy, vibrational or translational, will be sufficient to overcome the energy barrier.

Chemical reactions can be simulated by taking into consideration the relative position, such as the distances, between the atom involved in the reaction. In fact, the interaction between the atoms that cause their motion depend on their location and are described as a PES, a function relative to the coordinates of the atoms <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision coordinates; for the systems analysed in this report, there are only two coordinates: r1, distances between atoms of the reacting molecule and r2<ref name="intro1"/>, distance between the atoms of the product molecule. The path of the reaction can be mapped with a reaction trajectory. A reactive trajectory will pass through a saddle point of the PES, also know as transition state(TS<ref name="intro1"/>). On the contrary, an unreactive trajectory will roll back toward the reagents upon meeting the TS<ref name="intro1"/>.

To understand the reactions and successfully predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
The only information necessary is the behaviour of the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on the following assumptions:
* It is possible to separate the motion of the collision from the other motions of the TS<ref name="tunneling" />.
* Energy distributions is in accordance with the Maxwell-Boltzmann distributionc<ref name="tunneling" />.
* it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling" />.
* the reaction is treated classically and the quantum effects are ignored<ref name="tunneling" />.

=== Exercise 1 ===

For the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

'''Q1''': On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Inter-nuclear distances vs time for the transition state.]]
'''Q2''': The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen (A or C) is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the vibrational energy is zero due to the absence of bonds. The value was obtained by trial and error: the first distance chosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from a periodic vibration ( at 150 pm) to being stationary at 90.8 pm. This can also be observed in the “Inter-nuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

'''Q3''': A MEP and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. The dynamics calculation resulted in a longer distance between atom B and C once the reaction finished: the reaction rolls toward the products.

If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This is illustrated by the following plots:
* in the Inter-nuclear distance vs time plot, the initial value of AB is equal to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
* in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in different ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]][[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Inter-nuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in following table was obtained. Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation ended.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
[[File:forming_dist_01512921.png|thumb|left|Plot of the Inter-nuclear distances vs time. The reaction reaches the transition state.]]

'''Q4''': for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product is not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction proceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enough to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were formed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

'''Q5''':The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doesn't take into consideration the possibility of quantum tunnelling, as the conventional transition state theory is purely classical motion. <ref name="tunneling"/>
In fact, if the system tunnels through the PES, then the kinetic energy could be lower than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
'''Q1:'''

F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
* The reaction is exothermic as the energy of the reagents is higher that that of the products.

* Position of TS: AB= 74.5 pm and BC = 181pm .
* It was identified thanks to Hammond's postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds" />. In this case, the transition state is early, as the reaction is exothermic. Therefore, the TS will resemble the reagents and the separation between the hydrogen molecule will be smaller than that of HF
* Total energy-433.98 KJ mol-1.
* Energy of reagent:-560.592 KJ mol-1
[[File:energy_H2_01512921.png|thumb|right|Energy vs Time MEP. The transition state rolls to the reagent H2]]
* Activation energy is: -126.612
H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
* The reaction is endothermic as the energy of the reagents is lower that that of the products.
* Position of TS: HF = 95 pm, H2=250 pm
* Energy is -433.98 KJ mol-1
* Reagent energy: -434.012 KJ mol-1
* Activation energy: -0.032 KJ mol-1
[[File:energy_HF_01512921.png|thumb|left|Energy vs Time mep. The transition state rolls to the reagent HF]]

Strength of the bonds:
Strength of H2 = 436 KJ mol

Strength of HF = 569 KJ mol

When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

<nowiki> </nowiki>'''Q2,''' part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy" />. This can be confirmed by spectroscopic methods like infrared, as it would be possible to see overtones due to the transition from the first to the second vibrational excited state. A reactive trajectory was found at r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1. While the changing value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with high vibrational energy in the products. At 3.1, the reaction has a barrier recrossing, where the product forms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reaction is not successful. Barrier recrossing is also seen at 6.1, with a successful collision. At 5.1 g.mol-1.pm.fs-1, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb exactly in the middle between Ha and F, not showing which reaction's side is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1 and pHH=0.2 g.mol-1.pm.fs-1. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

'''Q2,''' part 2: FH + H ==> H2 +F

A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1. The skew plot of the reaction is shown.

[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

'''Q5''': The energy distribution of the products can be determined by locating the energy barrier on the reaction coordinate (The TS). For a reaction is A+ BC:
* an early barrier or TS will result in high vibrational energy in the products. In an exothermic reaction, the energy is released while AB distance is changing

* a late barrier results in low vibrational energy in the products. The energy is released after AB is formed and BC is changing, which corresponds to the formation of the products and the translational energy.
The position of the energy barrier would also help selecting what distribution of reactant energy is most likely to lead to a reaction.
* for an early TS, a molecule with high translational energy will be able to overcome the barrier, as all its motion along the reaction coordinates. On the other hand, a molecule with high vibration will not have enough energy to reach the barrier.
* for a late TS, the barrier will be overcome by vibrational energy rather than translational. In fact, a molecule with high translational energy will crash in the inner wall of the PES and bounce back<ref name="polanyi's" />.

=== Conclusions===
The investigation of the reaction dynamics for the three reactions examined in the lab concerned with the PES and the reaction trajectories.
Using the program, it was possible to identify the transition state of the reaction (using Hammond's postulate), the trajectories (both reactive and unreactive ones) and the activation energies (thank to the energy of the transition state and the of the reactants).
A comparison was made between the rate constant from the canonical transition state theory and the experimental one. It was noted how, due t the absence of tunnelling, the CTST would overestimate the rate of the reaction.
By analysing the transition state, it was possible to observe how its position influences which energy, translational or vibrational, is required for the reaction to happen. An early transition state can be overcome with translational energy, while a late one will prefer high vibrational energy.

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
<ref name="polanyi's">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 9, pp 272-274. </ref>
</references>

MRD:01512921

2020-05-22T18:03:55Z

Sp3418:

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and inter-nuclear distances affected the outcome of the reaction. It was also observed how the position of the transition state determines which type of energy, vibrational or translational, will be sufficient to overcome the energy barrier.

Chemical reactions can be simulated by taking into consideration the relative position, such as the distances, between the atom involved in the reaction. In fact, the interaction between the atoms that cause their motion depend on their location and are described as a PES, a function relative to the coordinates of the atoms <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision coordinates; for the systems analysed in this report, there are only two coordinates: r1, distances between atoms of the reacting molecule and r2<ref name="intro1"/>, distance between the atoms of the product molecule. The path of the reaction can be mapped with a reaction trajectory. A reactive trajectory will pass through a saddle point of the PES, also know as transition state(TS<ref name="intro1"/>). On the contrary, an unreactive trajectory will roll back toward the reagents upon meeting the TS<ref name="intro1"/>.

To understand the reactions and successfully predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
The only information necessary is the behaviour of the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on the following assumptions:
* It is possible to separate the motion of the collision from the other motions of the TS<ref name="tunneling" />.
* Energy distributions is in accordance with the Maxwell-Boltzmann distributionc<ref name="tunneling" />.
* it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling" />.
* the reaction is treated classically and the quantum effects are ignored<ref name="tunneling" />.

=== Exercise 1 ===

For the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

'''Q1''': On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Inter-nuclear distances vs time for the transition state.]]
'''Q2''': The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen (A or C) is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the vibrational energy is zero due to the absence of bonds. The value was obtained by trial and error: the first distance chosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from a periodic vibration ( at 150 pm) to being stationary at 90.8 pm. This can also be observed in the “Inter-nuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

'''Q3''': A MEP and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. The dynamics calculation resulted in a longer distance between atom B and C once the reaction finished: the reaction rolls toward the products.

If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This is illustrated by the following plots:
* in the Inter-nuclear distance vs time plot, the initial value of AB is equal to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
* in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in different ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]][[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Inter-nuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in following table was obtained. Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation ended.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
[[File:forming_dist_01512921.png|thumb|centre|Plot of the Inter-nuclear distances vs time. The reaction reaches the transition state.]]

'''Q4''': for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product is not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction proceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enough to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were formed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

'''Q5''':The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doesn't take into consideration the possibility of quantum tunnelling, as the conventional transition state theory is purely classical motion. <ref name="tunneling"/>
In fact, if the system tunnels through the PES, then the kinetic energy could be lower than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
'''Q1:'''

F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
* The reaction is exothermic as the energy of the reagents is higher that that of the products.

* Position of TS: AB= 74.5 pm and BC = 181pm .
* It was identified thanks to Hammond's postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds" />. In this case, the transition state is early, as the reaction is exothermic. Therefore, the TS will resemble the reagents and the separation between the hydrogen molecule will be smaller than that of HF
* Total energy-433.98 KJ mol-1.
* Energy of reagent:-560.592 KJ mol-1
[[File:energy_H2_01512921.png|thumb|right|Energy vs Time MEP. The transition state rolls to the reagent H2]]
* Activation energy is: -126.612
H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
* The reaction is endothermic as the energy of the reagents is lower that that of the products.
* Position of TS: HF = 95 pm, H2=250 pm
* Energy is -433.98 KJ mol-1
* Reagent energy: -434.012 KJ mol-1
* Activation energy: -0.032 KJ mol-1
[[File:energy_HF_01512921.png|thumb|left|Energy vs Time mep. The transition state rolls to the reagent HF]]

Strength of the bonds:
Strength of H2 = 436 KJ mol

Strength of HF = 569 KJ mol

When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

<nowiki> </nowiki>'''Q2,''' part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy" />. This can be confirmed by spectroscopic methods like infrared, as it would be possible to see overtones due to the transition from the first to the second vibrational excited state. A reactive trajectory was found at r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1. While the changing value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with high vibrational energy in the products. At 3.1, the reaction has a barrier recrossing, where the product forms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reaction is not successful. Barrier recrossing is also seen at 6.1, with a successful collision. At 5.1 g.mol-1.pm.fs-1, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb exactly in the middle between Ha and F, not showing which reaction's side is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1 and pHH=0.2 g.mol-1.pm.fs-1. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

'''Q2,''' part 2: FH + H ==> H2 +F

A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1. The skew plot of the reaction is shown.

[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

'''Q5''': The energy distribution of the products can be determined by locating the energy barrier on the reaction coordinate (The TS). For a reaction is A+ BC:
* an early barrier or TS will result in high vibrational energy in the products. In an exothermic reaction, the energy is released while AB distance is changing

* a late barrier results in low vibrational energy in the products. The energy is released after AB is formed and BC is changing, which corresponds to the formation of the products and the translational energy.
The position of the energy barrier would also help selecting what distribution of reactant energy is most likely to lead to a reaction.
* for an early TS, a molecule with high translational energy will be able to overcome the barrier, as all its motion along the reaction coordinates. On the other hand, a molecule with high vibration will not have enough energy to reach the barrier.
* for a late TS, the barrier will be overcome by vibrational energy rather than translational. In fact, a molecule with high translational energy will crash in the inner wall of the PES and bounce back<ref name="polanyi's" />.

=== Conclusions===
The investigation of the reaction dynamics for the three reactions examined in the lab concerned with the PES and the reaction trajectories.
Using the program, it was possible to identify the transition state of the reaction (using Hammond's postulate), the trajectories (both reactive and unreactive ones) and the activation energies (thank to the energy of the transition state and the of the reactants).
A comparison was made between the rate constant from the canonical transition state theory and the experimental one. It was noted how, due t the absence of tunnelling, the CTST would overestimate the rate of the reaction.
By analysing the transition state, it was possible to observe how its position influences which energy, translational or vibrational, is required for the reaction to happen. An early transition state can be overcome with translational energy, while a late one will prefer high vibrational energy.

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
<ref name="polanyi's">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 9, pp 272-274. </ref>
</references>

MRD:01512921

2020-05-22T18:01:15Z

Sp3418: /* Conclusions */

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and inter-nuclear distances affected the outcome of the reaction. It was also observed how the position of the transition state determines which type of energy, vibrational or translational, will be sufficient to overcome the energy barrier.

Chemical reactions can be simulated by taking into consideration the relative position, such as the distances, between the atom involved in the reaction. In fact, the interaction between the atoms that cause their motion depend on their location and are described as a PES, a function relative to the coordinates of the atoms <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision coordinates; for the systems analysed in this report, there are only two coordinates: r1, distances between atoms of the reacting molecule and r2<ref name="intro1"/>, distance between the atoms of the product molecule. The path of the reaction can be mapped with a reaction trajectory. A reactive trajectory will pass through a saddle point of the PES, also know as transition state(TS<ref name="intro1"/>). On the contrary, an unreactive trajectory will roll back toward the reagents upon meeting the TS<ref name="intro1"/>.

To understand the reactions and successfully predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
The only information necessary is the behaviour of the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on the following assumptions:
* It is possible to separate the motion of the collision from the other motions of the TS<ref name="tunneling" />.
* Energy distributions is in accordance with the Maxwell-Boltzmann distributionc<ref name="tunneling" />.
* it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling" />.
* the reaction is treated classically and the quantum effects are ignored<ref name="tunneling" />.

=== Exercise 1 ===

For the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Inter-nuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen (A or C) is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the vibrational energy is zero due to the absence of bonds. The value was obtained by trial and error: the first distance chosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from a periodic vibration ( at 150 pm) to being stationary at 90.8 pm. This can also be observed in the “Inter-nuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A MEP and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. The dynamics calculation resulted in a longer distance between atom B and C once the reaction finished: the reaction rolls toward the products.

If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This is illustrated by the following plots:

- in the Inter-nuclear distance vs time plot, the initial value of AB is equal to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in different ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]][[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Inter-nuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation ended. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Inter-nuclear distances vs time. The reaction reaches the transition state.]]

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product is not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction proceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enough to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were formed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doesn't take into consideration the possibility of quantum tunnelling, as the conventional transition state theory is purely classical motion. <ref name="tunneling"/>
In fact, if the system tunnels through the PES, then the kinetic energy could be lower than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:

F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
* The reaction is exothermic as the energy of the reagents is higher that that of the products.

* Position of TS: AB= 74.5 pm and BC = 181pm .
* It was identified thanks to Hammond's postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds" />. In this case, the transition state is early, as the reaction is exothermic. Therefore, the TS will resemble the reagents and the separation between the hydrogen molecule will be smaller than that of HF
* Total energy-433.98 KJ mol-1.
* Energy of reagent:-560.592 KJ mol-1
[[File:energy_H2_01512921.png|thumb|right|Energy vs Time MEP. The transition state rolls to the reagent H2]]
* Activation energy is: -126.612
H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
* The reaction is endothermic as the energy of the reagents is lower that that of the products.
* Position of TS: HF = 95 pm, H2=250 pm
* Energy is -433.98 KJ mol-1
* Reagent energy: -434.012 KJ mol-1
* Activation energy: -0.032 KJ mol-1
[[File:energy_HF_01512921.png|thumb|left|Energy vs Time mep. The transition state rolls to the reagent HF]]

Strength of the bonds:
Strength of H2 = 436 KJ mol

Strength of HF = 569 KJ mol

When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

<nowiki> </nowiki>Q2, part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy" />. This can be confirmed by spectroscopic methods like infrared, as it would be possible to see overtones due to the transition from the first to the second vibrational excited state. A reactive trajectory was found at r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1. While the changing value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with high vibrational energy in the products. At 3.1, the reaction has a barrier recrossing, where the product forms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reaction is not successful. Barrier recrossing is also seen at 6.1, with a successful collision. At 5.1 g.mol-1.pm.fs-1, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb exactly in the middle between Ha and F, not showing which reaction's side is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1 and pHH=0.2 g.mol-1.pm.fs-1. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

Q2, part 2: FH + H ==> H2 +F

A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1. The skew plot of the reaction is shown.
[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

Q5: The energy distribution of the products can be determined by locating the energy barrier on the reaction coordinate (The TS). For a reaction is A+ BC:
* an early barrier or TS will result in high vibrational energy in the products. In an exothermic reaction, the energy is released while AB distance is changing

* a late barrier results in low vibrational energy in the products. The energy is released after AB is formed and BC is changing, which corresponds to the formation of the products and the translational energy.
The position of the energy barrier would also help selecting what distribution of reactant energy is most likely to lead to a reaction.
* for an early TS, a molecule with high translational energy will be able to overcome the barrier, as all its motion along the reaction coordinates. On the other hand, a molecule with high vibration will not have enough energy to reach the barrier.
* for a late TS, the barrier will be overcome by vibrational energy rather than translational. In fact, a molecule with high translational energy will crash in the inner wall of the PES and bounce back<ref name="polanyi's" />.

=== Conclusions===
The investigation of the reaction dynamics for the three reactions examined in the lab concerned with the PES and the reaction trajectories.
Using the program, it was possible to identify the transition state of the reaction (using Hammond's postulate), the trajectories (both reactive and unreactive ones) and the activation energies (thank to the energy of the transition state and the of the reactants).
A comparison was made between the rate constant from the canonical transition state theory and the experimental one. It was noted how, due t the absence of tunnelling, the CTST would overestimate the rate of the reaction.
By analysing the transition state, it was possible to observe how its position influences which energy, translational or vibrational, is required for the reaction to happen. An early transition state can be overcome with translational energy, while a late one will prefer high vibrational energy.

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
<ref name="polanyi's">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 9, pp 272-274. </ref>
</references>

MRD:01512921

2020-05-22T18:00:44Z

Sp3418: /* Exercise 2 */

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and inter-nuclear distances affected the outcome of the reaction. It was also observed how the position of the transition state determines which type of energy, vibrational or translational, will be sufficient to overcome the energy barrier.

Chemical reactions can be simulated by taking into consideration the relative position, such as the distances, between the atom involved in the reaction. In fact, the interaction between the atoms that cause their motion depend on their location and are described as a PES, a function relative to the coordinates of the atoms <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision coordinates; for the systems analysed in this report, there are only two coordinates: r1, distances between atoms of the reacting molecule and r2<ref name="intro1"/>, distance between the atoms of the product molecule. The path of the reaction can be mapped with a reaction trajectory. A reactive trajectory will pass through a saddle point of the PES, also know as transition state(TS<ref name="intro1"/>). On the contrary, an unreactive trajectory will roll back toward the reagents upon meeting the TS<ref name="intro1"/>.

To understand the reactions and successfully predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
The only information necessary is the behaviour of the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on the following assumptions:
* It is possible to separate the motion of the collision from the other motions of the TS<ref name="tunneling" />.
* Energy distributions is in accordance with the Maxwell-Boltzmann distributionc<ref name="tunneling" />.
* it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling" />.
* the reaction is treated classically and the quantum effects are ignored<ref name="tunneling" />.

=== Exercise 1 ===

For the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Inter-nuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen (A or C) is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the vibrational energy is zero due to the absence of bonds. The value was obtained by trial and error: the first distance chosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from a periodic vibration ( at 150 pm) to being stationary at 90.8 pm. This can also be observed in the “Inter-nuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A MEP and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. The dynamics calculation resulted in a longer distance between atom B and C once the reaction finished: the reaction rolls toward the products.

If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This is illustrated by the following plots:

- in the Inter-nuclear distance vs time plot, the initial value of AB is equal to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in different ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]][[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Inter-nuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation ended. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Inter-nuclear distances vs time. The reaction reaches the transition state.]]

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product is not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction proceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enough to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were formed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doesn't take into consideration the possibility of quantum tunnelling, as the conventional transition state theory is purely classical motion. <ref name="tunneling"/>
In fact, if the system tunnels through the PES, then the kinetic energy could be lower than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:

F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
* The reaction is exothermic as the energy of the reagents is higher that that of the products.

* Position of TS: AB= 74.5 pm and BC = 181pm .
* It was identified thanks to Hammond's postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds" />. In this case, the transition state is early, as the reaction is exothermic. Therefore, the TS will resemble the reagents and the separation between the hydrogen molecule will be smaller than that of HF
* Total energy-433.98 KJ mol-1.
* Energy of reagent:-560.592 KJ mol-1
[[File:energy_H2_01512921.png|thumb|right|Energy vs Time MEP. The transition state rolls to the reagent H2]]
* Activation energy is: -126.612
H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
* The reaction is endothermic as the energy of the reagents is lower that that of the products.
* Position of TS: HF = 95 pm, H2=250 pm
* Energy is -433.98 KJ mol-1
* Reagent energy: -434.012 KJ mol-1
* Activation energy: -0.032 KJ mol-1
[[File:energy_HF_01512921.png|thumb|left|Energy vs Time mep. The transition state rolls to the reagent HF]]

Strength of the bonds:
Strength of H2 = 436 KJ mol

Strength of HF = 569 KJ mol

When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

<nowiki> </nowiki>Q2, part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy" />. This can be confirmed by spectroscopic methods like infrared, as it would be possible to see overtones due to the transition from the first to the second vibrational excited state. A reactive trajectory was found at r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1. While the changing value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with high vibrational energy in the products. At 3.1, the reaction has a barrier recrossing, where the product forms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reaction is not successful. Barrier recrossing is also seen at 6.1, with a successful collision. At 5.1 g.mol-1.pm.fs-1, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb exactly in the middle between Ha and F, not showing which reaction's side is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1 and pHH=0.2 g.mol-1.pm.fs-1. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

Q2, part 2: FH + H ==> H2 +F

A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1. The skew plot of the reaction is shown.
[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

Q5: The energy distribution of the products can be determined by locating the energy barrier on the reaction coordinate (The TS). For a reaction is A+ BC:
* an early barrier or TS will result in high vibrational energy in the products. In an exothermic reaction, the energy is released while AB distance is changing

* a late barrier results in low vibrational energy in the products. The energy is released after AB is formed and BC is changing, which corresponds to the formation of the products and the translational energy.
The position of the energy barrier would also help selecting what distribution of reactant energy is most likely to lead to a reaction.
* for an early TS, a molecule with high translational energy will be able to overcome the barrier, as all its motion along the reaction coordinates. On the other hand, a molecule with high vibration will not have enough energy to reach the barrier.
* for a late TS, the barrier will be overcome by vibrational energy rather than translational. In fact, a molecule with high translational energy will crash in the inner wall of the PES and bounce back<ref name="polanyi's" />.

=== Conclusions===
The investigation of the reaction dynamics for the three reactions examined in the lab concerned with the PES and the reaction trajectories.
Using the program, it was possible to identify the transition state of the reaction (using Hammond's postulate), the trajectories (both reactive and unreactive ones) and the activation energies (thank to the energy of the trasition state and thta of the reactants).
A comparison was made between the rate constant from the canonical transition state theory and the experimental one. It was noted how, due t the absence of tunneling, the CTST would overestimate the rate of the reaction.
By analysing the transition state, it was possible to observe how its position influences which energy, translational or vibrational, is required for the reaction to happen. An early transition state can be overcome with translational energy, while a late one will prefer high vibrational enegry.

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
<ref name="polanyi's">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 9, pp 272-274. </ref>
</references>

MRD:01512921

2020-05-22T17:57:43Z

Sp3418: /* Exercise 1 */

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and inter-nuclear distances affected the outcome of the reaction. It was also observed how the position of the transition state determines which type of energy, vibrational or translational, will be sufficient to overcome the energy barrier.

Chemical reactions can be simulated by taking into consideration the relative position, such as the distances, between the atom involved in the reaction. In fact, the interaction between the atoms that cause their motion depend on their location and are described as a PES, a function relative to the coordinates of the atoms <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision coordinates; for the systems analysed in this report, there are only two coordinates: r1, distances between atoms of the reacting molecule and r2<ref name="intro1"/>, distance between the atoms of the product molecule. The path of the reaction can be mapped with a reaction trajectory. A reactive trajectory will pass through a saddle point of the PES, also know as transition state(TS<ref name="intro1"/>). On the contrary, an unreactive trajectory will roll back toward the reagents upon meeting the TS<ref name="intro1"/>.

To understand the reactions and successfully predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
The only information necessary is the behaviour of the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on the following assumptions:
* It is possible to separate the motion of the collision from the other motions of the TS<ref name="tunneling" />.
* Energy distributions is in accordance with the Maxwell-Boltzmann distributionc<ref name="tunneling" />.
* it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling" />.
* the reaction is treated classically and the quantum effects are ignored<ref name="tunneling" />.

=== Exercise 1 ===

For the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Inter-nuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen (A or C) is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the vibrational energy is zero due to the absence of bonds. The value was obtained by trial and error: the first distance chosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from a periodic vibration ( at 150 pm) to being stationary at 90.8 pm. This can also be observed in the “Inter-nuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A MEP and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. The dynamics calculation resulted in a longer distance between atom B and C once the reaction finished: the reaction rolls toward the products.

If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This is illustrated by the following plots:

- in the Inter-nuclear distance vs time plot, the initial value of AB is equal to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in different ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]][[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Inter-nuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation ended. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Inter-nuclear distances vs time. The reaction reaches the transition state.]]

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enough kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product is not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction proceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enough to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were formed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doesn't take into consideration the possibility of quantum tunnelling, as the conventional transition state theory is purely classical motion. <ref name="tunneling"/>
In fact, if the system tunnels through the PES, then the kinetic energy could be lower than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:

F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
* The reaction is exothermic as the energy of the reagents is higher that that of the products.

* Position of TS: AB= 74.5 pm and BC = 181pm .
* It was identified thanks to hammonds postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds" />. In this case, the transition state is early, as the reaction is exothermic. Therefore, the TS will resemble the reagents and the separation between the hydrogen molecule will be smaller than that of HF
* Total energy-433.98 KJ mol-1.
* Energy of reagent:-560.592 KJ mol-1
[[File:energy_H2_01512921.png|thumb|right|Energy vs Time mep. The transition state rolls to the reagent H2]]
* Activation energy is: -126.612
H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
* The reaction is endothermic as the enrgy of the reagents is lower that that of the products.
* Position of TS: HF = 95 pm, H2=250 pm
* Energy is -433.98 KJ mol-1
* Reagent energy: -434.012 KJ mol-1
* Activation enegry: -0.032 KJ mol-1
[[File:energy_HF_01512921.png|thumb|left|Energy vs Time mep. The transition state rolls to the reagent HF]]

Strenght of the bonds:
Strength of H2 = 436 KJ mol

Strength of HF = 569 KJ mol

When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

<nowiki> </nowiki>Q2, part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy" />. This can be confirmed by spectroscopic methods like infrared, as it would be possible to see overtones due to the transition from the first to the second vibrational excited state. A reactive trajcetory was found at r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1. While varing the value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with hight vibrational energy in the products. At 3.1, the reaction has a barrier recrossing, where the product forms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reation is not successful. Barrier recrossing is alwo seen at 6.1, with a successful collision. At 5.1 g.mol-1.pm.fs-1, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb extacly in the middle between Ha and F, not showing which reaction's side is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1 and pHH=0.2 g.mol-1.pm.fs-1. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

Q2, part 2: FH + H ==> H2 +F

A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1. The skew plot of the reaction is shown.
[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

Q5: The energy distribution of the products can be determined by locating the energy barrier on the raction coordinate (The TS). For a reaction is A+ BC:
* an early barrier or TS will result in high vibrational energy in the products. In an exothermic reaction, the energy is released while AB distance is changing

* a late barrier results in low vibrational energy in the products. The energy is releaved after AB is formed and BC is changing, which corresponds to the formation of the products and corresponds to the translational energy.
The position of the energy barrier would also help sleecting what distribution o freactant energy is most likely to lead to a reaction.
* for an early TS, a molecule with high translational energy will be able to overcome the barrier, as it will thabe all its motion along the reaction coordinates. On the other hand, a molcule with high vibration will not have enough energy to reach the barrier.
* for a late TS, the barrier will be overcome by vibrational energy rather than translational. In fact, a molecule with high translational energy will crash in the inner wall of the PES and bounce back<ref name="polanyi's" />.

=== Conclusions===
The investigation of the reaction dynamics for the three reactions examined in the lab concerned with the PES and the reaction trajectories.
Using the program, it was possible to identify the transition state of the reaction (using Hammond's postulate), the trajectories (both reactive and unreactive ones) and the activation energies (thank to the energy of the trasition state and thta of the reactants).
A comparison was made between the rate constant from the canonical transition state theory and the experimental one. It was noted how, due t the absence of tunneling, the CTST would overestimate the rate of the reaction.
By analysing the transition state, it was possible to observe how its position influences which energy, translational or vibrational, is required for the reaction to happen. An early transition state can be overcome with translational energy, while a late one will prefer high vibrational enegry.

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
<ref name="polanyi's">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 9, pp 272-274. </ref>
</references>

MRD:01512921

2020-05-22T17:55:34Z

Sp3418: /* Introduction */

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and inter-nuclear distances affected the outcome of the reaction. It was also observed how the position of the transition state determines which type of energy, vibrational or translational, will be sufficient to overcome the energy barrier.

Chemical reactions can be simulated by taking into consideration the relative position, such as the distances, between the atom involved in the reaction. In fact, the interaction between the atoms that cause their motion depend on their location and are described as a PES, a function relative to the coordinates of the atoms <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision coordinates; for the systems analysed in this report, there are only two coordinates: r1, distances between atoms of the reacting molecule and r2<ref name="intro1"/>, distance between the atoms of the product molecule. The path of the reaction can be mapped with a reaction trajectory. A reactive trajectory will pass through a saddle point of the PES, also know as transition state(TS<ref name="intro1"/>). On the contrary, an unreactive trajectory will roll back toward the reagents upon meeting the TS<ref name="intro1"/>.

To understand the reactions and successfully predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
The only information necessary is the behaviour of the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on the following assumptions:
* It is possible to separate the motion of the collision from the other motions of the TS<ref name="tunneling" />.
* Energy distributions is in accordance with the Maxwell-Boltzmann distributionc<ref name="tunneling" />.
* it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling" />.
* the reaction is treated classically and the quantum effects are ignored<ref name="tunneling" />.

=== Exercise 1 ===

For the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Internuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen (A or C) is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the virational energy is zero due to the absence of bonds. The value was obtained by trial and error: the first distance choosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from a peridodic vibration ( at 150 pm) to being stationaty at 90.8 pm. This can also be observed in the “Internuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A mep and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. The dynamics calculation resulted in a longer distance between atom B and C once the reaction finished: the reaction rolls toward the products.

If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This is illustrated by the following plots:

- in the Internuclear distance vs time plot, the initial value of AB is equale to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in differetn ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]][[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Internuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation ended. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Internuclear distances vs time. The reaction reaches the transition state.]]

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product is not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction proceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enought to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were ormed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doen't take into consideration the possibility of quantum tunnelling, as the convetional transition state theory is purerly classical motion. <ref name="tunneling"/>
Infact, if the system tunnels throught the PES, then the kinetic energy could be lowere than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:

F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
* The reaction is exothermic as the energy of the reagents is higher that that of the products.

* Position of TS: AB= 74.5 pm and BC = 181pm .
* It was identified thanks to hammonds postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds" />. In this case, the transition state is early, as the reaction is exothermic. Therefore, the TS will resemble the reagents and the separation between the hydrogen molecule will be smaller than that of HF
* Total energy-433.98 KJ mol-1.
* Energy of reagent:-560.592 KJ mol-1
[[File:energy_H2_01512921.png|thumb|right|Energy vs Time mep. The transition state rolls to the reagent H2]]
* Activation energy is: -126.612
H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
* The reaction is endothermic as the enrgy of the reagents is lower that that of the products.
* Position of TS: HF = 95 pm, H2=250 pm
* Energy is -433.98 KJ mol-1
* Reagent energy: -434.012 KJ mol-1
* Activation enegry: -0.032 KJ mol-1
[[File:energy_HF_01512921.png|thumb|left|Energy vs Time mep. The transition state rolls to the reagent HF]]

Strenght of the bonds:
Strength of H2 = 436 KJ mol

Strength of HF = 569 KJ mol

When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

<nowiki> </nowiki>Q2, part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy" />. This can be confirmed by spectroscopic methods like infrared, as it would be possible to see overtones due to the transition from the first to the second vibrational excited state. A reactive trajcetory was found at r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1. While varing the value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with hight vibrational energy in the products. At 3.1, the reaction has a barrier recrossing, where the product forms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reation is not successful. Barrier recrossing is alwo seen at 6.1, with a successful collision. At 5.1 g.mol-1.pm.fs-1, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb extacly in the middle between Ha and F, not showing which reaction's side is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1 and pHH=0.2 g.mol-1.pm.fs-1. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

Q2, part 2: FH + H ==> H2 +F

A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1. The skew plot of the reaction is shown.
[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

Q5: The energy distribution of the products can be determined by locating the energy barrier on the raction coordinate (The TS). For a reaction is A+ BC:
* an early barrier or TS will result in high vibrational energy in the products. In an exothermic reaction, the energy is released while AB distance is changing

* a late barrier results in low vibrational energy in the products. The energy is releaved after AB is formed and BC is changing, which corresponds to the formation of the products and corresponds to the translational energy.
The position of the energy barrier would also help sleecting what distribution o freactant energy is most likely to lead to a reaction.
* for an early TS, a molecule with high translational energy will be able to overcome the barrier, as it will thabe all its motion along the reaction coordinates. On the other hand, a molcule with high vibration will not have enough energy to reach the barrier.
* for a late TS, the barrier will be overcome by vibrational energy rather than translational. In fact, a molecule with high translational energy will crash in the inner wall of the PES and bounce back<ref name="polanyi's" />.

=== Conclusions===
The investigation of the reaction dynamics for the three reactions examined in the lab concerned with the PES and the reaction trajectories.
Using the program, it was possible to identify the transition state of the reaction (using Hammond's postulate), the trajectories (both reactive and unreactive ones) and the activation energies (thank to the energy of the trasition state and thta of the reactants).
A comparison was made between the rate constant from the canonical transition state theory and the experimental one. It was noted how, due t the absence of tunneling, the CTST would overestimate the rate of the reaction.
By analysing the transition state, it was possible to observe how its position influences which energy, translational or vibrational, is required for the reaction to happen. An early transition state can be overcome with translational energy, while a late one will prefer high vibrational enegry.

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
<ref name="polanyi's">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 9, pp 272-274. </ref>
</references>

MRD:01512921

2020-05-22T17:52:45Z

Sp3418:

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and internuclear distances affected the outcome of the reaction. It was also observed how the position of the transition state determiens which type of energy, vibrational or translational, will be suffiect to overcome the energy barrier.

Chemical reactions can be simulated by taking into consideration the relative position, such as the distances, between the atom involved in the reaction. In fact, the interaction between the atoms that cause their motion depend on their location and are described as a PES, a functionrelative to the coordinates of the costituent atoms <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision coordinates; for the systems analysed in this report, there are only two coordinates: r1, distances between atoms of the reacting molecule and r2<ref name="intro1"/>, distance betweem the atoms of the product molecule. The path of the reaction can be mappes with a reaction trajectory. A reactiv trajectory will pass through a saddle point of the PES, also know as transition state(TS<ref name="intro1"/>). On the contrary, an unreactive trajectory will roll back toward the reagents upon meeting the TS<ref name="intro1"/>.

To understand the reactions and successfuly predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
The only information necessary is the behaviour of the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on the following assumptions:
* It is possible to separate the motion of the collision from the other motions of the TS<ref name="tunneling" />.
* Energy distributions is in accordacne with the Maxwell-Boltzmann distributionc<ref name="tunneling" />.
* it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling" />.
* the reaction is treated classically and the quantum effects are ignored<ref name="tunneling" />.

=== Exercise 1 ===

For the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Internuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen (A or C) is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the virational energy is zero due to the absence of bonds. The value was obtained by trial and error: the first distance choosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from a peridodic vibration ( at 150 pm) to being stationaty at 90.8 pm. This can also be observed in the “Internuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A mep and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. The dynamics calculation resulted in a longer distance between atom B and C once the reaction finished: the reaction rolls toward the products.

If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This is illustrated by the following plots:

- in the Internuclear distance vs time plot, the initial value of AB is equale to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in differetn ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]][[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Internuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation ended. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Internuclear distances vs time. The reaction reaches the transition state.]]

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product is not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction proceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enought to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were ormed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doen't take into consideration the possibility of quantum tunnelling, as the convetional transition state theory is purerly classical motion. <ref name="tunneling"/>
Infact, if the system tunnels throught the PES, then the kinetic energy could be lowere than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:

F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
* The reaction is exothermic as the energy of the reagents is higher that that of the products.

* Position of TS: AB= 74.5 pm and BC = 181pm .
* It was identified thanks to hammonds postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds" />. In this case, the transition state is early, as the reaction is exothermic. Therefore, the TS will resemble the reagents and the separation between the hydrogen molecule will be smaller than that of HF
* Total energy-433.98 KJ mol-1.
* Energy of reagent:-560.592 KJ mol-1
[[File:energy_H2_01512921.png|thumb|right|Energy vs Time mep. The transition state rolls to the reagent H2]]
* Activation energy is: -126.612
H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
* The reaction is endothermic as the enrgy of the reagents is lower that that of the products.
* Position of TS: HF = 95 pm, H2=250 pm
* Energy is -433.98 KJ mol-1
* Reagent energy: -434.012 KJ mol-1
* Activation enegry: -0.032 KJ mol-1
[[File:energy_HF_01512921.png|thumb|left|Energy vs Time mep. The transition state rolls to the reagent HF]]

Strenght of the bonds:
Strength of H2 = 436 KJ mol

Strength of HF = 569 KJ mol

When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

<nowiki> </nowiki>Q2, part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy" />. This can be confirmed by spectroscopic methods like infrared, as it would be possible to see overtones due to the transition from the first to the second vibrational excited state. A reactive trajcetory was found at r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1. While varing the value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with hight vibrational energy in the products. At 3.1, the reaction has a barrier recrossing, where the product forms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reation is not successful. Barrier recrossing is alwo seen at 6.1, with a successful collision. At 5.1 g.mol-1.pm.fs-1, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb extacly in the middle between Ha and F, not showing which reaction's side is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1 and pHH=0.2 g.mol-1.pm.fs-1. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

Q2, part 2: FH + H ==> H2 +F

A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1. The skew plot of the reaction is shown.
[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

Q5: The energy distribution of the products can be determined by locating the energy barrier on the raction coordinate (The TS). For a reaction is A+ BC:
* an early barrier or TS will result in high vibrational energy in the products. In an exothermic reaction, the energy is released while AB distance is changing

* a late barrier results in low vibrational energy in the products. The energy is releaved after AB is formed and BC is changing, which corresponds to the formation of the products and corresponds to the translational energy.
The position of the energy barrier would also help sleecting what distribution o freactant energy is most likely to lead to a reaction.
* for an early TS, a molecule with high translational energy will be able to overcome the barrier, as it will thabe all its motion along the reaction coordinates. On the other hand, a molcule with high vibration will not have enough energy to reach the barrier.
* for a late TS, the barrier will be overcome by vibrational energy rather than translational. In fact, a molecule with high translational energy will crash in the inner wall of the PES and bounce back<ref name="polanyi's" />.

=== Conclusions===
The investigation of the reaction dynamics for the three reactions examined in the lab concerned with the PES and the reaction trajectories.
Using the program, it was possible to identify the transition state of the reaction (using Hammond's postulate), the trajectories (both reactive and unreactive ones) and the activation energies (thank to the energy of the trasition state and thta of the reactants).
A comparison was made between the rate constant from the canonical transition state theory and the experimental one. It was noted how, due t the absence of tunneling, the CTST would overestimate the rate of the reaction.
By analysing the transition state, it was possible to observe how its position influences which energy, translational or vibrational, is required for the reaction to happen. An early transition state can be overcome with translational energy, while a late one will prefer high vibrational enegry.

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
<ref name="polanyi's">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 9, pp 272-274. </ref>
</references>

MRD:01512921

2020-05-22T17:47:40Z

Sp3418: /* Introduction */

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and internuclear distances affected the outcome of the reaction. It was also observed how the position of the transition state determiens which type of energy, vibrational or translational, will be suffiect to overcome the energy barrier.

Chemical reactions can be simulated by taking into consideration the relative position, such as the distances, between the atom involved in the reaction. In fact, the interaction between the atoms that cause their motion depend on their location and are described as a PES, a functionrelative to the coordinates of the costituent atoms <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision coordinates; for the systems analysed in this report, there are only two coordinates: r1, distances between atoms of the reacting molecule and r2<ref name="intro1"/>, distance betweem the atoms of the product molecule. The path of the reaction can be mappes with a reaction trajectory. A reactiv trajectory will pass through a saddle point of the PES, also know as transition state(TS). On the contrary, an unreactive trajectory will roll back toward the reagents upon meeting the TS.

To understand the reactions and successfuly predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
This is possible thanks to only few informations about the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on different assumptions:

* it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling" />.
* the reaction is treated classically and the quantum effects are ignored<ref name="tunneling" />.

=== Exercise 1 ===

For the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Internuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen (A or C) is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the virational energy is zero due to the absence of bonds. The value was obtained by trial and error: the first distance choosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from a peridodic vibration ( at 150 pm) to being stationaty at 90.8 pm. This can also be observed in the “Internuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A mep and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. The dynamics calculation resulted in a longer distance between atom B and C once the reaction finished: the reaction rolls toward the products.

If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This is illustrated by the following plots:

- in the Internuclear distance vs time plot, the initial value of AB is equale to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in differetn ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]][[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Internuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation ended. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Internuclear distances vs time. The reaction reaches the transition state.]]

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product is not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction proceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enought to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were ormed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doen't take into consideration the possibility of quantum tunnelling, as the convetional transition state theory is purerly classical motion. <ref name="tunneling"/>
Infact, if the system tunnels throught the PES, then the kinetic energy could be lowere than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:

F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
* The reaction is exothermic as the energy of the reagents is higher that that of the products.

* Position of TS: AB= 74.5 pm and BC = 181pm .
* It was identified thanks to hammonds postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds" />. In this case, the transition state is early, as the reaction is exothermic. Therefore, the TS will resemble the reagents and the separation between the hydrogen molecule will be smaller than that of HF
* Total energy-433.98 KJ mol-1.
* Energy of reagent:-560.592 KJ mol-1
[[File:energy_H2_01512921.png|thumb|right|Energy vs Time mep. The transition state rolls to the reagent H2]]
* Activation energy is: -126.612
H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
* The reaction is endothermic as the enrgy of the reagents is lower that that of the products.
* Position of TS: HF = 95 pm, H2=250 pm
* Energy is -433.98 KJ mol-1
* Reagent energy: -434.012 KJ mol-1
* Activation enegry: -0.032 KJ mol-1
[[File:energy_HF_01512921.png|thumb|left|Energy vs Time mep. The transition state rolls to the reagent HF]]

Strenght of the bonds:
Strength of H2 = 436 KJ mol

Strength of HF = 569 KJ mol

When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

<nowiki> </nowiki>Q2, part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy" />. This can be confirmed by spectroscopic methods like infrared, as it would be possible to see overtones due to the transition from the first to the second vibrational excited state. A reactive trajcetory was found at r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1. While varing the value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with hight vibrational energy in the products. At 3.1, the reaction has a barrier recrossing, where the product forms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reation is not successful. Barrier recrossing is alwo seen at 6.1, with a successful collision. At 5.1 g.mol-1.pm.fs-1, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb extacly in the middle between Ha and F, not showing which reaction's side is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1 and pHH=0.2 g.mol-1.pm.fs-1. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

Q2, part 2: FH + H ==> H2 +F

A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1. The skew plot of the reaction is shown.
[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

Q5: The energy distribution of the products can be determined by locating the energy barrier on the raction coordinate (The TS). For a reaction is A+ BC:
* an early barrier or TS will result in high vibrational energy in the products. In an exothermic reaction, the energy is released while AB distance is changing

* a late barrier results in low vibrational energy in the products. The energy is releaved after AB is formed and BC is changing, which corresponds to the formation of the products and corresponds to the translational energy.
The position of the energy barrier would also help sleecting what distribution o freactant energy is most likely to lead to a reaction.
* for an early TS, a molecule with high translational energy will be able to overcome the barrier, as it will thabe all its motion along the reaction coordinates. On the other hand, a molcule with high vibration will not have enough energy to reach the barrier.
* for a late TS, the barrier will be overcome by vibrational energy rather than translational. In fact, a molecule with high translational energy will crash in the inner wall of the PES and bounce back<ref name="polanyi's" />.

=== Conclusions===
The investigation of the reaction dynamics for the three reactions examined in the lab concerned with the PES and the reaction trajectories.
Using the program, it was possible to identify the transition state of the reaction (using Hammond's postulate), the trajectories (both reactive and unreactive ones) and the activation energies (thank to the energy of the trasition state and thta of the reactants).
A comparison was made between the rate constant from the canonical transition state theory and the experimental one. It was noted how, due t the absence of tunneling, the CTST would overestimate the rate of the reaction.
By analysing the transition state, it was possible to observe how its position influences which energy, translational or vibrational, is required for the reaction to happen. An early transition state can be overcome with translational energy, while a late one will prefer high vibrational enegry.

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
<ref name="polanyi's">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 9, pp 272-274. </ref>
</references>

MRD:01512921

2020-05-22T17:41:48Z

Sp3418: /* Introduction */

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and internuclear distances affected the outcome of the reaction. It was also observed how the position of the transition state determiens which type of energy, vibrational or translational, will be suffiect to overcome the energy barrier.

Chemical reactions can be simulated by taking into consideration the relative position, such as the distances, between the atom involved in the reaction. In fact, the interaction between the atoms that cause their motion depend on their location and are described as a PES, a functionrelative to the coordinates of the costituent atoms <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision coordinates; for the systems analysed in this report, there are only two coordinates: r1, distances between atoms of the reacting molecule and r2<ref name="intro1"/>, distance betweem the atoms of the product molecule. The path of the reaction can be mappes with a reaction trajectory. A reactiv trajectory will pass through a saddle point of the PES, also know as transition state(TS). On the contrary, an unreactive trajectory will roll back toward the reagents upon meeting the TS.

To understand the reactions and successfuly predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
This is possible thanks to only few informations about the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on different assumptions, of which the following are important for the purpose of this report:
- it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling"/>.
- The reaction is treated classically and the quantum effects are ignored<ref name="tunneling"/>.

=== Exercise 1 ===

For the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Internuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen (A or C) is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the virational energy is zero due to the absence of bonds. The value was obtained by trial and error: the first distance choosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from a peridodic vibration ( at 150 pm) to being stationaty at 90.8 pm. This can also be observed in the “Internuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A mep and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. The dynamics calculation resulted in a longer distance between atom B and C once the reaction finished: the reaction rolls toward the products.

If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This is illustrated by the following plots:

- in the Internuclear distance vs time plot, the initial value of AB is equale to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in differetn ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]][[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Internuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation ended. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Internuclear distances vs time. The reaction reaches the transition state.]]

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product is not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction proceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enought to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were ormed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doen't take into consideration the possibility of quantum tunnelling, as the convetional transition state theory is purerly classical motion. <ref name="tunneling"/>
Infact, if the system tunnels throught the PES, then the kinetic energy could be lowere than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:

F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
* The reaction is exothermic as the energy of the reagents is higher that that of the products.

* Position of TS: AB= 74.5 pm and BC = 181pm .
* It was identified thanks to hammonds postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds" />. In this case, the transition state is early, as the reaction is exothermic. Therefore, the TS will resemble the reagents and the separation between the hydrogen molecule will be smaller than that of HF
* Total energy-433.98 KJ mol-1.
* Energy of reagent:-560.592 KJ mol-1
[[File:energy_H2_01512921.png|thumb|right|Energy vs Time mep. The transition state rolls to the reagent H2]]
* Activation energy is: -126.612
H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
* The reaction is endothermic as the enrgy of the reagents is lower that that of the products.
* Position of TS: HF = 95 pm, H2=250 pm
* Energy is -433.98 KJ mol-1
* Reagent energy: -434.012 KJ mol-1
* Activation enegry: -0.032 KJ mol-1
[[File:energy_HF_01512921.png|thumb|left|Energy vs Time mep. The transition state rolls to the reagent HF]]

Strenght of the bonds:
Strength of H2 = 436 KJ mol

Strength of HF = 569 KJ mol

When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

<nowiki> </nowiki>Q2, part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy" />. This can be confirmed by spectroscopic methods like infrared, as it would be possible to see overtones due to the transition from the first to the second vibrational excited state. A reactive trajcetory was found at r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1. While varing the value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with hight vibrational energy in the products. At 3.1, the reaction has a barrier recrossing, where the product forms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reation is not successful. Barrier recrossing is alwo seen at 6.1, with a successful collision. At 5.1 g.mol-1.pm.fs-1, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb extacly in the middle between Ha and F, not showing which reaction's side is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1 and pHH=0.2 g.mol-1.pm.fs-1. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

Q2, part 2: FH + H ==> H2 +F

A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1. The skew plot of the reaction is shown.
[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

Q5: The energy distribution of the products can be determined by locating the energy barrier on the raction coordinate (The TS). For a reaction is A+ BC:
* an early barrier or TS will result in high vibrational energy in the products. In an exothermic reaction, the energy is released while AB distance is changing

* a late barrier results in low vibrational energy in the products. The energy is releaved after AB is formed and BC is changing, which corresponds to the formation of the products and corresponds to the translational energy.
The position of the energy barrier would also help sleecting what distribution o freactant energy is most likely to lead to a reaction.
* for an early TS, a molecule with high translational energy will be able to overcome the barrier, as it will thabe all its motion along the reaction coordinates. On the other hand, a molcule with high vibration will not have enough energy to reach the barrier.
* for a late TS, the barrier will be overcome by vibrational energy rather than translational. In fact, a molecule with high translational energy will crash in the inner wall of the PES and bounce back<ref name="polanyi's" />.

=== Conclusions===
The investigation of the reaction dynamics for the three reactions examined in the lab concerned with the PES and the reaction trajectories.
Using the program, it was possible to identify the transition state of the reaction (using Hammond's postulate), the trajectories (both reactive and unreactive ones) and the activation energies (thank to the energy of the trasition state and thta of the reactants).
A comparison was made between the rate constant from the canonical transition state theory and the experimental one. It was noted how, due t the absence of tunneling, the CTST would overestimate the rate of the reaction.
By analysing the transition state, it was possible to observe how its position influences which energy, translational or vibrational, is required for the reaction to happen. An early transition state can be overcome with translational energy, while a late one will prefer high vibrational enegry.

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
<ref name="polanyi's">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 9, pp 272-274. </ref>
</references>

MRD:01512921

2020-05-22T13:26:59Z

Sp3418: /* Reaction dynamics report */

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and internuclear distances affected the outcome of the reaction.

A function which is relative to the coordinates of the costituent atoms of a reaction is expressed as a Potential energy surface (PES) <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision trajectories; for the systems analysed in this report, there are only two coordinates: r1 and r2<ref name="intro1"/>.

To understand the reactions and successfuly predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
This is possible thanks to only few informations about the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on different assumptions, of which the following are important for the purpose of this report:
- it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling"/>.
- The reaction is treated classically and the quantum effects are ignored<ref name="tunneling"/>.

=== Exercise 1 ===

For the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Internuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen (A or C) is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the virational energy is zero due to the absence of bonds. The value was obtained by trial and error: the first distance choosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from a peridodic vibration ( at 150 pm) to being stationaty at 90.8 pm. This can also be observed in the “Internuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A mep and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. The dynamics calculation resulted in a longer distance between atom B and C once the reaction finished: the reaction rolls toward the products.

If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This is illustrated by the following plots:

- in the Internuclear distance vs time plot, the initial value of AB is equale to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in differetn ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]][[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Internuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation ended. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Internuclear distances vs time. The reaction reaches the transition state.]]

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product is not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction proceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enought to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were ormed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doen't take into consideration the possibility of quantum tunnelling, as the convetional transition state theory is purerly classical motion. <ref name="tunneling"/>
Infact, if the system tunnels throught the PES, then the kinetic energy could be lowere than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:

F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
* The reaction is exothermic as the energy of the reagents is higher that that of the products.

* Position of TS: AB= 74.5 pm and BC = 181pm .
* It was identified thanks to hammonds postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds" />. In this case, the transition state is early, as the reaction is exothermic. Therefore, the TS will resemble the reagents and the separation between the hydrogen molecule will be smaller than that of HF
* Total energy-433.98 KJ mol-1.
* Energy of reagent:-560.592 KJ mol-1
[[File:energy_H2_01512921.png|thumb|right|Energy vs Time mep. The transition state rolls to the reagent H2]]
* Activation energy is: -126.612
H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
* The reaction is endothermic as the enrgy of the reagents is lower that that of the products.
* Position of TS: HF = 95 pm, H2=250 pm
* Energy is -433.98 KJ mol-1
* Reagent energy: -434.012 KJ mol-1
* Activation enegry: -0.032 KJ mol-1
[[File:energy_HF_01512921.png|thumb|left|Energy vs Time mep. The transition state rolls to the reagent HF]]

Strenght of the bonds:
Strength of H2 = 436 KJ mol

Strength of HF = 569 KJ mol

When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

<nowiki> </nowiki>Q2, part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy" />. This can be confirmed by spectroscopic methods like infrared, as it would be possible to see overtones due to the transition from the first to the second vibrational excited state. A reactive trajcetory was found at r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1. While varing the value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with hight vibrational energy in the products. At 3.1, the reaction has a barrier recrossing, where the product forms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reation is not successful. Barrier recrossing is alwo seen at 6.1, with a successful collision. At 5.1 g.mol-1.pm.fs-1, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb extacly in the middle between Ha and F, not showing which reaction's side is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1 and pHH=0.2 g.mol-1.pm.fs-1. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

Q2, part 2: FH + H ==> H2 +F

A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1. The skew plot of the reaction is shown.
[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

Q5: The energy distribution of the products can be determined by locating the energy barrier on the raction coordinate (The TS). For a reaction is A+ BC:
* an early barrier or TS will result in high vibrational energy in the products. In an exothermic reaction, the energy is released while AB distance is changing

* a late barrier results in low vibrational energy in the products. The energy is releaved after AB is formed and BC is changing, which corresponds to the formation of the products and corresponds to the translational energy.
The position of the energy barrier would also help sleecting what distribution o freactant energy is most likely to lead to a reaction.
* for an early TS, a molecule with high translational energy will be able to overcome the barrier, as it will thabe all its motion along the reaction coordinates. On the other hand, a molcule with high vibration will not have enough energy to reach the barrier.
* for a late TS, the barrier will be overcome by vibrational energy rather than translational. In fact, a molecule with high translational energy will crash in the inner wall of the PES and bounce back<ref name="polanyi's" />.

=== Conclusions===
The investigation of the reaction dynamics for the three reactions examined in the lab concerned with the PES and the reaction trajectories.
Using the program, it was possible to identify the transition state of the reaction (using Hammond's postulate), the trajectories (both reactive and unreactive ones) and the activation energies (thank to the energy of the trasition state and thta of the reactants).
A comparison was made between the rate constant from the canonical transition state theory and the experimental one. It was noted how, due t the absence of tunneling, the CTST would overestimate the rate of the reaction.
By analysing the transition state, it was possible to observe how its position influences which energy, translational or vibrational, is required for the reaction to happen. An early transition state can be overcome with translational energy, while a late one will prefer high vibrational enegry.

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
<ref name="polanyi's">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 9, pp 272-274. </ref>
</references>

MRD:01512921

2020-05-22T13:26:41Z

Sp3418: /* Reaction dynamics report */

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and internuclear distances affected the outcome of the reaction.

A function which is relative to the coordinates of the costituent atoms of a reaction is expressed as a Potential energy surface (PES) <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision trajectories; for the systems analysed in this report, there are only two coordinates: r1 and r2<ref name="intro1"/>.

To understand the reactions and successfuly predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
This is possible thanks to only few informations about the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on different assumptions, of which the following are important for the purpose of this report:
- it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling"/>.
- The reaction is treated classically and the quantum effects are ignored<ref name="tunneling"/>.

=== Exercise 1 ===

For the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Internuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen (A or C) is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the virational energy is zero due to the absence of bonds. The value was obtained by trial and error: the first distance choosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from a peridodic vibration ( at 150 pm) to being stationaty at 90.8 pm. This can also be observed in the “Internuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A mep and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. The dynamics calculation resulted in a longer distance between atom B and C once the reaction finished: the reaction rolls toward the products.

If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This is illustrated by the following plots:

- in the Internuclear distance vs time plot, the initial value of AB is equale to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in differetn ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]][[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Internuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation ended. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Internuclear distances vs time. The reaction reaches the transition state.]]

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product is not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction proceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enought to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were ormed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doen't take into consideration the possibility of quantum tunnelling, as the convetional transition state theory is purerly classical motion. <ref name="tunneling"/>
Infact, if the system tunnels throught the PES, then the kinetic energy could be lowere than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:

F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
* The reaction is exothermic as the energy of the reagents is higher that that of the products.

* Position of TS: AB= 74.5 pm and BC = 181pm .
* It was identified thanks to hammonds postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds" />. In this case, the transition state is early, as the reaction is exothermic. Therefore, the TS will resemble the reagents and the separation between the hydrogen molecule will be smaller than that of HF
* Total energy-433.98 KJ mol-1.
* Energy of reagent:-560.592 KJ mol-1
[[File:energy_H2_01512921.png|thumb|right|Energy vs Time mep. The transition state rolls to the reagent H2]]
* Activation energy is: -126.612
H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
* The reaction is endothermic as the enrgy of the reagents is lower that that of the products.
* Position of TS: HF = 95 pm, H2=250 pm
* Energy is -433.98 KJ mol-1
* Reagent energy: -434.012 KJ mol-1
* Activation enegry: -0.032 KJ mol-1
[[File:energy_HF_01512921.png|thumb|left|Energy vs Time mep. The transition state rolls to the reagent HF]]

Strenght of the bonds:
Strength of H2 = 436 KJ mol

Strength of HF = 569 KJ mol

When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

<nowiki> </nowiki>Q2, part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy" />. This can be confirmed by spectroscopic methods like infrared, as it would be possible to see overtones due to the transition from the first to the second vibrational excited state. A reactive trajcetory was found at r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1. While varing the value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with hight vibrational energy in the products. At 3.1, the reaction has a barrier recrossing, where the product forms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reation is not successful. Barrier recrossing is alwo seen at 6.1, with a successful collision. At 5.1 g.mol-1.pm.fs-1, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb extacly in the middle between Ha and F, not showing which reaction's side is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1 and pHH=0.2 g.mol-1.pm.fs-1. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

Q2, part 2: FH + H ==> H2 +F

A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1. The skew plot of the reaction is shown.
[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

Q5: The energy distribution of the products can be determined by locating the energy barrier on the raction coordinate (The TS). For a reaction is A+ BC:
* an early barrier or TS will result in high vibrational energy in the products. In an exothermic reaction, the energy is released while AB distance is changing

* a late barrier results in low vibrational energy in the products. The energy is releaved after AB is formed and BC is changing, which corresponds to the formation of the products and corresponds to the translational energy.
The position of the energy barrier would also help sleecting what distribution o freactant energy is most likely to lead to a reaction.
* for an early TS, a molecule with high translational energy will be able to overcome the barrier, as it will thabe all its motion along the reaction coordinates. On the other hand, a molcule with high vibration will not have enough energy to reach the barrier.
* for a late TS, the barrier will be overcome by vibrational energy rather than translational. In fact, a molecule with high translational energy will crash in the inner wall of the PES and bounce back<ref name="polanyi's" />.

=== Conclusions===
The investigation of the reaction dynamics for the three reactions examined in the lab concerned with the PES and the reaction trajectories.
Using the program, it was possible to identify the transition state of the reaction (using Hammond's postulate), the trajectories (both reactive and unreactive ones) and the activation energies (thank to the energy of the trasition state and thta of the reactants).
A comparison was made between the rate constant from the canonical transition state theory and the experimental one. It was noted how, due t the absence of tunneling, the CTST would overestimate the rate of the reaction.
By analysing the transition state, it was possible to observe how its position influences which energy, translational or vibrational, is required for the reaction to happen. An early transition state can be overcome with translational energy, while a late one will prefer high vibrational enegry.

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
<ref name="polanyi's">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 9, pp 272-274. </ref>
</references>

MRD:01512921

2020-05-22T13:26:16Z

Sp3418: /* Conclusions */

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and internuclear distances affected the outcome of the reaction.

A function which is relative to the coordinates of the costituent atoms of a reaction is expressed as a Potential energy surface (PES) <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision trajectories; for the systems analysed in this report, there are only two coordinates: r1 and r2<ref name="intro1"/>.

To understand the reactions and successfuly predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
This is possible thanks to only few informations about the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on different assumptions, of which the following are important for the purpose of this report:
- it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling"/>.
- The reaction is treated classically and the quantum effects are ignored<ref name="tunneling"/>.

=== Exercise 1 ===

For the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Internuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen (A or C) is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the virational energy is zero due to the absence of bonds. The value was obtained by trial and error: the first distance choosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from a peridodic vibration ( at 150 pm) to being stationaty at 90.8 pm. This can also be observed in the “Internuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A mep and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. The dynamics calculation resulted in a longer distance between atom B and C once the reaction finished: the reaction rolls toward the products.

If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This is illustrated by the following plots:

- in the Internuclear distance vs time plot, the initial value of AB is equale to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in differetn ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]][[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Internuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation ended. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Internuclear distances vs time. The reaction reaches the transition state.]]

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product is not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction proceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enought to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were ormed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doen't take into consideration the possibility of quantum tunnelling, as the convetional transition state theory is purerly classical motion. <ref name="tunneling"/>
Infact, if the system tunnels throught the PES, then the kinetic energy could be lowere than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:

F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
* The reaction is exothermic as the energy of the reagents is higher that that of the products.

* Position of TS: AB= 74.5 pm and BC = 181pm .
* It was identified thanks to hammonds postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds" />. In this case, the transition state is early, as the reaction is exothermic. Therefore, the TS will resemble the reagents and the separation between the hydrogen molecule will be smaller than that of HF
* Total energy-433.98 KJ mol-1.
* Energy of reagent:-560.592 KJ mol-1
[[File:energy_H2_01512921.png|thumb|right|Energy vs Time mep. The transition state rolls to the reagent H2]]
* Activation energy is: -126.612
H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
* The reaction is endothermic as the enrgy of the reagents is lower that that of the products.
* Position of TS: HF = 95 pm, H2=250 pm
* Energy is -433.98 KJ mol-1
* Reagent energy: -434.012 KJ mol-1
* Activation enegry: -0.032 KJ mol-1
[[File:energy_HF_01512921.png|thumb|left|Energy vs Time mep. The transition state rolls to the reagent HF]]

Strenght of the bonds:
Strength of H2 = 436 KJ mol

Strength of HF = 569 KJ mol

When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

<nowiki> </nowiki>Q2, part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy" />. This can be confirmed by spectroscopic methods like infrared, as it would be possible to see overtones due to the transition from the first to the second vibrational excited state. A reactive trajcetory was found at r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1. While varing the value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with hight vibrational energy in the products. At 3.1, the reaction has a barrier recrossing, where the product forms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reation is not successful. Barrier recrossing is alwo seen at 6.1, with a successful collision. At 5.1 g.mol-1.pm.fs-1, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb extacly in the middle between Ha and F, not showing which reaction's side is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1 and pHH=0.2 g.mol-1.pm.fs-1. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

Q2, part 2: FH + H ==> H2 +F

A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1. The skew plot of the reaction is shown.
[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

Q5: The energy distribution of the products can be determined by locating the energy barrier on the raction coordinate (The TS). For a reaction is A+ BC:
* an early barrier or TS will result in high vibrational energy in the products. In an exothermic reaction, the energy is released while AB distance is changing

* a late barrier results in low vibrational energy in the products. The energy is releaved after AB is formed and BC is changing, which corresponds to the formation of the products and corresponds to the translational energy.
The position of the energy barrier would also help sleecting what distribution o freactant energy is most likely to lead to a reaction.
* for an early TS, a molecule with high translational energy will be able to overcome the barrier, as it will thabe all its motion along the reaction coordinates. On the other hand, a molcule with high vibration will not have enough energy to reach the barrier.
* for a late TS, the barrier will be overcome by vibrational energy rather than translational. In fact, a molecule with high translational energy will crash in the inner wall of the PES and bounce back<ref name="polanyi's" />.

=== Conclusions===
The investigation of the reaction dynamics for the three reactions examined in the lab concerned with the PES and the reaction trajectories.
Using the program, it was possible to identify the transition state of the reaction (using Hammond's postulate), the trajectories (both reactive and unreactive ones) and the activation energies (thank to the energy of the trasition state and thta of the reactants).
A comparison was made between the rate constant from the canonical transition state theory and the experimental one. It was noted how, due t the absence of tunneling, the CTST would overestimate the rate of the reaction.
By analysing the transition state, it was possible to observe how its position influences which energy, translational or vibrational, is required for the reaction to happen. An early transition state can be overcome with translational energy, while a late one will prefer high vibrational enegry.

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
<ref name="polanyi's">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 9, pp 272-274. </ref>
</references>

MRD:01512921

2020-05-22T13:25:39Z

Sp3418: /* Conclusions */

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and internuclear distances affected the outcome of the reaction.

A function which is relative to the coordinates of the costituent atoms of a reaction is expressed as a Potential energy surface (PES) <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision trajectories; for the systems analysed in this report, there are only two coordinates: r1 and r2<ref name="intro1"/>.

To understand the reactions and successfuly predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
This is possible thanks to only few informations about the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on different assumptions, of which the following are important for the purpose of this report:
- it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling"/>.
- The reaction is treated classically and the quantum effects are ignored<ref name="tunneling"/>.

=== Exercise 1 ===

For the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Internuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen (A or C) is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the virational energy is zero due to the absence of bonds. The value was obtained by trial and error: the first distance choosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from a peridodic vibration ( at 150 pm) to being stationaty at 90.8 pm. This can also be observed in the “Internuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A mep and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. The dynamics calculation resulted in a longer distance between atom B and C once the reaction finished: the reaction rolls toward the products.

If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This is illustrated by the following plots:

- in the Internuclear distance vs time plot, the initial value of AB is equale to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in differetn ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]][[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Internuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation ended. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Internuclear distances vs time. The reaction reaches the transition state.]]

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product is not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction proceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enought to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were ormed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doen't take into consideration the possibility of quantum tunnelling, as the convetional transition state theory is purerly classical motion. <ref name="tunneling"/>
Infact, if the system tunnels throught the PES, then the kinetic energy could be lowere than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:

F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
* The reaction is exothermic as the energy of the reagents is higher that that of the products.

* Position of TS: AB= 74.5 pm and BC = 181pm .
* It was identified thanks to hammonds postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds" />. In this case, the transition state is early, as the reaction is exothermic. Therefore, the TS will resemble the reagents and the separation between the hydrogen molecule will be smaller than that of HF
* Total energy-433.98 KJ mol-1.
* Energy of reagent:-560.592 KJ mol-1
[[File:energy_H2_01512921.png|thumb|right|Energy vs Time mep. The transition state rolls to the reagent H2]]
* Activation energy is: -126.612
H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
* The reaction is endothermic as the enrgy of the reagents is lower that that of the products.
* Position of TS: HF = 95 pm, H2=250 pm
* Energy is -433.98 KJ mol-1
* Reagent energy: -434.012 KJ mol-1
* Activation enegry: -0.032 KJ mol-1
[[File:energy_HF_01512921.png|thumb|left|Energy vs Time mep. The transition state rolls to the reagent HF]]

Strenght of the bonds:
Strength of H2 = 436 KJ mol

Strength of HF = 569 KJ mol

When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

<nowiki> </nowiki>Q2, part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy" />. This can be confirmed by spectroscopic methods like infrared, as it would be possible to see overtones due to the transition from the first to the second vibrational excited state. A reactive trajcetory was found at r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1. While varing the value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with hight vibrational energy in the products. At 3.1, the reaction has a barrier recrossing, where the product forms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reation is not successful. Barrier recrossing is alwo seen at 6.1, with a successful collision. At 5.1 g.mol-1.pm.fs-1, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb extacly in the middle between Ha and F, not showing which reaction's side is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1 and pHH=0.2 g.mol-1.pm.fs-1. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

Q2, part 2: FH + H ==> H2 +F

A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1. The skew plot of the reaction is shown.
[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

Q5: The energy distribution of the products can be determined by locating the energy barrier on the raction coordinate (The TS). For a reaction is A+ BC:
* an early barrier or TS will result in high vibrational energy in the products. In an exothermic reaction, the energy is released while AB distance is changing

* a late barrier results in low vibrational energy in the products. The energy is releaved after AB is formed and BC is changing, which corresponds to the formation of the products and corresponds to the translational energy.
The position of the energy barrier would also help sleecting what distribution o freactant energy is most likely to lead to a reaction.
* for an early TS, a molecule with high translational energy will be able to overcome the barrier, as it will thabe all its motion along the reaction coordinates. On the other hand, a molcule with high vibration will not have enough energy to reach the barrier.
* for a late TS, the barrier will be overcome by vibrational energy rather than translational. In fact, a molecule with high translational energy will crash in the inner wall of the PES and bounce back<ref name="polanyi's" />.

=== Conclusions===
The investigation of the reaction dynamics for the three reactions examined in the lab concerned with the PES and the reaction trajectories.
Using the program, it was possible to identify the transition state of the reaction (using Hammond's postulate), the trajectories (both reactive and unreactive ones) and the activation energies (thank to the energy of the trasition state and thta of the reactants).
A comparison was made between the rate constant from the canonical transition state theory and the experimental one. It was noted how, due t the absence of tunneling, the CTST would overestimate the rate of the reaction.
By analysing the transition state, it was possible to observe how its position influences which energy, translational or vibrational, is required for the reaction to happen. An early transition state would prefer

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
<ref name="polanyi's">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 9, pp 272-274. </ref>
</references>

MRD:01512921

2020-05-22T13:16:43Z

Sp3418:

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and internuclear distances affected the outcome of the reaction.

A function which is relative to the coordinates of the costituent atoms of a reaction is expressed as a Potential energy surface (PES) <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision trajectories; for the systems analysed in this report, there are only two coordinates: r1 and r2<ref name="intro1"/>.

To understand the reactions and successfuly predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
This is possible thanks to only few informations about the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on different assumptions, of which the following are important for the purpose of this report:
- it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling"/>.
- The reaction is treated classically and the quantum effects are ignored<ref name="tunneling"/>.

=== Exercise 1 ===

For the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Internuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen (A or C) is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the virational energy is zero due to the absence of bonds. The value was obtained by trial and error: the first distance choosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from a peridodic vibration ( at 150 pm) to being stationaty at 90.8 pm. This can also be observed in the “Internuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A mep and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. The dynamics calculation resulted in a longer distance between atom B and C once the reaction finished: the reaction rolls toward the products.

If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This is illustrated by the following plots:

- in the Internuclear distance vs time plot, the initial value of AB is equale to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in differetn ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]][[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Internuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation ended. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Internuclear distances vs time. The reaction reaches the transition state.]]

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product is not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction proceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enought to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were ormed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doen't take into consideration the possibility of quantum tunnelling, as the convetional transition state theory is purerly classical motion. <ref name="tunneling"/>
Infact, if the system tunnels throught the PES, then the kinetic energy could be lowere than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:

F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
* The reaction is exothermic as the energy of the reagents is higher that that of the products.

* Position of TS: AB= 74.5 pm and BC = 181pm .
* It was identified thanks to hammonds postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds" />. In this case, the transition state is early, as the reaction is exothermic. Therefore, the TS will resemble the reagents and the separation between the hydrogen molecule will be smaller than that of HF
* Total energy-433.98 KJ mol-1.
* Energy of reagent:-560.592 KJ mol-1
[[File:energy_H2_01512921.png|thumb|right|Energy vs Time mep. The transition state rolls to the reagent H2]]
* Activation energy is: -126.612
H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
* The reaction is endothermic as the enrgy of the reagents is lower that that of the products.
* Position of TS: HF = 95 pm, H2=250 pm
* Energy is -433.98 KJ mol-1
* Reagent energy: -434.012 KJ mol-1
* Activation enegry: -0.032 KJ mol-1
[[File:energy_HF_01512921.png|thumb|left|Energy vs Time mep. The transition state rolls to the reagent HF]]

Strenght of the bonds:
Strength of H2 = 436 KJ mol

Strength of HF = 569 KJ mol

When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

<nowiki> </nowiki>Q2, part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy" />. This can be confirmed by spectroscopic methods like infrared, as it would be possible to see overtones due to the transition from the first to the second vibrational excited state. A reactive trajcetory was found at r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=74 pm, r2=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1. While varing the value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with hight vibrational energy in the products. At 3.1, the reaction has a barrier recrossing, where the product forms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reation is not successful. Barrier recrossing is alwo seen at 6.1, with a successful collision. At 5.1 g.mol-1.pm.fs-1, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb extacly in the middle between Ha and F, not showing which reaction's side is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1 and pHH=0.2 g.mol-1.pm.fs-1. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

Q2, part 2: FH + H ==> H2 +F

A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1. The skew plot of the reaction is shown.
[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

Q5: The energy distribution of the products can be determined by locating the energy barrier on the raction coordinate (The TS). For a reaction is A+ BC:
* an early barrier or TS will result in high vibrational energy in the products. In an exothermic reaction, the energy is released while AB distance is changing

* a late barrier results in low vibrational energy in the products. The energy is releaved after AB is formed and BC is changing, which corresponds to the formation of the products and corresponds to the translational energy.
The position of the energy barrier would also help sleecting what distribution o freactant energy is most likely to lead to a reaction.
* for an early TS, a molecule with high translational energy will be able to overcome the barrier, as it will thabe all its motion along the reaction coordinates. On the other hand, a molcule with high vibration will not have enough energy to reach the barrier.
* for a late TS, the barrier will be overcome by vibrational energy rather than translational. In fact, a molecule with high translational energy will crash in the inner wall of the PES and bounce back<ref name="polanyi's" />.

=== Conclusions===

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
<ref name="polanyi's">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 9, pp 272-274. </ref>
</references>

MRD:01512921

2020-05-22T12:35:59Z

Sp3418:

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and internuclear distances affected the outcome of the reaction.

A function which is relative to the coordinates of the costituent atoms of a reaction is expressed as a Potential energy surface (PES) <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision trajectories; for the systems analysed in this report, there are only two coordinates: r1 and r2<ref name="intro1"/>.

To understand the reactions and successfuly predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
This is possible thanks to only few informations about the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on different assumptions, of which the following are important for the purpose of this report:
- it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling"/>.
- The reaction is treated classically and the quantum effects are ignored<ref name="tunneling"/>.

=== Exercise 1 ===

For the purpose of the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Internuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the virational energy is zero due to the absence of bonds. The value was obtained by trial an error: the first distance choosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from aperidodic vibration ( at 150 pm) to being stationaty at 90.8 pm. This can also be observed in the “Internuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A mep and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. TThe dynamics calculation resulted in a longer diatnce between atom B and C once the reaction finished: the reaction would roll toward the products.
If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This isllustrated by the following plots:

- in the Internuclear distance vs time plot, the initial value of AB is equale to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in differetn ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]].[[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Internuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation stopped. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Internuclear distances vs time. The reaction reaches the transition state.]].

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product are not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction preoceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enought to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were ormed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doen't take into consideration the possibility of quantum tunnelling, as the convetional transition state theory is purerly classical motion. <ref name="tunneling"/>
Infact, if the system tunnels throught the PES, then the kinetic energy could be lowere than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:
F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
The reaction is exothermic as the enrgy of the reagents is higher that that of the products.
Position of TS: AB= 74.5 and BC = 181. It was identified thank to hammonds postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds"/>. In this case, the transition state is early, as the reaction is exothermic. Theefore the TS will resemble the reagents, therefore the separation between the hydrogen molecule will be smaller than that of HF

Total energy-433.98.
Energy of reagent:-560.592 [[File:energy_H2_01512921.png|thumb|right|Energy vs Time mep. The transition state rolls to the reagent H2]]

Activation energy is: -126.612
H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
The reaction is endothermic as the enrgy of the reagents is lower that that of the products.
Position of TS: HF = 95 pm, H2=250 pm
Energy is -433.98
Reagent energy: -434.012
Activation enegry: -0.032 [[File:energy_HF_01512921.png|thumb|left|Energy vs Time mep. The transition state rolls to the reagent HF]]

Strength of H2 = 436 KJ mol
Strength of HF = 569 KJ mol
When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

Q2, part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy"/>. This can be confirmed by spectroscopic methods like infrared, as it would be possible to see overtones due to the transition from the first to the second vibrational excited state. A reactive trajcetory was found at r1=HH=74 pm, r2=HF=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=HH=74 pm, r2=HF=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1. While varing the value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with hight vibrational energy in the products. AT 3.1, the reaction has a barrier recrossing, where the product froms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reation is not successful. Barrier recrossing is alwo seen at 6.1, with a successful collision.
At 5.1 g.mol-1.pm.fs-1, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb extacly in the middle between Ha and F, not sowing wihich of the sides of the reaction is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1 and pHH=0.2 g.mol-1.pm.fs-1. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

Q2, part 2:
FH + H ==> H2 +F
A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1. The skew plot of the reaction is shown
[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

Q5: The energy distribution of the products can be determined by locating the energy barrier on the raction coordinate (The TS). For a reaction is A+ BC
• an early barrier or TS will result in high vibrational energy in the products. In an exothermic reaction, the energy is released while AB distance is changing
• a late barrier results in low vibrational energy in the products. The energy is releaved after AB is formed and BC is changing, which corresponds to the formation of the products and corresponds to the translational energy.
The position of the energy barrier would also help sleecting what distribution o freactant energy is most likely to lead to a reaction.
- for an early TS, a molecule with high translational energy will be able to overcome the barrier, as it will thabe all its motion along the reaction coordinates. On the other hand, a molcule with high vibration will not have enough energy to reach the barrier.
- for a late TS, the barrier will be overcome by vibrational energy rather than translational. In fact, a molecule with high translational energy will crash in the inner wall of the PES and bounce back<ref name="polanyi's"/>.

=== Conclusions===

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
<ref name="polanyi's">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 9, pp 272-274. </ref>
</references>

MRD:01512921

2020-05-22T12:34:21Z

Sp3418: /* Exercise 2 */

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and internuclear distances affected the outcome of the reaction.

A function which is relative to the coordinates of the costituent atoms of a reaction is expressed as a Potential energy surface (PES) <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision trajectories; for the systems analysed in this report, there are only two coordinates: r1 and r2<ref name="intro1"/>.

To understand the reactions and successfuly predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
This is possible thanks to only few informations about the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on different assumptions, of which the following are important for the purpose of this report:
- it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling"/>.
- The reaction is treated classically and the quantum effects are ignored<ref name="tunneling"/>.

=== Exercise 1 ===

For the purpose of the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Internuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the virational energy is zero due to the absence of bonds. The value was obtained by trial an error: the first distance choosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from aperidodic vibration ( at 150 pm) to being stationaty at 90.8 pm. This can also be observed in the “Internuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A mep and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. TThe dynamics calculation resulted in a longer diatnce between atom B and C once the reaction finished: the reaction would roll toward the products.
If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This isllustrated by the following plots:

- in the Internuclear distance vs time plot, the initial value of AB is equale to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in differetn ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]].[[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Internuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation stopped. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Internuclear distances vs time. The reaction reaches the transition state.]].

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product are not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction preoceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enought to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were ormed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doen't take into consideration the possibility of quantum tunnelling, as the convetional transition state theory is purerly classical motion. <ref name="tunneling"/>
Infact, if the system tunnels throught the PES, then the kinetic energy could be lowere than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:
F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
The reaction is exothermic as the enrgy of the reagents is higher that that of the products.
Position of TS: AB= 74.5 and BC = 181. It was identified thank to hammonds postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds"/>. In this case, the transition state is early, as the reaction is exothermic. Theefore the TS will resemble the reagents, therefore the separation between the hydrogen molecule will be smaller than that of HF

Total energy-433.98.
Energy of reagent:-560.592 [[File:energy_H2_01512921.png|thumb|right|Energy vs Time mep. The transition state rolls to the reagent H2]]

Activation energy is: -126.612
H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
The reaction is endothermic as the enrgy of the reagents is lower that that of the products.
Position of TS: HF = 95 pm, H2=250 pm
Energy is -433.98
Reagent energy: -434.012
Activation enegry: -0.032 [[File:energy_HF_01512921.png|thumb|left|Energy vs Time mep. The transition state rolls to the reagent HF]]

Strength of H2 = 436 KJ mol
Strength of HF = 569 KJ mol
When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

Q2, part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy"/>. This can be confirmed by spectroscopic methods like infrared, as it would be possible to see overtones due to the transition from the first to the second vibrational excited state. A reactive trajcetory was found at r1=HH=74 pm, r2=HF=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=HH=74 pm, r2=HF=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1. While varing the value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with hight vibrational energy in the products. AT 3.1, the reaction has a barrier recrossing, where the product froms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reation is not successful. Barrier recrossing is alwo seen at 6.1, with a successful collision.
At 5.1 g.mol-1.pm.fs-1, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb extacly in the middle between Ha and F, not sowing wihich of the sides of the reaction is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1 and pHH=0.2 g.mol-1.pm.fs-1. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

Q2, part 2:
FH + H ==> H2 +F
A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1. The skew plot of the reaction is shown
[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

Q5: The energy distribution of the products can be determined by locating the energy barrier on the raction coordinate (The TS). For a reaction is A+ BC
• an early barrier or TS will result in high vibrational energy in the products. In an exothermic reaction, the energy is released while AB distance is changing
• a late barrier results in low vibrational energy in the products. The energy is releaved after AB is formed and BC is changing, which corresponds to the formation of the products and corresponds to the translational energy.
The position of the energy barrier would also help sleecting what distribution o freactant energy is most likely to lead to a reaction.
- for an early TS, a molecule with high translational energy will be able to overcome the barrier, as it will thabe all its motion along the reaction coordinates. On the other hand, a molcule with high vibration will not have enough energy to reach the barrier.
- for a late TS, the barrier will be overcome by vibrational energy rather than translational. In fact, a molecule with high translational energy will crash in the inner wall of the PES and bounce back<ref name="polanyi's"/>.

=== Conclusions===

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
</references>

File:Energy H2 01512921.png

2020-05-22T11:27:01Z

Sp3418:

File:Energy HF 01512921.png

2020-05-22T11:26:32Z

Sp3418:

MRD:01512921

2020-05-21T17:23:39Z

Sp3418: /* Exercise 2 */

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and internuclear distances affected the outcome of the reaction.

A function which is relative to the coordinates of the costituent atoms of a reaction is expressed as a Potential energy surface (PES) <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision trajectories; for the systems analysed in this report, there are only two coordinates: r1 and r2<ref name="intro1"/>.

To understand the reactions and successfuly predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
This is possible thanks to only few informations about the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on different assumptions, of which the following are important for the purpose of this report:
- it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling"/>.
- The reaction is treated classically and the quantum effects are ignored<ref name="tunneling"/>.

=== Exercise 1 ===

For the purpose of the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Internuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the virational energy is zero due to the absence of bonds. The value was obtained by trial an error: the first distance choosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from aperidodic vibration ( at 150 pm) to being stationaty at 90.8 pm. This can also be observed in the “Internuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A mep and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. TThe dynamics calculation resulted in a longer diatnce between atom B and C once the reaction finished: the reaction would roll toward the products.
If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This isllustrated by the following plots:

- in the Internuclear distance vs time plot, the initial value of AB is equale to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in differetn ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]].[[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Internuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation stopped. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Internuclear distances vs time. The reaction reaches the transition state.]].

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product are not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction preoceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enought to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were ormed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doen't take into consideration the possibility of quantum tunnelling, as the convetional transition state theory is purerly classical motion. <ref name="tunneling"/>
Infact, if the system tunnels throught the PES, then the kinetic energy could be lowere than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:
F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
The reaction is exothermic as the enrgy of the reagents is higher that that of the products.
Position of TS: AB= 74.5 and BC = 181. It was identified thank to hammonds postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds"/>. In this case, the transition state is early, as the reaction is exothermic. Theefore the TS will resemble the reagents, therefore the separation between the hydrogen molecule. will be smaller than that of HF
in case the collision is at 116 degree angle, then the ts is at AB= 74 and BC= 300. and energy is -434.98
The activation energy is:
Total energy-433.98 and it's exclusively vibrational.
Energy of reagent::
Activation energy is:

H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
The reaction is endothermic as the enrgy of the reagents is lower that that of the products.
Position of TS

Strength of H2 = 436 KJ mol
Strength of HF = 569 KJ mol
When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

Q2, part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy"/>. This can be confirmed by spectroscopic methods like infrared. A reactive trajcetory was found at r1=HH=74 pm, r2=HF=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=HH=74 pm, r2=HF=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1. While varing the value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with hight vibrational energy in the products. AT 3.1, the reaction has a barrier recrossing, where the product froms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reation is not successful. Barrier recrossing is alwo seen at 6.1, with a successful collision.
At 5.1 g.mol-1.pm.fs-1, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb extacly in the middle between Ha and F, not sowing wihich of the sides of the reaction is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1 and pHH=0.2 g.mol-1.pm.fs-1. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

Q2, part 2:
F +H2 ==> HF + H.
A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1. The skew plot of the reaction is shown
[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

Q5: Experimental Chemical Dynamics (9.4.2)
Book Title: Chemical kinetics and dynamics
Book Author: Steinfeld, Jeffrey I.
Additional Person Name: Francisco, Joseph S.; Hase, William L.
Start page: 272 End page: 274

=== Conclusions===

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
</references>

MRD:01512921

2020-05-21T17:23:21Z

Sp3418: /* Exercise 2 */

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and internuclear distances affected the outcome of the reaction.

A function which is relative to the coordinates of the costituent atoms of a reaction is expressed as a Potential energy surface (PES) <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision trajectories; for the systems analysed in this report, there are only two coordinates: r1 and r2<ref name="intro1"/>.

To understand the reactions and successfuly predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
This is possible thanks to only few informations about the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on different assumptions, of which the following are important for the purpose of this report:
- it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling"/>.
- The reaction is treated classically and the quantum effects are ignored<ref name="tunneling"/>.

=== Exercise 1 ===

For the purpose of the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Internuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the virational energy is zero due to the absence of bonds. The value was obtained by trial an error: the first distance choosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from aperidodic vibration ( at 150 pm) to being stationaty at 90.8 pm. This can also be observed in the “Internuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A mep and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. TThe dynamics calculation resulted in a longer diatnce between atom B and C once the reaction finished: the reaction would roll toward the products.
If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This isllustrated by the following plots:

- in the Internuclear distance vs time plot, the initial value of AB is equale to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in differetn ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]].[[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Internuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation stopped. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Internuclear distances vs time. The reaction reaches the transition state.]].

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product are not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction preoceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enought to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were ormed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doen't take into consideration the possibility of quantum tunnelling, as the convetional transition state theory is purerly classical motion. <ref name="tunneling"/>
Infact, if the system tunnels throught the PES, then the kinetic energy could be lowere than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:
F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
The reaction is exothermic as the enrgy of the reagents is higher that that of the products.
Position of TS: AB= 74.5 and BC = 181. It was identified thank to hammonds postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds"/>. In this case, the transition state is early, as the reaction is exothermic. Theefore the TS will resemble the reagents, therefore the separation between the hydrogen molecule. will be smaller than that of HF
in case the collision is at 116 degree angle, then the ts is at AB= 74 and BC= 300. and energy is -434.98
The activation energy is:
Total energy-433.98 and it's exclusively vibrational.
Energy of reagent::
Activation energy is:

H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
The reaction is endothermic as the enrgy of the reagents is lower that that of the products.
Position of TS

Strength of H2 = 436 KJ mol
Strength of HF = 569 KJ mol
When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

Q2, part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy"/>. This can be confirmed by spectroscopic methods like infrared. A reactive trajcetory was found at r1=HH=74 pm, r2=HF=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1</sup.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=HH=74 pm, r2=HF=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1. While varing the value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with hight vibrational energy in the products. AT 3.1, the reaction has a barrier recrossing, where the product froms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reation is not successful. Barrier recrossing is alwo seen at 6.1, with a successful collision.
At 5.1 g.mol-1.pm.fs-1, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb extacly in the middle between Ha and F, not sowing wihich of the sides of the reaction is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1 and pHH=0.2 g.mol-1.pm.fs-1. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

Q2, part 2:
F +H2 ==> HF + H.
A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1. The skew plot of the reaction is shown
[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

Q5: Experimental Chemical Dynamics (9.4.2)
Book Title: Chemical kinetics and dynamics
Book Author: Steinfeld, Jeffrey I.
Additional Person Name: Francisco, Joseph S.; Hase, William L.
Start page: 272 End page: 274

=== Conclusions===

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
</references>

MRD:01512921

2020-05-21T17:22:43Z

Sp3418:

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and internuclear distances affected the outcome of the reaction.

A function which is relative to the coordinates of the costituent atoms of a reaction is expressed as a Potential energy surface (PES) <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision trajectories; for the systems analysed in this report, there are only two coordinates: r1 and r2<ref name="intro1"/>.

To understand the reactions and successfuly predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
This is possible thanks to only few informations about the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on different assumptions, of which the following are important for the purpose of this report:
- it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling"/>.
- The reaction is treated classically and the quantum effects are ignored<ref name="tunneling"/>.

=== Exercise 1 ===

For the purpose of the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Internuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the virational energy is zero due to the absence of bonds. The value was obtained by trial an error: the first distance choosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from aperidodic vibration ( at 150 pm) to being stationaty at 90.8 pm. This can also be observed in the “Internuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A mep and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. TThe dynamics calculation resulted in a longer diatnce between atom B and C once the reaction finished: the reaction would roll toward the products.
If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This isllustrated by the following plots:

- in the Internuclear distance vs time plot, the initial value of AB is equale to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in differetn ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]].[[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Internuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation stopped. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Internuclear distances vs time. The reaction reaches the transition state.]].

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product are not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction preoceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enought to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were ormed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doen't take into consideration the possibility of quantum tunnelling, as the convetional transition state theory is purerly classical motion. <ref name="tunneling"/>
Infact, if the system tunnels throught the PES, then the kinetic energy could be lowere than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:
F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
The reaction is exothermic as the enrgy of the reagents is higher that that of the products.
Position of TS: AB= 74.5 and BC = 181. It was identified thank to hammonds postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds"/>. In this case, the transition state is early, as the reaction is exothermic. Theefore the TS will resemble the reagents, therefore the separation between the hydrogen molecule. will be smaller than that of HF
in case the collision is at 116 degree angle, then the ts is at AB= 74 and BC= 300. and energy is -434.98
The activation energy is:
Total energy-433.98 and it's exclusively vibrational.
Energy of reagent::
Activation energy is:

H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
The reaction is endothermic as the enrgy of the reagents is lower that that of the products.
Position of TS

Strength of H2 = 436 KJ mol
Strength of HF = 569 KJ mol
When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

Q2, part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy"/>. This can be confirmed by spectroscopic methods like infrared. A reactive trajcetory was found at r1=HH=74 pm, r2=HF=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1</sup.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=HH=74 pm, r2=HF=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1. While varing the value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with hight vibrational energy in the products. AT 3.1, the reaction has a barrier recrossing, where the product froms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reation is not successful. Barrier recrossing is alwo seen at 6.1, with a successful collision.
At 5.1 g.mol-1.pm.fs-1</sup, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb extacly in the middle between Ha and F, not sowing wihich of the sides of the reaction is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1HH=0.2 g.mol-1.pm.fs-1</sup. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

Q2, part 2:
F +H2 ==> HF + H.
A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1</sup. The skew plot of the reaction is shown
[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

Q5: Experimental Chemical Dynamics (9.4.2)
Book Title: Chemical kinetics and dynamics
Book Author: Steinfeld, Jeffrey I.
Additional Person Name: Francisco, Joseph S.; Hase, William L.
Start page: 272 End page: 274

=== Conclusions===

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
</references>

MRD:01512921

2020-05-21T17:21:16Z

Sp3418:

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and internuclear distances affected the outcome of the reaction.

A function which is relative to the coordinates of the costituent atoms of a reaction is expressed as a Potential energy surface (PES) <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision trajectories; for the systems analysed in this report, there are only two coordinates: r1 and r2<ref name="intro1"/>.

To understand the reactions and successfuly predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
This is possible thanks to only few informations about the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on different assumptions, of which the following are important for the purpose of this report:
- it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling"/>.
- The reaction is treated classically and the quantum effects are ignored<ref name="tunneling"/>.

=== Exercise 1 ===

For the purpose of the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Internuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the virational energy is zero due to the absence of bonds. The value was obtained by trial an error: the first distance choosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from aperidodic vibration ( at 150 pm) to being stationaty at 90.8 pm. This can also be observed in the “Internuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A mep and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. TThe dynamics calculation resulted in a longer diatnce between atom B and C once the reaction finished: the reaction would roll toward the products.
If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This isllustrated by the following plots:

- in the Internuclear distance vs time plot, the initial value of AB is equale to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in differetn ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]].[[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Internuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation stopped. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Internuclear distances vs time. The reaction reaches the transition state.]].

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product are not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction preoceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enought to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were ormed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doen't take into consideration the possibility of quantum tunnelling, as the convetional transition state theory is purerly classical motion. <ref name="tunneling"/>
Infact, if the system tunnels throught the PES, then the kinetic energy could be lowere than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:
F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
The reaction is exothermic as the enrgy of the reagents is higher that that of the products.
Position of TS: AB= 74.5 and BC = 181. It was identified thank to hammonds postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds"/>. In this case, the transition state is early, as the reaction is exothermic. Theefore the TS will resemble the reagents, therefore the separation between the hydrogen molecule. will be smaller than that of HF
in case the collision is at 116 degree angle, then the ts is at AB= 74 and BC= 300. and energy is -434.98
The activation energy is:
Total energy-433.98 and it's exclusively vibrational.
Energy of reagent::
Activation energy is:

H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
The reaction is endothermic as the enrgy of the reagents is lower that that of the products.
Position of TS

Strength of H2 = 436 KJ mol
Strength of HF = 569 KJ mol
When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

Q2, part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy"/>. This can be confirmed by spectroscopic methods like infrared. A reactive trajcetory was found at r1=HH=74 pm, r2=HF=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1</sup.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=HH=74 pm, r2=HF=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1</sup. While varing the value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with hight vibrational energy in the products. AT 3.1, the reaction has a barrier recrossing, where the product froms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reation is not successful. Barrier recrossing is alwo seen at 6.1, with a successful collision.
At 5.1 g.mol-1.pm.fs-1</sup, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb extacly in the middle between Ha and F, not sowing wihich of the sides of the reaction is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1HH=0.2 g.mol-1.pm.fs-1</sup. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

Q2, part 2:
F +H2 ==> HF + H.
A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1</sup. The skew plot of the reaction is shown
[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

Q5: Experimental Chemical Dynamics (9.4.2)
Book Title: Chemical kinetics and dynamics
Book Author: Steinfeld, Jeffrey I.
Additional Person Name: Francisco, Joseph S.; Hase, William L.
Start page: 272 End page: 274

=== Conclusions===

=== References===
<references>
<ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
</references>

MRD:01512921

2020-05-21T17:20:51Z

Sp3418: /* Introduction */

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and internuclear distances affected the outcome of the reaction.

A function which is relative to the coordinates of the costituent atoms of a reaction is expressed as a Potential energy surface (PES) <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision trajectories; for the systems analysed in this report, there are only two coordinates: r1 and r2<ref name="intro1"/>.

To understand the reactions and successfuly predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
This is possible thanks to only few informations about the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on different assumptions, of which the following are important for the purpose of this report:
- it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling"/>.
- The reaction is treated classically and the quantum effects are ignored<ref name="tunneling"/>.

=== Exercise 1 ===

For the purpose of the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Internuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the virational energy is zero due to the absence of bonds. The value was obtained by trial an error: the first distance choosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from aperidodic vibration ( at 150 pm) to being stationaty at 90.8 pm. This can also be observed in the “Internuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A mep and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. TThe dynamics calculation resulted in a longer diatnce between atom B and C once the reaction finished: the reaction would roll toward the products.
If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This isllustrated by the following plots:

- in the Internuclear distance vs time plot, the initial value of AB is equale to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in differetn ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]].[[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Internuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation stopped. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Internuclear distances vs time. The reaction reaches the transition state.]].

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product are not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction preoceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enought to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were ormed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doen't take into consideration the possibility of quantum tunnelling, as the convetional transition state theory is purerly classical motion. <ref name="tunneling"/>
Infact, if the system tunnels throught the PES, then the kinetic energy could be lowere than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:
F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
The reaction is exothermic as the enrgy of the reagents is higher that that of the products.
Position of TS: AB= 74.5 and BC = 181. It was identified thank to hammonds postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds"/>. In this case, the transition state is early, as the reaction is exothermic. Theefore the TS will resemble the reagents, therefore the separation between the hydrogen molecule. will be smaller than that of HF
in case the collision is at 116 degree angle, then the ts is at AB= 74 and BC= 300. and energy is -434.98
The activation energy is:
Total energy-433.98 and it's exclusively vibrational.
Energy of reagent::
Activation energy is:

H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
The reaction is endothermic as the enrgy of the reagents is lower that that of the products.
Position of TS

Strength of H2 = 436 KJ mol
Strength of HF = 569 KJ mol
When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

Q2, part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy"/>. This can be confirmed by spectroscopic methods like infrared. A reactive trajcetory was found at r1=HH=74 pm, r2=HF=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1</sup.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=HH=74 pm, r2=HF=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1</sup. While varing the value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with hight vibrational energy in the products. AT 3.1, the reaction has a barrier recrossing, where the product froms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reation is not successful. Barrier recrossing is alwo seen at 6.1, with a successful collision.
At 5.1 g.mol-1.pm.fs-1</sup, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb extacly in the middle between Ha and F, not sowing wihich of the sides of the reaction is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1HH=0.2 g.mol-1.pm.fs-1</sup. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

Q2, part 2:
F +H2 ==> HF + H.
A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1</sup. The skew plot of the reaction is shown
[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

Q5: Experimental Chemical Dynamics (9.4.2)
Book Title: Chemical kinetics and dynamics
Book Author: Steinfeld, Jeffrey I.
Additional Person Name: Francisco, Joseph S.; Hase, William L.
Start page: 272 End page: 274

=== Conclusions===

=== References===
<references>
ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
</references>

MRD:01512921

2020-05-21T17:19:54Z

Sp3418: /* References */

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and internuclear distances affected the outcome of the reaction.

A function which is relative to the coordinates of the costituent atoms of a reaction is expressed as a Potential energy surface (PES) <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision trajectories; for the systems analysed in this report, there are only two coordinates: r1 and r2<ref name="intro1"/>.

To understand the reactions and successfuly predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
This is possible thanks to only few informations about the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on different assumptions, of which the following are important for the purpose of this report:
- it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling"/>.
- The reaction is treated classically and the quantum effects are ignored<ref name="tunneling"/>.

=== Exercise 1 ===

For the purpose of the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Internuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the virational energy is zero due to the absence of bonds. The value was obtained by trial an error: the first distance choosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from aperidodic vibration ( at 150 pm) to being stationaty at 90.8 pm. This can also be observed in the “Internuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A mep and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. TThe dynamics calculation resulted in a longer diatnce between atom B and C once the reaction finished: the reaction would roll toward the products.
If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This isllustrated by the following plots:

- in the Internuclear distance vs time plot, the initial value of AB is equale to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in differetn ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]].[[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Internuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation stopped. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Internuclear distances vs time. The reaction reaches the transition state.]].

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product are not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction preoceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enought to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were ormed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doen't take into consideration the possibility of quantum tunnelling, as the convetional transition state theory is purerly classical motion. <ref name="tunneling"/>
Infact, if the system tunnels throught the PES, then the kinetic energy could be lowere than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:
F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
The reaction is exothermic as the enrgy of the reagents is higher that that of the products.
Position of TS: AB= 74.5 and BC = 181. It was identified thank to hammonds postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds"/>. In this case, the transition state is early, as the reaction is exothermic. Theefore the TS will resemble the reagents, therefore the separation between the hydrogen molecule. will be smaller than that of HF
in case the collision is at 116 degree angle, then the ts is at AB= 74 and BC= 300. and energy is -434.98
The activation energy is:
Total energy-433.98 and it's exclusively vibrational.
Energy of reagent::
Activation energy is:

H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
The reaction is endothermic as the enrgy of the reagents is lower that that of the products.
Position of TS

Strength of H2 = 436 KJ mol
Strength of HF = 569 KJ mol
When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

Q2, part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy"/>. This can be confirmed by spectroscopic methods like infrared. A reactive trajcetory was found at r1=HH=74 pm, r2=HF=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1</sup.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=HH=74 pm, r2=HF=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1</sup. While varing the value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with hight vibrational energy in the products. AT 3.1, the reaction has a barrier recrossing, where the product froms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reation is not successful. Barrier recrossing is alwo seen at 6.1, with a successful collision.
At 5.1 g.mol-1.pm.fs-1</sup, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb extacly in the middle between Ha and F, not sowing wihich of the sides of the reaction is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1HH=0.2 g.mol-1.pm.fs-1</sup. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

Q2, part 2:
F +H2 ==> HF + H.
A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1</sup. The skew plot of the reaction is shown
[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

Q5: Experimental Chemical Dynamics (9.4.2)
Book Title: Chemical kinetics and dynamics
Book Author: Steinfeld, Jeffrey I.
Additional Person Name: Francisco, Joseph S.; Hase, William L.
Start page: 272 End page: 274

=== Conclusions===

=== References===
<references>
ref name="intro1">J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
</references>

MRD:01512921

2020-05-21T17:19:08Z

Sp3418: /* Reaction dynamics report */

== Reaction dynamics report ==
=== Introduction ===
During this lab, the potential energy surface (PES) of three different reactions were analysed, along with their trajectories. The PES were used to identify the transition state and observe how different momenta and internuclear distances affected the outcome of the reaction.

A function which is relative to the coordinates of the costituent atoms of a reaction is expressed as a Potential energy surface (PES) <ref name="intro1"/>.
The region of space of the PES is separated into the reactant, where the system is before reacting, and the product regions, where the system is after reacting<ref name="intro2"/>. The boundary between these two regions is the transition state<ref name="intro2"/>.
The PES allows to solve classical equation of motion for collision trajectories; for the systems analysed in this report, there are only two coordinates: r1 and r2<ref name="intro1"/>.

To understand the reactions and successfuly predict their rates, the conventional transition state theory is used<ref name="tunneling"/>.
This is possible thanks to only few informations about the potential energy surface near the transition state and the reactants<ref name="intro"/>.
This theory is based on different assumptions, of which the following are important for the purpose of this report:
- it's impossible for a system to revert back to the reagents once the energy barrier is overcome<ref name="tunneling"/>.
- The reaction is treated classically and the quantum effects are ignored<ref name="tunneling"/>.

=== Exercise 1 ===

For the purpose of the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Internuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the virational energy is zero due to the absence of bonds. The value was obtained by trial an error: the first distance choosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from aperidodic vibration ( at 150 pm) to being stationaty at 90.8 pm. This can also be observed in the “Internuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A mep and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. TThe dynamics calculation resulted in a longer diatnce between atom B and C once the reaction finished: the reaction would roll toward the products.
If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This isllustrated by the following plots:

- in the Internuclear distance vs time plot, the initial value of AB is equale to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in differetn ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]].[[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Internuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation stopped. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Internuclear distances vs time. The reaction reaches the transition state.]].

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product are not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction preoceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enought to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were ormed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doen't take into consideration the possibility of quantum tunnelling, as the convetional transition state theory is purerly classical motion. <ref name="tunneling"/>
Infact, if the system tunnels throught the PES, then the kinetic energy could be lowere than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:
F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
The reaction is exothermic as the enrgy of the reagents is higher that that of the products.
Position of TS: AB= 74.5 and BC = 181. It was identified thank to hammonds postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds"/>. In this case, the transition state is early, as the reaction is exothermic. Theefore the TS will resemble the reagents, therefore the separation between the hydrogen molecule. will be smaller than that of HF
in case the collision is at 116 degree angle, then the ts is at AB= 74 and BC= 300. and energy is -434.98
The activation energy is:
Total energy-433.98 and it's exclusively vibrational.
Energy of reagent::
Activation energy is:

H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
The reaction is endothermic as the enrgy of the reagents is lower that that of the products.
Position of TS

Strength of H2 = 436 KJ mol
Strength of HF = 569 KJ mol
When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

Q2, part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy"/>. This can be confirmed by spectroscopic methods like infrared. A reactive trajcetory was found at r1=HH=74 pm, r2=HF=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH= -2 g.mol-1.pm.fs-1</sup.[[File:skew_HF_01512921.png|thumb|right|Plot of HF formation. The HF bond has high vibrational energy]]

A calculation was set up with r1=HH=74 pm, r2=HF=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1</sup. While varing the value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with hight vibrational energy in the products. AT 3.1, the reaction has a barrier recrossing, where the product froms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reation is not successful. Barrier recrossing is alwo seen at 6.1, with a successful collision.
At 5.1 g.mol-1.pm.fs-1</sup, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb extacly in the middle between Ha and F, not sowing wihich of the sides of the reaction is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1HH=0.2 g.mol-1.pm.fs-1</sup. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

Q2, part 2:
F +H2 ==> HF + H.
A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1</sup. The skew plot of the reaction is shown
[[File:skew_HH_01512921.png|thumb|left|Skew plot of HH formation. The HH bond has high vibrational energy]]

Q5: Experimental Chemical Dynamics (9.4.2)
Book Title: Chemical kinetics and dynamics
Book Author: Steinfeld, Jeffrey I.
Additional Person Name: Francisco, Joseph S.; Hase, William L.
Start page: 272 End page: 274

=== Conclusions===

=== References===
<references>J.I Steinfeld, J.S. Francisco, W.L. HAse, Chemical kinetics and dynamics,Prentice-Hall, 2nd ed., 1989,Upper Saddle River, chap 8, pp 232-239. </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro"> B. Peters,Reaction Rate Theory and Rare Events Simulations, Elsevier, 2017, chap 10, pp.227-271 </ref>
<ref name="intro2">T. Bligaard, J.K. Nørskov,Chemical Bonding at Surfaces and Interfaces,A. Nilsson, L. G.M. Pettersson, J. K. Nørskov,Elsevier,2008, Chap. 4, pp. 255-321. </ref>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
</references>

File:Skew HF 01512921.png

2020-05-21T16:19:25Z

Sp3418:

File:Skew HH 01512921.png

2020-05-21T16:13:09Z

Sp3418:

MRD:01512921

2020-05-21T16:12:01Z

Sp3418: /* Exercise 2 */

== Reaction dynamics report ==
=== Introduction ===
==== The transition state theory====
=== Exercise 1 ===

For the purpose of the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Internuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the virational energy is zero due to the absence of bonds. The value was obtained by trial an error: the first distance choosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from aperidodic vibration ( at 150 pm) to being stationaty at 90.8 pm. This can also be observed in the “Internuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A mep and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. TThe dynamics calculation resulted in a longer diatnce between atom B and C once the reaction finished: the reaction would roll toward the products.
If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This isllustrated by the following plots:

- in the Internuclear distance vs time plot, the initial value of AB is equale to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in differetn ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]].[[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Internuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation stopped. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Internuclear distances vs time. The reaction reaches the transition state.]].

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product are not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction preoceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enought to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were ormed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doen't take into consideration the possibility of quantum tunnelling, as the convetional transition state theory is purerly classical motion. <ref name="tunneling"/>
Infact, if the system tunnels throught the PES, then the kinetic energy could be lowere than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:
F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
The reaction is exothermic as the enrgy of the reagents is higher that that of the products.
Position of TS: AB= 74.5 and BC = 181. It was identified thank to hammonds postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds"/>. In this case, the transition state is early, as the reaction is exothermic. Theefore the TS will resemble the reagents, therefore the separation between the hydrogen molecule. will be smaller than that of HF
in case the collision is at 116 degree angle, then the ts is at AB= 74 and BC= 300. and energy is -434.98
The activation energy is:
Total energy-433.98 and it's exclusively vibrational.
Energy of reagent::
Activation energy is:

H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
The reaction is endothermic as the enrgy of the reagents is lower that that of the products.
Position of TS

Strength of H2 = 436 KJ mol
Strength of HF = 569 KJ mol
When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

Q2, part 1:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF<ref name="released_energy"/>. This can be confirmed by spectroscopic methods like infrared.

A calculation was set up with r1=HH=74 pm, r2=HF=150 pm and the following momenta: pHF= -1.0 g.mol-1.pm.fs-1 and pHH between -6.1 and 6.1 g.mol-1.pm.fs-1</sup. While varing the value of the HF momenta, it was noticed that at pHF= -6.1 the atom HB bounced several time between the atom HA and F. Between -6.1 and -3.1, the transition state was still crossed more than once to go back to the reagents, but the number of times this happened decreased from value to value. From -3.1 to 3.1, the reaction was successful with hight vibrational energy in the products. AT 3.1, the reaction has a barrier recrossing, where the product froms only to roll back to the reagent and then a second time toward the product. At 4.1, there is a barrier recrossing but the reation is not successful. Barrier recrossing is alwo seen at 6.1, with a successful collision.
At 5.1 g.mol-1.pm.fs-1</sup, the reaction reaches the Ts and then has a few barrier recrossing. However, the simulation ends with Hb extacly in the middle between Ha and F, not sowing wihich of the sides of the reaction is preferred.

For the same initial conditions, the following changes were applied pHF=-1.6 g.mol-1.pm.fs-1HH=0.2 g.mol-1.pm.fs-1</sup. With these settings, the reaction was successful, with the excess energy released as vibrational energy in the HF bond.

Q2, part 2:
F +H2 ==> HF + H.
A reactive trajectory was obtained with the following set up r1=HF=74 pm, r2=HH=200 pm and the following momenta: pHF= 4.5 g.mol-1.pm.fs-1 and pHH= 0.5 g.mol-1.pm.fs-1</sup. The skew plot of the reaction is shown

=== Conclusions===
=== References===
<references>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
</references>

MRD:01512921

2020-05-21T15:30:35Z

Sp3418: /* References */

== Reaction dynamics report ==
=== Introduction ===
==== The transition state theory====
=== Exercise 1 ===

For the purpose of the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Internuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the virational energy is zero due to the absence of bonds. The value was obtained by trial an error: the first distance choosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from aperidodic vibration ( at 150 pm) to being stationaty at 90.8 pm. This can also be observed in the “Internuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A mep and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. TThe dynamics calculation resulted in a longer diatnce between atom B and C once the reaction finished: the reaction would roll toward the products.
If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This isllustrated by the following plots:

- in the Internuclear distance vs time plot, the initial value of AB is equale to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in differetn ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]].[[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Internuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation stopped. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Internuclear distances vs time. The reaction reaches the transition state.]].

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product are not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction preoceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enought to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were ormed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doen't take into consideration the possibility of quantum tunnelling, as the convetional transition state theory is purerly classical motion. <ref name="tunneling"/>
Infact, if the system tunnels throught the PES, then the kinetic energy could be lowere than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:
F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
The reaction is exothermic as the enrgy of the reagents is higher that that of the products.
Position of TS: AB= 74.5 and BC = 181. It was identified thank to hammonds postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds"/>. In this case, the transition state is early, as the reaction is exothermic. Theefore the TS will resemble the reagents, therefore the separation between the hydrogen molecule. will be smaller than that of HF
in case the collision is at 116 degree angle, then the ts is at AB= 74 and BC= 300. and energy is -434.98
The activation energy is:
Total energy-433.98 and it's exclusively vibrational.
Energy of reagent::
Activation energy is:

H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
The reaction is endothermic as the enrgy of the reagents is lower that that of the products.
Position of TS

Strength of H2 = 436 KJ mol
Strength of HF = 569 KJ mol
When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

Q2:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF. This can be confirmed by spectroscopic methods like infrared. <ref name="released_energy"/>

=== Conclusions===
=== References===
<references>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
<ref name="released_energy">K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 12, pp 460-471</ref>
</references>

MRD:01512921

2020-05-21T15:29:07Z

Sp3418: /* Exercise 2 */

== Reaction dynamics report ==
=== Introduction ===
==== The transition state theory====
=== Exercise 1 ===

For the purpose of the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Internuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the virational energy is zero due to the absence of bonds. The value was obtained by trial an error: the first distance choosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from aperidodic vibration ( at 150 pm) to being stationaty at 90.8 pm. This can also be observed in the “Internuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A mep and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. TThe dynamics calculation resulted in a longer diatnce between atom B and C once the reaction finished: the reaction would roll toward the products.
If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This isllustrated by the following plots:

- in the Internuclear distance vs time plot, the initial value of AB is equale to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in differetn ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]].[[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Internuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation stopped. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Internuclear distances vs time. The reaction reaches the transition state.]].

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product are not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction preoceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enought to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were ormed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doen't take into consideration the possibility of quantum tunnelling, as the convetional transition state theory is purerly classical motion. <ref name="tunneling"/>
Infact, if the system tunnels throught the PES, then the kinetic energy could be lowere than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:
F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
The reaction is exothermic as the enrgy of the reagents is higher that that of the products.
Position of TS: AB= 74.5 and BC = 181. It was identified thank to hammonds postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds"/>. In this case, the transition state is early, as the reaction is exothermic. Theefore the TS will resemble the reagents, therefore the separation between the hydrogen molecule. will be smaller than that of HF
in case the collision is at 116 degree angle, then the ts is at AB= 74 and BC= 300. and energy is -434.98
The activation energy is:
Total energy-433.98 and it's exclusively vibrational.
Energy of reagent::
Activation energy is:

H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
The reaction is endothermic as the enrgy of the reagents is lower that that of the products.
Position of TS

Strength of H2 = 436 KJ mol
Strength of HF = 569 KJ mol
When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

Q2:
F +H2 ==> HF + H is an example of mixed energy release, where a high amount of the released energy is converted into vibrational energy of HF. This can be confirmed by spectroscopic methods like infrared. <ref name="released_energy"/>

=== Conclusions===
=== References===
<references>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
</references>

MRD:01512921

2020-05-21T14:57:11Z

Sp3418: /* Exercise 2 */

== Reaction dynamics report ==
=== Introduction ===
==== The transition state theory====
=== Exercise 1 ===

For the purpose of the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Internuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the virational energy is zero due to the absence of bonds. The value was obtained by trial an error: the first distance choosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from aperidodic vibration ( at 150 pm) to being stationaty at 90.8 pm. This can also be observed in the “Internuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A mep and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. TThe dynamics calculation resulted in a longer diatnce between atom B and C once the reaction finished: the reaction would roll toward the products.
If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This isllustrated by the following plots:

- in the Internuclear distance vs time plot, the initial value of AB is equale to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in differetn ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]].[[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Internuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation stopped. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Internuclear distances vs time. The reaction reaches the transition state.]].

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product are not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction preoceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enought to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were ormed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doen't take into consideration the possibility of quantum tunnelling, as the convetional transition state theory is purerly classical motion. <ref name="tunneling"/>
Infact, if the system tunnels throught the PES, then the kinetic energy could be lowere than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:
F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
The reaction is exothermic as the enrgy of the reagents is higher that that of the products.
Position of TS: AB= 74.5 and BC = 181. It was identified thank to hammonds postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds"/>. In this case, the transition state is early, as the reaction is exothermic. Theefore the TS will resemble the reagents, therefore the separation between the hydrogen molecule. will be smaller than that of HF
in case the collision is at 116 degree angle, then the ts is at AB= 74 and BC= 300. and energy is -434.98
The activation energy is:
Total energy-433.98 and it's exclusively vibrational.
Energy of reagent::
Activation energy is:

H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
The reaction is endothermic as the enrgy of the reagents is lower that that of the products.
Position of TS

Strength of H2 = 436 KJ mol
Strength of HF = 569 KJ mol
When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

=== Conclusions===
=== References===
<references>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
</references>

MRD:01512921

2020-05-21T12:23:15Z

Sp3418: /* Reaction dynamics report */

== Reaction dynamics report ==
=== Introduction ===
==== The transition state theory====
=== Exercise 1 ===

For the purpose of the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Internuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the virational energy is zero due to the absence of bonds. The value was obtained by trial an error: the first distance choosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from aperidodic vibration ( at 150 pm) to being stationaty at 90.8 pm. This can also be observed in the “Internuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A mep and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. TThe dynamics calculation resulted in a longer diatnce between atom B and C once the reaction finished: the reaction would roll toward the products.
If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This isllustrated by the following plots:

- in the Internuclear distance vs time plot, the initial value of AB is equale to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in differetn ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]].[[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Internuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation stopped. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Internuclear distances vs time. The reaction reaches the transition state.]].

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product are not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction preoceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enought to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were ormed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doen't take into consideration the possibility of quantum tunnelling, as the convetional transition state theory is purerly classical motion. <ref name="tunneling"/>
Infact, if the system tunnels throught the PES, then the kinetic energy could be lowere than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:
F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
The reaction is exothermic as the enrgy of the reagents is higher that that of the products.
Position of TS: AB= 74.5 and BC = 181. It was identified thank to hammonds postulate: the position of the transition state determines if it will more closely resemble the products or the reagents<ref name="hammonds"/>. In this case, the transition state is early, as the reaction is exothermic. Theefore the TS will resemble the reagents, therefore the separation between the hydrogen molecule. will be smaller than that of HF

H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
The reaction is endothermic as the enrgy of the reagents is lower that that of the products.
Position of TS

Strength of H2 = 436 KJ mol
Strength of HF = 569 KJ mol
When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

=== Conclusions===
=== References===
<references>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123</ref>
<ref name="hammonds"> Roman F. Nalewajski, Elżbieta Broniatowska, Information distance approach to Hammond postulate, Chemical Physics Letters, Volume 376, Issues 1–2, 2003, Pages 33-39</ref>
</references>

MRD:01512921

2020-05-21T12:21:49Z

Sp3418: /* Exercise 2 */

== Reaction dynamics report ==
=== Introduction ===
==== The transition state theory====
=== Exercise 1 ===

For the purpose of the exercise, A + BC ==> AB + C is mirrored by H + H2 ==> H2 + H. Therefore, AB= r2 and BC=r1

Q1: On a potential energy surface diagram, the transition state is defined as the saddle point, which causes the first derivative of the potential (the slope) to be zero. To test whether the point found is a saddle point or a local minimum, the second partial derivative test can be used. The test takes into consideration the determinant, D, of a Hessian matrix, a 2x2 matrix of partial derivatives of the function, which is generated by the program. If the determinant is positive, the point is either a maximum or a minimum. If the determinant is negative, then the point is a saddle point<ref name="second derivative test"/>.
[[File:Ts_01512921.png|thumb|centre|Plot of the Internuclear distances vs time for the transition state.]]
Q2: The best estimate of the transition state position ('''rts''') is at AB=BC=90.8 pm. By having equal distances, neither hydrogen is favoured in forming a bond with hydrogen B. It was identified by observing the forces on the single atoms: they all approached zero, as the virational energy is zero due to the absence of bonds. The value was obtained by trial an error: the first distance choosen was 150 pm, as it's the distance between atom A and C at the start of the reaction divided by two. The forces resulted to be quite negative (-1.759), so the value was lowered until eventually they reached zero. From the animation window, it was possible to observe how the atoms went from aperidodic vibration ( at 150 pm) to being stationaty at 90.8 pm. This can also be observed in the “Internuclear Distances vs Time” plot, where the distances between the atoms are constant in time. .

Q3: A mep and a dynamics calculation for AB= 90.8 and BC= 91.8 were run. TThe dynamics calculation resulted in a longer diatnce between atom B and C once the reaction finished: the reaction would roll toward the products.
If the values are exchange, AB= 91.8 and BC=90.8, then the transition state rolls back to the initial reagent and the molecule AB is not formed. This isllustrated by the following plots:

- in the Internuclear distance vs time plot, the initial value of AB is equale to that of BC. However, as time increases, the distance between A and B increases while that of B and C gets smaller.
- in the momenta vs time plot, the initial values are the same. After a small amount of time, the momenta decreases and then increase in differetn ways. The molecule BC presents a vibrating momentum, while the momentum of A-B increases until it reaches a plateau when they are quite far. [[File:Not_forming_mom_01512921.png|thumb|right|Plot of the momenta vs time. The transition state rolls back to the reagents.]].[[File:Not_forming_dist_01512921.png|thumb|left|Plot of the Internuclear distances vs time. The transition state rolls back to the reagents.]]

With AB= 91.8 and BC=90.8 and the dynamic set up, the data in Table 1 was obtained.
{| class="wikitable" border="1"
|+ Distance and momenta values at t=50 sec
! !! distances !! momenta
|-
| r2 || 352 || 5
|-
| r1 || 75 || 3.2
|}
Using this values ( the final values of the reaction), a new calculation was performed. This time, the result was the two reactant getting closer together to reach the transition state, where the calculation stopped. [[File:forming_dist_01512921.png|thumb|centre|Plot of the Internuclear distances vs time. The reaction reaches the transition state.]].

Q4: for the initial position of r1= 74 pm and r2= 200 pm, the following table was obtained using the momenta given.
{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! 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 || -414.1 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_1_01512921.png|150px]]
|-
| -3.1 || -4.1 || -419.9 || no || the momenta do not have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_2_01512921.png|150px]]
|-
| -3.1 || -5.1 || -413.8 || yes || the momenta have enougth kinetic energy to overcome the activation barrier || [[File:Trajectory_3_01512921.png|150px]]
|-
| -5.1 || -10.1 || -357.3 || no || The system crosses the transition state but, instead of forming a new bond, the product bounces back to the transition state and eventually the product are not formed. || [[File:Trajectory_4_01512921.png|150px]]
|-
| -5.1 || -10.6 || -349.5 || yes || The reaction preoceeds as the case above, but in this case the product is formed || [[File:Trajectory_5_01512921.png|150px]]
|}

For a successful reaction, kinetic energy possessed by the reagents has to be enought to overcome the saddle point. The momenta were varied so that the molecules had different kinetic and vibrational energy, in order to observe if the product were ormed and if they were formed in a vibrational mode. For the first three reactions, if only one of the reagent was in the momenta range proven successful by previous calculations ( -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1), then the reaction was successful <ref name="atkins"/> . For the last two example, these are cases of barrier crossing <ref name="barrier corssing"/>.

Q5:The conventional transition state theory assumes that as long as there is enough kinetic energy to overcome the energy barrier, then the reaction will proceed and it's not possible to recross the barrier<ref name="tunneling"/>. However, the theory doen't take into consideration the possibility of quantum tunnelling, as the convetional transition state theory is purerly classical motion. <ref name="tunneling"/>
Infact, if the system tunnels throught the PES, then the kinetic energy could be lowere than the one needed to reach the TS as the system can go through it, therefore and the rate constant form the CTST KTST is overestimated <ref name="tunneling"/> compared to that of the program which is used in this exercise (it takes into consideration barrier crossing).

=== Exercise 2 ===
Q1:
F+ H2 ==> FH + H, where AB=r1=H2 and HF=BC=r2
The reaction is exothermic as the enrgy of the reagents is higher that that of the products.
Position of TS: AB= 74.5 and BC = 181. It was identified thank to hammonds postulate: the position of the transition state determines if it will more closely resemble the products or the reagents. In this case, the transition state is early, as the reaction is exothermic. Theefore the TS will resemble the reagents, therefore the separation between the hydrogen molecule. will be smaller than that of HF

H + HF ==> H2 +F where AB=r1=HF and H2=BC=r2
The reaction is endothermic as the enrgy of the reagents is lower that that of the products.
Position of TS

Strength of H2 = 436 KJ mol
Strength of HF = 569 KJ mol
When breaking a strong bond to make a weaker bond, more energy is required and the reaction is endothermic. Therefore the formation of H2 from HF and H is endothermic, while the formation of HF is exothermic.

=== Conclusions===
=== References===
<references>
<ref name="second derivative test">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 2, pg 103-105 </ref>
<ref name="atkins"> Atkins, P. W., and Julio De Paula, Atkins' Physical chemistry. Oxford: Oxford University Press, 2006, chapter 18, pg 807-808</ref>
<ref name="barrier corssing">W. Kaplan, B. J. Berne, G. Ciccotti, D. F. Coker, Maxima and Minima with Applications: Practical Optimization and Duality, John Wiley & Sons, 1998, chap 1, pg 3-23. </ref>
<ref name="tunneling"> K.J. Laidler,Chemical kinetics, Harper & Row, 3rd ed., 1987, London, chap 4, pp 88-123
</references>

MRD:01512921

2020-05-21T11:29:54Z

Sp3418: /* References */

MRD:01512921

2020-05-21T11:23:20Z

Sp3418: /* Exercise 1 */

MRD:01512921

2020-05-20T19:02:36Z

Sp3418: /* Reaction dynamics report */

MRD:01512921

2020-05-20T19:01:57Z

Sp3418: /* Reaction dynamics report */

MRD:01512921

2020-05-20T19:00:54Z

Sp3418: /* Reaction dynamics report */

MRD:01512921

2020-05-20T18:59:50Z

Sp3418: /* Reaction dynamics report */

MRD:01512921

2020-05-20T18:59:18Z

Sp3418: /* References */

MRD:01512921

2020-05-20T18:53:46Z

Sp3418: /* Exercise 1 */

File:Trajectory 5 01512921.png

2020-05-20T18:33:23Z

Sp3418: