As6918:

MRD:01349769

2020-05-20T02:02:45Z

As6918:

== EXERCISE 1: H + H2 system ==
{{fontcolor1|blue|On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?}}

The transition state is defined as the maximum on the minimum energy path linking reactants and the products on a potential energy surface (PES) diagram. Consequently, if one starts a trajectory exactly at the transition state, with no initial momentum, it will remain there forever. However, if one changes the geometry by a small amount in the direction of the products it will roll towards the products (and similarly for the reactants). One way of locating the transition state is to start trajectories near the transition state and see whether they "roll" towards the reactants or products. For this system, a transition state can be mathematically defined to be where ∂V('''r1, r2''')/∂'''r1 '''= ∂V('''r1,r2''')/∂'''r2 '''= 0 '''and '''where the point in question is a maximum . To distinguish between a maximum and a local minimum, one can use the second partial derivative test [https://mathworld.wolfram.com/SecondDerivativeTest.html] . If H > 0 and ∂V2('''r1, r2''')/∂'''r1'''2 < 0 then a maximum is present which would be the transition state. For a local minimum, the condition is H > 0 and ∂V2('''r1, r2''')/∂'''r1'''2r > 0, which indicates a minimum in potential energy with respect to reaction coordinates (∂V2('''r1, r2''')/∂'''r2'''2 can also be used - the choice is arbitrary and gives the same result). Alternatively, inspection of a PES diagram for this reaction will show two minima and a maximum for potential energy, thus providing a visual distinction as well for this system.

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

Calculations showed that the best estimate is '''rts'''= 91pm. The reasoning here is that with this value, the potential energy of the transition state is at its lowest relative to other positions and the most realistic transition state is naturally going to be the one of lowest energy, since this make the reaction more kinetically feasible. Furthermore, given that partial bonds are present at a transition state, either through the process of forming or breaking, it's no surprise that the distance between H atoms would be greater than the equilibrium bond length of 74pm for dihydrogen. '''Figure 1''' shows an internuclear distances vs time plot for a trajectory originating near the stated transition state position of 91pm. Here, the "rolling" phenomenon is seen as the trajectory slowly moves towards the products and the distance, BC, decreases. Further, at smaller values of transition state position, a slow moving trajectory isn't seen, rather a very fast trajectory is observed - which can be a consequence of the repulsive energy present as distances between atoms shorten.

[[File:Q2_MRD_01349769.png|alt=Alt|Figure 1 showing a plot of internuclear distance against time for initial conditions of no momentum, AB = 91.1pm and BC = 90pm|465x465px|centre|thumb]]

'''Comment on how the ''mep'' and the trajectory you just calculated differ.*'''

The mep trajectory does not have any vibration occurring in the molecule during the trajectory, since momentum/velocity is set to zero in each step; this means that the total energy will decrease when the trajectory moves towards the products as the potential energy decreases while the kinetic energy remains 0 throughout the trajectory. As a further consequence, no conversion is possible between the two forms of energy as a result of this. These points can be seen in '''Figure 2''' which shows the contour plot for the mep for the same initial conditions. The motion of the trajectory is straight, owing to kinetic energy being removed in each step, thus there's no vibration in the molecule.

[[File:Q3_MRD_01349769.png|alt=Alt|Figure 2 showing the contour plot for the mep trajectory with same initial conditions (AB = 91.1pm, BC = 91pm, no momentum).|465x465px|centre|thumb]]

'''Figure 3 '''shows the contour plot for the dynamic trajectory for the same initial conditions. Here, the trajectory isn't straight, but wavy and so oscillating; this is due to changes in potential energy as some is converted to kinetic energy. This is in stark contrast to the mep trajectory.

[[File:Q3_MRD2_01349769.png|alt=Alt|Figure 3 showing a contour plot of the dynamic trajectory for the same initial conditions (AB = 91.1pm, BC = 91pm, no momentum).|465x465px|centre|thumb]]

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || || || ||
|-
| -3.1 || -4.1 || || || ||
|-
| -3.1 || -5.1 || || || ||
|-
| -5.1 || -10.1 || || || ||
|-
| -5.1 || -10.6 || || || ||
|}

MRD:01349769

2020-05-20T01:57:02Z

As6918:

== EXERCISE 1: H + H2 system ==
'''On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state is defined as the maximum on the minimum energy path linking reactants and the products on a potential energy surface (PES) diagram. Consequently, if one starts a trajectory exactly at the transition state, with no initial momentum, it will remain there forever. However, if one changes the geometry by a small amount in the direction of the products it will roll towards the products (and similarly for the reactants). One way of locating the transition state is to start trajectories near the transition state and see whether they "roll" towards the reactants or products. For this system, a transition state can be mathematically defined to be where ∂V('''r1, r2''')/∂'''r1 '''= ∂V('''r1,r2''')/∂'''r2 '''= 0 '''and '''where the point in question is a maximum . To distinguish between a maximum and a local minimum, one can use the second partial derivative test [https://mathworld.wolfram.com/SecondDerivativeTest.html] . If H > 0 and ∂V2('''r1, r2''')/∂'''r1'''2 < 0 then a maximum is present which would be the transition state. For a local minimum, the condition is H > 0 and ∂V2('''r1, r2''')/∂'''r1'''2r > 0, which indicates a minimum in potential energy with respect to reaction coordinates (∂V2('''r1, r2''')/∂'''r2'''2 can also be used - the choice is arbitrary and gives the same result). Alternatively, inspection of a PES diagram for this reaction will show two minima and a maximum for potential energy, thus providing a visual distinction as well for this system.

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

Calculations showed that the best estimate is '''rts'''= 91pm. The reasoning here is that with this value, the potential energy of the transition state is at its lowest relative to other positions and the most realistic transition state is naturally going to be the one of lowest energy, since this make the reaction more kinetically feasible. Furthermore, given that partial bonds are present at a transition state, either through the process of forming or breaking, it's no surprise that the distance between H atoms would be greater than the equilibrium bond length of 74pm for dihydrogen. '''Figure 1''' shows an internuclear distances vs time plot for a trajectory originating near the stated transition state position of 91pm. Here, the "rolling" phenomenon is seen as the trajectory slowly moves towards the products and the distance, BC, decreases. Further, at smaller values of transition state position, a slow moving trajectory isn't seen, rather a very fast trajectory is observed - which can be a consequence of the repulsive energy present as distances between atoms shorten.

[[File:Q2_MRD_01349769.png|alt=Alt|Figure 1 showing a plot of internuclear distance against time for initial conditions of no momentum, AB = 91.1pm and BC = 90pm|465x465px|centre|thumb]]

'''Comment on how the ''mep'' and the trajectory you just calculated differ.*'''

The mep trajectory does not have any vibration occurring in the molecule during the trajectory, since momentum/velocity is set to zero in each step; this means that the total energy will decrease when the trajectory moves towards the products as the potential energy decreases while the kinetic energy remains 0 throughout the trajectory. As a further consequence, no conversion is possible between the two forms of energy as a result of this. These points can be seen in '''Figure 2''' which shows the contour plot for the mep for the same initial conditions. The motion of the trajectory is straight, owing to kinetic energy being removed in each step, thus there's no vibration in the molecule.

[[File:Q3_MRD_01349769.png|alt=Alt|Figure 2 showing the contour plot for the mep trajectory with same initial conditions (AB = 91.1pm, BC = 91pm, no momentum).|465x465px|centre|thumb]]

'''Figure 3 '''shows the contour plot for the dynamic trajectory for the same initial conditions. Here, the trajectory isn't straight, but wavy and so oscillating; this is due to changes in potential energy as some is converted to kinetic energy. This is in stark contrast to the mep trajectory.

[[File:Q3_MRD2_01349769.png|alt=Alt|Figure 3 showing a contour plot of the dynamic trajectory for the same initial conditions (AB = 91.1pm, BC = 91pm, no momentum).|465x465px|centre|thumb]]

MRD:01349769

2020-05-20T00:52:51Z

As6918:

== EXERCISE 1: H + H2 system ==
'''On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state is defined as the maximum on the minimum energy path linking reactants and the products on a potential energy surface (PES) diagram. This point can be mathematically defined as ∂V('''ri...''')/∂'''ri'''=0 (the gradient of the potential is zero and the point is a maximum). Consequently, if one starts a trajectory exactly at the transition state, with no initial momentum, it will remain there forever. However, if one changes the geometry by a small amount in the direction of the products it will roll towards the products (and similarly for the reactants). Thus a transition state can be identified by starting the trajectories near the transition state and observing whether move towards the reactants or products. For this system, It can be distinguished from a local minimum by using the second derivative test [https://mathworld.wolfram.com/SecondDerivativeTest.html] . If D > 0 and ∂V2('''ri...''')/∂'''ri'''2 < 0 then a maximum is present which would be the transition state. For a local minimum, the condition is D > 0 and ∂V2('''ri...''')/∂'''ri'''2 > 0, which indicates a minimum in potential energy with respect to reaction coordinates. Alternatively, inspection of a PES diagram for this reaction will show two minima and a maximum for potential energy, thus providing a visual distinction too for this system.

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

Calculations showed that the best estimate is '''rts'''= 91pm. The reasoning here is that with this value, the potential energy of the transition state is at its lowest relative to other positions and the most realistic transition state is naturally going to be the one of lowest energy, since this make the reaction more kinetically feasible. Furthermore, given that partial bonds are present at a transition state, either through the process of forming or breaking, it's no surprise that the distance between H atoms would be greater than the equilibrium bond length of 74pm for dihydrogen. '''Figure 1''' shows an internuclear distances vs time plot for a trajectory originating near the stated transition state position of 91pm. Here, the "rolling" phenomenon is seen as the trajectory slowly moves towards the products and the distance, BC, decreases. Further, at smaller values of transition state position, a slow moving trajectory isn't seen, rather a very fast trajectory is observed - which can be a consequence of the repulsive energy present as distances between atoms shorten.

[[File:Q2_MRD_01349769.png|alt=Alt|Figure 1 showing a plot of internuclear distance against time for initial conditions of no momentum, AB = 91.1pm and BC = 90pm|465x465px|centre|thumb]]

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

The mep trajectory does not have any vibration occurring in the molecule during the trajectory, since momentum/velocity is set to zero in each step; this means that the total energy will decrease when the trajectory moves towards the products as the potential energy decreases while the kinetic energy remains 0 throughout the trajectory. As a further consequence, no conversion is possible between the two forms of energy as a result of this. These points can be seen in '''Figure 2''' which shows the contour plot for the mep for the same initial conditions. The motion of the trajectory is straight, owing to kinetic energy being removed in each step, thus there's no vibration in the molecule.

[[File:Q3_MRD_01349769.png|alt=Alt|Figure 2 showing the contour plot for the mep trajectory with same initial conditions (AB = 91.1pm, BC = 91pm, no momentum).|465x465px|centre|thumb]]

'''Figure 3 '''shows the contour plot for the dynamic trajectory for the same initial conditions. Here, the trajectory isn't straight, but wavy and so oscillating; this is due to changes in potential energy as some is converted to kinetic energy. This is in stark contrast to the mep trajectory.

[[File:Q3_MRD2_01349769.png|alt=Alt|Figure 3 showing a contour plot of the dynamic trajectory for the same initial conditions (AB = 91.1pm, BC = 91pm, no momentum).|465x465px|centre|thumb]]

MRD:01349769

2020-05-19T22:33:03Z

As6918:

== EXERCISE 1: H + H2 system ==
'''On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

The transition state is defined as the maximum on the minimum energy path linking reactants and the products on a potential energy surface (PES) diagram. This point can be mathematically defined as ∂V('''ri...''')/∂'''ri'''=0 (the gradient of the potential is zero and the point is a maximum). Consequently, if one starts a trajectory exactly at the transition state, with no initial momentum, it will remain there forever. However, if one changes the geometry by a small amount in the direction of the products it will roll towards the products (and similarly for the reactants). Thus a transition state can be identified by starting the trajectories near the transition state and observing whether move towards the reactants or products. For this system, It can be distinguished from a local minimum by using the second derivative test [https://mathworld.wolfram.com/SecondDerivativeTest.html] . If D > 0 and ∂V2('''ri...''')/∂'''ri'''2 < 0 then a maximum is present which would be the transition state. For a local minimum, the condition is D > 0 and ∂V2('''ri...''')/∂'''ri'''2 > 0, which indicates a minimum in potential energy with respect to reaction coordinates. Alternatively, inspection of a PES diagram for this reaction will show two minima and a maximum for potential energy, thus providing a visual distinction too for this system.

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

Calculations showed that the best estimate is '''rts'''= 91pm. The reasoning here is that with this value, the potential energy of the transition state is at its lowest relative to other positions and the most realistic transition state is naturally going to be the one of lowest energy, since this make the reaction more kinetically feasible. Furthermore, given that partial bonds are present at a transition state, either through the process of forming or breaking, it's no surprise that the distance between H atoms would be greater than the equilibrium bond length of 74pm for dihydrogen. Figure 1 shows a internuclear vs time plot for a trajectory near the transition state coordinates. Relative to shorter values of the transition state,

[[File:Q2_MRD_01349769.png|alt=Alt|Figure 1 showing a plot of internuclear distance against time for initial conditions of no momentum, AB = 91.1pm and BC = 90pm|465x465px|centre|thumb]]Figure 1 shows an internuclear distances vs time plot for a trajectory originating near the stated transition state position of 91pm. Here, the "rolling" phenomenon is seen as the trajectory slowly moves towards the products and the distance, BC, decreases. Further, at smaller values of transition state position, a slow moving trajectory isn't seen, rather a very fast trajectory is observed - which can be a consequence of the repulsive energy present as distances between atoms shorten.

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

The mep trajectory does not have any vibration occurring in the molecule during the trajectory, since momentum/velocity is set to zero in each step; this means as a result that the total energy will decrease when the trajectory moves towards the products as the potential energy decreases while the kinetic energy remains 0 throughout the trajectory.

[[File:Q3_MRD_01349769.png|alt=Alt|Figure 1 showing a plot of internuclear distance against time for initial conditions of no momentum, AB = 91.1pm and BC = 90pm|465x465px|centre|thumb]]Figure 1 shows an internuclear distances vs time plot for a trajectory originating near the stated transition state position of 91pm. Here, the "rolling" phenomenon is seen as the trajectory slowly moves towards the products and the distance, BC, decreases. Further, at smaller values of transition state position, a slow moving trajectory isn't seen, rather a very fast trajectory is observed - which can be a consequence of the repulsive energy present as distances between atoms shorten.

[[File:Q3_MRD2_01349769.png|alt=Alt|Figure 1 showing a plot of internuclear distance against time for initial conditions of no momentum, AB = 91.1pm and BC = 90pm|465x465px|centre|thumb]]Figure 1 shows an internuclear distances vs time plot for a trajectory originating near the stated transition state position of 91pm. Here, the "rolling" phenomenon is seen as the trajectory slowly moves towards the products and the distance, BC, decreases. Further, at smaller values of transition state position, a slow moving trajectory isn't seen, rather a very fast trajectory is observed - which can be a consequence of the repulsive energy present as distances between atoms shorten.