MRD:MAlberts

2020-05-08T15:40:13Z

Maa218: /* Report the activation energy for both reactions. */

== Exercise 1 ==
[[File:MAlberts Saddle point.png|thumb|''Figure 1: Saddle point''|450x450px]]
=== 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? ===
<math></math>
The transition state on a potential energy surface is defined as a first order saddle point. A saddle point is mathematically defined in cartesian space as:

<math>F:U \rightarrow \R</math> with <math>U \subseteq \R^2</math>

'''Condition 1: ''' <math>\frac{\partial F(\overrightarrow{r})}{\partial \overrightarrow{r}} = 0</math>

'''Condition 2''': <math>\frac{\partial ^2 F(r_1)}{\partial ^2 r_1} > 0</math> and <math>\frac{\partial ^2 F(r_2)}{\partial ^2 r_2} < 0</math> with <math>r_1</math> and <math>r_2</math> being interchangable.

Visually this means that the saddle point is the point where the function F has a minimum along one direction and a maximum along another orthogonal to the first one.

In the context of a potential energy surface diagram, the transition state is the saddle point. It can be thought of as the maximum of the minimum energy path. This, by definition, has to be the saddle point as the minimum energy path runs along the minima of the potential energy surface and thus adheres to the definition of there being a minimum and maximum in the same point. In the case of a local minima or maxima condition, 1 would also be fulfilled. Condition 2 is needed to distinguish between a local minimum or maxima and the saddle point as for a minimum both derivatives would be positive while for a maximum they would both be negative.

Another way to identify a saddle point is to form the Hess matrix. If the Hess matrix is indefinite the point is a saddle point.

=== Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===
The transition state is defined as a saddle point on the potential energy surface. Thus if the molecule and the atom it collides with have no momentum and they are exactly at the transition state they will remain there forever. The transition state was found by setting the momentums to zero and as this system is symmetric <math>r_1</math> equal to <math>r_2</math>. Then the values of <math>r_1</math> and <math>r_2</math> were varied while observing the contour plot and the forces along <math>r_1</math> and <math>r_2</math>. For the position to be the transition state, no movement can be observed in the contour plot and the forces need to be equal to zero as the atoms are static and don't move. Additionally, verify the position of the transition state the internuclear distance vs time was examined. At the transition state, this plot should show two flat lines as there is no movement at the transition state.

The transition state was found to be at 90.772pm. This is corroborated by the forces along <math>r_1</math> and <math>r_2</math>being zero (see figure 2), the contour plot showing no movement (see figure 3) and the internuclear distance vs time plot showed two flat lines (see Figure 4).
[[File:MAlberts Q2 contour plot.png|left|thumb|350x350px|Figure 2: Contour plot at the transition state showing no movement.]]
[[File:MAlberts Q2 Internuclear distance vs time.png|thumb|350x350px|Figure 4: Internuclear distance vs time plot showing no movement.]]
[[File:MAlberts Q2 Forces.png|thumb|508x508px|Figure 3: Screenshot of the ''lepsgui.py'' control panel showing that the forces are equal to zero.|centre]]
<math></math><math></math>

The transition state was found to be at 90.772pm. This is corroborated by the forces along <math>r_1</math> and <math>r_2</math>being zero (see figure 2), the contour plot showing no movement (see figure 3) and the internuclear distance vs time plot showed two flat lines (see Figure 4).

=== Comment on how the ''mep ''and the trajectory you just calculated differ. ===
The minimum energy path describes the path that follows all the minima of the potential energy surface. In the simulation with the minimum energy path the molecule follows the minima at a velocity approaching zero (internuclear velocity vs time remains at zero). Additionally, there are no vibrations in this simulation and the total energy decreases. The molecule starts close to the transition state and the moves at a velocity near zero to a position in the product where it seems to stop and remain.

For the dynamic simulation, the molecules do not follow a straight path but as indicated by the oscillatory behaviour of the internuclear distance show vibrations. In addition, the total energy does not decrease as compared to the ''mep'' and the velocity is not at zero but the molecule continues at a constant oscillating velocity.
[[File:MAlberts Q3.png|thumb|800x800px|Figure 5: The path calculated by the dynamic simulation on the right and ''mep'' path on the left|centre]]

===Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable"
!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
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|In this collision, the Ha-Hb collides with Hc. This breaks the bond between Ha-Hband forms a new one between Ha and Hc. Initially, the Ha-Hb molecule does not show any vibration indicated by the path not oscillating. However, the newly formed molecule consisting of Ha and Hc shows oscillation indicating that the new bond vibrates.
|[[File:MAlberts Q4 1.png|frameless]]
|-
|-3.1
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|In this collision, Ha-Hb collides again with Hc. However, in contrast to the case above no reaction occurs and Hc is repelled. The bond between Ha-Hb vibrates as indicated by the oscillating path. This is caused by the increased momentum of Ha-Hb in contrast to reaction one. On the other hand, the momentum of Hc is decreased which also causes a decrease in the kinetic energy which is the reason that the activation energy barrier can't be overcome and no reaction occurs.
|[[File:MAlberts Q4 2.png|frameless]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Reaction three keeps the same momentum from reaction one but increases the momentum of Hc. This causes an increase in the kinetic energy and thus Hc can surmount the energy barrier and the reaction can proceed. Just with reaction two the bond between Ha-Hb vibrates and as seen in reaction one the newly formed bond between Hc and Ha vibrates more strongly than the original bond.
|[[File:MAlberts Q4 3.png|frameless]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|In this reaction, both Ha-Hb and Hc are given a high momentum and thus have a high kinetic energy. However when they collide no reaction happens. At first the bond length between Ha-Hb lengthens and it seems like the reaction is going to proceed. However then due to very strong vibrations the newly formed bond between Hb-Hc breaks and the one between Ha-Hb is reformed. This indicates that even if the molecules have enough kinetic energy to overcome the activation energy the vibrations still play a large role.
|[[File:MAlberts Q4 4.png|frameless]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|For reaction five the kinetic energy for Hc is increased again. However, in contrast to reaction four the reaction proceeds in this case. Here the Ha-Hb molecule seems to bounce back from the potential energy surface three times and then be reflected in such a way that it overcomes the activation energy boundary. The newly formed bond between Hc and Hb shows strong vibrations.
|[[File:MAlberts Q4 5.png|frameless]]
|}
The conclusion that can be drawn from this table is that even if the molecules have enough kinetic energy to overcome the activation energy boundary a reaction does not necessarily have to happen. If one molecule has a high excess of kinetic energy it can happen that the newly formed bond is torn apart by strong vibrations.

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

Transition state theory makes four essential assumptions. The first one is that all molecules that move into the transition state are going to form products or if they enter from the product sides that all molecules reform reactants. This means that reactants that e.g move into the transition state but do not cross it are counted as products even though they remain products. An example of reactants moving into the transition state but not forming products can be seen above in reaction four. Another, assumption of transition state theory is that no quantum effects occur. One of these effects tunnelling where a molecule that according to the theory would not have enough energy to form products still does. This effect is called tunnelling as it travels through the potential energy surface.

In general transition state theory tends to overestimate the reaction rate values compared to experimentally obtained ones. This is mostly due to the fact that it is assumed that all molecules that enter the transition state cross it.
==Exercise 2==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===
The energetics of both reactions were evaluated using the surface plot feature of lepsgui as shown below:
[[File:MAlberts Q6 Combined.png|centre|thumb|800x800px|Figure 7: Potential energy surface diagram of the reaction of H2 + F and HF + H.]]

As can be seen in the potential energy surface diagram for the reaction of H2 + F the reactants are higher in energy than the products. Thus the reaction and between H2 and F is exothermic. In this reaction, the bond between H2 is broken and a new one between HF is formed. As the reaction is exothermic the energy required to break the bond in H2 has to be smaller than the energy released by the formation of the HF bond. Thus the HF bond is stronger than the of H2 bond.

The reverse is true in the reaction of HF and H. Here the reactants are higher in energy than the products and the reaction is endothermic. Thus energy released by the formation of the H2 bond is smaller than the energy required to break the bond in HF.

===Locate the approximate position of the transition state.===
The transition state was determined using the same methods as in question 1.2. However, as in this reaction, not all atoms are the same, the transition state will not be found at a point where <math>r_1</math> and <math>r_2</math>are equal.
[[File:MAlberts Q7 Combined.png|centre|thumb|960x960px|Figure 8: The transition state of the reaction can be seen above.]]
As can be seen above the transition state was found at were <math>r_{HF} = 181.1pm</math> and <math>r_{HH} = 74.5pm</math>. This is most probably not the exact transition state as the forces along the BC axis are not equal to zero.

===Report the activation energy for both reactions.===
The activation energy can be interpreted as the energy required to reach the transition state. Thus the energy of the system at the transition state was determined and then compared to the energy of the reactants for both reactions. The energy for the transition state was found to be -433.98 kJ/mol. The energy of the reactants was by setting the momentum to zero and adjusting the bond distances so that the molecule was at a local minimum (the bond length of the H2 or HF molecule) in the potential energy surface and very far removed from the F or H with which it is going to react.

'''H2 + F:''' The energy of the reactants was found to be -435.099 kJ/mol with H2 bond distance being 74 pm and 500 pm for the H2-F distance. Thus the activation energy is 1.119 kJ/mol.

'''HF + H:''' The energy of the reactants was found to -560.402 kJ/mol with HF bond distance being 91 pm and 500 for the HF-H distance. Thus the activation energy is 126.422 kJ/mol.

===In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===

===Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.===

MRD:MAlberts

2020-05-08T15:24:32Z

Maa218: /* Report the activation energy for both reactions. */

== Exercise 1 ==
[[File:MAlberts Saddle point.png|thumb|''Figure 1: Saddle point''|450x450px]]
=== 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? ===
<math></math>
The transition state on a potential energy surface is defined as a first order saddle point. A saddle point is mathematically defined in cartesian space as:

<math>F:U \rightarrow \R</math> with <math>U \subseteq \R^2</math>

'''Condition 1: ''' <math>\frac{\partial F(\overrightarrow{r})}{\partial \overrightarrow{r}} = 0</math>

'''Condition 2''': <math>\frac{\partial ^2 F(r_1)}{\partial ^2 r_1} > 0</math> and <math>\frac{\partial ^2 F(r_2)}{\partial ^2 r_2} < 0</math> with <math>r_1</math> and <math>r_2</math> being interchangable.

Visually this means that the saddle point is the point where the function F has a minimum along one direction and a maximum along another orthogonal to the first one.

In the context of a potential energy surface diagram, the transition state is the saddle point. It can be thought of as the maximum of the minimum energy path. This, by definition, has to be the saddle point as the minimum energy path runs along the minima of the potential energy surface and thus adheres to the definition of there being a minimum and maximum in the same point. In the case of a local minima or maxima condition, 1 would also be fulfilled. Condition 2 is needed to distinguish between a local minimum or maxima and the saddle point as for a minimum both derivatives would be positive while for a maximum they would both be negative.

Another way to identify a saddle point is to form the Hess matrix. If the Hess matrix is indefinite the point is a saddle point.

=== Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===
The transition state is defined as a saddle point on the potential energy surface. Thus if the molecule and the atom it collides with have no momentum and they are exactly at the transition state they will remain there forever. The transition state was found by setting the momentums to zero and as this system is symmetric <math>r_1</math> equal to <math>r_2</math>. Then the values of <math>r_1</math> and <math>r_2</math> were varied while observing the contour plot and the forces along <math>r_1</math> and <math>r_2</math>. For the position to be the transition state, no movement can be observed in the contour plot and the forces need to be equal to zero as the atoms are static and don't move. Additionally, verify the position of the transition state the internuclear distance vs time was examined. At the transition state, this plot should show two flat lines as there is no movement at the transition state.

The transition state was found to be at 90.772pm. This is corroborated by the forces along <math>r_1</math> and <math>r_2</math>being zero (see figure 2), the contour plot showing no movement (see figure 3) and the internuclear distance vs time plot showed two flat lines (see Figure 4).
[[File:MAlberts Q2 contour plot.png|left|thumb|350x350px|Figure 2: Contour plot at the transition state showing no movement.]]
[[File:MAlberts Q2 Internuclear distance vs time.png|thumb|350x350px|Figure 4: Internuclear distance vs time plot showing no movement.]]
[[File:MAlberts Q2 Forces.png|thumb|508x508px|Figure 3: Screenshot of the ''lepsgui.py'' control panel showing that the forces are equal to zero.|centre]]
<math></math><math></math>

The transition state was found to be at 90.772pm. This is corroborated by the forces along <math>r_1</math> and <math>r_2</math>being zero (see figure 2), the contour plot showing no movement (see figure 3) and the internuclear distance vs time plot showed two flat lines (see Figure 4).

=== Comment on how the ''mep ''and the trajectory you just calculated differ. ===
The minimum energy path describes the path that follows all the minima of the potential energy surface. In the simulation with the minimum energy path the molecule follows the minima at a velocity approaching zero (internuclear velocity vs time remains at zero). Additionally, there are no vibrations in this simulation and the total energy decreases. The molecule starts close to the transition state and the moves at a velocity near zero to a position in the product where it seems to stop and remain.

For the dynamic simulation, the molecules do not follow a straight path but as indicated by the oscillatory behaviour of the internuclear distance show vibrations. In addition, the total energy does not decrease as compared to the ''mep'' and the velocity is not at zero but the molecule continues at a constant oscillating velocity.
[[File:MAlberts Q3.png|thumb|800x800px|Figure 5: The path calculated by the dynamic simulation on the right and ''mep'' path on the left|centre]]

===Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable"
!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
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|In this collision, the Ha-Hb collides with Hc. This breaks the bond between Ha-Hband forms a new one between Ha and Hc. Initially, the Ha-Hb molecule does not show any vibration indicated by the path not oscillating. However, the newly formed molecule consisting of Ha and Hc shows oscillation indicating that the new bond vibrates.
|[[File:MAlberts Q4 1.png|frameless]]
|-
|-3.1
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|In this collision, Ha-Hb collides again with Hc. However, in contrast to the case above no reaction occurs and Hc is repelled. The bond between Ha-Hb vibrates as indicated by the oscillating path. This is caused by the increased momentum of Ha-Hb in contrast to reaction one. On the other hand, the momentum of Hc is decreased which also causes a decrease in the kinetic energy which is the reason that the activation energy barrier can't be overcome and no reaction occurs.
|[[File:MAlberts Q4 2.png|frameless]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Reaction three keeps the same momentum from reaction one but increases the momentum of Hc. This causes an increase in the kinetic energy and thus Hc can surmount the energy barrier and the reaction can proceed. Just with reaction two the bond between Ha-Hb vibrates and as seen in reaction one the newly formed bond between Hc and Ha vibrates more strongly than the original bond.
|[[File:MAlberts Q4 3.png|frameless]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|In this reaction, both Ha-Hb and Hc are given a high momentum and thus have a high kinetic energy. However when they collide no reaction happens. At first the bond length between Ha-Hb lengthens and it seems like the reaction is going to proceed. However then due to very strong vibrations the newly formed bond between Hb-Hc breaks and the one between Ha-Hb is reformed. This indicates that even if the molecules have enough kinetic energy to overcome the activation energy the vibrations still play a large role.
|[[File:MAlberts Q4 4.png|frameless]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|For reaction five the kinetic energy for Hc is increased again. However, in contrast to reaction four the reaction proceeds in this case. Here the Ha-Hb molecule seems to bounce back from the potential energy surface three times and then be reflected in such a way that it overcomes the activation energy boundary. The newly formed bond between Hc and Hb shows strong vibrations.
|[[File:MAlberts Q4 5.png|frameless]]
|}
The conclusion that can be drawn from this table is that even if the molecules have enough kinetic energy to overcome the activation energy boundary a reaction does not necessarily have to happen. If one molecule has a high excess of kinetic energy it can happen that the newly formed bond is torn apart by strong vibrations.

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

Transition state theory makes four essential assumptions. The first one is that all molecules that move into the transition state are going to form products or if they enter from the product sides that all molecules reform reactants. This means that reactants that e.g move into the transition state but do not cross it are counted as products even though they remain products. An example of reactants moving into the transition state but not forming products can be seen above in reaction four. Another, assumption of transition state theory is that no quantum effects occur. One of these effects tunnelling where a molecule that according to the theory would not have enough energy to form products still does. This effect is called tunnelling as it travels through the potential energy surface.

In general transition state theory tends to overestimate the reaction rate values compared to experimentally obtained ones. This is mostly due to the fact that it is assumed that all molecules that enter the transition state cross it.
==Exercise 2==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===
The energetics of both reactions were evaluated using the surface plot feature of lepsgui as shown below:
[[File:MAlberts Q6 Combined.png|centre|thumb|800x800px|Figure 7: Potential energy surface diagram of the reaction of H2 + F and HF + H.]]

As can be seen in the potential energy surface diagram for the reaction of H2 + F the reactants are higher in energy than the products. Thus the reaction and between H2 and F is exothermic. In this reaction, the bond between H2 is broken and a new one between HF is formed. As the reaction is exothermic the energy required to break the bond in H2 has to be smaller than the energy released by the formation of the HF bond. Thus the HF bond is stronger than the of H2 bond.

The reverse is true in the reaction of HF and H. Here the reactants are higher in energy than the products and the reaction is endothermic. Thus energy released by the formation of the H2 bond is smaller than the energy required to break the bond in HF.

===Locate the approximate position of the transition state.===
The transition state was determined using the same methods as in question 1.2. However, as in this reaction, not all atoms are the same, the transition state will not be found at a point where <math>r_1</math> and <math>r_2</math>are equal.
[[File:MAlberts Q7 Combined.png|centre|thumb|960x960px|Figure 8: The transition state of the reaction can be seen above.]]
As can be seen above the transition state was found at were <math>r_{HF} = 181.1pm</math> and <math>r_{HH} = 74.5pm</math>. This is most probably not the exact transition state as the forces along the BC axis are not equal to zero.

===Report the activation energy for both reactions.===
The activation energy can be interpreted as the energy required to reach the transition state. Thus the energy of the system at the transition state was determined and then compared to the energy of the reactants for both reactions. The energy for the transition state was found to be

===In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===

===Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.===

MRD:MAlberts

2020-05-08T13:56:26Z

Maa218: /* Locate the approximate position of the transition state. */

== Exercise 1 ==
[[File:MAlberts Saddle point.png|thumb|''Figure 1: Saddle point''|450x450px]]
=== 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? ===
<math></math>
The transition state on a potential energy surface is defined as a first order saddle point. A saddle point is mathematically defined in cartesian space as:

<math>F:U \rightarrow \R</math> with <math>U \subseteq \R^2</math>

'''Condition 1: ''' <math>\frac{\partial F(\overrightarrow{r})}{\partial \overrightarrow{r}} = 0</math>

'''Condition 2''': <math>\frac{\partial ^2 F(r_1)}{\partial ^2 r_1} > 0</math> and <math>\frac{\partial ^2 F(r_2)}{\partial ^2 r_2} < 0</math> with <math>r_1</math> and <math>r_2</math> being interchangable.

Visually this means that the saddle point is the point where the function F has a minimum along one direction and a maximum along another orthogonal to the first one.

In the context of a potential energy surface diagram, the transition state is the saddle point. It can be thought of as the maximum of the minimum energy path. This, by definition, has to be the saddle point as the minimum energy path runs along the minima of the potential energy surface and thus adheres to the definition of there being a minimum and maximum in the same point. In the case of a local minima or maxima condition, 1 would also be fulfilled. Condition 2 is needed to distinguish between a local minimum or maxima and the saddle point as for a minimum both derivatives would be positive while for a maximum they would both be negative.

Another way to identify a saddle point is to form the Hess matrix. If the Hess matrix is indefinite the point is a saddle point.

=== Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===
The transition state is defined as a saddle point on the potential energy surface. Thus if the molecule and the atom it collides with have no momentum and they are exactly at the transition state they will remain there forever. The transition state was found by setting the momentums to zero and as this system is symmetric <math>r_1</math> equal to <math>r_2</math>. Then the values of <math>r_1</math> and <math>r_2</math> were varied while observing the contour plot and the forces along <math>r_1</math> and <math>r_2</math>. For the position to be the transition state, no movement can be observed in the contour plot and the forces need to be equal to zero as the atoms are static and don't move. Additionally, verify the position of the transition state the internuclear distance vs time was examined. At the transition state, this plot should show two flat lines as there is no movement at the transition state.

The transition state was found to be at 90.772pm. This is corroborated by the forces along <math>r_1</math> and <math>r_2</math>being zero (see figure 2), the contour plot showing no movement (see figure 3) and the internuclear distance vs time plot showed two flat lines (see Figure 4).
[[File:MAlberts Q2 contour plot.png|left|thumb|350x350px|Figure 2: Contour plot at the transition state showing no movement.]]
[[File:MAlberts Q2 Internuclear distance vs time.png|thumb|350x350px|Figure 4: Internuclear distance vs time plot showing no movement.]]
[[File:MAlberts Q2 Forces.png|thumb|508x508px|Figure 3: Screenshot of the ''lepsgui.py'' control panel showing that the forces are equal to zero.|centre]]
<math></math><math></math>

The transition state was found to be at 90.772pm. This is corroborated by the forces along <math>r_1</math> and <math>r_2</math>being zero (see figure 2), the contour plot showing no movement (see figure 3) and the internuclear distance vs time plot showed two flat lines (see Figure 4).

=== Comment on how the ''mep ''and the trajectory you just calculated differ. ===
The minimum energy path describes the path that follows all the minima of the potential energy surface. In the simulation with the minimum energy path the molecule follows the minima at a velocity approaching zero (internuclear velocity vs time remains at zero). Additionally, there are no vibrations in this simulation and the total energy decreases. The molecule starts close to the transition state and the moves at a velocity near zero to a position in the product where it seems to stop and remain.

For the dynamic simulation, the molecules do not follow a straight path but as indicated by the oscillatory behaviour of the internuclear distance show vibrations. In addition, the total energy does not decrease as compared to the ''mep'' and the velocity is not at zero but the molecule continues at a constant oscillating velocity.
[[File:MAlberts Q3.png|thumb|800x800px|Figure 5: The path calculated by the dynamic simulation on the right and ''mep'' path on the left|centre]]

===Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable"
!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
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|In this collision, the Ha-Hb collides with Hc. This breaks the bond between Ha-Hband forms a new one between Ha and Hc. Initially, the Ha-Hb molecule does not show any vibration indicated by the path not oscillating. However, the newly formed molecule consisting of Ha and Hc shows oscillation indicating that the new bond vibrates.
|[[File:MAlberts Q4 1.png|frameless]]
|-
|-3.1
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|In this collision, Ha-Hb collides again with Hc. However, in contrast to the case above no reaction occurs and Hc is repelled. The bond between Ha-Hb vibrates as indicated by the oscillating path. This is caused by the increased momentum of Ha-Hb in contrast to reaction one. On the other hand, the momentum of Hc is decreased which also causes a decrease in the kinetic energy which is the reason that the activation energy barrier can't be overcome and no reaction occurs.
|[[File:MAlberts Q4 2.png|frameless]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Reaction three keeps the same momentum from reaction one but increases the momentum of Hc. This causes an increase in the kinetic energy and thus Hc can surmount the energy barrier and the reaction can proceed. Just with reaction two the bond between Ha-Hb vibrates and as seen in reaction one the newly formed bond between Hc and Ha vibrates more strongly than the original bond.
|[[File:MAlberts Q4 3.png|frameless]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|In this reaction, both Ha-Hb and Hc are given a high momentum and thus have a high kinetic energy. However when they collide no reaction happens. At first the bond length between Ha-Hb lengthens and it seems like the reaction is going to proceed. However then due to very strong vibrations the newly formed bond between Hb-Hc breaks and the one between Ha-Hb is reformed. This indicates that even if the molecules have enough kinetic energy to overcome the activation energy the vibrations still play a large role.
|[[File:MAlberts Q4 4.png|frameless]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|For reaction five the kinetic energy for Hc is increased again. However, in contrast to reaction four the reaction proceeds in this case. Here the Ha-Hb molecule seems to bounce back from the potential energy surface three times and then be reflected in such a way that it overcomes the activation energy boundary. The newly formed bond between Hc and Hb shows strong vibrations.
|[[File:MAlberts Q4 5.png|frameless]]
|}
The conclusion that can be drawn from this table is that even if the molecules have enough kinetic energy to overcome the activation energy boundary a reaction does not necessarily have to happen. If one molecule has a high excess of kinetic energy it can happen that the newly formed bond is torn apart by strong vibrations.

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

Transition state theory makes four essential assumptions. The first one is that all molecules that move into the transition state are going to form products or if they enter from the product sides that all molecules reform reactants. This means that reactants that e.g move into the transition state but do not cross it are counted as products even though they remain products. An example of reactants moving into the transition state but not forming products can be seen above in reaction four. Another, assumption of transition state theory is that no quantum effects occur. One of these effects tunnelling where a molecule that according to the theory would not have enough energy to form products still does. This effect is called tunnelling as it travels through the potential energy surface.

In general transition state theory tends to overestimate the reaction rate values compared to experimentally obtained ones. This is mostly due to the fact that it is assumed that all molecules that enter the transition state cross it.
==Exercise 2==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===
The energetics of both reactions were evaluated using the surface plot feature of lepsgui as shown below:
[[File:MAlberts Q6 Combined.png|centre|thumb|800x800px|Figure 7: Potential energy surface diagram of the reaction of H2 + F and HF + H.]]

As can be seen in the potential energy surface diagram for the reaction of H2 + F the reactants are higher in energy than the products. Thus the reaction and between H2 and F is exothermic. In this reaction, the bond between H2 is broken and a new one between HF is formed. As the reaction is exothermic the energy required to break the bond in H2 has to be smaller than the energy released by the formation of the HF bond. Thus the HF bond is stronger than the of H2 bond.

The reverse is true in the reaction of HF and H. Here the reactants are higher in energy than the products and the reaction is endothermic. Thus energy released by the formation of the H2 bond is smaller than the energy required to break the bond in HF.

===Locate the approximate position of the transition state.===
The transition state was determined using the same methods as in question 1.2. However, as in this reaction, not all atoms are the same, the transition state will not be found at a point where <math>r_1</math> and <math>r_2</math>are equal.
[[File:MAlberts Q7 Combined.png|centre|thumb|960x960px|Figure 8: The transition state of the reaction can be seen above.]]
As can be seen above the transition state was found at were <math>r_{HF} = 181.1pm</math> and <math>r_{HH} = 74.5pm</math>. This is most probably not the exact transition state as the forces along the BC axis are not equal to zero.

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

===In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===

===Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.===

MRD:MAlberts

2020-05-08T13:55:04Z

Maa218: /* Locate the approximate position of the transition state. */

== Exercise 1 ==
[[File:MAlberts Saddle point.png|thumb|''Figure 1: Saddle point''|450x450px]]
=== 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? ===
<math></math>
The transition state on a potential energy surface is defined as a first order saddle point. A saddle point is mathematically defined in cartesian space as:

<math>F:U \rightarrow \R</math> with <math>U \subseteq \R^2</math>

'''Condition 1: ''' <math>\frac{\partial F(\overrightarrow{r})}{\partial \overrightarrow{r}} = 0</math>

'''Condition 2''': <math>\frac{\partial ^2 F(r_1)}{\partial ^2 r_1} > 0</math> and <math>\frac{\partial ^2 F(r_2)}{\partial ^2 r_2} < 0</math> with <math>r_1</math> and <math>r_2</math> being interchangable.

Visually this means that the saddle point is the point where the function F has a minimum along one direction and a maximum along another orthogonal to the first one.

In the context of a potential energy surface diagram, the transition state is the saddle point. It can be thought of as the maximum of the minimum energy path. This, by definition, has to be the saddle point as the minimum energy path runs along the minima of the potential energy surface and thus adheres to the definition of there being a minimum and maximum in the same point. In the case of a local minima or maxima condition, 1 would also be fulfilled. Condition 2 is needed to distinguish between a local minimum or maxima and the saddle point as for a minimum both derivatives would be positive while for a maximum they would both be negative.

Another way to identify a saddle point is to form the Hess matrix. If the Hess matrix is indefinite the point is a saddle point.

=== Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===
The transition state is defined as a saddle point on the potential energy surface. Thus if the molecule and the atom it collides with have no momentum and they are exactly at the transition state they will remain there forever. The transition state was found by setting the momentums to zero and as this system is symmetric <math>r_1</math> equal to <math>r_2</math>. Then the values of <math>r_1</math> and <math>r_2</math> were varied while observing the contour plot and the forces along <math>r_1</math> and <math>r_2</math>. For the position to be the transition state, no movement can be observed in the contour plot and the forces need to be equal to zero as the atoms are static and don't move. Additionally, verify the position of the transition state the internuclear distance vs time was examined. At the transition state, this plot should show two flat lines as there is no movement at the transition state.

The transition state was found to be at 90.772pm. This is corroborated by the forces along <math>r_1</math> and <math>r_2</math>being zero (see figure 2), the contour plot showing no movement (see figure 3) and the internuclear distance vs time plot showed two flat lines (see Figure 4).
[[File:MAlberts Q2 contour plot.png|left|thumb|350x350px|Figure 2: Contour plot at the transition state showing no movement.]]
[[File:MAlberts Q2 Internuclear distance vs time.png|thumb|350x350px|Figure 4: Internuclear distance vs time plot showing no movement.]]
[[File:MAlberts Q2 Forces.png|thumb|508x508px|Figure 3: Screenshot of the ''lepsgui.py'' control panel showing that the forces are equal to zero.|centre]]
<math></math><math></math>

The transition state was found to be at 90.772pm. This is corroborated by the forces along <math>r_1</math> and <math>r_2</math>being zero (see figure 2), the contour plot showing no movement (see figure 3) and the internuclear distance vs time plot showed two flat lines (see Figure 4).

=== Comment on how the ''mep ''and the trajectory you just calculated differ. ===
The minimum energy path describes the path that follows all the minima of the potential energy surface. In the simulation with the minimum energy path the molecule follows the minima at a velocity approaching zero (internuclear velocity vs time remains at zero). Additionally, there are no vibrations in this simulation and the total energy decreases. The molecule starts close to the transition state and the moves at a velocity near zero to a position in the product where it seems to stop and remain.

For the dynamic simulation, the molecules do not follow a straight path but as indicated by the oscillatory behaviour of the internuclear distance show vibrations. In addition, the total energy does not decrease as compared to the ''mep'' and the velocity is not at zero but the molecule continues at a constant oscillating velocity.
[[File:MAlberts Q3.png|thumb|800x800px|Figure 5: The path calculated by the dynamic simulation on the right and ''mep'' path on the left|centre]]

===Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable"
!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
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|In this collision, the Ha-Hb collides with Hc. This breaks the bond between Ha-Hband forms a new one between Ha and Hc. Initially, the Ha-Hb molecule does not show any vibration indicated by the path not oscillating. However, the newly formed molecule consisting of Ha and Hc shows oscillation indicating that the new bond vibrates.
|[[File:MAlberts Q4 1.png|frameless]]
|-
|-3.1
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|In this collision, Ha-Hb collides again with Hc. However, in contrast to the case above no reaction occurs and Hc is repelled. The bond between Ha-Hb vibrates as indicated by the oscillating path. This is caused by the increased momentum of Ha-Hb in contrast to reaction one. On the other hand, the momentum of Hc is decreased which also causes a decrease in the kinetic energy which is the reason that the activation energy barrier can't be overcome and no reaction occurs.
|[[File:MAlberts Q4 2.png|frameless]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Reaction three keeps the same momentum from reaction one but increases the momentum of Hc. This causes an increase in the kinetic energy and thus Hc can surmount the energy barrier and the reaction can proceed. Just with reaction two the bond between Ha-Hb vibrates and as seen in reaction one the newly formed bond between Hc and Ha vibrates more strongly than the original bond.
|[[File:MAlberts Q4 3.png|frameless]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|In this reaction, both Ha-Hb and Hc are given a high momentum and thus have a high kinetic energy. However when they collide no reaction happens. At first the bond length between Ha-Hb lengthens and it seems like the reaction is going to proceed. However then due to very strong vibrations the newly formed bond between Hb-Hc breaks and the one between Ha-Hb is reformed. This indicates that even if the molecules have enough kinetic energy to overcome the activation energy the vibrations still play a large role.
|[[File:MAlberts Q4 4.png|frameless]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|For reaction five the kinetic energy for Hc is increased again. However, in contrast to reaction four the reaction proceeds in this case. Here the Ha-Hb molecule seems to bounce back from the potential energy surface three times and then be reflected in such a way that it overcomes the activation energy boundary. The newly formed bond between Hc and Hb shows strong vibrations.
|[[File:MAlberts Q4 5.png|frameless]]
|}
The conclusion that can be drawn from this table is that even if the molecules have enough kinetic energy to overcome the activation energy boundary a reaction does not necessarily have to happen. If one molecule has a high excess of kinetic energy it can happen that the newly formed bond is torn apart by strong vibrations.

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

Transition state theory makes four essential assumptions. The first one is that all molecules that move into the transition state are going to form products or if they enter from the product sides that all molecules reform reactants. This means that reactants that e.g move into the transition state but do not cross it are counted as products even though they remain products. An example of reactants moving into the transition state but not forming products can be seen above in reaction four. Another, assumption of transition state theory is that no quantum effects occur. One of these effects tunnelling where a molecule that according to the theory would not have enough energy to form products still does. This effect is called tunnelling as it travels through the potential energy surface.

In general transition state theory tends to overestimate the reaction rate values compared to experimentally obtained ones. This is mostly due to the fact that it is assumed that all molecules that enter the transition state cross it.
==Exercise 2==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===
The energetics of both reactions were evaluated using the surface plot feature of lepsgui as shown below:
[[File:MAlberts Q6 Combined.png|centre|thumb|800x800px|Figure 7: Potential energy surface diagram of the reaction of H2 + F and HF + H.]]

As can be seen in the potential energy surface diagram for the reaction of H2 + F the reactants are higher in energy than the products. Thus the reaction and between H2 and F is exothermic. In this reaction, the bond between H2 is broken and a new one between HF is formed. As the reaction is exothermic the energy required to break the bond in H2 has to be smaller than the energy released by the formation of the HF bond. Thus the HF bond is stronger than the of H2 bond.

The reverse is true in the reaction of HF and H. Here the reactants are higher in energy than the products and the reaction is endothermic. Thus energy released by the formation of the H2 bond is smaller than the energy required to break the bond in HF.

===Locate the approximate position of the transition state.===
The transition state was determined using the same methods as in question 1.2. However, as in this reaction, not all atoms are the same, the transition state will not be found at a point where <math>r_1</math> and <math>r_2</math>are equal.
[[File:MAlberts Q7 Combined.png|centre|thumb|960x960px|Figure 8: The transition state of the reaction can be seen above.]]
As can be seen above the transition state was found at ere <math>r_1</math> 181.1pm and 74.5pm. This is most probably not the exact transition state as the forces along the BC axis are not equal to zero.

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

===In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===

===Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.===

File:MAlberts Q7 Combined.png

2020-05-08T13:51:52Z

2020-05-06T11:29:10Z

Maa218: /* Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? */

MRD:MAlberts

2020-05-06T11:13:24Z

MRD:MAlberts

2020-05-06T11:02:07Z

MRD:MAlberts

2020-05-06T11:01:39Z

MRD:MAlberts

2020-05-06T10:03:02Z

Maa218: /* Molecular Reaction Dynamics */

MRD:MAlberts

2020-05-06T10:00:17Z

= Molecular Reaction Dynamics =
== Exercise 1 ==
[[File:MAlberts Saddle point.png|thumb|''Figure 1: Saddle point''|450x450px]]
=== 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? ===
<math></math>
The transition state on a potential energy surface is defined as the saddle point. A saddle point is mathematically defined in cartesian space as:

<math>F:U \rightarrow \R</math> with <math>U \subseteq \R^2</math>

'''Condition 1: ''' <math>\frac{\partial F(\overrightarrow{r})}{\partial \overrightarrow{r}} = 0</math>

'''Condition 2''': <math>\frac{\partial ^2 F(r_1)}{\partial ^2 r_1} > 0</math> and <math>\frac{\partial ^2 F(r_2)}{\partial ^2 r_2} < 0</math> with <math>r_1</math> and <math>r_2</math> being interchangable.

Visually this means that the saddle point is the point where the function F has a minimum along one direction and a maximum along another orthogonal to the first one.

In the context of a potential energy surface diagram, the transition state is the saddle point. It can be thought of as the maximum of the minimum energy path. This, by definition, has to be the saddle point as the minimum energy path runs along the minima of the potential energy surface and thus adheres to the definition of there being a minimum and maximum in the same point. In the case of a local minima or maxima condition, 1 would also be fulfilled. Condition 2 is needed to distinguish between a local minimum or maxima and the saddle point as for a minimum both derivatives would be positive while for a maximum they would both be negative.

=== Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===
The transition state is defined as a saddle point on the potential energy surface. Thus if the molecule and the atom it collides with have no momentum and they are exactly at the transition state they will remain there forever. The transition state was found by setting the momentums to zero and as this system is symmetric <math>r_1</math> equal to <math>r_2</math>. Then the values of <math>r_1</math> and <math>r_2</math> were varied while observing the contour plot and the forces along <math>r_1</math> and <math>r_2</math>. For the position to be the transition state, no movement can be observed in the contour plot and the forces need to be equal to zero as the atoms are static and don't move. Additionally, verify the position of the transition state the internuclear distance vs time was examined. At the transition state, this plot should show two flat lines as there is no movement at the transition state.

The transition state was found to be at 90.772pm. This is corroborated by the forces along <math>r_1</math> and <math>r_2</math>being zero (see figure 2), the contour plot showing no movement (see figure 3) and the internuclear distance vs time plot showed two flat lines (see Figure 4):
[[File:MAlberts Q2 Internuclear distance vs time.png|thumb|419x419px|Figure 4: Internuclear distance vs time plot showing no movement.]]
[[File:MAlberts Q2 contour plot.png|left|thumb|419x419px|Figure 2: Contour plot at the transition state showing no movement.]]
[[File:MAlberts Q2 Forces.png|thumb|608x608px|Figure 3: Screenshot of the ''lepsgui.py'' control panel showing that the forces are equal to zero.|centre]]
<math></math>

=== Comment on how the ''mep ''and the trajectory you just calculated differ. ===
The minimum energy path describes the path that follows all the minima of the potential energy surface. In the simulation with the minimum energy path the molecule follows the minima at a velocity approaching zero (internuclear velocity vs time remains at zero). Additionally, there are no vibrations in this simulation and the total energy decreases. The molecule starts close to the transition state and the moves at a velocity near zero to a position in the product where it seems to stop and remain.

For the dynamic simulation the

===Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable"
!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
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|
|[[File:MAlberts Q4 1.png|frameless]]
|-
|-3.1
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|
|[[File:MAlberts Q4 2.png|frameless]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|
|[[File:MAlberts Q4 3.png|frameless]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|
|[[File:MAlberts Q4 4.png|frameless]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|
|[[File:MAlberts Q4 5.png|frameless]]
|}

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

2020-05-06T09:09:31Z

Maa218: /* 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. */

MRD:MAlberts

2020-05-06T09:07:46Z

Maa218: /* 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. */