File:Contour norxn.png

2019-05-16T14:32:57Z

Mt217:

MRD:MToth96

2019-05-16T14:32:35Z

Mt217:

== '''Collison of H2 and H'''==

=== Q1 Potential Energy Surface ===

[[File:Surface_Plot.PNG]] .
On a potential energy surface (''see Figure 1''), the transition state is found and defined as the point where it is true that ∂V(ri)/∂ri=0, and ∂V2(ri)/∂ri2>0.
Empirically, the gradient of the surface is hence equal to zero. Looking at the surface plot, the kinetic energy surface, as a function of AB distance has
a curvature that describes the potential energy as a function of distance. The transition state is at its minimum, as the distances are in balance.
Also, the trajectory (see black line) has to have its maximum value at the transition state, as we are expecting an energy maximum at this point.
The transition state is the highest point along the trajectory. At this point, AB and BC are also equal. After the second differentiation, a local maximum
is found if the derived values is positive. It is distinguished from other local minima by implementing the following criterium: ∂V/dq1 = 0 and ∂V/dq2 = 0,
where q1 is the tangent to the minimum and q2 is orthogonal to this as well as to the reaction path. Hence, ∂V2/dq12 < 0 and ∂V2/dq22 > 0 for that point.

=== Q2 Trajectories, location TS ===

When the transition state is reached, both the AB and BC distances are equal. Atoms show no vibration and therefore the graph displays no vibration but they are static. The optimal distance was found to be 0.908 A, at which both AB and BC distances were equal. Also, the momentum for both p1 and p2 were set to zero.
The intermolecular distanced vs Time plot is shown below. The aim was to minimise oscillation and obtain flat lines with no corresponding gradients. No vibration means no change in distance and the vibration of the H and H2 molecule is restricted, a restriction in vibrational freedom means that the TS is reached. Moreover, the BC and AB distances are equal and the corresponding lines are on top of each other.

[[File:Diastance_vs_time_h2.PNG]] . [[File:Animation_h2_static.PNG]]

=== Q3 MEP and dynamic trajectory ===

[[File:Surface_mep.PNG]]

The trajectory is displayed as the black line along the bottom of the surface. When the MEP calculation is used, the line starts from the transition state but is only displayed for a shirt length. Whereas, the dynamic calculation gives a much longer, complete black trajectory (see figure). The program tries to find the steepest descent along the potential energy surface to obtained a minimum. The MEP method is being reset to zero after every step, whilst the dynamic calculation the data (kinetic energy) is being carried further to the next step. Hence, the calculation is being carried on regardless but with MEP the potential energy stays at its minimum. Thus, an oscillating line is observed with the dynamic, and a local maximum with the MEP is observed.

[[File:Surface_dynamic.PNG]]

Values are being changed from r1=ts+0.01, r2=ts and r1=ts and r2=ts+0.01

[[File:normal.PNG]] [[File:Changed_distance.PNG]]

In both cases, it was observed that the lines corresponding to the AB and BC distances were exchanged (as expected). When the numbers got changed, the transition state was approached by the opposite direction - as a symmetrical system is being considered - and the values are exactly the same but opposite. Hence, the same plot was displayed in the opposite direction.

[[File:Normal_momentum.PNG]] [[File:Changed.PNG]]

=== Q4 Reactive and unreactive trajectories ===

{| class="wikitable" border=1
! p1 !! p2 !! Etot (kcal.mol-1) !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Yes || A-B oscillates, B and C bond formation || [[File:-1.25,-2.5_p1,p2.png]]
|-
| -1.5 || -2.0 || -100.456 || No || A-B oscillates, no reaction || [[File:-1.5,-2.0.png]]
|-
| -1.5 || -2.5 || -98.956 || Yes || A-B oscillates, B and C bond formation || [[File:-1.5,-2.5.png]]
|-
| -2.5 || -5.0 || -84.956 || No || A-B oscillates intensely, B and C bond formation is expected, but it bounces off and C remains a single atom || [[File:-2.5,-5.0.png]]
|-
| -2.5 || -5.2 || -83.416 || Yes || A-B oscillates intensely, very close transition state, B and C bond formation || [[File:-2.5,-5.2.png]]
|}
'''(Table 1)'''

From the table above, it was concluded that the E total of the system depends on the momentum. Moreover, it shows correlation . with the kinetic energy of the system, as the potential enegy is constant, only the kinetic energy is chening. Moreover, in terms of reaxtivity, the system's susceptibility to undergo a reaction is also dpeendent of the momentum and not just of the enegy. If there is a large momentum difference between the single atom and the H2 moelceud, especially when the incoming atoms momentum is marginally lower, than the reaction will not happen.

=== Q5 Transition State Theory ===

Transition State theory is a way of describing chemical reactions as such that they are being a result the atoms and molecules being part of the reaction are being subjected to continous chhange in potential enegies and positions. The transistion stae is being described as the position along the reaction path where the enegy is at its maximum. The activation energy corrwespongin to a specific reaction and to its reactants necessary to initiate the collison leading to the products can be calculated as the differenc between the transition state and the enegy of the reactants. The state of being at the transition state is being defined as the activated complex which is in equilibrium with the reactants and hence its rpoerties can be approximated using Hammonds postulate. Moreover, the rate of reaction - formation of the finla products can be approximated from the concentration of these activated complexes and their ability to form the products.

Furthermore, the trsnsition state theory can also be applied tot eh quasi-equilibrium state. The reactants are in equailibrium with the transition state complex. This means the an equalibtrium is always established between those two states, even if the reactant and proudtc are not in euqialibrium and fo exaple the reaction is driven hardly to one side. This section of the theroy can be imagined mathemmaticlly throguht the Vant Hoff equation. Again, this equatioj desbribes an equalibrium state which ois revesible and is temperature dependent.

\frac{d\ln K}{dT} = \frac{\Delta U}{RT^{2}} ; K - eq. constant, T - temeprature in K, U - internal energy, R - gas constant

In order to get the tempereature dependence of reaction rates, the Vant Hoff reaction is derived, and tje Arrhenius equation is residued,

Integration of this equation leads to Equation 2, otherwise known as the Arrhenius equation. The arrhenius equation shows the temperature dpeendence of reaction rates and allows the calculation of axtivation enegy and reaction rates.

k = Ae^{-E_a/RT} ; k - rate constant, T - temperature in K, A - exponential factor, Ea - activation energy, R - gas constant

According to the theory, the last reaction wheer it was -2.5 and -5.0 should go to completion. However, the exact opposite was observed. As a conclusion, it is suggested that the rate constant is lower in real life than the theory would suggest and therefore the rate conctant prevents the reaction from going to competion. (energy barrier is sufficient to reatin the AB moleculed whn is it sibjected to collision with C)

== '''System of F-H-H'''==

=== Q1 Classification of F + H2 and H + HF reactions ===

[[File:FH_E_surface.png]]

Atoms A and B are H-s, whereas C was F. The surface plot shows that the rectants (on the right ha d side) are higher in energy than the procuct on the left hamd side. Therefore the reaction:

F + H2 → HF + H

is exothermic (negative sign, accompanoed by a decrease in energy). Hence the reaction:

H + HF → H2 + F

must be endothermic, as this is the correspoing backward reaction. In that case, the reactants will be lower in energy and the products will be higher. Additional enegy is required to go through the activaion enegy barrier.

[[File:HF_surfce_endothemic.png]]

The first reaciton is exothemic, (bond making is exothemric) whoch emns that the final prodcts will be in a lower enegy state tha the rectants. Hence, the HF bond will be stron ger and lower in enegy than the HH bond. This is due to the ionic nature of the HF bond due to a large differnnce in electronegativity.

also no momentum was used

=== Q2 Position of transition state ===

The transition state is when the distance are in such configuration that the reaxtion pathway does no exist and would only show a single point. This can be shown wither on the controur plot (no reaction trajctory) or on the surface plot, on whoch only a single dot will be displayed. The ttsnsition state was obtained at H-H distance: 0.745 A, and H-F distance: 1.808 A.

no momentum

[[File:Surface_plot_norxn.png]]

[[File:Example.jpg]]

=== Q3 Activation enegyies for both reactions ===

For the reaction H2 + F → HF + H, the transition state lies at -103.742 kcal.mol-1 and the reactants lie at -103.911 kcal.mol-1. Therefore, the activation energy of this reaction is 0.2 kcal.mol-1. For the reverse reaction, the reactants lie at -133.864 kcal.mol-1, so the activation energy is 30 kcal.mol-1. These energies were established using Energy vs Time plots for extremes of distances and the transition state.

=== Q4 Mechasism of the release of the reaction energy ===

The mechanism is the conversion between translational and vibrational energy. Upon collision, translational is converted into vibrational energy. For the exothermic reaction, energy should be released whereas for the endothermic reaction, energy should be absorbed. Vibrational energy can be confirmed experimentally using infrared spectroscopy. Translational energy can be measured by using bomb calorimetry which measures kinetic energy using an isochoric process.

By inputting rHH = 0.74, with a momentum pFH = -0.5, several values of pHH in the range -3 to 3 were explored. rHF was set to 2.3.

The reaction only appears to be successful for momentum values between 0 and -2 and between 2.85 and 2.95. This indicates there is no obvious trend between the momentum of the H-H bond and the success of the reaction. The success of the reaction between 2.85 and 2.95 violates Polanyi's rules.

Mt 1.pngFinalmomm.png
A set of initial conditions that results in a reactive trajectory for H2 + F → HF + H is as follows:

=== Q5 Distribution of enegy between different modes ===

Polanyi's empirical rules involve the effect of the transition state on the importance of different modes[2]. For an early transition state, such as in an exothermic reaction, translational energy is more important than vibrational energy in terms of the effect on the efficiency of the reaction. On the other hand, endothermic reactions, with a late transition state, place a greater importance on vibrational energy.

As F + H2 → HF + H is an exothermic reaction, its early transition state suggests greater importance of translational energy in affecting the efficiency of the reaction. This translational energy is reflected in the H-F momentum, whereas vibrational energy is reflected in the H-H momentum. The theory predicts that as H-F momentum increases, the reaction goes more to completion whereas increasing the H-H momentum has less of an effect. This is proven as if we increase the momentum of the H-H bond from -1 to -4, the reaction becomes unsuccessful, as seen in Fig.11. The lack of trend in the table above, Table 2, also perpetuates the agreement with Polanyi's rules as no trend is observed. The results between 2.85 and 2.95 violate Polanyi's rules, however. Whereas, as the reverse reaction is an endothermic reaction, if we decrease the momentum of the H-F bond from -1 to -0.5, the reaction becomes unsuccessful, as seen in Fig.12.

2019-05-16T14:19:09Z

Mt217:

MRD:MToth96

2019-05-16T14:18:56Z

Mt217:

== '''Collison of H2 and H'''==

=== Q1 Potential Energy Surface ===

[[File:Surface_Plot.PNG]] .
On a potential energy surface (''see Figure 1''), the transition state is found and defined as the point where it is true that ∂V(ri)/∂ri=0, and ∂V2(ri)/∂ri2>0.
Empirically, the gradient of the surface is hence equal to zero. Looking at the surface plot, the kinetic energy surface, as a function of AB distance has
a curvature that describes the potential energy as a function of distance. The transition state is at its minimum, as the distances are in balance.
Also, the trajectory (see black line) has to have its maximum value at the transition state, as we are expecting an energy maximum at this point.
The transition state is the highest point along the trajectory. At this point, AB and BC are also equal. After the second differentiation, a local maximum
is found if the derived values is positive. It is distinguished from other local minima by implementing the following criterium: ∂V/dq1 = 0 and ∂V/dq2 = 0,
where q1 is the tangent to the minimum and q2 is orthogonal to this as well as to the reaction path. Hence, ∂V2/dq12 < 0 and ∂V2/dq22 > 0 for that point.

=== Q2 Trajectories, location TS ===

When the transition state is reached, both the AB and BC distances are equal. Atoms show no vibration and therefore the graph displays no vibration but they are static. The optimal distance was found to be 0.908 A, at which both AB and BC distances were equal. Also, the momentum for both p1 and p2 were set to zero.
The intermolecular distanced vs Time plot is shown below. The aim was to minimise oscillation and obtain flat lines with no corresponding gradients. No vibration means no change in distance and the vibration of the H and H2 molecule is restricted, a restriction in vibrational freedom means that the TS is reached. Moreover, the BC and AB distances are equal and the corresponding lines are on top of each other.

[[File:Diastance_vs_time_h2.PNG]] . [[File:Animation_h2_static.PNG]]

=== Q3 MEP and dynamic trajectory ===

[[File:Surface_mep.PNG]]

The trajectory is displayed as the black line along the bottom of the surface. When the MEP calculation is used, the line starts from the transition state but is only displayed for a shirt length. Whereas, the dynamic calculation gives a much longer, complete black trajectory (see figure). The program tries to find the steepest descent along the potential energy surface to obtained a minimum. The MEP method is being reset to zero after every step, whilst the dynamic calculation the data (kinetic energy) is being carried further to the next step. Hence, the calculation is being carried on regardless but with MEP the potential energy stays at its minimum. Thus, an oscillating line is observed with the dynamic, and a local maximum with the MEP is observed.

[[File:Surface_dynamic.PNG]]

Values are being changed from r1=ts+0.01, r2=ts and r1=ts and r2=ts+0.01

[[File:normal.PNG]] [[File:Changed_distance.PNG]]

In both cases, it was observed that the lines corresponding to the AB and BC distances were exchanged (as expected). When the numbers got changed, the transition state was approached by the opposite direction - as a symmetrical system is being considered - and the values are exactly the same but opposite. Hence, the same plot was displayed in the opposite direction.

[[File:Normal_momentum.PNG]] [[File:Changed.PNG]]

=== Q4 Reactive and unreactive trajectories ===

{| class="wikitable" border=1
! p1 !! p2 !! Etot (kcal.mol-1) !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Yes || A-B oscillates, B and C bond formation || [[File:-1.25,-2.5_p1,p2.png]]
|-
| -1.5 || -2.0 || -100.456 || No || A-B oscillates, no reaction || [[File:-1.5,-2.0.png]]
|-
| -1.5 || -2.5 || -98.956 || Yes || A-B oscillates, B and C bond formation || [[File:-1.5,-2.5.png]]
|-
| -2.5 || -5.0 || -84.956 || No || A-B oscillates intensely, B and C bond formation is expected, but it bounces off and C remains a single atom || [[File:-2.5,-5.0.png]]
|-
| -2.5 || -5.2 || -83.416 || Yes || A-B oscillates intensely, very close transition state, B and C bond formation || [[File:-2.5,-5.2.png]]
|}
'''(Table 1)'''

From the table above, it was concluded that the E total of the system depends on the momentum. Moreover, it shows correlation . with the kinetic energy of the system, as the potential enegy is constant, only the kinetic energy is chening. Moreover, in terms of reaxtivity, the system's susceptibility to undergo a reaction is also dpeendent of the momentum and not just of the enegy. If there is a large momentum difference between the single atom and the H2 moelceud, especially when the incoming atoms momentum is marginally lower, than the reaction will not happen.

=== Q5 Transition State Theory ===

Transition State theory is a way of describing chemical reactions as such that they are being a result the atoms and molecules being part of the reaction are being subjected to continous chhange in potential enegies and positions. The transistion stae is being described as the position along the reaction path where the enegy is at its maximum. The activation energy corrwespongin to a specific reaction and to its reactants necessary to initiate the collison leading to the products can be calculated as the differenc between the transition state and the enegy of the reactants. The state of being at the transition state is being defined as the activated complex which is in equilibrium with the reactants and hence its rpoerties can be approximated using Hammonds postulate. Moreover, the rate of reaction - formation of the finla products can be approximated from the concentration of these activated complexes and their ability to form the products.

Furthermore, the trsnsition state theory can also be applied tot eh quasi-equilibrium state. The reactants are in equailibrium with the transition state complex. This means the an equalibtrium is always established between those two states, even if the reactant and proudtc are not in euqialibrium and fo exaple the reaction is driven hardly to one side. This section of the theroy can be imagined mathemmaticlly throguht the Vant Hoff equation. Again, this equatioj desbribes an equalibrium state which ois revesible and is temperature dependent.

\frac{d\ln K}{dT} = \frac{\Delta U}{RT^{2}} ; K - eq. constant, T - temeprature in K, U - internal energy, R - gas constant

In order to get the tempereature dependence of reaction rates, the Vant Hoff reaction is derived, and tje Arrhenius equation is residued,

Integration of this equation leads to Equation 2, otherwise known as the Arrhenius equation. The arrhenius equation shows the temperature dpeendence of reaction rates and allows the calculation of axtivation enegy and reaction rates.

k = Ae^{-E_a/RT} ; k - rate constant, T - temperature in K, A - exponential factor, Ea - activation energy, R - gas constant

According to the theory, the last reaction wheer it was -2.5 and -5.0 should go to completion. However, the exact opposite was observed. As a conclusion, it is suggested that the rate constant is lower in real life than the theory would suggest and therefore the rate conctant prevents the reaction from going to competion. (energy barrier is sufficient to reatin the AB moleculed whn is it sibjected to collision with C)

== '''System of F-H-H'''==

=== Q1 Classification of F + H2 and H + HF reactions ===

[[File:FH_E_surface.png]]

Atoms A and B are H-s, whereas C was F. The surface plot shows that the rectants (on the right ha d side) are higher in energy than the procuct on the left hamd side. Therefore the reaction:

F + H2 → HF + H

is exothermic (negative sign, accompanoed by a decrease in energy). Hence the reaction:

H + HF → H2 + F

must be endothermic, as this is the correspoing backward reaction. In that case, the reactants will be lower in energy and the products will be higher. Additional enegy is required to go through the activaion enegy barrier.

[[File:Example.jpg]]

By looking at the potential, energy surfaces, F + H2 → HF + H is an exothermic reaction as the products are lower in energy than the reactants, as seen in Fig.9. The backwards reaction must be endothermic. This can be seen by looking at the plot for H + HF → H2 + F which concurs that the reaction is endothermic because the products are higher in energy than the reactants so energy is required from the surroundings to overcome the potential energy barrier. Therefore, the reverse reaction must be exothermic.

Since F + H2 → H + HF is an exothermic reaction and bond making is an exothermic process, the H-F bond must be stronger than the H-H bond. This can be attributed to its ionic nature as the electronegativity difference between H and F is large, leading to a more polarised bond.

=== Q2 Position of transition state ===

The transition state H-H distance is 0.745 Å and the H-F distance is 1.808 Å. This can be shown using a contour plot, Fig.10, as there is no reaction pathway shown.

=== Q3 Activation enegyies for both reactions ===

For the reaction H2 + F → HF + H, the transition state lies at -103.742 kcal.mol-1 and the reactants lie at -103.911 kcal.mol-1. Therefore, the activation energy of this reaction is 0.2 kcal.mol-1. For the reverse reaction, the reactants lie at -133.864 kcal.mol-1, so the activation energy is 30 kcal.mol-1. These energies were established using Energy vs Time plots for extremes of distances and the transition state.

=== Q4 Mechasism of the release of the reaction energy ===

The mechanism is the conversion between translational and vibrational energy. Upon collision, translational is converted into vibrational energy. For the exothermic reaction, energy should be released whereas for the endothermic reaction, energy should be absorbed. Vibrational energy can be confirmed experimentally using infrared spectroscopy. Translational energy can be measured by using bomb calorimetry which measures kinetic energy using an isochoric process.

By inputting rHH = 0.74, with a momentum pFH = -0.5, several values of pHH in the range -3 to 3 were explored. rHF was set to 2.3.

The reaction only appears to be successful for momentum values between 0 and -2 and between 2.85 and 2.95. This indicates there is no obvious trend between the momentum of the H-H bond and the success of the reaction. The success of the reaction between 2.85 and 2.95 violates Polanyi's rules.

Mt 1.pngFinalmomm.png
A set of initial conditions that results in a reactive trajectory for H2 + F → HF + H is as follows:

=== Q5 Distribution of enegy between different modes ===

Polanyi's empirical rules involve the effect of the transition state on the importance of different modes[2]. For an early transition state, such as in an exothermic reaction, translational energy is more important than vibrational energy in terms of the effect on the efficiency of the reaction. On the other hand, endothermic reactions, with a late transition state, place a greater importance on vibrational energy.

As F + H2 → HF + H is an exothermic reaction, its early transition state suggests greater importance of translational energy in affecting the efficiency of the reaction. This translational energy is reflected in the H-F momentum, whereas vibrational energy is reflected in the H-H momentum. The theory predicts that as H-F momentum increases, the reaction goes more to completion whereas increasing the H-H momentum has less of an effect. This is proven as if we increase the momentum of the H-H bond from -1 to -4, the reaction becomes unsuccessful, as seen in Fig.11. The lack of trend in the table above, Table 2, also perpetuates the agreement with Polanyi's rules as no trend is observed. The results between 2.85 and 2.95 violate Polanyi's rules, however. Whereas, as the reverse reaction is an endothermic reaction, if we decrease the momentum of the H-F bond from -1 to -0.5, the reaction becomes unsuccessful, as seen in Fig.12.

2019-05-16T13:11:01Z

Mt217: Mt217 uploaded a new version of File:-2.5,-5.2.png

File:-2.5,-5.0.png

2019-05-16T13:10:47Z

Mt217:

2019-05-16T13:07:06Z

Mt217:

MRD:MToth96

2019-05-15T23:01:48Z

Mt217:

MRD:MToth96

2019-05-15T22:44:22Z

Mt217: