01485762jy

2020-05-19T03:09:47Z

Jy8418: /* References */

== '''EXERCISE 1: H + H2 system''' ==

==== Question 1 (Dynamics from the transition state region): ====
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?

==== Answer 1: ====
On a potential energy surface diagram, the transition state is defined mathematically as following:
δV/δr1 = δV/δr2 = 0, where V is the chemical potential, r1 and r2 are the distances between atoms A & B and atoms B & C respectively.
[[File:path1.jpg|thumb|frame|right|Figure 1-The minimised energy pathway]]
On a reaction co-ordinate diagram, where the minimum energy path (the line in bold in Figure 1) linking reactants and reactants is plotted, the transition state is the maximum (the one located at the least negative potential energy value). (Figure 1) On the potential energy surface diagram, it is a saddle point.

Both the saddle point and the local minimum have the first derivative equal to zero, but they are distinguished by the following method. For the collision between three homogeneous atoms, the transition state is when rAB = rBC. Other turning points are classified as local maximum/minimum (depends on how you 'rotate' the diagram). Mathematically, for collisions between heterogeneous atoms, the term fxx(x0,y0)*fyy(x0,y0)−f2xy(x0,y0) has to be calculated, where fxx is the second derivative with respect to x, fyy to y and fxy is to differentiate with respect to x then to y. If the term is less than zero, then the coordinate a saddle point. If the term is greater than zero, then the coordinate is a minimum/maximum. For a local minimum, Its second derivative (fxx or fyy) has to be greater than zero.

==== Question 2 (Trajectories from r1 = r2: locating the transition state): ====
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.

==== Answer 2: ====
The best estimate of the transition state position (rts) is rAB=rBC=90.8 pm. The possible range of transition state is between 85 pm to 110 pm from the surface plot, so the values were tested in this range, keeping rAB=rBC and momentum zero. This result was verified by plotting the “Internuclear Distances vs Time” graph. Two straight horizontal lines at the same height were obtained, indicating the constant distances between A & B and B & C. There is no oscillations on the ridge.

[[File:Rts.png|thumb|frame|right|Figure 2-The "Internuclear distances vs Time" graph]]

==== Question 3 (Calculating the reaction path): ====
Comment on how the ''mep'' and the trajectory you just calculated differ.

==== Answer 3: ====
From the “Internuclear Distances vs Time” graphs generated by two methods, it is found that the distance between atoms A and B is higher in the Dynamics method and tend to infinity, whereas the distance generated by MEP tends to converge at a lower value. Therefore, it can be deduced that MEP is based on the conservation of energy whereas Dynamics ignores that. Also, there are some oscillations in distance B-C when using Dynamics, whereas MEP does not. So MEP ignores the motion of atoms and their inertia.

==== Question 4 (Reactive and unreactive trajectories): ====
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?

==== Answer 4: ====
{| 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
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Reactive
|The H2 molecule(HA-HB) moves towards HC atom and the kinetic barrier is overcome, forming a new H2 molecule(HB-HC). The newly formed molecule vibrates away and HA also moves back.
|
[[File:1cp.png|thumb]]

|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The vibrating H2 molecule(HA-HB) moves towards Hc atom, but the energy was too low to pass the transition state region, so no reaction takes place; the H2 molecule(HA-HB) and HC vibrate backwards.
|
[[File:2cp.png|thumb]]

|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The vibrating H2 molecule(HA-HB) approaches HC atom and form a new H2 molecule(HB-HC). The new molecule vibrates away.
|
[[File:3cp.png|thumb]]

|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-359.277</nowiki>
|Unreactive
|The H2 molecule(HA-HB) moves fast towards HC. The transition state region was crossed. Due to the high momentum, HB bounces between HA and HC. Finally, HB reforms bonds with HA and the molecule vibrates backwards.
|
[[File:4cp.png|thumb]]

|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Reactive
|The H2 molecule(HA-HB) moves fast towards HC. The transition state region was crossed. Due to the high momentum, HB bounces between HA and HC. Finally, HB forms new bonds with HC and the new molecule(HB-HC) vibrates backwards.
|
[[File:5cp.png|thumb]]

|}
In conclusion, the momentum of the reactants is crucial in determining the success of reactions. If the momentum is too small, the energy barrier cannot be overcome and the trajectory is unreactive. If the momentum is too large, the transition state region can certainly be crossed, but the atoms may bounce around and the outcome is unpredictable, products are likely to be formed when one momentum is more than double of the other. From the above experimental data, the 'safest' p1 is around -3.1 g.mol-1.pm.fs-1 and p2 is -5.1 g.mol-1.pm.fs-1.

==== Question 5 (Transition State Theory): ====
Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

==== Answer 5: ====
The transition state theory (TST) predicts a higher rate value than the actual experimental value gives. TST assumes that all the trajectories with the energy path along the reaction coordinates greater than the activation energy will be reactive, by ignoring the fact that atoms may 'bounce' around the transition state region. By looking at example 4 in the above table, atoms have enough energy but the reaction failed. In this case, TST failed. Also, even though the reaction can proceed under sufficient energy, the time taken for 'bouncing around' also reduce the actual rate constant (e.g. example 5), making TST overestimate the rate constant.

== '''EXERCISE 2: F - H - H system''' ==

==== Question 6 (PES inspection): ====
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?

==== Answer 6: ====
The reaction between F and H2 is exothermic and the reaction between H and HF is endothermic.

On the surface plot, the reactants H2 and F lie in a higher energy level than the products, so the reaction is exothermic. (Figure 3, with reactants on RHS and products on LHS).
[[File:exooo.png|thumb|frame|Figure 3- energy pathway for an exothermic reaction]]
The reactants H and HF lie in lower energy level (more negative in value) than the products, so it's endothermic. (Figure 4, with reactants on LHS and products on RHS)

[[File:endoo.png|thumb|frame|Figure 4- energy pathway for an endothermic reaction]]

{| class="wikitable"
!
!Bond dissociation energy at 298 K (kJ/mol)
|-
|H-H
|436
|-
|H-F
|569
|}
For reaction between F and H2, H-F bond is newly formed and H-H is broken. It has a higher bond energy than H-H does, indicating more energy released, so the reaction is exothermic.

The reverse process takes place for reaction between HF and H, H-F is broken and H-H is formed. Therefore, more energy is absorbed to break the strong H-bond H-F and less energy released, so it is an endothermic reaction.

==== Question 7 (PES inspection): ====
Locate the approximate position of the transition state.

==== Answer 7: ====
Different combinations of numbers were tried for distances until there is only one spot on the contour spot and the graph of "internuclear distance vs time" shows straight horizontal lines (Figure 5 & 6). For the reaction 1--H2+F, the position of transition state is at rH-H=74.4 pm and rH-F=182 pm.
[[File:tshf.png|thumb|frame|none|Figure 5- "Internuclear distance vs time" graph for reaction 1]]
For reaction 2--HF+H, the position of transition state is at the same position, where rH-H=74.4 pm and rH-F=182 pm. This is because both the two reactions involve interactions between H-H and H-F, with only the differences in bond breaking/forming.
[[File:hh.png|thumb|frame|none|Figure 6- "Internuclear distance vs time" graph for reaction 2]]

==== Question 8 (PES inspection): ====
Report the activation energy for both reactions.

==== Answer 8: ====
The activation energy is calculated as the difference between the potential energy of the transition state and the reactant. The respective energies are determined by plotting "Energy vs Time" graphs. As the transition state for both reactions are at the same position with the same structure of F--H--H, the two reactions have the same potential energy for the transition state, which is -433.980 kJmol-1 (Figure 7). For the potential energy of the reactants, it is assumed that the reactant species are not interacting, so the distance B-C is set near infinity and the most negative value of potential energy on the graph is taken as the potential energy of the reactants.

[[File:tsenergy.png|thumb|frame|none|Figure 7-potential energy for the transition state]]
For reaction 1--H2+F, the potential energy for the reactants is -435.073 kJmol-1 (Figure 8), so the activation energy is 1.093 kJmol-1.

[[File:di8.png|thumb|frame|none|Figure 8-potential energy for the reactants 1]]
For reaction 2--HF+H, the potential energy for the reactants is -562.459 kJmol-1(Figure 9), so the activation energy is 127.386 kJmol-1.

[[File:di9.png|thumb|frame|none|Figure 9-potential energy for the reactants 2]]

==== Question 9 (Reaction dynamics): ====
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.

==== Answer 9: ====
The distance AB is set as 66 pm and distance BC is 210 pm. The momentum of AB is -9.1 g.mol-1.pm.fs-1 and that of C is -5.8 g.mol-1.pm.fs-1. Under this condition, the H2 molecule vibrates towards F atom, after bouncing around a few times, the HB atom forms bonds with F, and HF vibrates away.

In this process, based on the conservation of energy, the potential energy is converted to kinetic energy. The kinetic energy consists of vibrational and translational energy. The vibrational energy can be detected through infrared spectroscopy. To measure the heat change, a colorimeter can also be used to determine whether a reaction is endothermic or exothermic, by how much amount.

==== Question 10 (Reaction dynamics): ====
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.

==== Answer 10: ====
For all reactions, A B are the atoms in the molecule and C is the atom for colliding. When pAB is set far more than pBC (pBC is also given a low value), the system is said to have a high vibrational energy (and a low translational energy). Oppositely, when pAB is set far less than pBC (pAB is also given a low value), the system is said to have a high translational energy (and a low vibrational energy).

For reaction 1--H2+F, when the positions are set at transition state of rAB=74 pm and rBC= 182 pm, pAB=8 g.mol-1.pm.fs-1 and pBC=-1 g.mol-1.pm.fs-1, a reactive trajectory (Figure 10) is obtained. On the surface plot, at such high vibrational energy, the transition state was found closer to the reactants--early transition state.
[[File:1vibb.png|thumb|frame|none|Figure 10-The trajectory for reaction 1 under high vibrational energy]]
For the same reaction, when the positions are set at transition state of rAB=74 pm and rBC= 182 pm, pAB=-1 g.mol-1.pm.fs-1 and pBC=-14.6 g.mol-1.pm.fs-1, a reactive trajectory (Figure 11) is obtained. On the surface plot, at such high translational energy, the transition state was found closer to the products--late transition state.
[[File:1trans.png|thumb|frame|none|Figure 11-The trajectory for reaction 1 under high translational energy]]
For reaction 2--HF+H, when the positions are set at transition state of rAB=182 pm and rBC= 74 pm, pAB=-12.5 g.mol-1.pm.fs-1 and pBC=-1 g.mol-1.pm.fs-1, a reactive trajectory (Figure 12) is obtained. On the surface plot, at such high vibrational energy, the transition state was found closer to the reactants--early transition state.
[[File:2vibb.png|thumb|frame|none|Figure 12-The trajectory for reaction 2 under high vibrational energy]]
For the same reaction, when the positions are set at transition state of rAB=182 pm and rBC= 74 pm, pAB=-0.5 g.mol-1.pm.fs-1 and pBC=-9.8 g.mol-1.pm.fs-1, a reactive trajectory (Figure 13) is obtained. On the surface plot, at such high translational energy, the transition state was found closer to the products--late transition state.
[[File:2trans.png|thumb|frame|none|Figure 13-The trajectory for reaction 2 under high translational energy]]
Therefore, high vibrational energy and low translational energy indicate the presence the early transition state; high translational energy and low vibrational energy favour the late transition state.

When investigating for the reactive trajectories, it is found that if the vibrational energy is high, the reaction can be more efficient (early transition state). According to Polanyi's principle,

Ea=E0+αΔH

where Ea is the activation energy, α is the position of transition state along the reaction coordinates. An early transition state (at high vibrational energy) indicates a small value of α, thus lower activation energy, making the reaction more efficient.

<nowiki>*</nowiki>In figures 10-11, the reactants are on RHS and products on LHS; for figures 12-13, the reactants are on LHS and products on RHS.

== '''References''' ==
<nowiki>https://aip.scitation.org/doi/pdf/10.1063/1.1672005</nowiki>

<nowiki>https://pubs.rsc.org/en/content/articlehtml/2016/sc/c6sc01066k</nowiki>

01485762jy

2020-05-19T02:43:47Z

Jy8418: /* Answer 10: */

== '''EXERCISE 1: H + H2 system''' ==

==== Question 1 (Dynamics from the transition state region): ====
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?

==== Answer 1: ====
On a potential energy surface diagram, the transition state is defined mathematically as following:
δV/δr1 = δV/δr2 = 0, where V is the chemical potential, r1 and r2 are the distances between atoms A & B and atoms B & C respectively.
[[File:path1.jpg|thumb|frame|right|Figure 1-The minimised energy pathway]]
On a reaction co-ordinate diagram, where the minimum energy path (the line in bold in Figure 1) linking reactants and reactants is plotted, the transition state is the maximum (the one located at the least negative potential energy value). (Figure 1) On the potential energy surface diagram, it is a saddle point.

Both the saddle point and the local minimum have the first derivative equal to zero, but they are distinguished by the following method. For the collision between three homogeneous atoms, the transition state is when rAB = rBC. Other turning points are classified as local maximum/minimum (depends on how you 'rotate' the diagram). Mathematically, for collisions between heterogeneous atoms, the term fxx(x0,y0)*fyy(x0,y0)−f2xy(x0,y0) has to be calculated, where fxx is the second derivative with respect to x, fyy to y and fxy is to differentiate with respect to x then to y. If the term is less than zero, then the coordinate a saddle point. If the term is greater than zero, then the coordinate is a minimum/maximum. For a local minimum, Its second derivative (fxx or fyy) has to be greater than zero.

==== Question 2 (Trajectories from r1 = r2: locating the transition state): ====
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.

==== Answer 2: ====
The best estimate of the transition state position (rts) is rAB=rBC=90.8 pm. The possible range of transition state is between 85 pm to 110 pm from the surface plot, so the values were tested in this range, keeping rAB=rBC and momentum zero. This result was verified by plotting the “Internuclear Distances vs Time” graph. Two straight horizontal lines at the same height were obtained, indicating the constant distances between A & B and B & C. There is no oscillations on the ridge.

[[File:Rts.png|thumb|frame|right|Figure 2-The "Internuclear distances vs Time" graph]]

==== Question 3 (Calculating the reaction path): ====
Comment on how the ''mep'' and the trajectory you just calculated differ.

==== Answer 3: ====
From the “Internuclear Distances vs Time” graphs generated by two methods, it is found that the distance between atoms A and B is higher in the Dynamics method and tend to infinity, whereas the distance generated by MEP tends to converge at a lower value. Therefore, it can be deduced that MEP is based on the conservation of energy whereas Dynamics ignores that. Also, there are some oscillations in distance B-C when using Dynamics, whereas MEP does not. So MEP ignores the motion of atoms and their inertia.

==== Question 4 (Reactive and unreactive trajectories): ====
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?

==== Answer 4: ====
{| 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
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Reactive
|The H2 molecule(HA-HB) moves towards HC atom and the kinetic barrier is overcome, forming a new H2 molecule(HB-HC). The newly formed molecule vibrates away and HA also moves back.
|
[[File:1cp.png|thumb]]

|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The vibrating H2 molecule(HA-HB) moves towards Hc atom, but the energy was too low to pass the transition state region, so no reaction takes place; the H2 molecule(HA-HB) and HC vibrate backwards.
|
[[File:2cp.png|thumb]]

|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The vibrating H2 molecule(HA-HB) approaches HC atom and form a new H2 molecule(HB-HC). The new molecule vibrates away.
|
[[File:3cp.png|thumb]]

|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-359.277</nowiki>
|Unreactive
|The H2 molecule(HA-HB) moves fast towards HC. The transition state region was crossed. Due to the high momentum, HB bounces between HA and HC. Finally, HB reforms bonds with HA and the molecule vibrates backwards.
|
[[File:4cp.png|thumb]]

|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Reactive
|The H2 molecule(HA-HB) moves fast towards HC. The transition state region was crossed. Due to the high momentum, HB bounces between HA and HC. Finally, HB forms new bonds with HC and the new molecule(HB-HC) vibrates backwards.
|
[[File:5cp.png|thumb]]

|}
In conclusion, the momentum of the reactants is crucial in determining the success of reactions. If the momentum is too small, the energy barrier cannot be overcome and the trajectory is unreactive. If the momentum is too large, the transition state region can certainly be crossed, but the atoms may bounce around and the outcome is unpredictable, products are likely to be formed when one momentum is more than double of the other. From the above experimental data, the 'safest' p1 is around -3.1 g.mol-1.pm.fs-1 and p2 is -5.1 g.mol-1.pm.fs-1.

==== Question 5 (Transition State Theory): ====
Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

==== Answer 5: ====
The transition state theory (TST) predicts a higher rate value than the actual experimental value gives. TST assumes that all the trajectories with the energy path along the reaction coordinates greater than the activation energy will be reactive, by ignoring the fact that atoms may 'bounce' around the transition state region. By looking at example 4 in the above table, atoms have enough energy but the reaction failed. In this case, TST failed. Also, even though the reaction can proceed under sufficient energy, the time taken for 'bouncing around' also reduce the actual rate constant (e.g. example 5), making TST overestimate the rate constant.

== '''EXERCISE 2: F - H - H system''' ==

==== Question 6 (PES inspection): ====
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?

==== Answer 6: ====
The reaction between F and H2 is exothermic and the reaction between H and HF is endothermic.

On the surface plot, the reactants H2 and F lie in a higher energy level than the products, so the reaction is exothermic. (Figure 3, with reactants on RHS and products on LHS).
[[File:exooo.png|thumb|frame|Figure 3- energy pathway for an exothermic reaction]]
The reactants H and HF lie in lower energy level (more negative in value) than the products, so it's endothermic. (Figure 4, with reactants on LHS and products on RHS)

[[File:endoo.png|thumb|frame|Figure 4- energy pathway for an endothermic reaction]]

{| class="wikitable"
!
!Bond dissociation energy at 298 K (kJ/mol)
|-
|H-H
|436
|-
|H-F
|569
|}
For reaction between F and H2, H-F bond is newly formed and H-H is broken. It has a higher bond energy than H-H does, indicating more energy released, so the reaction is exothermic.

The reverse process takes place for reaction between HF and H, H-F is broken and H-H is formed. Therefore, more energy is absorbed to break the strong H-bond H-F and less energy released, so it is an endothermic reaction.

==== Question 7 (PES inspection): ====
Locate the approximate position of the transition state.

==== Answer 7: ====
Different combinations of numbers were tried for distances until there is only one spot on the contour spot and the graph of "internuclear distance vs time" shows straight horizontal lines (Figure 5 & 6). For the reaction 1--H2+F, the position of transition state is at rH-H=74.4 pm and rH-F=182 pm.
[[File:tshf.png|thumb|frame|none|Figure 5- "Internuclear distance vs time" graph for reaction 1]]
For reaction 2--HF+H, the position of transition state is at the same position, where rH-H=74.4 pm and rH-F=182 pm. This is because both the two reactions involve interactions between H-H and H-F, with only the differences in bond breaking/forming.
[[File:hh.png|thumb|frame|none|Figure 6- "Internuclear distance vs time" graph for reaction 2]]

==== Question 8 (PES inspection): ====
Report the activation energy for both reactions.

==== Answer 8: ====
The activation energy is calculated as the difference between the potential energy of the transition state and the reactant. The respective energies are determined by plotting "Energy vs Time" graphs. As the transition state for both reactions are at the same position with the same structure of F--H--H, the two reactions have the same potential energy for the transition state, which is -433.980 kJmol-1 (Figure 7). For the potential energy of the reactants, it is assumed that the reactant species are not interacting, so the distance B-C is set near infinity and the most negative value of potential energy on the graph is taken as the potential energy of the reactants.

[[File:tsenergy.png|thumb|frame|none|Figure 7-potential energy for the transition state]]
For reaction 1--H2+F, the potential energy for the reactants is -435.073 kJmol-1 (Figure 8), so the activation energy is 1.093 kJmol-1.

[[File:di8.png|thumb|frame|none|Figure 8-potential energy for the reactants 1]]
For reaction 2--HF+H, the potential energy for the reactants is -562.459 kJmol-1(Figure 9), so the activation energy is 127.386 kJmol-1.

[[File:di9.png|thumb|frame|none|Figure 9-potential energy for the reactants 2]]

==== Question 9 (Reaction dynamics): ====
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.

==== Answer 9: ====
The distance AB is set as 66 pm and distance BC is 210 pm. The momentum of AB is -9.1 g.mol-1.pm.fs-1 and that of C is -5.8 g.mol-1.pm.fs-1. Under this condition, the H2 molecule vibrates towards F atom, after bouncing around a few times, the HB atom forms bonds with F, and HF vibrates away.

In this process, based on the conservation of energy, the potential energy is converted to kinetic energy. The kinetic energy consists of vibrational and translational energy. The vibrational energy can be detected through infrared spectroscopy. To measure the heat change, a colorimeter can also be used to determine whether a reaction is endothermic or exothermic, by how much amount.

==== Question 10 (Reaction dynamics): ====
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.

==== Answer 10: ====
For all reactions, A B are the atoms in the molecule and C is the atom for colliding. When pAB is set far more than pBC (pBC is also given a low value), the system is said to have a high vibrational energy (and a low translational energy). Oppositely, when pAB is set far less than pBC (pAB is also given a low value), the system is said to have a high translational energy (and a low vibrational energy).

For reaction 1--H2+F, when the positions are set at transition state of rAB=74 pm and rBC= 182 pm, pAB=8 g.mol-1.pm.fs-1 and pBC=-1 g.mol-1.pm.fs-1, a reactive trajectory (Figure 10) is obtained. On the surface plot, at such high vibrational energy, the transition state was found closer to the reactants--early transition state.
[[File:1vibb.png|thumb|frame|none|Figure 10-The trajectory for reaction 1 under high vibrational energy]]
For the same reaction, when the positions are set at transition state of rAB=74 pm and rBC= 182 pm, pAB=-1 g.mol-1.pm.fs-1 and pBC=-14.6 g.mol-1.pm.fs-1, a reactive trajectory (Figure 11) is obtained. On the surface plot, at such high translational energy, the transition state was found closer to the products--late transition state.
[[File:1trans.png|thumb|frame|none|Figure 11-The trajectory for reaction 1 under high translational energy]]
For reaction 2--HF+H, when the positions are set at transition state of rAB=182 pm and rBC= 74 pm, pAB=-12.5 g.mol-1.pm.fs-1 and pBC=-1 g.mol-1.pm.fs-1, a reactive trajectory (Figure 12) is obtained. On the surface plot, at such high vibrational energy, the transition state was found closer to the reactants--early transition state.
[[File:2vibb.png|thumb|frame|none|Figure 12-The trajectory for reaction 2 under high vibrational energy]]
For the same reaction, when the positions are set at transition state of rAB=182 pm and rBC= 74 pm, pAB=-0.5 g.mol-1.pm.fs-1 and pBC=-9.8 g.mol-1.pm.fs-1, a reactive trajectory (Figure 13) is obtained. On the surface plot, at such high translational energy, the transition state was found closer to the products--late transition state.
[[File:2trans.png|thumb|frame|none|Figure 13-The trajectory for reaction 2 under high translational energy]]
Therefore, high vibrational energy and low translational energy indicate the presence the early transition state; high translational energy and low vibrational energy favour the late transition state.

When investigating for the reactive trajectories, it is found that if the vibrational energy is high, the reaction can be more efficient (early transition state). According to Polanyi's principle,

Ea=E0+αΔH

where Ea is the activation energy, α is the position of transition state along the reaction coordinates. An early transition state (at high vibrational energy) indicates a small value of α, thus lower activation energy, making the reaction more efficient.

<nowiki>*</nowiki>In figures 10-11, the reactants are on RHS and products on LHS; for figures 12-13, the reactants are on LHS and products on RHS.

== '''References''' ==
<nowiki>https://aip.scitation.org/doi/pdf/10.1063/1.1672005</nowiki>

01485762jy

2020-05-19T02:32:43Z

Jy8418: /* Answer 10: */

== '''EXERCISE 1: H + H2 system''' ==

==== Question 1 (Dynamics from the transition state region): ====
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?

==== Answer 1: ====
On a potential energy surface diagram, the transition state is defined mathematically as following:
δV/δr1 = δV/δr2 = 0, where V is the chemical potential, r1 and r2 are the distances between atoms A & B and atoms B & C respectively.
[[File:path1.jpg|thumb|frame|right|Figure 1-The minimised energy pathway]]
On a reaction co-ordinate diagram, where the minimum energy path (the line in bold in Figure 1) linking reactants and reactants is plotted, the transition state is the maximum (the one located at the least negative potential energy value). (Figure 1) On the potential energy surface diagram, it is a saddle point.

Both the saddle point and the local minimum have the first derivative equal to zero, but they are distinguished by the following method. For the collision between three homogeneous atoms, the transition state is when rAB = rBC. Other turning points are classified as local maximum/minimum (depends on how you 'rotate' the diagram). Mathematically, for collisions between heterogeneous atoms, the term fxx(x0,y0)*fyy(x0,y0)−f2xy(x0,y0) has to be calculated, where fxx is the second derivative with respect to x, fyy to y and fxy is to differentiate with respect to x then to y. If the term is less than zero, then the coordinate a saddle point. If the term is greater than zero, then the coordinate is a minimum/maximum. For a local minimum, Its second derivative (fxx or fyy) has to be greater than zero.

==== Question 2 (Trajectories from r1 = r2: locating the transition state): ====
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.

==== Answer 2: ====
The best estimate of the transition state position (rts) is rAB=rBC=90.8 pm. The possible range of transition state is between 85 pm to 110 pm from the surface plot, so the values were tested in this range, keeping rAB=rBC and momentum zero. This result was verified by plotting the “Internuclear Distances vs Time” graph. Two straight horizontal lines at the same height were obtained, indicating the constant distances between A & B and B & C. There is no oscillations on the ridge.

[[File:Rts.png|thumb|frame|right|Figure 2-The "Internuclear distances vs Time" graph]]

==== Question 3 (Calculating the reaction path): ====
Comment on how the ''mep'' and the trajectory you just calculated differ.

==== Answer 3: ====
From the “Internuclear Distances vs Time” graphs generated by two methods, it is found that the distance between atoms A and B is higher in the Dynamics method and tend to infinity, whereas the distance generated by MEP tends to converge at a lower value. Therefore, it can be deduced that MEP is based on the conservation of energy whereas Dynamics ignores that. Also, there are some oscillations in distance B-C when using Dynamics, whereas MEP does not. So MEP ignores the motion of atoms and their inertia.

==== Question 4 (Reactive and unreactive trajectories): ====
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?

==== Answer 4: ====
{| 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
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Reactive
|The H2 molecule(HA-HB) moves towards HC atom and the kinetic barrier is overcome, forming a new H2 molecule(HB-HC). The newly formed molecule vibrates away and HA also moves back.
|
[[File:1cp.png|thumb]]

|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The vibrating H2 molecule(HA-HB) moves towards Hc atom, but the energy was too low to pass the transition state region, so no reaction takes place; the H2 molecule(HA-HB) and HC vibrate backwards.
|
[[File:2cp.png|thumb]]

|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The vibrating H2 molecule(HA-HB) approaches HC atom and form a new H2 molecule(HB-HC). The new molecule vibrates away.
|
[[File:3cp.png|thumb]]

|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-359.277</nowiki>
|Unreactive
|The H2 molecule(HA-HB) moves fast towards HC. The transition state region was crossed. Due to the high momentum, HB bounces between HA and HC. Finally, HB reforms bonds with HA and the molecule vibrates backwards.
|
[[File:4cp.png|thumb]]

|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Reactive
|The H2 molecule(HA-HB) moves fast towards HC. The transition state region was crossed. Due to the high momentum, HB bounces between HA and HC. Finally, HB forms new bonds with HC and the new molecule(HB-HC) vibrates backwards.
|
[[File:5cp.png|thumb]]

|}
In conclusion, the momentum of the reactants is crucial in determining the success of reactions. If the momentum is too small, the energy barrier cannot be overcome and the trajectory is unreactive. If the momentum is too large, the transition state region can certainly be crossed, but the atoms may bounce around and the outcome is unpredictable, products are likely to be formed when one momentum is more than double of the other. From the above experimental data, the 'safest' p1 is around -3.1 g.mol-1.pm.fs-1 and p2 is -5.1 g.mol-1.pm.fs-1.

==== Question 5 (Transition State Theory): ====
Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

==== Answer 5: ====
The transition state theory (TST) predicts a higher rate value than the actual experimental value gives. TST assumes that all the trajectories with the energy path along the reaction coordinates greater than the activation energy will be reactive, by ignoring the fact that atoms may 'bounce' around the transition state region. By looking at example 4 in the above table, atoms have enough energy but the reaction failed. In this case, TST failed. Also, even though the reaction can proceed under sufficient energy, the time taken for 'bouncing around' also reduce the actual rate constant (e.g. example 5), making TST overestimate the rate constant.

== '''EXERCISE 2: F - H - H system''' ==

==== Question 6 (PES inspection): ====
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?

==== Answer 6: ====
The reaction between F and H2 is exothermic and the reaction between H and HF is endothermic.

On the surface plot, the reactants H2 and F lie in a higher energy level than the products, so the reaction is exothermic. (Figure 3, with reactants on RHS and products on LHS).
[[File:exooo.png|thumb|frame|Figure 3- energy pathway for an exothermic reaction]]
The reactants H and HF lie in lower energy level (more negative in value) than the products, so it's endothermic. (Figure 4, with reactants on LHS and products on RHS)

[[File:endoo.png|thumb|frame|Figure 4- energy pathway for an endothermic reaction]]

{| class="wikitable"
!
!Bond dissociation energy at 298 K (kJ/mol)
|-
|H-H
|436
|-
|H-F
|569
|}
For reaction between F and H2, H-F bond is newly formed and H-H is broken. It has a higher bond energy than H-H does, indicating more energy released, so the reaction is exothermic.

The reverse process takes place for reaction between HF and H, H-F is broken and H-H is formed. Therefore, more energy is absorbed to break the strong H-bond H-F and less energy released, so it is an endothermic reaction.

==== Question 7 (PES inspection): ====
Locate the approximate position of the transition state.

==== Answer 7: ====
Different combinations of numbers were tried for distances until there is only one spot on the contour spot and the graph of "internuclear distance vs time" shows straight horizontal lines (Figure 5 & 6). For the reaction 1--H2+F, the position of transition state is at rH-H=74.4 pm and rH-F=182 pm.
[[File:tshf.png|thumb|frame|none|Figure 5- "Internuclear distance vs time" graph for reaction 1]]
For reaction 2--HF+H, the position of transition state is at the same position, where rH-H=74.4 pm and rH-F=182 pm. This is because both the two reactions involve interactions between H-H and H-F, with only the differences in bond breaking/forming.
[[File:hh.png|thumb|frame|none|Figure 6- "Internuclear distance vs time" graph for reaction 2]]

==== Question 8 (PES inspection): ====
Report the activation energy for both reactions.

==== Answer 8: ====
The activation energy is calculated as the difference between the potential energy of the transition state and the reactant. The respective energies are determined by plotting "Energy vs Time" graphs. As the transition state for both reactions are at the same position with the same structure of F--H--H, the two reactions have the same potential energy for the transition state, which is -433.980 kJmol-1 (Figure 7). For the potential energy of the reactants, it is assumed that the reactant species are not interacting, so the distance B-C is set near infinity and the most negative value of potential energy on the graph is taken as the potential energy of the reactants.

[[File:tsenergy.png|thumb|frame|none|Figure 7-potential energy for the transition state]]
For reaction 1--H2+F, the potential energy for the reactants is -435.073 kJmol-1 (Figure 8), so the activation energy is 1.093 kJmol-1.

[[File:di8.png|thumb|frame|none|Figure 8-potential energy for the reactants 1]]
For reaction 2--HF+H, the potential energy for the reactants is -562.459 kJmol-1(Figure 9), so the activation energy is 127.386 kJmol-1.

[[File:di9.png|thumb|frame|none|Figure 9-potential energy for the reactants 2]]

==== Question 9 (Reaction dynamics): ====
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.

==== Answer 9: ====
The distance AB is set as 66 pm and distance BC is 210 pm. The momentum of AB is -9.1 g.mol-1.pm.fs-1 and that of C is -5.8 g.mol-1.pm.fs-1. Under this condition, the H2 molecule vibrates towards F atom, after bouncing around a few times, the HB atom forms bonds with F, and HF vibrates away.

In this process, based on the conservation of energy, the potential energy is converted to kinetic energy. The kinetic energy consists of vibrational and translational energy. The vibrational energy can be detected through infrared spectroscopy. To measure the heat change, a colorimeter can also be used to determine whether a reaction is endothermic or exothermic, by how much amount.

==== Question 10 (Reaction dynamics): ====
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.

==== Answer 10: ====
For all reactions, A B are the atoms in the molecule and C is the atom for colliding. When pAB is set far more than pBC (pBC is also given a low value), the system is said to have a high vibrational energy (and a low translational energy). Oppositely, when pAB is set far less than pBC (pAB is also given a low value), the system is said to have a high translational energy (and a low vibrational energy).

For reaction 1--H2+F, when the positions are set at transition state of rAB=74 pm and rBC= 182 pm, pAB=8 g.mol-1.pm.fs-1 and pBC=-1 g.mol-1.pm.fs-1, a reactive trajectory (Figure 10) is obtained. On the surface plot, at such high vibrational energy, the transition state was found closer to the reactants--early transition state.
[[File:1vibb.png|thumb|frame|none|Figure 10-The trajectory for reaction 1 under high vibrational energy]]
For the same reaction, when the positions are set at transition state of rAB=74 pm and rBC= 182 pm, pAB=-1 g.mol-1.pm.fs-1 and pBC=-14.6 g.mol-1.pm.fs-1, a reactive trajectory (Figure 11) is obtained. On the surface plot, at such high translational energy, the transition state was found closer to the products--late transition state.
[[File:1trans.png|thumb|frame|none|Figure 11-The trajectory for reaction 1 under high translational energy]]
For reaction 2--HF+H, when the positions are set at transition state of rAB=182 pm and rBC= 74 pm, pAB=-12.5 g.mol-1.pm.fs-1 and pBC=-1 g.mol-1.pm.fs-1, a reactive trajectory (Figure 12) is obtained. On the surface plot, at such high vibrational energy, the transition state was found closer to the reactants--early transition state.
[[File:2vibb.png|thumb|frame|none|Figure 12-The trajectory for reaction 2 under high vibrational energy]]
For the same reaction, when the positions are set at transition state of rAB=182 pm and rBC= 74 pm, pAB=-0.5 g.mol-1.pm.fs-1 and pBC=-9.8 g.mol-1.pm.fs-1, a reactive trajectory (Figure 13) is obtained. On the surface plot, at such high translational energy, the transition state was found closer to the products--late transition state.
[[File:2trans.png|thumb|frame|none|Figure 13-The trajectory for reaction 2 under high translational energy]]
Therefore, high vibrational energy and low translational energy indicate the presence the early transition state; high translational energy and low vibrational energy favour the late transition state.

When investigating for the reactive trajectories, it is found that if the vibrational energy is high, the reaction can be more efficient (early transition state). According to Polanyi's principle,

Ea=E0+αΔH

where Ea is the activation energy, α is the position of transition state along the reaction coordinates. An early transition state (at high vibrational energy) indicates a small value of α, thus lower activation energy, making the reaction more efficient.

<nowiki>*</nowiki>In all the figures 10-13, the reactants are on LHS and products on RHS.

== '''References''' ==
<nowiki>https://aip.scitation.org/doi/pdf/10.1063/1.1672005</nowiki>

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