ChemWiki - User contributions [en]

MRD:01552413

2020-05-22T21:55:24Z

Msi3018:

'''''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?'''''[[File:Gradient_123.png|400x400px|thumb|Figure 1|none]]The transition state is defined as the maximum point on the minimum energy path linking reactants and the products hence can be mathematically defined as ∂V(ri)/∂ri=0 which represents that the gradient of the potential is zero. It can be identified by starting a trajectory near the transition state and see whether they roll towards the reactants or products. This can be done several times with slightly varying trajectories in order to accurately identify the transition state. The transition state can be distinguished from the local minimum of the potential energy surface by taking the second derivative and checking whether ∂2V(ri)/∂r2i > 0 or ∂2V(ri)/∂r2i < 0 . ∂2V(ri)/∂r2i > 0 corresponds to the local minimum whereas ∂2V([[ri|ri]])/∂r2i < 0 corresponds to the maximum ( 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.'''''[[File:Surface_Plot_01552413.png|495x495px|thumb|Figure 3]]
[[File:TS_distance_plot.png|480x480px|thumb|Figure 2|none]]Since the H + H2 surface is symmetric, the transition state corresponds to where r1=r2 and hence Figure 2 shows the intersection point at 20 fs with a distance of 90.8 pm. Hence AB=BC= 90.8 pm. Figure 3 shows this point (saddle point) is at the diagonal intersect further providing evidence the transition state is this distance and also the force of the bonds is 0 kJ/mol/pm.

'''''Comment on how the mep and the trajectory you just calculated differ'''''
[[File:Dynamic_01552413.png|443x443px|thumb|Figure 5]][[File:MEP_01552413.png|464x464px|thumb|Figure 4|none]]The trajectories, was found using r1 = rts+δ, r2 = rts where δ = 1 hence r1 = 91.8 pm and r2 = 90.8 pm and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1 . Figure 4 shows a mep plot which doesn't provide a realistic account of the motion of atoms during a reaction as the vibration of the atoms aren't taken into account. However, Figure 5 showing the dynamic plot does include the atomic vibrations hence provides a more realistic account of the motion of atoms.

'''''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"
|-
|-
|-
|-
|-
|-
|}
{| class="wikitable"
|
{| 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
|-414.280
|Reactive
|Figure 5 shows a reactive trajectory as the kinetic energy is large enough to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. After the formation of AB and as rBC increases the molecule vibrates.
|[[File:Counter_01552413_1.png|400x400px|thumb|Figure 6]]

|-
|-3.1
|-4.1
|-420.081
|Unreactive
|Figure 6 shows an unreactive trajectory as the kinetic energy is not large enough to overcome the activation barrier and hence the collision doesn't have enough energy to break the BC bond and form the new AB bond. Compared to Figure 5 the momentum of the incoming A atom is reduced from -5.1 to -4.1 hence decreasing the kinetic energy and so the reactants are reformed
|[[File:Counter_01552413_2.png|400x400px|thumb|Figure 7]]
|-
|-3.1
|-5.1
|-413.978
|Reactive
|Figure 7 shows a reactive trajectory as the kinetic energy is large enough to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. After the formation of AB and as rBC increases the molecule vibrates. Compared to Figure 6 the momentum of the incoming A atom is increased from -4.1 to -5.1 hence increasing the kinetic energy and so the product forms.
|[[File:Counter_01552413_3.png|400x400px|thumb|Figure 8]]
|-
|-5.1
|-10.1
|-357.274
|Unreactive
|Figure 8 shows an unreactive trajectory. Even though there is enough kinetic energy to overcome the activation barrier the product AB starts to form but then the system reverts back to the reactants. This is barrier recrossing, where due to large kinetic energies there's a lot of extra vibrational energy which causes the trajectory to recross the barrier and therefore reform the reactants.
|[[File:Counter_01552413_4.png|400x400px|thumb|Figure 9]]
|-
|-5.1
|-10.6
|-349.476
|Reactive
|Figure 9 shows a reactive trajectory. There is enough kinetic energy to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. Similarly to Figure 7 barrier recrossing occurs where the system reverts back to the reactants but since atom A has higher energy -10.6 vs -10.1 (Figure 8), it's able to revert back to the product via a higher potential energy.
|[[File:Counter_01552413_5.png|400x400px|thumb|Figure 10]]
|}
|}
'''''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''''

The transition state theory (TST) states that a molecular collision that leads to reaction must pass through an intermediate state known as the transition state essentially where A-B is partially formed and B-C is partially broken as discussed previously. It assumes once a transition state has been achieved the transition state structure does not collapse back to the reactants. However this does not fit with experimental data as shown in Figure 9 the transition state collapses back to the reactants. Therefore TST overestimates the rate of this bimolecular reaction, since it assumes all reactions go to completion, but experimentally this violates the assumption. Furthermore, the TST is also based on the assumption that atomic nuclei behave according to the classic mechanics, hence it ignores Quantum Tunnelling which leads to the underestimation of the TST, but since this is a small effect as compared to barrier recrossing, the TST is mainly an overestimate of rate. The TST also fails for some reactions at high temperatures due to the more complex motions of molecules or at very low temperatures due to the quantum tunneling.

'''''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? Locate the approximate position of the transition state.and the value of Activation energy'''''
[[File:ATS_01552413.png|left|452x452px|thumb|Figure 11]]
[[File:TS1_01552413.png|448x448px|centre|thumb|Figure 12]]

This reaction is exothermic as H-F is stronger than H-H. Even though H-H has stronger covalent character than H-F as the overlap of the 1s orbitals is greater than the overlap of 1s and 2p orbitals of H-F, the H-F molecule is polar due to the very electronegative fluorine atom, hence it has a greater ionic character so forms stronger bonds. H-H is non-polar and hence has no ionic character. The transition state was found by changing the distances of H-H and H-F till the force = 0 kJ/mol/pm. This occurred at HF= 181.1 pm and HH= 74.5 pm. This provides evidence of an early transition sate hence an exothermic process occurs. From this, the activation energy was obtained from the mep plot of energy vs time as shown in Figure 10 and so the activation energy = 121.76 kJ/mol
{| class="wikitable"
!p= -6.1
!p=-4.1
!p=-2.1
!p=0
|-
|[[File:-6.1_01552413.png|left|250x250px]]
|[[File:-4.1_01552413.png|250x250px|centre]]
|[[File:-2.1_01552413.png|left|250x250px]]
|[[File:0_01552413.png|left|250x250px]]
|}

{| class="wikitable"
!p=2.1
!p=4.1
!p=6.1
|-
|[[File:2.1_01552413.png|left|250x250px]]
|[[File:4.1_01552413.png|left|250x250px]]
|[[File:6.1_01552413.png|left|250x250px]]
|}
The table above shows the contour diagrams at changing momentums of H-H. When p=-4.1, p=2.1, and p=4.1 the reaction proceeds to the products (reactive trajectories)whereas for p=-6.1, p= -2.1 p=0 and p=6.1 after reaching the transition state, it reverts back to the reactants hence these are unreactive trajectories. After reaching the TS for the reactive trajectories there is greater vibrational energy as the H-F bond is stronger than the H-H bond and hence the force constant will be higher therefore the vibrational energy will be larger. This further shows the difference between experimental results and the transition state theory as when the transition state has reached, the product (H-F) doesn't always form as it reverts back to the reactants.[[File:Momentaa_01552413.png|491x491px|right|thumb|Figure 13]][[File:Standard_01552413.png|498x498px|thumb|Figure 14|none]]

The mechanism of energy release can be interpreted, in terms of vibrational energy and translational energy as shown in Figure 13. At t = 0 fs H-F has no vibrational energy whereas H-H does, as shown by the oscillations on the graph. As the reaction proceeds H-F vibrates more and H-H vibrates less and eventually converges at p=3.12 g.mol-1.pm.fs-1 as no oscillations occur hence the energy is purely translational. H-F has greater oscillations than H-H as the H-F bond is stronger and so the force constant is greater so vibrational energy will be greater. For greater accuracy, IR spectroscopy can be used to confirm this as H-F is has a dipole moment so it is IR active, unlike H2 which is IR inactive hence transition between the energy levels is forbidden. Since H-F molecules can be excited, overtones are produced which suggests vibrational energy is present in the product.

'''''HF + H'''''

[[File:HF1_01552413.png|488x488px|thumb|Figure 15]][[File:HF2_01552413.png|500x500px|thumb|Figure 16|none]]

the bond distances found for the reactive trajectory was rHF = 74.49 pm and rHH= 181 pm with p = -3 and -5.1 repectively.

This reaction will be endothermic as the reactant H-F is lower in energy than the product H-H since H-F is a stronger bond than H-H as discussed previously. Via Hammond's postulate, the reactants are lower in energy compared to the products, hence a late transition state is observed. The transition state is located as previously discussed, was H-H = 74.12 pm and HF = 181.5 pm. The activation energy = 145.7 kJ/mol as shown in Figure 14.

'''''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.'''''

Te effect of translational and vibrational energies on the reaction efficiency and position of the TS can be explained by Polanyi's empirical rules. For an exothermic F + H2 reaction, the reaction barrier is usually located in the entrance valley of the reaction, that is, an early barrier. According to Polanyi's rules, reactant translational energy is then more effective than vibration to surmount the barrier to reaction, thus, accelerating the reaction rate. For an endothermic, H + HF, process a late transition barrier occurs, and hence the vibrational energy is more effective than translation to surmount the reaction barrier, thus, accelerating the reaction rate. '''''1'''''

'''''References'''''
# Yeston, J. (2012). Stretching the Polanyi Rules. ''Science'', 338(6113), pp.1397–1397.
# Schramm, V.L. (2007). Enzymatic Transition State Theory and Transition State Analogue Design. ''Journal of Biological Chemistry'', 282(39), pp.28297–28300.
# Levine, R.D. (1990). The steric factor in transition state theory and in collison theory. ''Chemical Physics Letters'', 175(4), pp.331–337. ‌ ‌
‌

MRD:01552413

2020-05-22T21:51:58Z

Msi3018:

2020-05-22T17:57:36Z

Msi3018:

MRD:01552413

2020-05-22T17:52:19Z

Msi3018:

'''''On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''''

The transition state is defined s the maximum on the minimum energy path linking reactants and the products hence can be mathematically defined as ∂V(ri)/∂ri=0 which represents that the gradient of the potential is zero).. The transition state can be identified by statring a trajectory near the transition state and see whether they roll towards the reactants or products. This can be done several times with slightly varying trajectories in order to accurately identify the transition state. The transition state can be distinguished from the local minimum of the potential energy surface by taking the second derivitive and checking whether ∂2V(ri)/∂r2i > 0 or ∂2V(ri)/∂r2i < 0 . ∂2V(ri)/∂r2i > 0 corresponds to the local minimum whereas ∂2V([[ri|ri]])/∂r2i < 0 corresponds to the maximum ( 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.'''''

Since the H + H2 surface is symmetric, the transition state corresponds to where r1=r2 and hence Figure 1 shows the intersection point at 20 fs with a distance of 90.8 pm. Hence AB=BC= 90.8 pm. Figure 2 shows this point (saddle point) is at the turning point further providing evidence the transition state is a this distance TS force =0 as no vibrational energy[[File:TS_distance_plot.png|400x400px|thumb|left|Internuclear distance against time plot.]][[File:Surface_Plot_01552413.png|400x400px]]

Comment on how the ''mep'' and the trajectory you just calculated differ.
[[File:MEP_01552413.png|400x400px]]
[[File:Dynamic_01552413.png|400x400px]]

The trajectories, mep, was found using r1 = rts+δ, r2 = rts where δ = 1 hence r1 = 91.8 pm and r2 = 90.8 pm and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1 . Figure 2 shows an mep plot doesn't provide a realistic account of the motion of atoms during a reaction as the vibration of the atoms aren't taken into account. However Figure 3 showing the dynamic plot does include the atomic vibrations hence provides a more realistic account of the motion of atoms.

'''''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"
|-
|-
|-
|-
|-
|-
|}
{| class="wikitable"
|
{| 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
|-414.280
|Reactive
|Figure 4 shows a reactive trajectory as the kinetic energy is large enough to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. After formation of AB and as rBC increases the molecule vibrates
|[[File:Counter_01552413_1.png|400x400px]]

|-
|-3.1
|-4.1
|-420.081
|Unreactive
|Figure 5 shows an unreactive trajectory as the kinetic energy is not large enough to overcome the activation barrier and hence the collision doesn't have enough energy to break the BC bond and form the new AB bond. Compared to Figure 4 the the momentum of the incoming A atom is reduced from -5.1 to -4.1 hence decreasing the kinetic energy and so the reactants are reformed
|[[File:Counter_01552413_2.png|400x400px]]
|-
|-3.1
|-5.1
|-413.978
|Reactive
|Figure 6 shows a reactive trajectory as the kinetic energy is large enough to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. After formation of AB and as rBC increases the molecule vibrates. Compared to Figure 5 the momentum of the incoming A atom is increased from -4.1 to -5.1 hence increasing the kinetic energy and so the product form.
|[[File:Counter_01552413_3.png|400x400px]]
|-
|-5.1
|-10.1
|-357.274
|Unreactive
|Figure 7 shows an unreactive trajectory. Even though there is enough kinetic energy to overcome the activation barrieran the product AB starts to form but then the system reverts back to the reactants. This is is barrier recrossing, where due to large kinetic energies there's a lot of extra vibrational energy which causes the trajectory to recross the barrier and therefore reform the reactants.
|[[File:Counter_01552413_4.png|400x400px]]
|-
|-5.1
|-10.6
|-349.476
|Reactive
|Figure 8 shows a reactive trajectory. There is enough kinetic energy to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. Similarly to Figure 7 barrier recrossing occurs where the system reverts back to the reactants but since atom A has higher energy -10.6 vs -10.1, it's able to revert back to the product via a higher potential energy
|[[File:Counter_01552413_5.png|400x400px]]
|}
|}
'''''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''''

The theory states that a molecular collision that leads to reaction must pass through an intermediate state known as the transition state essentially where A-B is partially formed and B-C is partially broken as discussed previously. It assumes once a transition state has been achieved the '''transition state''' structure does not collapse back to the reactants. However this does not fit with experimental data as shown in Figure 7 the transition state collapses back to the reactants.The Eyring Equation is based on transition state theory and is used to describe the relationship between reaction rate and temperature,both via ''Rcl'' and the partition functions.When the intermediates are very short-living, so that the Boltzmann distribution22 of energies is not reached before the process continues to the next step.

3. The transition state theory also fails for some reactions at high temperatures due to the more complex motions of molecules or at very low temperatures due to the quantum tunneling.

Furthermore, the TST is also based on the assumption that atomic nuclei behave according to the classic mechanics. It is assumed that unless atoms or molecules collide with enough energy to form the transition structure (activated complex), the reaction does not occur. However, according to the quantum mechanics, for any barrier with a finite amount of energy, there is a possibility that particles can still tunnel across the barrier.

symmetric sa wells are same energy reactant and product neither exothermic nor endothermic. TS force =0 as no vibrational energy
[[File:TST_equation_01552413.PNG|400x400px]]

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? Locate the approximate position of the transition state.and the value of Activation energy
[[File:ATS_01552413.png|left|452x452px]]
[[File:TS1_01552413.png|411x411px|centre]]

This reaction is exothermic as H-F is stronger than H-H. Even though H-H has stronger covalent character than H-F as the overlap of the 1s orbitals is greater than the overlap of 1s and 2p orbitals of H-F. Since H-F is polar due to the very electronegative fluorine atom, it has greater ionic character so forms stronger bonds. H-H is non polar and hence has no ionic character. The transition state was found by changing the distances of H-H and H-F til the force = 0 kJ/mol/pm. This occured at HF= 181.1 and HH= 74.5. This provides evidence of an early transition sate hence an exothermic process occurs. From this the activation energy was obtained from the mep plot of energy vs time as shown in figure and so the activation energy = 121.76 kJ/mol

aph we can see the reactants are higher in energy compared to the products, we have an early transition state hence this is an exothermic reaction. Hence H2 syste
{| class="wikitable"
!p= -6.1
!p=-4.1
!p=-2.1
!p=0
|-
|[[File:-6.1_01552413.png|left|250x250px]]
|[[File:-4.1_01552413.png|250x250px|centre]]
|[[File:-2.1_01552413.png|left|250x250px]]
|[[File:0_01552413.png|left|250x250px]]
|}

{| class="wikitable"
!p=2.1
!p=4.1
!p=6.1
|-
|[[File:2.1_01552413.png|left|250x250px]]
|[[File:4.1_01552413.png|left|250x250px]]
|[[File:6.1_01552413.png|left|250x250px]]
|}
These show that when p=-4.1, p=2.1 and p=4.1 the reaction
proceeds to the products whereas in the for p=-6.1, p= -2.1 p=0 and p=6.1 after
reaching the transition state, it reverts back to the reactants hence these are
unreactive trajectories. After reaching the TS for the reactive trajectories
there is greater vibrational energy as the H-F bond is stronger than the H-H
bond and hence the force constant will be higher. hence the vibrational energy will be larger. This further shows the difference between experimental results and the transition state theory as when the transition state is reached the product (H-F) doesn't always form.[[File:Momentaa_01552413.png|464x464px|right]][[File:Standard_01552413.png|left|400x400px]]

This plot shows the overall energy of the system is reduced as the momentum is reduced hence it's an unreactive trajectory as when the transition state is reached, it reverts back to the reactants.

The reactive trajectory where the momentum is 4.1 for fluorine atom. The mechanism of energy release can be interpreted, in terms of vibrational energy and translational energy as shown in Figure. At t = 0 fs H-F has no vibrational energy whereas H-H does, as shown by the oscillations on the graph. As the reaction proceeds H-F vibrates more and H-H vibrates less and eventually converges at we see there is a convergence at around p=3.12 as no oscillations occur hence the energy is purely translational. H-F has greater oscillations as the H-F bond is stronger and so the force constant is greater so vibrational energy will be greater. For greater accuracy, IR spectroscopy can be used to confirm this as H-F is has a dipole moment so is IR active, unlike H2 which is IR inactive since transition between the energy levels is forbidden. Since H-F molecules can be excited, overtones are produced which suggests vibrational energy is present in the product.

HF + H

[[File:HF1_01552413.png|464x464px|right]][[File:HF2_01552413.png|left|400x400px]]

the bond distances found for the reactive trajectory was rHF = 74.49 pm and rHH= 181 with p = -3 and -5.1 repectively

This reaction will be endothermic as the reactant H-F is lower in energy than the product H-H, since H-F is a stronger bond than H-H as discussed previpously Via Hammond's postulate the reactants are lower in energy compared to the products, hence a late transition state is observed. The transition state is located as previously discussed, was H-H = 74.12 pm and HF = 181.5 pm. The activation energy = 145.7 kJ/mol as shown in Figure.

File:HF1 01552413.png

2020-05-22T17:51:13Z

Msi3018:

2020-05-22T16:58:14Z

Msi3018:

MRD:01552413

2020-05-22T16:57:09Z

Msi3018:

'''''On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''''

The transition state is defined s the maximum on the minimum energy path linking reactants and the products hence can be mathematically defined as ∂V(ri)/∂ri=0 which represents that the gradient of the potential is zero).. The transition state can be identified by statring a trajectory near the transition state and see whether they roll towards the reactants or products. This can be done several times with slightly varying trajectories in order to accurately identify the transition state. The transition state can be distinguished from the local minimum of the potential energy surface by taking the second derivitive and checking whether ∂2V(ri)/∂r2i > 0 or ∂2V(ri)/∂r2i < 0 . ∂2V(ri)/∂r2i > 0 corresponds to the local minimum whereas ∂2V([[ri|ri]])/∂r2i < 0 corresponds to the maximum ( 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.'''''

Since the H + H2 surface is symmetric, the transition state corresponds to where r1=r2 and hence Figure 1 shows the intersection point at 20 fs with a distance of 90.8 pm. Hence AB=BC= 90.8 pm. Figure 2 shows this point (saddle point) is at the turning point further providing evidence the transition state is a this distance TS force =0 as no vibrational energy[[File:TS_distance_plot.png|400x400px|thumb|left|Internuclear distance against time plot.]][[File:Surface_Plot_01552413.png|400x400px]]

Comment on how the ''mep'' and the trajectory you just calculated differ.
[[File:MEP_01552413.png|400x400px]]
[[File:Dynamic_01552413.png|400x400px]]

The trajectories, mep, was found using r1 = rts+δ, r2 = rts where δ = 1 hence r1 = 91.8 pm and r2 = 90.8 pm and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1 . Figure 2 shows an mep plot doesn't provide a realistic account of the motion of atoms during a reaction as the vibration of the atoms aren't taken into account. However Figure 3 showing the dynamic plot does include the atomic vibrations hence provides a more realistic account of the motion of atoms.

'''''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"
|-
|-
|-
|-
|-
|-
|}
{| class="wikitable"
|
{| 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
|-414.280
|Reactive
|Figure 4 shows a reactive trajectory as the kinetic energy is large enough to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. After formation of AB and as rBC increases the molecule vibrates
|[[File:Counter_01552413_1.png|400x400px]]

|-
|-3.1
|-4.1
|-420.081
|Unreactive
|Figure 5 shows an unreactive trajectory as the kinetic energy is not large enough to overcome the activation barrier and hence the collision doesn't have enough energy to break the BC bond and form the new AB bond. Compared to Figure 4 the the momentum of the incoming A atom is reduced from -5.1 to -4.1 hence decreasing the kinetic energy and so the reactants are reformed
|[[File:Counter_01552413_2.png|400x400px]]
|-
|-3.1
|-5.1
|-413.978
|Reactive
|Figure 6 shows a reactive trajectory as the kinetic energy is large enough to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. After formation of AB and as rBC increases the molecule vibrates. Compared to Figure 5 the momentum of the incoming A atom is increased from -4.1 to -5.1 hence increasing the kinetic energy and so the product form.
|[[File:Counter_01552413_3.png|400x400px]]
|-
|-5.1
|-10.1
|-357.274
|Unreactive
|Figure 7 shows an unreactive trajectory. Even though there is enough kinetic energy to overcome the activation barrieran the product AB starts to form but then the system reverts back to the reactants. This is is barrier recrossing, where due to large kinetic energies there's a lot of extra vibrational energy which causes the trajectory to recross the barrier and therefore reform the reactants.
|[[File:Counter_01552413_4.png|400x400px]]
|-
|-5.1
|-10.6
|-349.476
|Reactive
|Figure 8 shows a reactive trajectory. There is enough kinetic energy to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. Similarly to Figure 7 barrier recrossing occurs where the system reverts back to the reactants but since atom A has higher energy -10.6 vs -10.1, it's able to revert back to the product via a higher potential energy
|[[File:Counter_01552413_5.png|400x400px]]
|}
|}
'''''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''''

The theory states that a molecular collision that leads to reaction must pass through an intermediate state known as the transition state essentially where A-B is partially formed and B-C is partially broken as discussed previously. It assumes once a transition state has been achieved the '''transition state''' structure does not collapse back to the reactants. However this does not fit with experimental data as shown in Figure 7 the transition state collapses back to the reactants.The Eyring Equation is based on transition state theory and is used to describe the relationship between reaction rate and temperature,both via ''Rcl'' and the partition functions.When the intermediates are very short-living, so that the Boltzmann distribution22 of energies is not reached before the process continues to the next step.

3. The transition state theory also fails for some reactions at high temperatures due to the more complex motions of molecules or at very low temperatures due to the quantum tunneling.

Furthermore, the TST is also based on the assumption that atomic nuclei behave according to the classic mechanics. It is assumed that unless atoms or molecules collide with enough energy to form the transition structure (activated complex), the reaction does not occur. However, according to the quantum mechanics, for any barrier with a finite amount of energy, there is a possibility that particles can still tunnel across the barrier.

symmetric sa wells are same energy reactant and product neither exothermic nor endothermic. TS force =0 as no vibrational energy
[[File:TST_equation_01552413.PNG|400x400px]]

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? Locate the approximate position of the transition state.and the value of Activation energy
[[File:ATS_01552413.png|left|452x452px]]
[[File:TS1_01552413.png|411x411px|centre]]

aph we can see the reactants are higher in energy compared to the products, we have an early transition state hence this is an exothermic reaction. Hence H2 syste

[[File:Surface_01552413.png|left|400x400px]]

{| class="wikitable"
!p= -6.1
!p=-4.1
!p=-2.1
!p=0
|-
|[[File:-6.1_01552413.png|left|250x250px]]
|[[File:-4.1_01552413.png|250x250px|centre]]
|[[File:-2.1_01552413.png|left|250x250px]]
|[[File:0_01552413.png|left|250x250px]]
|}

{| class="wikitable"
!p=2.1
!p=4.1
!p=6.1
|-
|[[File:2.1_01552413.png|left|250x250px]]
|[[File:4.1_01552413.png|left|250x250px]]
|[[File:6.1_01552413.png|left|250x250px]]
|}
These show that when p=-4.1, p=2.1 and p=4.1 the reaction
proceeds to the products whereas in the for p=-6.1, p= -2.1 p=0 and p=6.1 after
reaching the transition state, it reverts back to the reactants hence these are
unreactive trajectories. After reaching the TS for the reactive trajectories
there is greater vibrational energy as the H-F bond is stronger than the H-H
bond and hence the force constant will be higher. hence the vibrational energy will be larger. This further shows the difference between experimental results and the transition state theory as when the transition state is reached the product (H-F) doesn't always form.

[[File:Standard_01552413.png|left|400x400px]]This plot shows the overall energy of the system is reduced as the momentum is reduced hence it's an unreactive trajectory as when the transition state is reached, it reverts back to the reactants.
[[File:Momenta_01552413.png|left|400x400px]]The reactive trajectory where the momentum is 4.1 for fluorine atom. The mechanism of energy release can be interpreted, in terms of vibrational energy and translational energy as shown in Figure. At t = 0 fs H-F has no vibrational energy whereas H-H does, as shown by the oscillations on the graph. As the reaction proceeds H-F vibrates more and H-H vibrates less and eventually converges at we see there is a convergence at around p=3.12 as no oscillations occur hence the energy is purely translational. H-F has greater oscillations as the H-F bond is stronger and so the force constant is greater so vibrational energy will be greater. For greater accuracy, IR spectroscopy can be used to confirm this as H-F is has a dipole moment so is IR active, unlike H2 which is IR inactive since transition between the energy levels is forbidden. Since H-F molecules can be excited, overtones are produced which suggests vibrational energy is present in the product.

therefore higher vibrational exited states will be populated, giving evidence of overtones. This therefore suggests that the energy has been distributed as vibrational energy for the products and translational energy to the H atom.

MRD:01552413

2020-05-22T16:31:50Z

Msi3018:

'''''On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''''

The transition state is defined s the maximum on the minimum energy path linking reactants and the products hence can be mathematically defined as ∂V(ri)/∂ri=0 which represents that the gradient of the potential is zero).. The transition state can be identified by statring a trajectory near the transition state and see whether they roll towards the reactants or products. This can be done several times with slightly varying trajectories in order to accurately identify the transition state. The transition state can be distinguished from the local minimum of the potential energy surface by taking the second derivitive and checking whether ∂2V(ri)/∂r2i > 0 or ∂2V(ri)/∂r2i < 0 . ∂2V(ri)/∂r2i > 0 corresponds to the local minimum whereas ∂2V([[ri|ri]])/∂r2i < 0 corresponds to the maximum ( 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.'''''

Since the H + H2 surface is symmetric, the transition state corresponds to where r1=r2 and hence Figure 1 shows the intersection point at 20 fs with a distance of 90.8 pm. Hence AB=BC= 90.8 pm. Figure 2 shows this point (saddle point) is at the turning point further providing evidence the transition state is a this distance TS force =0 as no vibrational energy[[File:TS_distance_plot.png|400x400px|thumb|left|Internuclear distance against time plot.]][[File:Surface_Plot_01552413.png|400x400px]]

Comment on how the ''mep'' and the trajectory you just calculated differ.
[[File:MEP_01552413.png|400x400px]]
[[File:Dynamic_01552413.png|400x400px]]

The trajectories, mep, was found using r1 = rts+δ, r2 = rts where δ = 1 hence r1 = 91.8 pm and r2 = 90.8 pm and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1 . Figure 2 shows an mep plot doesn't provide a realistic account of the motion of atoms during a reaction as the vibration of the atoms aren't taken into account. However Figure 3 showing the dynamic plot does include the atomic vibrations hence provides a more realistic account of the motion of atoms.

'''''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"
|-
|-
|-
|-
|-
|-
|}
{| class="wikitable"
|
{| 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
|-414.280
|Reactive
|Figure 4 shows a reactive trajectory as the kinetic energy is large enough to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. After formation of AB and as rBC increases the molecule vibrates
|[[File:Counter_01552413_1.png|400x400px]]

|-
|-3.1
|-4.1
|-420.081
|Unreactive
|Figure 5 shows an unreactive trajectory as the kinetic energy is not large enough to overcome the activation barrier and hence the collision doesn't have enough energy to break the BC bond and form the new AB bond. Compared to Figure 4 the the momentum of the incoming A atom is reduced from -5.1 to -4.1 hence decreasing the kinetic energy and so the reactants are reformed
|[[File:Counter_01552413_2.png|400x400px]]
|-
|-3.1
|-5.1
|-413.978
|Reactive
|Figure 6 shows a reactive trajectory as the kinetic energy is large enough to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. After formation of AB and as rBC increases the molecule vibrates. Compared to Figure 5 the momentum of the incoming A atom is increased from -4.1 to -5.1 hence increasing the kinetic energy and so the product form.
|[[File:Counter_01552413_3.png|400x400px]]
|-
|-5.1
|-10.1
|-357.274
|Unreactive
|Figure 7 shows an unreactive trajectory. Even though there is enough kinetic energy to overcome the activation barrieran the product AB starts to form but then the system reverts back to the reactants. This is is barrier recrossing, where due to large kinetic energies there's a lot of extra vibrational energy which causes the trajectory to recross the barrier and therefore reform the reactants.
|[[File:Counter_01552413_4.png|400x400px]]
|-
|-5.1
|-10.6
|-349.476
|Reactive
|Figure 8 shows a reactive trajectory. There is enough kinetic energy to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. Similarly to Figure 7 barrier recrossing occurs where the system reverts back to the reactants but since atom A has higher energy -10.6 vs -10.1, it's able to revert back to the product via a higher potential energy
|[[File:Counter_01552413_5.png|400x400px]]
|}
|}
'''''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''''

The theory states that a molecular collision that leads to reaction must pass through an intermediate state known as the transition state essentially where A-B is partially formed and B-C is partially broken as discussed previously. It assumes once a transition state has been achieved the '''transition state''' structure does not collapse back to the reactants. However this does not fit with experimental data as shown in Figure 7 the transition state collapses back to the reactants.The Eyring Equation is based on transition state theory and is used to describe the relationship between reaction rate and temperature,both via ''Rcl'' and the partition functions.When the intermediates are very short-living, so that the Boltzmann distribution22 of energies is not reached before the process continues to the next step.

3. The transition state theory also fails for some reactions at high temperatures due to the more complex motions of molecules or at very low temperatures due to the quantum tunneling.

Furthermore, the TST is also based on the assumption that atomic nuclei behave according to the classic mechanics. It is assumed that unless atoms or molecules collide with enough energy to form the transition structure (activated complex), the reaction does not occur. However, according to the quantum mechanics, for any barrier with a finite amount of energy, there is a possibility that particles can still tunnel across the barrier.

symmetric sa wells are same energy reactant and product neither exothermic nor endothermic. TS force =0 as no vibrational energy
[[File:TST_equation_01552413.PNG|400x400px]]

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? Locate the approximate position of the transition state.and the value of Activation energy
[[File:ATS_01552413.png|left|452x452px]]
[[File:TS1_01552413.png|411x411px|centre]]

aph we can see the reactants are higher in energy compared to the products, we have an early transition state hence this is an exothermic reaction. Hence H2 syste

[[File:Surface_01552413.png|left|400x400px]]

{| class="wikitable"
!p= -6.1
!p=-4.1
!p=-2.1
!p=0
|-
|[[File:-6.1_01552413.png|left|250x250px]]
|[[File:-4.1_01552413.png|250x250px|centre]]
|[[File:-2.1_01552413.png|left|250x250px]]
|[[File:0_01552413.png|left|250x250px]]
|}

{| class="wikitable"
!p=2.1
!p=4.1
!p=6.1
|-
|[[File:2.1_01552413.png|left|250x250px]]
|[[File:4.1_01552413.png|left|250x250px]]
|[[File:6.1_01552413.png|left|250x250px]]
|}
These show that when p=-4.1, p=2.1 and p=4.1 the reaction
proceeds to the products whereas in the for p=-6.1, p= -2.1 p=0 and p=6.1 after
reaching the transition state, it reverts back to the reactants hence these are
unreactive trajectories. After reaching the TS for the reactive trajectories
there is greater vibrational energy as the H-F bond is stronger than the H-H
bond and hence the force constant will be higher. hence the vibrational energy will be larger. This further shows the difference between experimental results and the transition state theory as when the transition state is reached the product (H-F) doesn't always form.

[[File:Standard_01552413.png|left|400x400px]]This plot shows the overall energy of the system is reduced as the momentum is reduced hence it's an unreactive trajectory as when the transition state is reached, it reverts back to the reactants.
[[File:Momenta_01552413.png|left|400x400px]]The reactive trajectory where the momentum is 4.1 for fluorine atom. The mechanism of energy release can be interpreted, in terms of vibrational energy and translational energy as shown in Figure. At t = 0 fs H-F has no vibrational energy whereas H-H does, as shown by the oscillations on the graph. As the reaction proceeds we see there is a convergence at around -7.28, after which translational energy of the reactants is transferred to vibrational energy, and vice versa. This is can be seen where the BC line plateaus, suggesting it has all translational energy now. This however could be more accurately confirmed via experimental analytical techniques such as IR spectroscopy. Since one of the selection rules state that only molecules which experience a change in dipole will be IR-active, and as the H-H reactant is not IR active, we should not see any vibrational bands in IR, i.e. they populate the ground state only. The product (H-F) however does possess a dipole moment, and therefore higher vibrational exited states will be populated, giving evidence of overtones. This therefore suggests that the energy has been distributed as vibrational energy for the products and translational energy to the H atom.

MRD:01552413

2020-05-22T16:16:35Z

Msi3018:

'''''On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''''

The transition state is defined s the maximum on the minimum energy path linking reactants and the products hence can be mathematically defined as ∂V(ri)/∂ri=0 which represents that the gradient of the potential is zero).. The transition state can be identified by statring a trajectory near the transition state and see whether they roll towards the reactants or products. This can be done several times with slightly varying trajectories in order to accurately identify the transition state. The transition state can be distinguished from the local minimum of the potential energy surface by taking the second derivitive and checking whether ∂2V(ri)/∂r2i > 0 or ∂2V(ri)/∂r2i < 0 . ∂2V(ri)/∂r2i > 0 corresponds to the local minimum whereas ∂2V([[ri|ri]])/∂r2i < 0 corresponds to the maximum ( 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.'''''

Since the H + H2 surface is symmetric, the transition state corresponds to where r1=r2 and hence Figure 1 shows the intersection point at 20 fs with a distance of 90.8 pm. Hence AB=BC= 90.8 pm. Figure 2 shows this point (saddle point) is at the turning point further providing evidence the transition state is a this distance TS force =0 as no vibrational energy[[File:TS_distance_plot.png|400x400px|thumb|left|Internuclear distance against time plot.]][[File:Surface_Plot_01552413.png|400x400px]]

Comment on how the ''mep'' and the trajectory you just calculated differ.
[[File:MEP_01552413.png|400x400px]]
[[File:Dynamic_01552413.png|400x400px]]

The trajectories, mep, was found using r1 = rts+δ, r2 = rts where δ = 1 hence r1 = 91.8 pm and r2 = 90.8 pm and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1 . Figure 2 shows an mep plot doesn't provide a realistic account of the motion of atoms during a reaction as the vibration of the atoms aren't taken into account. However Figure 3 showing the dynamic plot does include the atomic vibrations hence provides a more realistic account of the motion of atoms.

'''''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"
|-
|-
|-
|-
|-
|-
|}
{| class="wikitable"
|
{| 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
|-414.280
|Reactive
|Figure 4 shows a reactive trajectory as the kinetic energy is large enough to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. After formation of AB and as rBC increases the molecule vibrates
|[[File:Counter_01552413_1.png|400x400px]]

|-
|-3.1
|-4.1
|-420.081
|Unreactive
|Figure 5 shows an unreactive trajectory as the kinetic energy is not large enough to overcome the activation barrier and hence the collision doesn't have enough energy to break the BC bond and form the new AB bond. Compared to Figure 4 the the momentum of the incoming A atom is reduced from -5.1 to -4.1 hence decreasing the kinetic energy and so the reactants are reformed
|[[File:Counter_01552413_2.png|400x400px]]
|-
|-3.1
|-5.1
|-413.978
|Reactive
|Figure 6 shows a reactive trajectory as the kinetic energy is large enough to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. After formation of AB and as rBC increases the molecule vibrates. Compared to Figure 5 the momentum of the incoming A atom is increased from -4.1 to -5.1 hence increasing the kinetic energy and so the product form.
|[[File:Counter_01552413_3.png|400x400px]]
|-
|-5.1
|-10.1
|-357.274
|Unreactive
|Figure 7 shows an unreactive trajectory. Even though there is enough kinetic energy to overcome the activation barrieran the product AB starts to form but then the system reverts back to the reactants. This is is barrier recrossing, where due to large kinetic energies there's a lot of extra vibrational energy which causes the trajectory to recross the barrier and therefore reform the reactants.
|[[File:Counter_01552413_4.png|400x400px]]
|-
|-5.1
|-10.6
|-349.476
|Reactive
|Figure 8 shows a reactive trajectory. There is enough kinetic energy to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. Similarly to Figure 7 barrier recrossing occurs where the system reverts back to the reactants but since atom A has higher energy -10.6 vs -10.1, it's able to revert back to the product via a higher potential energy
|[[File:Counter_01552413_5.png|400x400px]]
|}
|}
'''''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''''

The theory states that a molecular collision that leads to reaction must pass through an intermediate state known as the transition state essentially where A-B is partially formed and B-C is partially broken as discussed previously. It assumes once a transition state has been achieved the '''transition state''' structure does not collapse back to the reactants. However this does not fit with experimental data as shown in Figure 7 the transition state collapses back to the reactants.The Eyring Equation is based on transition state theory and is used to describe the relationship between reaction rate and temperature,both via ''Rcl'' and the partition functions.When the intermediates are very short-living, so that the Boltzmann distribution22 of energies is not reached before the process continues to the next step.

3. The transition state theory also fails for some reactions at high temperatures due to the more complex motions of molecules or at very low temperatures due to the quantum tunneling.

Furthermore, the TST is also based on the assumption that atomic nuclei behave according to the classic mechanics. It is assumed that unless atoms or molecules collide with enough energy to form the transition structure (activated complex), the reaction does not occur. However, according to the quantum mechanics, for any barrier with a finite amount of energy, there is a possibility that particles can still tunnel across the barrier.

symmetric sa wells are same energy reactant and product neither exothermic nor endothermic. TS force =0 as no vibrational energy
[[File:TST_equation_01552413.PNG|400x400px]]

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? Locate the approximate position of the transition state.and the value of Activation energy
[[File:ATS_01552413.png|left|452x452px]]
[[File:TS1_01552413.png|411x411px|centre]]

Locate the approximate position of the transition state.

Initial conditions: rHH= 74 pm pHH = -5.1 g.mol-1.pm.fs-1 rHF= 180 pm pHF = -1 g.mol-1.pm.fs-1

Since Hammond's postulate states that for an exothermic reaction, the reactants resemble the transition state and so the transition state bond length resembles the bond length of the reactant(H-H=74 pm). Hence by adjusting the bond lengths it was determined that at H-H = So by adjusting the H-F distance, keeping the H-H bond length consistent at 74 pm until the forces along both bonds are close to zero gives us the transition state where AB= 74 pm, BC=181.3 74.5 pm and H-F= 181.3 pm, the force was +0.000kJ/mol/pm for both H-F and H-H. The saddle point is shown in Figure as a red cross..

This reaction is exothermic as H-F is stronger than H-H. Evenr though in H2 the overlap between the 1s orbital is stronger than the overlap of the 2p and 1s orbital in HF. However as HF bond is very polar due to Fluorine being very electronegative hence has a large ionic componenet to the bond strenght. H2 however is non polar so has no ionic contribution this can be shown by Figure 10 which shows the product energy (HF) is lower than the reactant energy (H2) hence exothemic. Hence the activation energy can be determined via an mep plot for energy vs time at the saddle point. Hence the activation energy is 123.3 kJ/mol-1. The Surface plot ( Figure 11) shows the H-H system higher in energy due to a weaker bond and hence the reaction has an early transition state as stated by Hammonds postulate the transition state resembles the reactants and products depending on how close it is in energy to either of them. Since from the graph we can see the reactants are higher in energy compared to the products, we have an early transition state hence this is an exothermic reaction. Hence H2 syste

[[File:Surface_01552413.png|left|400x400px]]

{| class="wikitable"
!p= -6.1
!p=-4.1
!p=-2.1
!p=0
|-
|[[File:-6.1_01552413.png|left|250x250px]]
|[[File:-4.1_01552413.png|250x250px|centre]]
|[[File:-2.1_01552413.png|left|250x250px]]
|[[File:0_01552413.png|left|250x250px]]
|}

{| class="wikitable"
!p=2.1
!p=4.1
!p=6.1
|-
|[[File:2.1_01552413.png|left|250x250px]]
|[[File:4.1_01552413.png|left|250x250px]]
|[[File:6.1_01552413.png|left|250x250px]]
|}
These show that when p=-4.1, p=2.1 and p=4.1 the reaction
proceeds to the products whereas in the for p=-6.1, p= -2.1 p=0 and p=6.1 after
reaching the transition state, it reverts back to the reactants hence these are
unreactive trajectories. After reaching the TS for the reactive trajectories
there is greater vibrational energy as the H-F bond is stronger than the H-H
bond and hence the force constant will be higher. hence the vibrational energy will be larger. This further shows the difference between experimental results and the transition state theory as when the transition state is reached the product (H-F) doesn't always form.

[[File:Standard_01552413.png|left|400x400px]]This plot shows the overall energy of the system is reduced as the momentum is reduced hence it's an unreactive trajectory as when the transition state is reached, it reverts back to the reactants.
[[File:Momenta_01552413.png|left|400x400px]]

2020-05-22T14:07:21Z

Msi3018:

MRD:01552413

2020-05-22T14:02:49Z

Msi3018:

'''''On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''''

The transition state is defined s the maximum on the minimum energy path linking reactants and the products hence can be mathematically defined as ∂V(ri)/∂ri=0 which represents that the gradient of the potential is zero).. The transition state can be identified by statring a trajectory near the transition state and see whether they roll towards the reactants or products. This can be done several times with slightly varying trajectories in order to accurately identify the transition state. The transition state can be distinguished from the local minimum of the potential energy surface by taking the second derivitive and checking whether ∂2V(ri)/∂r2i > 0 or ∂2V(ri)/∂r2i < 0 . ∂2V(ri)/∂r2i > 0 corresponds to the local minimum whereas ∂2V([[ri|ri]])/∂r2i < 0 corresponds to the maximum ( 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.'''''

Since the H + H2 surface is symmetric, the transition state corresponds to where r1=r2 and hence Figure 1 shows the intersection point at 20 fs with a distance of 90.8 pm. Hence AB=BC= 90.8 pm. Figure 2 shows this point (saddle point) is at the turning point further providing evidence the transition state is a this distance TS force =0 as no vibrational energy[[File:TS_distance_plot.png|400x400px|thumb|left|Internuclear distance against time plot.]][[File:Surface_Plot_01552413.png|400x400px]]

Comment on how the ''mep'' and the trajectory you just calculated differ.
[[File:MEP_01552413.png|400x400px]]
[[File:Dynamic_01552413.png|400x400px]]

The trajectories, mep, was found using r1 = rts+δ, r2 = rts where δ = 1 hence r1 = 91.8 pm and r2 = 90.8 pm and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1 . Figure 2 shows an mep plot doesn't provide a realistic account of the motion of atoms during a reaction as the vibration of the atoms aren't taken into account. However Figure 3 showing the dynamic plot does include the atomic vibrations hence provides a more realistic account of the motion of atoms.

'''''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"
|-
|-
|-
|-
|-
|-
|}
{| class="wikitable"
|
{| 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
|-414.280
|Reactive
|Figure 4 shows a reactive trajectory as the kinetic energy is large enough to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. After formation of AB and as rBC increases the molecule vibrates
|[[File:Counter_01552413_1.png|400x400px]]

|-
|-3.1
|-4.1
|-420.081
|Unreactive
|Figure 5 shows an unreactive trajectory as the kinetic energy is not large enough to overcome the activation barrier and hence the collision doesn't have enough energy to break the BC bond and form the new AB bond. Compared to Figure 4 the the momentum of the incoming A atom is reduced from -5.1 to -4.1 hence decreasing the kinetic energy and so the reactants are reformed
|[[File:Counter_01552413_2.png|400x400px]]
|-
|-3.1
|-5.1
|-413.978
|Reactive
|Figure 6 shows a reactive trajectory as the kinetic energy is large enough to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. After formation of AB and as rBC increases the molecule vibrates. Compared to Figure 5 the momentum of the incoming A atom is increased from -4.1 to -5.1 hence increasing the kinetic energy and so the product form.
|[[File:Counter_01552413_3.png|400x400px]]
|-
|-5.1
|-10.1
|-357.274
|Unreactive
|Figure 7 shows an unreactive trajectory. Even though there is enough kinetic energy to overcome the activation barrieran the product AB starts to form but then the system reverts back to the reactants. This is is barrier recrossing, where due to large kinetic energies there's a lot of extra vibrational energy which causes the trajectory to recross the barrier and therefore reform the reactants.
|[[File:Counter_01552413_4.png|400x400px]]
|-
|-5.1
|-10.6
|-349.476
|Reactive
|Figure 8 shows a reactive trajectory. There is enough kinetic energy to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. Similarly to Figure 7 barrier recrossing occurs where the system reverts back to the reactants but since atom A has higher energy -10.6 vs -10.1, it's able to revert back to the product via a higher potential energy
|[[File:Counter_01552413_5.png|400x400px]]
|}
|}
'''''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''''

The theory states that a molecular collision that leads to reaction must pass through an intermediate state known as the transition state essentially where A-B is partially formed and B-C is partially broken as discussed previously. It assumes once a transition state has been achieved the '''transition state''' structure does not collapse back to the reactants. However this does not fit with experimental data as shown in Figure 7 the transition state collapses back to the reactants.The Eyring Equation is based on transition state theory and is used to describe the relationship between reaction rate and temperature,both via ''Rcl'' and the partition functions.When the intermediates are very short-living, so that the Boltzmann distribution22 of energies is not reached before the process continues to the next step.

3. The transition state theory also fails for some reactions at high temperatures due to the more complex motions of molecules or at very low temperatures due to the quantum tunneling.

Furthermore, the TST is also based on the assumption that atomic nuclei behave according to the classic mechanics. It is assumed that unless atoms or molecules collide with enough energy to form the transition structure (activated complex), the reaction does not occur. However, according to the quantum mechanics, for any barrier with a finite amount of energy, there is a possibility that particles can still tunnel across the barrier.

symmetric sa wells are same energy reactant and product neither exothermic nor endothermic. TS force =0 as no vibrational energy
[[File:TST_equation_01552413.PNG|400x400px]]

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?
[[File:ATS_01552413.png|left|400x400px]]
Locate the approximate position of the transition state.

Initial conditions: rHH= 74 pm pHH = -5.1 g.mol-1.pm.fs-1 rHF= 180 pm pHF = -1 g.mol-1.pm.fs-1

This reaction is exothermic as H-F is stronger than H-H. Evenr though in H2 the overlap between the 1s orbital is stronger than the overlap of the 2p and 1s orbital in HF. However as HF bond is very polar due to Fluorine being very electronegative hence has a large ionic componenet to the bond strenght. H2 however is non polar so has no ionic contribution this can be shown by Figure 10 which shows the product energy (HF) is lower than the reactant energy (H2) hence exothemic. Hence the activation energy is 123.3 kJ/mol-1 determined via the energy vs time for an mep plot. The Surface plot ( Figure 11) shows the H-H system hoigher in energy due to a weaker bond and hence the reaction has an early transition state as stated by Hammonds postulate the transition state resembles the reactants and products depending on how close it is in energy to either of them. Since from the graph we can see the reactants are higher in energy compared to the products, we have an early transition state hence this is an exothermic reaction. Hence H2 syste

[[File:Surface_01552413.png|left|400x400px]]

[[File:6.1_01552413.png|left|250x250px]]

{| class="wikitable"
!p= -6.1
!p=-4.1
!p=-2.1
!p=0
|-
|[[File:-6.1_01552413.png|left|250x250px]]
|[[File:-4.1_01552413.png|250x250px|centre]]
|[[File:-2.1_01552413.png|left|250x250px]]
|[[File:0_01552413.png|left|250x250px]]
|}

{| class="wikitable"
!p=2.1
!p=4.1
!p=6.1
|-
|[[File:2.1_01552413.png|left|250x250px]]
|[[File:4.1_01552413.png|left|250x250px]]
|[[File:6.1_01552413.png|left|250x250px]]
|}
These show that when p=-4.1, p=2.1 and p=4.1 the reaction
proceeds to the products whereas in the for p=-6.1, p= -2.1 p=0 and p=6.1 after
reaching the transition state, it reverts back to the reactants hence these are
unreactive trajectories. After reaching the TS for the reactive trajectories
there is greater vibrational energy as the H-F bond is stronger than the H-H
bond and hence the force constant will be higher. hence the vibrational energy will be larger. This further shows the difference between experimental results and the transition state theory as when the transition state is reached the product (H-F) doesn't always form.

[[File:Standard_01552413.png|left|400x400px]]This plot shows the overall energy of the system is reduced as the momentum is reduced hence it's an unreactive trajectory as when the transition state is reached, it reverts back to the reactants.

MRD:01552413

2020-05-22T12:59:57Z

Msi3018:

'''''On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''''

The transition state is defined s the maximum on the minimum energy path linking reactants and the products hence can be mathematically defined as ∂V(ri)/∂ri=0 which represents that the gradient of the potential is zero).. The transition state can be identified by statring a trajectory near the transition state and see whether they roll towards the reactants or products. This can be done several times with slightly varying trajectories in order to accurately identify the transition state. The transition state can be distinguished from the local minimum of the potential energy surface by taking the second derivitive and checking whether ∂2V(ri)/∂r2i > 0 or ∂2V(ri)/∂r2i < 0 . ∂2V(ri)/∂r2i > 0 corresponds to the local minimum whereas ∂2V([[ri|ri]])/∂r2i < 0 corresponds to the maximum ( 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.'''''

Since the H + H2 surface is symmetric, the transition state corresponds to where r1=r2 and hence Figure 1 shows the intersection point at 20 fs with a distance of 90.8 pm. Hence AB=BC= 90.8 pm. Figure 2 shows this point (saddle point) is at the turning point further providing evidence the transition state is a this distance TS force =0 as no vibrational energy[[File:TS_distance_plot.png|400x400px|thumb|left|Internuclear distance against time plot.]][[File:Surface_Plot_01552413.png|400x400px]]

Comment on how the ''mep'' and the trajectory you just calculated differ.
[[File:MEP_01552413.png|400x400px]]
[[File:Dynamic_01552413.png|400x400px]]

The trajectories, mep, was found using r1 = rts+δ, r2 = rts where δ = 1 hence r1 = 91.8 pm and r2 = 90.8 pm and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1 . Figure 2 shows an mep plot doesn't provide a realistic account of the motion of atoms during a reaction as the vibration of the atoms aren't taken into account. However Figure 3 showing the dynamic plot does include the atomic vibrations hence provides a more realistic account of the motion of atoms.

'''''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"
|-
|-
|-
|-
|-
|-
|}
{| class="wikitable"
|
{| 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
|-414.280
|Reactive
|Figure 4 shows a reactive trajectory as the kinetic energy is large enough to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. After formation of AB and as rBC increases the molecule vibrates
|[[File:Counter_01552413_1.png|400x400px]]

|-
|-3.1
|-4.1
|-420.081
|Unreactive
|Figure 5 shows an unreactive trajectory as the kinetic energy is not large enough to overcome the activation barrier and hence the collision doesn't have enough energy to break the BC bond and form the new AB bond. Compared to Figure 4 the the momentum of the incoming A atom is reduced from -5.1 to -4.1 hence decreasing the kinetic energy and so the reactants are reformed
|[[File:Counter_01552413_2.png|400x400px]]
|-
|-3.1
|-5.1
|-413.978
|Reactive
|Figure 6 shows a reactive trajectory as the kinetic energy is large enough to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. After formation of AB and as rBC increases the molecule vibrates. Compared to Figure 5 the momentum of the incoming A atom is increased from -4.1 to -5.1 hence increasing the kinetic energy and so the product form.
|[[File:Counter_01552413_3.png|400x400px]]
|-
|-5.1
|-10.1
|-357.274
|Unreactive
|Figure 7 shows an unreactive trajectory. Even though there is enough kinetic energy to overcome the activation barrieran the product AB starts to form but then the system reverts back to the reactants. This is is barrier recrossing, where due to large kinetic energies there's a lot of extra vibrational energy which causes the trajectory to recross the barrier and therefore reform the reactants.
|[[File:Counter_01552413_4.png|400x400px]]
|-
|-5.1
|-10.6
|-349.476
|Reactive
|Figure 8 shows a reactive trajectory. There is enough kinetic energy to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. Similarly to Figure 7 barrier recrossing occurs where the system reverts back to the reactants but since atom A has higher energy -10.6 vs -10.1, it's able to revert back to the product via a higher potential energy
|[[File:Counter_01552413_5.png|400x400px]]
|}
|}
'''''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''''

The theory states that a molecular collision that leads to reaction must pass through an intermediate state known as the transition state essentially where A-B is partially formed and B-C is partially broken as discussed previously. It assumes once a transition state has been achieved the '''transition state''' structure does not collapse back to the reactants. However this does not fit with experimental data as shown in Figure 7 the transition state collapses back to the reactants.The Eyring Equation is based on transition state theory and is used to describe the relationship between reaction rate and temperature,both via ''Rcl'' and the partition functions.When the intermediates are very short-living, so that the Boltzmann distribution22 of energies is not reached before the process continues to the next step.

3. The transition state theory also fails for some reactions at high temperatures due to the more complex motions of molecules or at very low temperatures due to the quantum tunneling.

Furthermore, the TST is also based on the assumption that atomic nuclei behave according to the classic mechanics. It is assumed that unless atoms or molecules collide with enough energy to form the transition structure (activated complex), the reaction does not occur. However, according to the quantum mechanics, for any barrier with a finite amount of energy, there is a possibility that particles can still tunnel across the barrier.

symmetric sa wells are same energy reactant and product neither exothermic nor endothermic. TS force =0 as no vibrational energy
[[File:TST_equation_01552413.PNG|400x400px]]

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?
[[File:ATS_01552413.png|left|400x400px]]
Locate the approximate position of the transition state.

Initial conditions: rHH= 74 pm pHH = -5.1 g.mol-1.pm.fs-1 rHF= 180 pm pHF = -1 g.mol-1.pm.fs-1

This reaction is exothermic as H-F is stronger than H-H. Evenr though in H2 the overlap between the 1s orbital is stronger than the overlap of the 2p and 1s orbital in HF. However as HF bond is very polar due to Fluorine being very electronegative hence has a large ionic componenet to the bond strenght. H2 however is non polar so has no ionic contribution this can be shown by Figure 10 which shows the product energy (HF) is lower than the reactant energy (H2) hence exothemic. Hence the activation energy is 123.3 kJ/mol-1 determined via the energy vs time for an mep plot. The Surface plot ( Figure 11) shows the H-H system hoigher in energy due to a weaker bond and hence the reaction has an early transition state as stated by Hammonds postulate the transition state resembles the reactants and products depending on how close it is in energy to either of them. Since from the graph we can see the reactants are higher in energy compared to the products, we have an early transition state hence this is an exothermic reaction. Hence H2 syste

[[File:Surface_01552413.png|left|400x400px]]

[[File:6.1_01552413.png|left|250x250px]]

{| class="wikitable"
!p= -6.1
!p=-4.1
!p=-2.1
!p=0
|-
|[[File:-6.1_01552413.png|left|250x250px]]
|[[File:-4.1_01552413.png|250x250px|centre]]
|[[File:-2.1_01552413.png|left|250x250px]]
|[[File:0_01552413.png|left|250x250px]]
|}

{| class="wikitable"
!p=2.1
!p=4.1
!p=6.1
|-
|[[File:2.1_01552413.png|left|250x250px]]
|[[File:4.1_01552413.png|left|250x250px]]
|[[File:6.1_01552413.png|left|250x250px]]
|}
These show that when p=-4.1, p=2.1 and p=4.1 the reaction
proceeds to the products whereas in the for p=-6.1, p= -2.1 p=0 and p=6.1 after
reaching the transition state, it reverts back to the reactants hence these are
unreactive trajectories. After reaching the TS for the reactive trajectories
there is greater vibrational energy as the H-F bond is stronger than the H-H
bond and hence the force constant will be higher. hence the vibrational energy will be larger. This further shows the difference between experimental results and the transition state theory as when the transition state is reached the product (H-F) doesn't always form.

[[File:Standard_01552413.png|left|400x400px]]

2020-05-22T12:37:14Z

Msi3018:

'''''On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''''

The transition state is defined s the maximum on the minimum energy path linking reactants and the products hence can be mathematically defined as ∂V(ri)/∂ri=0 which represents that the gradient of the potential is zero).. The transition state can be identified by statring a trajectory near the transition state and see whether they roll towards the reactants or products. This can be done several times with slightly varying trajectories in order to accurately identify the transition state. The transition state can be distinguished from the local minimum of the potential energy surface by taking the second derivitive and checking whether ∂2V(ri)/∂r2i > 0 or ∂2V(ri)/∂r2i < 0 . ∂2V(ri)/∂r2i > 0 corresponds to the local minimum whereas ∂2V([[ri|ri]])/∂r2i < 0 corresponds to the maximum ( 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.'''''

Since the H + H2 surface is symmetric, the transition state corresponds to where r1=r2 and hence Figure 1 shows the intersection point at 20 fs with a distance of 90.8 pm. Hence AB=BC= 90.8 pm. Figure 2 shows this point (saddle point) is at the turning point further providing evidence the transition state is a this distance TS force =0 as no vibrational energy[[File:TS_distance_plot.png|400x400px|thumb|left|Internuclear distance against time plot.]][[File:Surface_Plot_01552413.png|400x400px]]

Comment on how the ''mep'' and the trajectory you just calculated differ.
[[File:MEP_01552413.png|400x400px]]
[[File:Dynamic_01552413.png|400x400px]]

The trajectories, mep, was found using r1 = rts+δ, r2 = rts where δ = 1 hence r1 = 91.8 pm and r2 = 90.8 pm and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1 . Figure 2 shows an mep plot doesn't provide a realistic account of the motion of atoms during a reaction as the vibration of the atoms aren't taken into account. However Figure 3 showing the dynamic plot does include the atomic vibrations hence provides a more realistic account of the motion of atoms.

'''''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"
|-
|-
|-
|-
|-
|-
|}
{| class="wikitable"
|
{| 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
|-414.280
|Reactive
|Figure 4 shows a reactive trajectory as the kinetic energy is large enough to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. After formation of AB and as rBC increases the molecule vibrates
|[[File:Counter_01552413_1.png|400x400px]]

|-
|-3.1
|-4.1
|-420.081
|Unreactive
|Figure 5 shows an unreactive trajectory as the kinetic energy is not large enough to overcome the activation barrier and hence the collision doesn't have enough energy to break the BC bond and form the new AB bond. Compared to Figure 4 the the momentum of the incoming A atom is reduced from -5.1 to -4.1 hence decreasing the kinetic energy and so the reactants are reformed
|[[File:Counter_01552413_2.png|400x400px]]
|-
|-3.1
|-5.1
|-413.978
|Reactive
|Figure 6 shows a reactive trajectory as the kinetic energy is large enough to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. After formation of AB and as rBC increases the molecule vibrates. Compared to Figure 5 the momentum of the incoming A atom is increased from -4.1 to -5.1 hence increasing the kinetic energy and so the product form.
|[[File:Counter_01552413_3.png|400x400px]]
|-
|-5.1
|-10.1
|-357.274
|Unreactive
|Figure 7 shows an unreactive trajectory. Even though there is enough kinetic energy to overcome the activation barrieran the product AB starts to form but then the system reverts back to the reactants. This is is barrier recrossing, where due to large kinetic energies there's a lot of extra vibrational energy which causes the trajectory to recross the barrier and therefore reform the reactants.
|[[File:Counter_01552413_4.png|400x400px]]
|-
|-5.1
|-10.6
|-349.476
|Reactive
|Figure 8 shows a reactive trajectory. There is enough kinetic energy to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. Similarly to Figure 7 barrier recrossing occurs where the system reverts back to the reactants but since atom A has higher energy -10.6 vs -10.1, it's able to revert back to the product via a higher potential energy
|[[File:Counter_01552413_5.png|400x400px]]
|}
|}
'''''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''''

The theory states that a molecular collision that leads to reaction must pass through an intermediate state known as the transition state essentially where A-B is partially formed and B-C is partially broken as discussed previously. It assumes once a transition state has been achieved the '''transition state''' structure does not collapse back to the reactants. However this does not fit with experimental data as shown in Figure 7 the transition state collapses back to the reactants.The Eyring Equation is based on transition state theory and is used to describe the relationship between reaction rate and temperature,both via ''Rcl'' and the partition functions.When the intermediates are very short-living, so that the Boltzmann distribution22 of energies is not reached before the process continues to the next step.

3. The transition state theory also fails for some reactions at high temperatures due to the more complex motions of molecules or at very low temperatures due to the quantum tunneling.

Furthermore, the TST is also based on the assumption that atomic nuclei behave according to the classic mechanics. It is assumed that unless atoms or molecules collide with enough energy to form the transition structure (activated complex), the reaction does not occur. However, according to the quantum mechanics, for any barrier with a finite amount of energy, there is a possibility that particles can still tunnel across the barrier.

symmetric sa wells are same energy reactant and product neither exothermic nor endothermic. TS force =0 as no vibrational energy
[[File:TST_equation_01552413.PNG|400x400px]]

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?
[[File:ATS_01552413.png|left|400x400px]]
Locate the approximate position of the transition state.

Initial conditions: rHH= 74 pm pHH = -5.1 g.mol-1.pm.fs-1 rHF= 180 pm pHF = -1 g.mol-1.pm.fs-1

This reaction is exothermic as H-F is stronger than H-H. Evenr though in H2 the overlap between the 1s orbital is stronger than the overlap of the 2p and 1s orbital in HF. However as HF bond is very polar due to Fluorine being very electronegative hence has a large ionic componenet to the bond strenght. H2 however is non polar so has no ionic contribution this can be shown by Figure 10 which shows the product energy (HF) is lower than the reactant energy (H2) hence exothemic. Hence the activation energy is 123.3 kJ/mol-1 determined via the energy vs time for an mep plot. The Surface plot ( Figure 11) shows the H-H system hoigher in energy due to a weaker bond and hence the reaction has an early transition state as stated by Hammonds postulate the transition state resembles the reactants and products depending on how close it is in energy to either of them. Since from the graph we can see the reactants are higher in energy compared to the products, we have an early transition state hence this is an exothermic reaction. Hence H2 syste

[[File:Surface_01552413.png|left|400x400px]]

[[File:6.1_01552413.png|left|250x250px]]
{| class="wikitable"
!p=2.1
!p=4.1
!p=6.1
|-
|
|
|
|}
{| class="wikitable"
!p= -6.1
!p=-4.1
!p=-2.1
!p=0
!p=2.1
!p=4.1
!p=6.1
|-
|[[File:-6.1_01552413.png|left|250x250px]]
|[[File:-4.1_01552413.png|250x250px|centre]]
|[[File:-2.1_01552413.png|left|250x250px]]
|[[File:0_01552413.png|left|250x250px]]
|[[File:2.1_01552413.png|left|250x250px]]
|[[File:4.1_01552413.png|left|250x250px]]
|[[File:6.1_01552413.png|left|250x250px]]
|}