ChemWiki - User contributions [en]

Kjo3118

2020-05-15T20:49:15Z

Kjo3118:

The answers to the highlighted questions have been underlined.

== EXERCISE 1: H+H2 ==
[[File:TS intermolecular dist vs time kjo3118.png|thumb|Figure 1. The internuclear distance vs time graph for the best estimate TS. Notice how both internuclear distances are almost constsnt, with only small osscillations.|587x587px]][[File:Kjo3118 rcn dynandmep.png|thumb|Figure 2. The genarated reaction path (right) as well as a trajectory generated using the dymanic calculation (left). Notie the osscilations on the left graph and the lack thereof on the right.|790x790px]]
In this exercise, a system comprised of only hydrogen atoms has been used. First the transition state has been located. The transition state is at a '''saddle point on the potential energy surface''', so it is a maximum along the reaction coordinate, but a minimum in the orthogonal direction. In this case '''both''' partial derivatives of the potential with respect to RAB and RBC are equal to 0. To distinguish a saddle ponit from a local minimum (which would also have both first derivatives equal to 0), you would have to take a look at the second derivatives: at a saddle ponit (TS) one would be positive and one would be negative; at a local minimum both second derivatives would be positive. To locate the position of the transition state first a consideration of geometry was made: since the system is totally symmetric, the transition state is also symmetric and so RAB=RBC . My best estimate is '''90.77 pm'''. For RAB(0)=RBC(0)=90.77 pm and both initial momenta equal to zero the molecule oscillates so little from the initial position that the internuclear diatances remain almost constant (Figure 1.), for these conditions the '''initial forces are very close to 0''' (both 0.001 kJ.mol-1.pm-1).

Figure 2 shows the reaction path genereated using the MEP calculation type as well as the trajectory genareted using the dynamic calculation. In this specific case they are very simmilar, the only difference is that the mep trajcetory just follows the valley of lowest potential, with not vibrational energy, as the total kinetic energy is set to 0 at each step, wheras the dynamically generated trajectory also includes vibrational osscillations.'''<nowiki/>'''
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|YES
|The system crosses the transition state and goes on to form the products. The vibrational osscillations are larger in products, meaning some energy was converted into vibrational.
|[[File:Kjo3118 traj1.png|thumb|357x357px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.1</nowiki>
|NO
|The system approaches the transition state, but it never really reaches it, it 'rolls' back into the reactants chanel.
|[[File:Kjo3118 traj2.png|thumb|359x359px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.0</nowiki>
|YES
|The system crosses the transition state and goes on to form the products. The vibrational osscillations are larger in products, meaning some energy was converted into vibrational.
|[[File:Kjo3118 traj3.png|thumb|360x360px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.3</nowiki>
|NO
|This case is interesting, the system crosses the transition state to form the products, however due to the very large vibrational oscillations the system then re-crosses the barrier to reforn the reactants and overall the system 'rolls' back into the reactants chanel.
|[[File:Kjo3118 traj4.png|thumb|367x367px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.5</nowiki>
|YES
|In this case barrier re-crossing is also seen, after crossing the transition state for the first time the system goes back to the reactants (barrier re-crossing), however then the system crosses the transition state again to make the products.
|[[File:Kjo3118 traj5.png|thumb|367x367px]]
|}
From the table you can conclude that barrier re-crossing can take place and that not all collisions with energy larger than the activation energy will result in a sucessfull collision. Meaning that the system can pass through the transition state to form the products, but then go back to reform the reactants. A good example of '''barrier re-crossing is the second to last row in the table above'''.

Transition state theory (TST) predicts reaction rates, however due to the main assumptions made it '''overestimates''' reaction rates as compared to experimental values. This is due to the assumption made, stating that any collision with kinetic energy greater than the activation energy will result in a reaction and formation of the products. However as shown in the table above (particularly the second to last row), barrier re-cressing can cause a collision to reform the reactants and therfore not result in a succesful reaction. This means that in reality the proportion of succesfull collisions is lower than the one predicted by transition state theory and thus transition state theory overestimates the reaction rates. It is important to meantion an opposing effect, that would cause transition state theory to underestimate reaction rates: quantum tunelling. Quantum tunnelling allows the energetic barrier to be crossed even if the reactants do not have sufficent kinetic energy. This would increase the proportion of sucesfull collisions, however the tunelling phenomenon is not as prevalent as barrier re-crossing, especially for heavier systems.

== EXERCISE 2: H+HF/F+H2 ==
[[File:Surf kjo318 1.png|thumb|500x500px|Figure 3. The potential surface plot for the reaction: F + H2 -> HF + H. The products are of lower energy than reactants, therfore the reaction is exothermic.]]
For this part of the exercise the F-H-H system was used.

'''F + H2 -> HF + H '''is an exothermic reaction, since the energy drops from reactants to products, Figure 3.
[[File:Surf kjo318 2.png|thumb|500x500px|Figure 4. The potential surface plot for the reaction: HF + H -> F + H2. The products are of higher energy than reactants, therfore the reaction is endothermic.]]
'''HF + H -> F + H2 '''is the exact 'opposite' reaction, and by that logic it has to be endothermic. Looking at the surface plot confirms this, Figure 4.

Since in both reaction the only bonds being made/broken are H-H and H-F bonds, the exo-/endothermicity of the reactions can be used to establish the relative bond strengths. Looking at Figure 3, the products have 1 H-F bond and the reactants have only 1 H-H bond. The products are lower in energy, suggesting that the H-F bond is more stable and therfore stronger.
[[File:Kjo3118 int fhf.png|thumb|500x500px|Figure 5. The intermolecular distance vs time plot for the F-H-H system, starting with zero momentum, at the estimated position of the transition state. Notice the only small oscillations fro mthe initial position.]]
To find the transition state of these reactions (which are the same since the reactions are the opposite of each other), first Hammond's postulate was used: since '''F + H2 -> HF + H '''is an exothermic reaction the transition state will be an 'early' transition state, that will be simmilar in structure to the reactants. That meanst the transition state will have a significantly higher F-H distance, while the H-H distance will be simmilar to the H-H bond lenght (around 75 pm). Using this information the transition state was estimated to be at RF-H = 181 pm and RH-H = 74 pm, by finding a position where, given initial momenta equal to 0, the trajectory only oscillates slightly from the initial positions and the initial forces are close to 0, Figure 5.

Next the activation energies were found, by subtracting the potential energy of the reactnts/products from the potential energy of the transition state, the values were found to be:

'''F + H2 -> HF + H '''activation energy was found to be +1.12 kJ.mol-1

'''HF + H -> F + H2 '''activation energy was found to be +126.70 kJ.mol-1[[File:Kjo3118 rcn2 path.png|thumb|1013x1013px|Figure 6. The MEP pathway (left) as well as the energy profile (right), of the F-H-H system starting close to the transition state, but displaced towards H2 + F. Notice the slow potential (here equal to total) energy drop from the transitions state to the products.]][[File:Kjo3118 rcn1 path.png|thumb|1010x1010px|Figure 7. The MEP pathway (left) as well as the energy profile (right), of the F-H-H system starting close to the transition state, but displaced towards HF + H. Notice the fast potential (here equal to total) energy drop from the transitions state to the products.]][[File:Kjo3118 hf reactive.png|thumb|Figure 8. THe reactive trajectory found for the F-H-H system.|463x463px]]The reactant and product potential energies were found by running a MEP calculations displaced slightly from the approximate position of the transition state towoards either '''HF + H''' or '''F + H2''', which allowed the trajectory to follow into the valley of either reactants or porducts. The energies vs time found by these calculations have initially changed and then reached a stable value - the potential energy of reactants/products. The results of the calculations can be found in Figure 6 and Figure 7.
A reactive pathway (Figure 8) was found for the initial conditions: RF-H = RAB=200 pm; RH-H =RBC=75 pm; pAB= -2 g.mol-1.pm.fs-1; pBC= 0 g.mol-1.pm.fs-1

In an exothermic reaction the reactants start with a higher potential energy and move on to a lower potential energy in the products. Since the energy has to be conserved it has to be converted into kinetic energy in two forms: either translational or vibrational. Experimantally this could be confirmed by monitoring the rection '''F + H2 -> HF + H''':
* The translational componet could be monitored using colorimetry, using a bomb colorimeter;
* The rotational kinetic energy could be measured using IR, where overtones can be observed when some molecules are promoted to higher vibrational levels, as potential energy is converted into rotational kinetic energy. As these relax back to the groud state the overtones lower in intensity, whlie the main peak gains intensity.
'''Polanyi's rules'''

Polanyi using his work has composed some simple rules for reaction trajectories<ref>J. POLANYI, ''Science'', 1987, '''236''', 680-690.</ref>:
* For a reaction with a late transition state (an endothermic reaction) a trajectory that has a higher proportion of its kinetic energy in the form of vibrational energy is more likely to result in a sucesful reaction;
* For a reaction with an early transition state (an exothermic one) the opposite is true: a trajectory with more of its kinetic energy in the form of translational energy is favoured.
Some simulations showing these rules are shown in the table below:
{| class="wikitable"
!Reaction type
!Transition state
!Reaction used in simulation
!Vibrational trajectory
!Translational trajectory
|-
|Endothermic
|Late
|'''HF + H -> F + H2'''
|[[File:Kjo3118 late vib.png|thumb|403x403px|The system starts with most of its kintic energy in the form of vibrational energy. As predictd by the rules, this is a reactive pathway.]]
|[[File:Kjo3118 late transla.png|thumb|348x348px|The system starts with most of its kinetic energy in the form of translational energy. As predicted by the rules, this is not a reactive pathway.]]
|-
|Exothermic
|Early
|'''F + H2 -> HF + H'''
|[[File:Kjo3118 early vib.png|thumb|409x409px|The system starts with most of its kintic energy in the form of vibrational energy. As predicted by the rules, this is not a reactive pathway.]]
|[[File:Kjo3118 early transla.png|thumb|371x371px|The system starts with most of its kintic energy in the form of translational energy. As predictd by the rules, this is a reactive pathway.]]
|}
<references />

Kjo3118

2020-05-15T20:46:24Z

Kjo3118:

The answers to the highlighted questions have been underlined.

== EXERCISE 1: H+H2 ==
[[File:TS intermolecular dist vs time kjo3118.png|thumb|Figure 1. The internuclear distance vs time graph for the best estimate TS. Notice how both internuclear distances are almost constsnt, with only small osscillations.|343x343px]][[File:Kjo3118 rcn dynandmep.png|thumb|Figure 2. The genarated reaction path (right) as well as a trajectory generated using the dymanic calculation (left). Notie the osscilations on the left graph and the lack thereof on the right.|587x587px]]
In this exercise, a system comprised of only hydrogen atoms has been used. First the transition state has been located. The transition state is at a '''saddle point on the potential energy surface''', so it is a maximum along the reaction coordinate, but a minimum in the orthogonal direction. In this case '''both''' partial derivatives of the potential with respect to RAB and RBC are equal to 0. To distinguish a saddle ponit from a local minimum (which would also have both first derivatives equal to 0), you would have to take a look at the second derivatives: at a saddle ponit (TS) one would be positive and one would be negative; at a local minimum both second derivatives would be positive. To locate the position of the transition state first a consideration of geometry was made: since the system is totally symmetric, the transition state is also symmetric and so RAB=RBC . My best estimate is '''90.77 pm'''. For RAB(0)=RBC(0)=90.77 pm and both initial momenta equal to zero the molecule oscillates so little from the initial position that the internuclear diatances remain almost constant (Figure 1.), for these conditions the '''initial forces are very close to 0''' (both 0.001 kJ.mol-1.pm-1).

Figure 2 shows the reaction path genereated using the MEP calculation type as well as the trajectory genareted using the dynamic calculation. In this specific case they are very simmilar, the only difference is that the mep trajcetory just follows the valley of lowest potential, with not vibrational energy, as the total kinetic energy is set to 0 at each step, wheras the dynamically generated trajectory also includes vibrational osscillations.'''<nowiki/>'''
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|YES
|The system crosses the transition state and goes on to form the products. The vibrational osscillations are larger in products, meaning some energy was converted into vibrational.
|[[File:Kjo3118 traj1.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.1</nowiki>
|NO
|The system approaches the transition state, but it never really reaches it, it 'rolls' back into the reactants chanel.
|[[File:Kjo3118 traj2.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.0</nowiki>
|YES
|The system crosses the transition state and goes on to form the products. The vibrational osscillations are larger in products, meaning some energy was converted into vibrational.
|[[File:Kjo3118 traj3.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.3</nowiki>
|NO
|This case is interesting, the system crosses the transition state to form the products, however due to the very large vibrational oscillations the system then re-crosses the barrier to reforn the reactants and overall the system 'rolls' back into the reactants chanel.
|[[File:Kjo3118 traj4.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.5</nowiki>
|YES
|In this case barrier re-crossing is also seen, after crossing the transition state for the first time the system goes back to the reactants (barrier re-crossing), however then the system crosses the transition state again to make the products.
|[[File:Kjo3118 traj5.png|thumb]]
|}
From the table you can conclude that barrier re-crossing can take place and that not all collisions with energy larger than the activation energy will result in a sucessfull collision. Meaning that the system can pass through the transition state to form the products, but then go back to reform the reactants. A good example of '''barrier re-crossing is the second to last row in the table above'''.

Transition state theory (TST) predicts reaction rates, however due to the main assumptions made it '''overestimates''' reaction rates as compared to experimental values. This is due to the assumption made, stating that any collision with kinetic energy greater than the activation energy will result in a reaction and formation of the products. However as shown in the table above (particularly the second to last row), barrier re-cressing can cause a collision to reform the reactants and therfore not result in a succesful reaction. This means that in reality the proportion of succesfull collisions is lower than the one predicted by transition state theory and thus transition state theory overestimates the reaction rates. It is important to meantion an opposing effect, that would cause transition state theory to underestimate reaction rates: quantum tunelling. Quantum tunnelling allows the energetic barrier to be crossed even if the reactants do not have sufficent kinetic energy. This would increase the proportion of sucesfull collisions, however the tunelling phenomenon is not as prevalent as barrier re-crossing, especially for heavier systems.

== EXERCISE 2: H+HF/F+H2 ==
[[File:Surf kjo318 1.png|thumb|433x433px|Figure 3. The potential surface plot for the reaction: F + H2 -> HF + H. The products are of lower energy than reactants, therfore the reaction is exothermic.]]
For this part of the exercise the F-H-H system was used.

'''F + H2 -> HF + H '''is an exothermic reaction, since the energy drops from reactants to products, Figure 3.
[[File:Surf kjo318 2.png|thumb|435x435px|Figure 4. The potential surface plot for the reaction: HF + H -> F + H2. The products are of higher energy than reactants, therfore the reaction is endothermic.]]
'''HF + H -> F + H2 '''is the exact 'opposite' reaction, and by that logic it has to be endothermic. Looking at the surface plot confirms this, Figure 4.

Since in both reaction the only bonds being made/broken are H-H and H-F bonds, the exo-/endothermicity of the reactions can be used to establish the relative bond strengths. Looking at Figure 3, the products have 1 H-F bond and the reactants have only 1 H-H bond. The products are lower in energy, suggesting that the H-F bond is more stable and therfore stronger.
[[File:Kjo3118 int fhf.png|thumb|375x375px|Figure 5. The intermolecular distance vs time plot for the F-H-H system, starting with zero momentum, at the estimated position of the transition state. Notice the only small oscillations fro mthe initial position.]]
To find the transition state of these reactions (which are the same since the reactions are the opposite of each other), first Hammond's postulate was used: since '''F + H2 -> HF + H '''is an exothermic reaction the transition state will be an 'early' transition state, that will be simmilar in structure to the reactants. That meanst the transition state will have a significantly higher F-H distance, while the H-H distance will be simmilar to the H-H bond lenght (around 75 pm). Using this information the transition state was estimated to be at RF-H = 181 pm and RH-H = 74 pm, by finding a position where, given initial momenta equal to 0, the trajectory only oscillates slightly from the initial positions and the initial forces are close to 0, Figure 5.

Next the activation energies were found, by subtracting the potential energy of the reactnts/products from the potential energy of the transition state, the values were found to be:

'''F + H2 -> HF + H '''activation energy was found to be +1.12 kJ.mol-1

'''HF + H -> F + H2 '''activation energy was found to be +126.70 kJ.mol-1[[File:Kjo3118 rcn2 path.png|thumb|915x915px|Figure 6. The MEP pathway (left) as well as the energy profile (right), of the F-H-H system starting close to the transition state, but displaced towards H2 + F. Notice the slow potential (here equal to total) energy drop from the transitions state to the products.]][[File:Kjo3118 rcn1 path.png|thumb|921x921px|Figure 7. The MEP pathway (left) as well as the energy profile (right), of the F-H-H system starting close to the transition state, but displaced towards HF + H. Notice the fast potential (here equal to total) energy drop from the transitions state to the products.]][[File:Kjo3118 hf reactive.png|thumb|Figure 8. THe reactive trajectory found for the F-H-H system.|345x345px]]The reactant and product potential energies were found by running a MEP calculations displaced slightly from the approximate position of the transition state towoards either '''HF + H''' or '''F + H2''', which allowed the trajectory to follow into the valley of either reactants or porducts. The energies vs time found by these calculations have initially changed and then reached a stable value - the potential energy of reactants/products. The results of the calculations can be found in Figure 6 and Figure 7.
A reactive pathway (Figure 8) was found for the initial conditions: RF-H = RAB=200 pm; RH-H =RBC=75 pm; pAB= -2 g.mol-1.pm.fs-1; pBC= 0 g.mol-1.pm.fs-1

In an exothermic reaction the reactants start with a higher potential energy and move on to a lower potential energy in the products. Since the energy has to be conserved it has to be converted into kinetic energy in two forms: either translational or vibrational. Experimantally this could be confirmed by monitoring the rection '''F + H2 -> HF + H''':
* The translational componet could be monitored using colorimetry, using a bomb colorimeter;
* The rotational kinetic energy could be measured using IR, where overtones can be observed when some molecules are promoted to higher vibrational levels, as potential energy is converted into rotational kinetic energy. As these relax back to the groud state the overtones lower in intensity, whlie the main peak gains intensity.
'''Polanyi's rules'''

Polanyi using his work has composed some simple rules for reaction trajectories<ref>J. POLANYI, ''Science'', 1987, '''236''', 680-690.</ref>:
* For a reaction with a late transition state (an endothermic reaction) a trajectory that has a higher proportion of its kinetic energy in the form of vibrational energy is more likely to result in a sucesful reaction;
* For a reaction with an early transition state (an exothermic one) the opposite is true: a trajectory with more of its kinetic energy in the form of translational energy is favoured.
Some simulations showing these rules are shown in the table below:
{| class="wikitable"
!Reaction type
!Transition state
!Reaction used in simulation
!Vibrational trajectory
!Translational trajectory
|-
|Endothermic
|Late
|'''HF + H -> F + H2'''
|[[File:Kjo3118 late vib.png|thumb|403x403px|The system starts with most of its kintic energy in the form of vibrational energy. As predictd by the rules, this is a reactive pathway.]]
|[[File:Kjo3118 late transla.png|thumb|348x348px|The system starts with most of its kinetic energy in the form of translational energy. As predicted by the rules, this is not a reactive pathway.]]
|-
|Exothermic
|Early
|'''F + H2 -> HF + H'''
|[[File:Kjo3118 early vib.png|thumb|409x409px|The system starts with most of its kintic energy in the form of vibrational energy. As predicted by the rules, this is not a reactive pathway.]]
|[[File:Kjo3118 early transla.png|thumb|371x371px|The system starts with most of its kintic energy in the form of translational energy. As predictd by the rules, this is a reactive pathway.]]
|}
<references />

Kjo3118

2020-05-15T20:44:04Z

Kjo3118:

The answers to the highlighted questions have been underlined.

== EXERCISE 1: H+H2 ==
[[File:TS intermolecular dist vs time kjo3118.png|thumb|Figure 1. The internuclear distance vs time graph for the best estimate TS. Notice how both internuclear distances are almost constsnt, with only small osscillations.|343x343px]][[File:Kjo3118 rcn dynandmep.png|thumb|Figure 2. The genarated reaction path (right) as well as a trajectory generated using the dymanic calculation (left). Notie the osscilations on the left graph and the lack thereof on the right.|587x587px]]
In this exercise, a system comprised of only hydrogen atoms has been used. First the transition state has been located. The transition state is at a '''saddle point on the potential energy surface''', so it is a maximum along the reaction coordinate, but a minimum in the orthogonal direction. In this case '''both''' partial derivatives of the potential with respect to RAB and RBC are equal to 0. To distinguish a saddle ponit from a local minimum (which would also have both first derivatives equal to 0), you would have to take a look at the second derivatives: at a saddle ponit (TS) one would be positive and one would be negative; at a local minimum both second derivatives would be positive. To locate the position of the transition state first a consideration of geometry was made: since the system is totally symmetric, the transition state is also symmetric and so RAB=RBC . My best estimate is '''90.77 pm'''. For RAB(0)=RBC(0)=90.77 pm and both initial momenta equal to zero the molecule oscillates so little from the initial position that the internuclear diatances remain almost constant (Figure 1.), for these conditions the '''initial forces are very close to 0''' (both 0.001 kJ.mol-1.pm-1).

Figure 2 shows the reaction path genereated using the MEP calculation type as well as the trajectory genareted using the dynamic calculation. In this specific case they are very simmilar, the only difference is that the mep trajcetory just follows the valley of lowest potential, with not vibrational energy, as the total kinetic energy is set to 0 at each step, wheras the dynamically generated trajectory also includes vibrational osscillations.'''<nowiki/>'''
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|YES
|The system crosses the transition state and goes on to form the products. The vibrational osscillations are larger in products, meaning some energy was converted into vibrational.
|[[File:Kjo3118 traj1.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.1</nowiki>
|NO
|The system approaches the transition state, but it never really reaches it, it 'rolls' back into the reactants chanel.
|[[File:Kjo3118 traj2.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.0</nowiki>
|YES
|The system crosses the transition state and goes on to form the products. The vibrational osscillations are larger in products, meaning some energy was converted into vibrational.
|[[File:Kjo3118 traj3.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.3</nowiki>
|NO
|This case is interesting, the system crosses the transition state to form the products, however due to the very large vibrational oscillations the system then re-crosses the barrier to reforn the reactants and overall the system 'rolls' back into the reactants chanel.
|[[File:Kjo3118 traj4.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.5</nowiki>
|YES
|In this case barrier re-crossing is also seen, after crossing the transition state for the first time the system goes back to the reactants (barrier re-crossing), however then the system crosses the transition state again to make the products.
|[[File:Kjo3118 traj5.png|thumb]]
|}
From the table you can conclude that barrier re-crossing can take place and that not all collisions with energy larger than the activation energy will result in a sucessfull collision. Meaning that the system can pass through the transition state to form the products, but then go back to reform the reactants. A good example of '''barrier re-crossing is the second to last row in the table above'''.

Transition state theory (TST) predicts reaction rates, however due to the main assumptions made it '''overestimates''' reaction rates as compared to experimental values. This is due to the assumption made, stating that any collision with kinetic energy greater than the activation energy will result in a reaction and formation of the products. However as shown in the table above (particularly the second to last row), barrier re-cressing can cause a collision to reform the reactants and therfore not result in a succesful reaction. This means that in reality the proportion of succesfull collisions is lower than the one predicted by transition state theory and thus transition state theory overestimates the reaction rates. It is important to meantion an opposing effect, that would cause transition state theory to underestimate reaction rates: quantum tunelling. TST is a classical theory and it does not include any quantum effects. Quantum tunnelling allows the energetic barrier to be crossed even if the reactants do not have sufficent kinetic energy. This would increase the proportion of sucesfull collisions, however the tunelling phenomenon is not as prevalent as barrier re-crossing, especially for heavier systems.

== EXERCISE 2: H+HF/F+H2 ==
[[File:Surf kjo318 1.png|thumb|433x433px|Figure 3. The potential surface plot for the reaction: F + H2 -> HF + H. The products are of lower energy than reactants, therfore the reaction is exothermic.]]
For this part of the exercise the F-H-H system was used.

'''F + H2 -> HF + H '''is an exothermic reaction, since the energy drops from reactants to products, Figure 3.
[[File:Surf kjo318 2.png|thumb|435x435px|Figure 4. The potential surface plot for the reaction: HF + H -> F + H2. The products are of higher energy than reactants, therfore the reaction is endothermic.]]
'''HF + H -> F + H2 '''is the exact 'opposite' reaction, and by that logic it has to be endothermic. Looking at the surface plot confirms this, Figure 4.

Since in both reaction the only bonds being made/broken are H-H and H-F bonds, the exo-/endothermicity of the reactions can be used to establish the relative bond strengths. Looking at Figure 3, the products have 1 H-F bond and the reactants have only 1 H-H bond. The products are lower in energy, suggesting that the H-F bond is more stable and therfore stronger.
[[File:Kjo3118 int fhf.png|thumb|375x375px|Figure 5. The intermolecular distance vs time plot for the F-H-H system, starting with zero momentum, at the estimated position of the transition state. Notice the only small oscillations fro mthe initial position.]]
To find the transition state of these reactions (which are the same since the reactions are the opposite of each other), first Hammond's postulate was used: since '''F + H2 -> HF + H '''is an exothermic reaction the transition state will be an 'early' transition state, that will be simmilar in structure to the reactants. That meanst the transition state will have a significantly higher F-H distance, while the H-H distance will be simmilar to the H-H bond lenght (around 75 pm). Using this information the transition state was estimated to be at RF-H = 181 pm and RH-H = 74 pm, by finding a position where, given initial momenta equal to 0, the trajectory only oscillates slightly from the initial positions and the initial forces are close to 0, Figure 5.

Next the activation energies were found, by subtracting the potential energy of the reactnts/products from the potential energy of the transition state, the values were found to be:

'''F + H2 -> HF + H '''activation energy was found to be +1.12 kJ.mol-1

'''HF + H -> F + H2 '''activation energy was found to be +126.70 kJ.mol-1[[File:Kjo3118 rcn2 path.png|thumb|915x915px|Figure 6. The MEP pathway (left) as well as the energy profile (right), of the F-H-H system starting close to the transition state, but displaced towards H2 + F. Notice the slow potential (here equal to total) energy drop from the transitions state to the products.]][[File:Kjo3118 rcn1 path.png|thumb|921x921px|Figure 7. The MEP pathway (left) as well as the energy profile (right), of the F-H-H system starting close to the transition state, but displaced towards HF + H. Notice the fast potential (here equal to total) energy drop from the transitions state to the products.]][[File:Kjo3118 hf reactive.png|thumb|Figure 8. THe reactive trajectory found for the F-H-H system.|345x345px]]The reactant and product potential energies were found by running a MEP calculations displaced slightly from the approximate position of the transition state towoards either '''HF + H''' or '''F + H2''', which allowed the trajectory to follow into the valley of either reactants or porducts. The energies vs time found by these calculations have initially changed and then reached a stable value - the potential energy of reactants/products. The results of the calculations can be found in Figure 6 and Figure 7.
A reactive pathway (Figure 8) was found for the initial conditions: RF-H = RAB=200 pm; RH-H =RBC=75 pm; pAB= -2 g.mol-1.pm.fs-1; pBC= 0 g.mol-1.pm.fs-1

In an exothermic reaction the reactants start with a higher potential energy and move on to a lower potential energy in the products. Since the energy has to be conserved it has to be converted into kinetic energy in two forms: either translational or vibrational. Experimantally this could be confirmed by monitoring the rection '''F + H2 -> HF + H''':
* The translational componet could be monitored using colorimetry, using a bomb colorimeter;
* The rotational kinetic energy could be measured using IR, where overtones can be observed when some molecules are promoted to higher vibrational levels, as potential energy is converted into rotational kinetic energy. As these relax back to the groud state the overtones lower in intensity, whlie the main peak gains intensity.
'''Polanyi's rules'''

Polanyi using his work has composed some simple rules for reaction trajectories<ref>J. POLANYI, ''Science'', 1987, '''236''', 680-690.</ref>:
* For a reaction with a late transition state (an endothermic reaction) a trajectory that has a higher proportion of its kinetic energy in the form of vibrational energy is more likely to result in a sucesful reaction;
* For a reaction with an early transition state (an exothermic one) the opposite is true: a trajectory with more of its kinetic energy in the form of translational energy is favoured.
Some simulations showing these rules are shown in the table below:
{| class="wikitable"
!Reaction type
!Transition state
!Reaction used in simulation
!Vibrational trajectory
!Translational trajectory
|-
|Endothermic
|Late
|'''HF + H -> F + H2'''
|[[File:Kjo3118 late vib.png|thumb|403x403px|The system starts with most of its kintic energy in the form of vibrational energy. As predictd by the rules, this is a reactive pathway.]]
|[[File:Kjo3118 late transla.png|thumb|348x348px|The system starts with most of its kinetic energy in the form of translational energy. As predicted by the rules, this is not a reactive pathway.]]
|-
|Exothermic
|Early
|'''F + H2 -> HF + H'''
|[[File:Kjo3118 early vib.png|thumb|409x409px|The system starts with most of its kintic energy in the form of vibrational energy. As predicted by the rules, this is not a reactive pathway.]]
|[[File:Kjo3118 early transla.png|thumb|371x371px|The system starts with most of its kintic energy in the form of translational energy. As predictd by the rules, this is a reactive pathway.]]
|}
<references />

File:Kjo3118 late vib.png

2020-05-15T20:38:02Z

Kjo3118: sa

File:Kjo3118 late transla.png

2020-05-15T20:37:50Z

Kjo3118: d

File:Kjo3118 early vib.png

2020-05-15T20:37:35Z

Kjo3118: s

2020-05-15T13:01:28Z

Kjo3118: r

Kjo3118

2020-05-15T12:46:22Z

Kjo3118:

== EXERCISE 1: H+H2 ==
[[File:TS intermolecular dist vs time kjo3118.png|thumb|Figure 1. The internuclear distance vs time graph for the best estimate TS. Notice how both internuclear distances are almost constsnt, with only small osscillations.|343x343px]][[File:Kjo3118 rcn dynandmep.png|thumb|Figure 2. The genarated reaction path (right) as well as a trajectory generated using the dymanic calculation (left). Notie the osscilations on the left graph and the lack thereof on the right.|587x587px]]
In this exercise, a system comprised of only hydrogen atoms has been used. First the transition state has been located. The transition state is at a '''saddle point on the potential energy surface''', so it is a maximum along the reaction coordinate, but a minimum in the orthogonal direction. In this case '''both''' partial derivatives of the potential with respect to RAB and RBC are equal to 0. To distinguish a saddle ponit from a local minimum (which would also have both first derivatives equal to 0), you would have to take a look at the second derivatives: at a saddle ponit (TS) one would be positive and one would be negative; at a local minimum both second derivatives would be positive. To locate the position of the transition state first a consideration of geometry was made: since the system is totally symmetric, the transition state is also symmetric and so RAB=RBC . My best estimate is '''90.77 pm'''. For RAB(0)=RBC(0)=90.77 pm and both initial momenta equal to zero the molecule oscillates so little from the initial position that the internuclear diatances remain almost constant (Figure 1.), for these conditions the '''initial forces are very close to 0''' (both 0.001 kJ.mol-1.pm-1).

Figure 2 shows the reaction path genereated using the MEP calculation type as well as the trajectory genareted using the dynamic calculation. They are very simmilar, the only difference is that the mep trajcetory just follows the valley of lowest potential, wheras the dynamically generated trajectory also includes vibrational osscillations.'''<nowiki/>'''
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|YES
|The system crosses the transition state and goes on to form the products. The vibrational osscillations are larger in products, meaning some energy was converted into vibrational.
|[[File:Kjo3118 traj1.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.1</nowiki>
|NO
|The system approaches the transition state, but it never really reaches it, it 'rolls' back into the reactants chanel.
|[[File:Kjo3118 traj2.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.0</nowiki>
|YES
|The system crosses the transition state and goes on to form the products. The vibrational osscillations are larger in products, meaning some energy was converted into vibrational.
|[[File:Kjo3118 traj3.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.3</nowiki>
|NO
|This case is interesting, the system crosses the transition state to form the products, however due to the very large vibrational oscillations the system then re-crosses the barrier to reforn the reactants and overall the system 'rolls' back into the reactants chanel.
|[[File:Kjo3118 traj4.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.5</nowiki>
|YES
|In this case barrier re-crossing is also seen, after crossing the transition state for the first time the system goes back to the reactants (barrier re-crossing), however then the system crosses the transition state again to make the products.
|[[File:Kjo3118 traj5.png|thumb]]
|}
From the table you can conclude that barrier re-crossing can take place and that not all collisions with energy larger than the activation energy will result in a sucessfull collision. Meaning that the system can pass through the transition state to form the products, but then go back to reform the reactants. A good example of '''barrier re-crossing is the second to last row in the table above'''.

Transition state theory (TST) predicts reaction rates, however due to the main assumptions made it '''overestimates''' reaction rates as compared to experimental values. This is due to the assumption made, stating that any collision with kinetic energy greater than the activation energy will result in a reaction and formation of the products. However as shown in the table above (particularly the second to last row), barrier re-cressing can cause a collision to reform the reactants and therfore not result in a succesful reaction. This means that in reality the proportion of succesfull collisions is lower than the one predicted by transition state theory and thus transition state theory overestimates the reaction rates. It is important to meantion an opposing effect, that would cause transition state theory to underestimate reaction rates: quantum tunelling. TST is a classical theory and it does not include any quantum effects. Quantum tunnelling allows the energetic barrier to be crossed even if the reactants do not have sufficent kinetic energy. This would increase the proportion of sucesfull collisions, however the tunelling phenomenon is not as prevalent as barrier re-crossing, especially for heavier systems.

== EXERCISE 2: H+HF/F+H2 ==
[[File:Surf kjo318 1.png|thumb|433x433px|Figure 3. The potential surface plot for the reaction: F + H2 -> HF + H. The products are of lower energy than reactants, therfore the reaction is exothermic.]]
For this part of the exercise the F-H-H system was used.

'''F + H2 -> HF + H '''is an exothermic reaction, since the energy drops from reactants to products, Figure 3.
[[File:Surf kjo318 2.png|thumb|435x435px|Figure 4. The potential surface plot for the reaction: HF + H -> F + H2. The products are of higher energy than reactants, therfore the reaction is endothermic.]]
'''HF + H -> F + H2 '''is the exact 'opposite' reaction, and by that logic it has to be endothermic. Looking at the surface plot confirms this, Figure 4.

Since in both reaction the only bonds being made/broken are H-H and H-F bonds, the exo-/endothermicity of the reactions can be used to establish the relative bond strengths. Looking at Figure 3, the products have 1 H-F bond and the reactants have only 1 H-H bond. The products are lower in energy, suggesting that the H-F bond is more stable and therfore stronger.
[[File:Kjo3118 int fhf.png|thumb|375x375px|Figure 5. The intermolecular distance vs time plot for the F-H-H system, starting with zero momentum, at the estimated position of the transition state. Notice the only small oscillations fro mthe initial position.]]
To find the transition state of these reactions (which are the same since the reactions are the opposite of each other), first Hammond's postulate was used: since '''F + H2 -> HF + H '''is an exothermic reaction the transition state will be an 'early' transition state, that will be simmilar in structure to the reactants. That meanst the transition state will have a significantly higher F-H distance, while the H-H distance will be simmilar to the H-H bond lenght (around 75 pm). Using this information the transition state was estimated to be at RF-H = 181 pm and RH-H = 74 pm, by finding a position where, given initial momenta equal to 0, the trajectory only oscillates slightly from the initial positions and the initial forces are close to 0, Figure 5.

Next the activation energies were found, by subtracting the potential energy of the reactnts/products from the potential energy of the transition state, the values were found to be:

'''F + H2 -> HF + H activation energy was found to be +1.12 kJ.mol-1'''

'''HF + H -> F + H2 activation energy was found to be +126.70 kJ.mol-1'''
[[File:Kjo3118 rcn1 path.png|thumb|921x921px|Figure 7. The MEP pathway (left) as well as the energy profile (right), of the F-H-H system starting close to the transition state, but displaced towards HF + H. Notice the fast potential (here equal to total) energy drop from the transitions state to the products.]]The reactant and product potential energies were found by running a MEP calculations displaced slightly from the approximate position of the transition state towoards either '''HF + H''' or '''F + H2''', which allowed the trajectory to follow into the valley of either reactants or porducts. The energies vs time found by these calculations have initially changed and then reached a stable value - the potential energy of reactants/products. The results of the calculations can be found in Figure 6.[[File:Kjo3118 rcn2 path.png|left|thumb|915x915px|Figure 6. The MEP pathway (left) as well as the energy profile (right), of the F-H-H system starting close to the transition state, but displaced towards H2 + F. Notice the slow potential (here equal to total) energy drop from the transitions state to the products.]]

Kjo3118

2020-05-15T12:22:01Z

Kjo3118:

== EXERCISE 1: H+H2 ==
[[File:TS intermolecular dist vs time kjo3118.png|thumb|Figure 1. The internuclear distance vs time graph for the best estimate TS. Notice how both internuclear distances are almost constsnt, with only small osscillations.|343x343px]][[File:Kjo3118 rcn dynandmep.png|thumb|Figure 2. The genarated reaction path (right) as well as a trajectory generated using the dymanic calculation (left). Notie the osscilations on the left graph and the lack thereof on the right.|587x587px]]
In this exercise, a system comprised of only hydrogen atoms has been used. First the transition state has been located. The transition state is at a '''saddle point on the potential energy surface''', so it is a maximum along the reaction coordinate, but a minimum in the orthogonal direction. In this case '''both''' partial derivatives of the potential with respect to RAB and RBC are equal to 0. To distinguish a saddle ponit from a local minimum (which would also have both first derivatives equal to 0), you would have to take a look at the second derivatives: at a saddle ponit (TS) one would be positive and one would be negative; at a local minimum both second derivatives would be positive. To locate the position of the transition state first a consideration of geometry was made: since the system is totally symmetric, the transition state is also symmetric and so RAB=RBC . My best estimate is '''90.77 pm'''. For RAB(0)=RBC(0)=90.77 pm and both initial momenta equal to zero the molecule oscillates so little from the initial position that the internuclear diatances remain almost constant (Figure 1.), for these conditions the '''initial forces are very close to 0''' (both 0.001 kJ.mol-1.pm-1).

Figure 2 shows the reaction path genereated using the MEP calculation type as well as the trajectory genareted using the dynamic calculation. They are very simmilar, the only difference is that the mep trajcetory just follows the valley of lowest potential, wheras the dynamically generated trajectory also includes vibrational osscillations.'''<nowiki/>'''
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|YES
|The system crosses the transition state and goes on to form the products. The vibrational osscillations are larger in products, meaning some energy was converted into vibrational.
|[[File:Kjo3118 traj1.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.1</nowiki>
|NO
|The system approaches the transition state, but it never really reaches it, it 'rolls' back into the reactants chanel.
|[[File:Kjo3118 traj2.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.0</nowiki>
|YES
|The system crosses the transition state and goes on to form the products. The vibrational osscillations are larger in products, meaning some energy was converted into vibrational.
|[[File:Kjo3118 traj3.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.3</nowiki>
|NO
|This case is interesting, the system crosses the transition state to form the products, however due to the very large vibrational oscillations the system then re-crosses the barrier to reforn the reactants and overall the system 'rolls' back into the reactants chanel.
|[[File:Kjo3118 traj4.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.5</nowiki>
|YES
|In this case barrier re-crossing is also seen, after crossing the transition state for the first time the system goes back to the reactants (barrier re-crossing), however then the system crosses the transition state again to make the products.
|[[File:Kjo3118 traj5.png|thumb]]
|}
From the table you can conclude that barrier re-crossing can take place and that not all collisions with energy larger than the activation energy will result in a sucessfull collision. Meaning that the system can pass through the transition state to form the products, but then go back to reform the reactants. A good example of '''barrier re-crossing is the second to last row in the table above'''.

Transition state theory (TST) predicts reaction rates, however due to the main assumptions made it '''overestimates''' reaction rates as compared to experimental values. This is due to the assumption made, stating that any collision with kinetic energy greater than the activation energy will result in a reaction and formation of the products. However as shown in the table above (particularly the second to last row), barrier re-cressing can cause a collision to reform the reactants and therfore not result in a succesful reaction. This means that in reality the proportion of succesfull collisions is lower than the one predicted by transition state theory and thus transition state theory overestimates the reaction rates. It is important to meantion an opposing effect, that would cause transition state theory to underestimate reaction rates: quantum tunelling. TST is a classical theory and it does not include any quantum effects. Quantum tunnelling allows the energetic barrier to be crossed even if the reactants do not have sufficent kinetic energy. This would increase the proportion of sucesfull collisions, however the tunelling phenomenon is not as prevalent as barrier re-crossing, especially for heavier systems.

== EXERCISE 2: H+HF/F+H2 ==
[[File:Surf kjo318 1.png|thumb|433x433px|Figure 3. The potential surface plot for the reaction: F + H2 -> HF + H. The products are of lower energy than reactants, therfore the reaction is exothermic.]]
'''F + H2 -> HF + H '''is an exothermic reaction, since the energy drops from reactants to products, Figure 3.
[[File:Surf kjo318 2.png|thumb|435x435px|Figure 4. The potential surface plot for the reaction: HF + H -> F + H2. The products are of higher energy than reactants, therfore the reaction is endothermic.]]
'''HF + H -> F + H2 '''is the exact 'opposite' reaction, and by that logic it has to be endothermic. Looking at the surface plot confirms this, Figure 4.

Since in both reaction the only bonds being made/broken are H-H and H-F bonds, the exo-/endothermicity of the reactions can be used to establish the relative bond strengths. Looking at Figure 3, the products have 1 H-F bond and the reactants have only 1 H-H bond. The products are lower in energy, suggesting that the H-F bond is more stable and therfore stronger.
[[File:Kjo3118 int fhf.png|thumb|375x375px|Figure 5. The intermolecular distance vs time plot for the F-H-H system, starting with zero momentum, at the estimated position of the transition state. Notice the only small oscillations fro mthe initial position.]]
To find the transition state of these reactions (which are the same since the reactions are the opposite of each other), first Hammond's postulate was used: since '''F + H2 -> HF + H '''is an exothermic reaction the transition state will be an 'early' transition state, that will be simmilar in structure to the reactants. That meanst the transition state will have a significantly higher F-H distance, while the H-H distance will be simmilar to the H-H bond lenght (around 75 pm). Using this information the transition state was estimated to be at RF-H = 181 pm and RH-H = 74 pm, by finding a position where, given initial momenta equal to 0, the trajectory only oscillates slightly from the initial positions and the initial forces are close to 0, Figure 5.

Next the activation energies were found, by subtracting the potential energy of the reactnts/products from the potential energy of the transition state, the values were found to be:

'''F + H2 -> HF + H activation energy was found to be +1.12 kJ.mol-1'''

'''HF + H -> F + H2 activation energy was found to be +126.70 kJ.mol-1'''
[[File:Kjo3118 rcn1 path.png|thumb|921x921px|Figure 7. The MEP pathway (left) as well as the energy profile (right), of the F-H-H system starting close to the transition state, but displaced towards HF + H. Notice the fast potential (here equal to total) energy drop from the transitions state to the products.]]The reactant and product potential energies were found by running a MEP calculations displaced slightly from the approximate position of the transition state towoards either '''HF + H''' or '''F + H2''', which allowed the trajectory to follow into the valley of either reactants or porducts. The energies vs time found by these calculations have initially changed and then reached a stable value - the potential energy of reactants/products. The results of the calculations can be found in Figure 6.[[File:Kjo3118 rcn2 path.png|left|thumb|915x915px|Figure 6. The MEP pathway (left) as well as the energy profile (right), of the F-H-H system starting close to the transition state, but displaced towards H2 + F. Notice the slow potential (here equal to total) energy drop from the transitions state to the products.]]

Kjo3118

2020-05-15T12:19:06Z

Kjo3118:

== EXERCISE 1: H+H2 ==
[[File:TS intermolecular dist vs time kjo3118.png|thumb|Figure 1. The internuclear distance vs time graph for the best estimate TS. Notice how both internuclear distances are almost constsnt, with only small osscillations.|343x343px]][[File:Kjo3118 rcn dynandmep.png|thumb|Figure 2. The genarated reaction path (right) as well as a trajectory generated using the dymanic calculation (left). Notie the osscilations on the left graph and the lack thereof on the right.|587x587px]]
In this exercise, a system comprised of only hydrogen atoms has been used. First the transition state has been located. The transition state is at a '''saddle point on the potential energy surface''', so it is a maximum along the reaction coordinate, but a minimum in the orthogonal direction. In this case '''both''' partial derivatives of the potential with respect to RAB and RBC are equal to 0. To distinguish a saddle ponit from a local minimum (which would also have both first derivatives equal to 0), you would have to take a look at the second derivatives: at a saddle ponit (TS) one would be positive and one would be negative; at a local minimum both second derivatives would be positive. To locate the position of the transition state first a consideration of geometry was made: since the system is totally symmetric, the transition state is also symmetric and so RAB=RBC . My best estimate is '''90.77 pm'''. For RAB(0)=RBC(0)=90.77 pm and both initial momenta equal to zero the molecule oscillates so little from the initial position that the internuclear diatances remain almost constant (Figure 1.), for these conditions the '''initial forces are very close to 0''' (both 0.001 kJ.mol-1.pm-1).

Figure 2 shows the reaction path genereated using the MEP calculation type as well as the trajectory genareted using the dynamic calculation. They are very simmilar, the only difference is that the mep trajcetory just follows the valley of lowest potential, wheras the dynamically generated trajectory also includes vibrational osscillations.'''<nowiki/>'''
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|YES
|The system crosses the transition state and goes on to form the products. The vibrational osscillations are larger in products, meaning some energy was converted into vibrational.
|[[File:Kjo3118 traj1.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.1</nowiki>
|NO
|The system approaches the transition state, but it never really reaches it, it 'rolls' back into the reactants chanel.
|[[File:Kjo3118 traj2.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.0</nowiki>
|YES
|The system crosses the transition state and goes on to form the products. The vibrational osscillations are larger in products, meaning some energy was converted into vibrational.
|[[File:Kjo3118 traj3.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.3</nowiki>
|NO
|This case is interesting, the system crosses the transition state to form the products, however due to the very large vibrational oscillations the system then re-crosses the barrier to reforn the reactants and overall the system 'rolls' back into the reactants chanel.
|[[File:Kjo3118 traj4.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.5</nowiki>
|YES
|In this case barrier re-crossing is also seen, after crossing the transition state for the first time the system goes back to the reactants (barrier re-crossing), however then the system crosses the transition state again to make the products.
|[[File:Kjo3118 traj5.png|thumb]]
|}
From the table you can conclude that barrier re-crossing can take place and that not all collisions with energy larger than the activation energy will result in a sucessfull collision. Meaning that the system can pass through the transition state to form the products, but then go back to reform the reactants. A good example of '''barrier re-crossing is the second to last row in the table above'''.

Transition state theory (TST) predicts reaction rates, however due to the main assumptions made it '''overestimates''' reaction rates as compared to experimental values. This is due to the assumption made, stating that any collision with kinetic energy greater than the activation energy will result in a reaction and formation of the products. However as shown in the table above (particularly the second to last row), barrier re-cressing can cause a collision to reform the reactants and therfore not result in a succesful reaction. This means that in reality the proportion of succesfull collisions is lower than the one predicted by transition state theory and thus transition state theory overestimates the reaction rates. It is important to meantion an opposing effect, that would cause transition state theory to underestimate reaction rates: quantum tunelling. TST is a classical theory and it does not include any quantum effects. Quantum tunnelling allows the energetic barrier to be crossed even if the reactants do not have sufficent kinetic energy. This would increase the proportion of sucesfull collisions, however the tunelling phenomenon is not as prevalent as barrier re-crossing, especially for heavier systems.

== EXERCISE 2: H+HF/F+H2 ==
[[File:Surf kjo318 1.png|thumb|433x433px|Figure 3. The potential surface plot for the reaction: F + H2 -> HF + H. The products are of lower energy than reactants, therfore the reaction is exothermic.]]
'''F + H2 -> HF + H '''is an exothermic reaction, since the energy drops from reactants to products, Figure 3.
[[File:Surf kjo318 2.png|thumb|435x435px|Figure 4. The potential surface plot for the reaction: HF + H -> F + H2. The products are of higher energy than reactants, therfore the reaction is endothermic.]]
'''HF + H -> F + H2 '''is the exact 'opposite' reaction, and by that logic it has to be endothermic. Looking at the surface plot confirms this, Figure 4.

Since in both reaction the only bonds being made/broken are H-H and H-F bonds, the exo-/endothermicity of the reactions can be used to establish the relative bond strengths. Looking at Figure 3, the products have 1 H-F bond and the reactants have only 1 H-H bond. The products are lower in energy, suggesting that the H-F bond is more stable and therfore stronger.
[[File:Kjo3118 int fhf.png|thumb|375x375px|Figure 5. The intermolecular distance vs time plot for the F-H-H system, starting with zero momentum, at the estimated position of the transition state. Notice the only small oscillations fro mthe initial position.]]
To find the transition state of these reactions (which are the same since the reactions are the opposite of each other), first Hammond's postulate was used: since '''F + H2 -> HF + H '''is an exothermic reaction the transition state will be an 'early' transition state, that will be simmilar in structure to the reactants. That meanst the transition state will have a significantly higher F-H distance, while the H-H distance will be simmilar to the H-H bond lenght (around 75 pm). Using this information the transition state was estimated to be at RF-H = 181 pm and RH-H = 74 pm, by finding a position where, given initial momenta equal to 0, the trajectory only oscillates slightly from the initial positions and the initial forces are close to 0, Figure 5.

Next the activation energies were found, by subtracting the potential energy of the reactnts/products from the potential energy of the transition state, the values were found to be:

'''F + H2 -> HF + H activation energy was found to be '''

'''HF + H -> F + H2 activation energy was found to be'''
[[File:Kjo3118 rcn1 path.png|thumb|921x921px|Figure 7. The MEP pathway (left) as well as the energy profile (right), of the F-H-H system starting close to the transition state, but displaced towards HF + H. Notice the fast potential (here equal to total) energy drop from the transitions state to the products.]]The reactant and product potential energies were found by running a MEP calculations displaced slightly from the approximate position of the transition state towoards either '''HF + H''' or '''F + H2''', which allowed the trajectory to follow into the valley of either reactants or porducts. The energies vs time found by these calculations have initially changed and then reached a stable value - the potential energy of reactants/products. The results of the calculations can be found in Figure 6.[[File:Kjo3118 rcn2 path.png|left|thumb|915x915px|Figure 6. The MEP pathway (left) as well as the energy profile (right), of the F-H-H system starting close to the transition state, but displaced towards H2 + F. Notice the slow potential (here equal to total) energy drop from the transitions state to the products.]]

Kjo3118

2020-05-15T12:12:50Z

Kjo3118:

== EXERCISE 1: H+H2 ==
[[File:TS intermolecular dist vs time kjo3118.png|thumb|Figure 1. The internuclear distance vs time graph for the best estimate TS. Notice how both internuclear distances are almost constsnt, with only small osscillations.|343x343px]][[File:Kjo3118 rcn dynandmep.png|thumb|Figure 2. The genarated reaction path (right) as well as a trajectory generated using the dymanic calculation (left). Notie the osscilations on the left graph and the lack thereof on the right.|587x587px]]
In this exercise, a system comprised of only hydrogen atoms has been used. First the transition state has been located. The transition state is at a '''saddle point on the potential energy surface''', so it is a maximum along the reaction coordinate, but a minimum in the orthogonal direction. In this case '''both''' partial derivatives of the potential with respect to RAB and RBC are equal to 0. To distinguish a saddle ponit from a local minimum (which would also have both first derivatives equal to 0), you would have to take a look at the second derivatives: at a saddle ponit (TS) one would be positive and one would be negative; at a local minimum both second derivatives would be positive. To locate the position of the transition state first a consideration of geometry was made: since the system is totally symmetric, the transition state is also symmetric and so RAB=RBC . My best estimate is '''90.77 pm'''. For RAB(0)=RBC(0)=90.77 pm and both initial momenta equal to zero the molecule oscillates so little from the initial position that the internuclear diatances remain almost constant (Figure 1.), for these conditions the '''initial forces are very close to 0''' (both 0.001 kJ.mol-1.pm-1).

Figure 2 shows the reaction path genereated using the MEP calculation type as well as the trajectory genareted using the dynamic calculation. They are very simmilar, the only difference is that the mep trajcetory just follows the valley of lowest potential, wheras the dynamically generated trajectory also includes vibrational osscillations.'''<nowiki/>'''
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|YES
|The system crosses the transition state and goes on to form the products. The vibrational osscillations are larger in products, meaning some energy was converted into vibrational.
|[[File:Kjo3118 traj1.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.1</nowiki>
|NO
|The system approaches the transition state, but it never really reaches it, it 'rolls' back into the reactants chanel.
|[[File:Kjo3118 traj2.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.0</nowiki>
|YES
|The system crosses the transition state and goes on to form the products. The vibrational osscillations are larger in products, meaning some energy was converted into vibrational.
|[[File:Kjo3118 traj3.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.3</nowiki>
|NO
|This case is interesting, the system crosses the transition state to form the products, however due to the very large vibrational oscillations the system then re-crosses the barrier to reforn the reactants and overall the system 'rolls' back into the reactants chanel.
|[[File:Kjo3118 traj4.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.5</nowiki>
|YES
|In this case barrier re-crossing is also seen, after crossing the transition state for the first time the system goes back to the reactants (barrier re-crossing), however then the system crosses the transition state again to make the products.
|[[File:Kjo3118 traj5.png|thumb]]
|}
From the table you can conclude that barrier re-crossing can take place and that not all collisions with energy larger than the activation energy will result in a sucessfull collision. Meaning that the system can pass through the transition state to form the products, but then go back to reform the reactants. A good example of '''barrier re-crossing is the second to last row in the table above'''.

Transition state theory (TST) predicts reaction rates, however due to the main assumptions made it '''overestimates''' reaction rates as compared to experimental values. This is due to the assumption made, stating that any collision with kinetic energy greater than the activation energy will result in a reaction and formation of the products. However as shown in the table above (particularly the second to last row), barrier re-cressing can cause a collision to reform the reactants and therfore not result in a succesful reaction. This means that in reality the proportion of succesfull collisions is lower than the one predicted by transition state theory and thus transition state theory overestimates the reaction rates. It is important to meantion an opposing effect, that would cause transition state theory to underestimate reaction rates: quantum tunelling. TST is a classical theory and it does not include any quantum effects. Quantum tunnelling allows the energetic barrier to be crossed even if the reactants do not have sufficent kinetic energy. This would increase the proportion of sucesfull collisions, however the tunelling phenomenon is not as prevalent as barrier re-crossing, especially for heavier systems.

== EXERCISE 2: H+HF/F+H2 ==
[[File:Surf kjo318 1.png|thumb|433x433px|Figure 3. The potential surface plot for the reaction: F + H2 -> HF + H. The products are of lower energy than reactants, therfore the reaction is exothermic.]]
'''F + H2 -> HF + H '''is an exothermic reaction, since the energy drops from reactants to products, Figure 3.
[[File:Surf kjo318 2.png|thumb|435x435px|Figure 4. The potential surface plot for the reaction: HF + H -> F + H2. The products are of higher energy than reactants, therfore the reaction is endothermic.]]
'''HF + H -> F + H2 '''is the exact 'opposite' reaction, and by that logic it has to be endothermic. Looking at the surface plot confirms this, Figure 4.

Since in both reaction the only bonds being made/broken are H-H and H-F bonds, the exo-/endothermicity of the reactions can be used to establish the relative bond strengths. Looking at Figure 3, the products have 1 H-F bond and the reactants have only 1 H-H bond. The products are lower in energy, suggesting that the H-F bond is more stable and therfore stronger.
[[File:Kjo3118 int fhf.png|thumb|375x375px|Figure 5. The intermolecular distance vs time plot for the F-H-H system, starting with zero momentum, at the estimated position of the transition state. Notice the only small oscillations fro mthe initial position.]]
To find the transition state of these reactions (which are the same since the reactions are the opposite of each other), first Hammond's postulate was used: since '''F + H2 -> HF + H '''is an exothermic reaction the transition state will be an 'early' transition state, that will be simmilar in structure to the reactants. That meanst the transition state will have a significantly higher F-H distance, while the H-H distance will be simmilar to the H-H bond lenght (around 75 pm). Using this information the transition state was estimated to be at RF-H = 181 pm and RH-H = 74 pm, by finding a position where, given initial momenta equal to 0, the trajectory only oscillates slightly from the initial positions and the initial forces are close to 0, Figire 5.

Next the activation energies were found, by subtracting the potential energy of the reactnts/products from the potential energy of the transition state, the values were found to be:

'''F + H2 -> HF + H'''

'''HF + H -> F + H2'''
[[File:Kjo3118 rcn1 path.png|thumb|921x921px|Figure 7. The MEP pathway (left) as well as the energy profile (right), of the F-H-H system starting close to the transition state, but displaced towards HF + H. Notice the fast potential (here equal to total) energy drop from the transitions state to the products.]]
[[File:Kjo3118 rcn2 path.png|left|thumb|915x915px|Figure 6. The MEP pathway (left) as well as the energy profile (right), of the F-H-H system starting close to the transition state, but displaced towards H2 + F. Notice the slow potential (here equal to total) energy drop from the transitions state to the products.]]
The reactant and product potential energies were found by running a MEP calculations displaced slightly from the approximate position of the transition state towoards either '''HF + H''' or '''F + H2''', which allowed the trajectory to follow into the valley of either reactants or porducts. The energies vs time found by these calculations have initially changed and then reached a stable value - the potential energy of reactants/products. The results of the calculations can be found in Figure 6.

File:Kjo3118 rcn2 path.png

2020-05-15T11:58:51Z

Kjo3118: ab

File:Kjo3118 rcn1 path.png

2020-05-15T11:58:00Z

Kjo3118: aa

File:Kjo3118 int fhf.png

2020-05-15T11:14:27Z

Kjo3118: aa

2020-05-14T10:13:43Z

Kjo3118: F H2

F H2

Kjo3118

2020-05-12T12:59:18Z

Kjo3118:

[[File:TS intermolecular dist vs time kjo3118.png|thumb|Figure 1. The internuclear distance vs time graph for the best estimate TS. Notice how both internuclear distances are almost constsnt, with only small osscillations.]]'''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 at a '''saddle point''', so it is a maximum along the reaction coordinate, but a minimum in the orthogonal direction. In this case '''both''' partial derivatives of the potential with respect to RAB and RBC are 0. To distinguish a saddle ponit from a local minimum (which would also have both first derivatives equal to 0), you would have to take a look at the second derivatives: at a saddle ponit (TS) one would be positive and one would be negative; at a local minimum both second derivatives would be positive.

'''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.'''

My best estimate is '''90.77 pm'''. For RAB(0)=RBC(0)=90.77 pm and both initial momenta equal to zero the molecule oscillates so little from the initial position that the internuclear diatances remain almost constant (Figure 1.), for these conditions the '''initial forces are very close to 0''' (both 0.001 kJ.mol-1.pm-1).
[[File:Kjo3118 rcn dynandmep.png|thumb|Figure 2. The genarate reaction path (right) as well as a trajectory generated using the dymanic calculation (left). Notie the osscilations on the left graph and the lack thereof on the right.|587x587px]]
'''Comment on how the ''mep'' and the trajectory you just calculated differ.'''

They are very simmilar, the only difference is that the mep trajcetory just follows the valley of lowest potential, wheras the dynamically generated trajectory also includes vibrational osscillations.

'''Complete the table above by adding the total
energy, whether the trajectory is reactive or unreactive, and provide a
plot of the trajectory and a small description for what happens along
the trajectory. What can you conclude from the table?'''
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|YES
|
|[[File:Kjo3118 traj1.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.1</nowiki>
|NO
|
|[[File:Kjo3118 traj2.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.0</nowiki>
|YES
|
|[[File:Kjo3118 traj3.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.3</nowiki>
|NO
|
|[[File:Kjo3118 traj4.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.5</nowiki>
|YES
|
|[[File:Kjo3118 traj5.png|thumb]]
|}
From the table you can conclude that for '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1

File:TS intermolecular dist vs time kjo3118.png

2020-05-12T12:56:02Z

Kjo3118: dsa

dsa

File:TS intermolecular dist vs time.png

2020-05-12T12:54:22Z

Kjo3118: aa

Kjo3118

2020-05-12T12:46:00Z

Kjo3118

2020-05-12T11:12:50Z

Kjo3118:

File:Kjo3118 p1.png

2020-05-12T11:10:42Z

Kjo3118:

Kjo3118

2020-05-12T10:41:26Z

Kjo3118: Created page with "'''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..."

Rep:Mod:kjoreport

2019-03-08T14:12:38Z

Kjo3118: /* Method and obtained data */

==NH3==
===3D molecule model===
<jmol><jmolApplet>
<title> Ammonia molecule </title>
<color>black</color>
<size>500</size>
<uploadedFileContents>Kjo3118nh3opt.LOG</uploadedFileContents>
</jmolApplet></jmol>

===Method and obtained data===
{| class="wikitable" border="1"
|-
!colspan="2" style="text-align: center;"| Ammonia data
|-
| Point group || C3V
|-
| Calculation method || RB3LYP
|-
| Basis set || 6-31G(d,p)
|-
| N-H bond length || 1.018Å
|-
| H-N-H bond angle || 105.7°
|-
| Final energy || -56.55777a.u.
|-
| RMS gradient || 0.0003244a.u.
|}

===File and 'Item' table===
This is the .LOG file comtaining all the relevant data:[[File:KJO3118_CO2_OPTIMISED.LOG]], and the 'Item' table from this file is:
<pre>
Item Value Threshold Converged?
Maximum Force 0.000024 0.000450 YES
RMS Force 0.000017 0.000300 YES
Maximum Displacement 0.000021 0.001800 YES
RMS Displacement 0.000015 0.001200 YES
</pre>

===Vibrations===
====Vibrational modes====
[[File:kjo3118_vib_tab.png|350px]]

====Vibrational modes table====
According to the 3N-6 rule for non-linear molecules, there is an expected number of 6 vibrational modes for a NH3 molecule. 3 of these are bending vibrations (one mode with wavenumber 1090cm-1 and two degenerate modes 1694cm-1), and 3 are stretching vibrations (one mode with wavenumber 3461cm-1 and two degenerate modes 3590cm-1). The first mode shown in the table (wavenumber 1090cm-1) is known as the 'umbrella' mode, and the mode with wavenumber 3461cm-1 is highly symmetric. These modes are shown in the table below:
{| class="wikitable" border="1"
|-
!colspan="7" style="text-align: center;"| Vibrational data
|-
|||colspan="3" style="text-align: center;"| Bending modes
|-
| Wavenumber/cm-1 || 1090 || 1694 || 1694
|-
| Intensity/arbitrary || 145 || 14 || 14
|-
| Symmetry || A1 || E || E
|-
| Visualisation ||[[File:kjo3118_1vib.png|180px]]||[[File:kjo3118_2vib.png|180px]]||[[File:kjo3118_3vib.png|180px]]
|-
|||colspan="3" style="text-align: center;"| Stretching modes
|-
| Wavenumber/cm-1|| 3461 || 3590 || 3590
|-
| Intensity/arbitrary|| 1.06 || 0.27 || 0.27
|-
| Symmetry|| A1 || E || E
|-
| Visualisation||[[File:kjo3118_4vib.png|180px]]||[[File:kjo3118_5vib.png|180px]]||[[File:kjo3118_6vib.png|180px]]
|}

====IR spectrum====
In an IR spectrum of gaseous ammonia 4 peaks are visible (starting from the left):
*first one due to the 'umbrella' bending mode;
*second peak due to the other two degenerate, asymmetric bending modes;
*third, very faint peak, due to the 3461 cm-1 stretch;
*fourth, again very faint peak, due to both degenerate 3590cm-1 modes.
[[File:kjo3118_ir_spec_nh3.png|800px]]

===Charge distribution in NH3===
In a NH3 molecule, the nitrogen is more electronegative than hydrogen (3.04 for Nitrogen<ref>Mark Winter, U. (2019). WebElements Periodic Table » Nitrogen » electronegativity. [online] Webelements.com. Available at: https://www.webelements.com/nitrogen/electronegativity.html [Accessed 8 Mar. 2019].</ref> vs 2.20 for Hydrogen <ref>Mark Winter, U. (2019). WebElements Periodic Table » Hydrogen » electronegativity. [online] Webelements.com. Available at: https://www.webelements.com/hydrogen/electronegativity.html [Accessed 8 Mar. 2019].</ref> on the Pauling scale). Therfore the N-H bonds are polarised, with a slight negative charge on the nitrogen atom, and a partial positive charge on each hydrogen atom:
*[[File:kjo3118_nh3_charge.png|400px]]

==H2 and N2==
===3D molecule models===
<jmol><jmolApplet>
<title> Nitrogen molecule </title>
<color>black</color>
<size>380</size>
<uploadedFileContents>KJO3118_N2_OPTIMISED.LOG</uploadedFileContents>
</jmolApplet></jmol>

<jmol><jmolApplet>
<title> Hydrogen molecule </title>
<color>black</color>
<size>380</size>
<uploadedFileContents>KJO3118_H2_OPTIMISED.LOG</uploadedFileContents>
</jmolApplet></jmol>

===Method and obtained data===
{| class="wikitable" border="1"
|-
!colspan="3" style="text-align: center;"| Obtained data
|-
|||Nitrogen||Hydrogen
|-
| Point group || D∞h||D∞h
|-
| Calculation method || RB3LYP || RB3LYP
|-
| Basis set || 6-31G(d,p) || 6-31G(d,p)
|-
| Bond length || 1.106Å || 0.743Å
|-
| Final energy || -109.52413a.u. ||-1.17854a.u.
|-
| RMS gradient || 0.0000006a.u. ||0.00000007a.u.
|}
Bond angles are not included, since a di-atomic molecule cannot have a bond angle (for an angle, mathematically, 3 points are required). However, this angle can be taken to be 180.0°.

===Files and 'Item' tables===
This is the .LOG file comtaining all the relevant data for nitrogen:[[File:KJO3118_N2_OPTIMISED.LOG]], and the .LOG file for hydrogen: [[File:KJO3118_H2_OPTIMISED.LOG]].
The 'Item' table for nitrogen:
<pre>
Item Value Threshold Converged?
Maximum Force 0.000001 0.000450 YES
RMS Force 0.000001 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000000 0.001200 YES
</pre>
And for hydrogen:
<pre>
Item Value Threshold Converged?
Maximum Force 0.000000 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000001 0.001200 YES
</pre>

===Vibrations===
====N2 vibrational modes====
[[File:Kjo3118_n2_vib_win.png|350px]]

====N2 vibrational modes table====
According to the 3N-5 rule for linear molecules, only one vibrational mode is present in a N2 molecule. It is a stretching vibration mode:
{| class="wikitable" border="1"
|-
!colspan="7" style="text-align: center;"| Vibrational data
|-
||| Stretching mode
|-
| Wavenumber/cm-1 || 2457
|-
| Intensity/arbitrary || 0
|-
| Symmetry || SGG
|-
| Visualisation ||[[File:kjo3118_n2_img.png|180px]]
|}

====H2 vibrational modes====
[[File:Kjo3118_h2_vib_win.png|350px]]

====H2 vibrational modes table====
According to the 3N-5 rule for linear molecules, only one vibrational mode is present in a H2 molecule. It is a stretching vibration mode:
{| class="wikitable" border="1"
|-
!colspan="7" style="text-align: center;"| Vibrational data
|-
||| Stretching mode
|-
| Wavenumber/cm-1 || 4466
|-
| Intensity/arbitrary || 0
|-
| Symmetry || SGG
|-
| Visualisation ||[[File:kjo3118_h2_img.png|180px]]
|}

===Charge distribution===
Both N2 and H2 are symmetric, 100% covalent molecules, therefore there is no permanent charge separation on the atoms:
====N2====
[[File:kjo3118_n2_charge.png|350px]]
====H2====
[[File:kjo3118_h2_charge.png|350px]]
===N2 in a transition metal complex===
*[[File:kjo_complex.jpg|500px]]

I found [[https://www.ccdc.cam.ac.uk/structures/Search?Ccdcid=doctum&DatabaseToSearch=Published DOCTUM]], a Cobalt metal complex in the CDC, using conquest. The N-N bond length in the complex was found to be 1.112Å, which is very close to the computed and modelled value of 1.106Å. However, it is experimentally measure to be slightly longer. This is likely because in the calculation only an N2 molecule is considered, with no other species in the surroundings. In the case of the complex, there would be a large delocalised electron system on the complex, making it very likely that some electron density would be placed into the π* antibonding orbitals of N2, increasing the bond length. This could be confirmed by checking the Raman absorption spectrum: for the longer bond, its stiffness would be less, therefore showing an absorption peak at a lower wavenumber<ref>Cavigliasso, G., Wilson, L., McAlpine, S., Attar, M., Stranger, R. and Yates, B. (2010). Activation and cleavage of the N–N bond in side-on bound [L2M-NN-ML2] (L = NH2, NMe2, NiPr2, C5H5, C5Me4H) dinitrogen complexes of transition metals from groups 4 through 9. Dalton Transactions, 39(19), p.4529.</ref>.

===Energetics of the Harbor process===
N2+3H2->2NH3
{| class="wikitable" border="1"
|-
!colspan="7" style="text-align: center;"| Energies table
|-
| E(NH3)=-56.55777 a.u.
|-
| 2*E(NH3)=-113.11554 a.u.
|-
| E(N2)=-109.52413 a.u.
|-
| E(H2)=-1.17854 a.u.
|-
| 3*E(H2)=-3.53562 a.u.
|-
| ΔE=2*E(NH3)-[E(N2)+3*E(H2)]=-0.05579 a.u.= -146.5 kJ/mol
|}
Since the reaction is exothermic, the products are thermodynamically more stable, at a lower energy.

==CO2-chosen molecule==
===3D molecule model===
<jmol><jmolApplet>
<title> Carbon dioxide molecule </title>
<color>black</color>
<size>380</size>
<uploadedFileContents>KJO3118_CO2_OPTIMISED.LOG</uploadedFileContents>
</jmolApplet></jmol>

===Method and obtained data===
{| class="wikitable" border="1"
|-
!colspan="2" style="text-align: center;"| Ammonia data
|-
| Point group || D∞h
|-
| Calculation method || RB3LYP
|-
| Basis set || 6-31G(d,p)
|-
| C=O bond length || 1.169Å
|-
| O-C-O bond angle || 180.0°
|-
| Final energy || -188.58094a.u.
|-
| RMS gradient || 0.00001154a.u.
|}

===File and 'Item' table===
This is the .LOG file comtaining all the relevant data:[[File:Kjo3118nh3opt.LOG]], and the 'Item' table from this file is:
<pre>
Item Value Threshold Converged?
Maximum Force 0.000004 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000072 0.001800 YES
RMS Displacement 0.000035 0.001200 YES
</pre>

===Vibrations===

====Vibrational modes====
[[File:kjo3118_co2_vib_tab.png|350px]]

====Vibrational modes table====
According to the 3N-5 rule for linear molecules, there is an expected number of 4 vibrational modes for a CO2 molecule. 2 of these are bending vibrations, these are degenaerate with a wavenumber of 640cm-1, and 2 are stretching vibrations (one mode with wavenumber 1372cm-1 and one with 2436cm-1). These modes are shown in the table below:
{| class="wikitable" border="1"
|-
!colspan="5" style="text-align: center;"| Vibrational data
|-
|||colspan="2" style="text-align: center;"| Bending modes
|-
| Wavenumber/cm-1 || 640 || 640
|-
| Intensity/arbitrary || 30 || 30
|-
| Symmetry || PIU || PIU
|-
| Visualisation ||[[File:kjo3118_co2_v1.gif|180px]]||[[File:kjo3118_co2_v2.gif|180px]]
|-
|||colspan="2" style="text-align: center;"| Stretching modes
|-
| Wavenumber/cm-1|| 1372 || 2436
|-
| Intensity/arbitrary || 0 || 546
|-
| Symmetry|| SGG || SGG
|-
| Visualisation||[[File:kjo3118_co2_v3.gif|180px]]||[[File:kjo3118_co2_v4.gif|180px]]
|}
The images are .gif files. MAKE SURE YOU CLICK ON THEM TO VIEW THE ANIMATION.

====IR spectrum====
In an IR spectrum of gaseous carbon dioxide 2 peaks are visible (starting from the left):
*first one due to the two degenerate bending modes, at 640cm-1;
*second peak due to the one IR active asymmetric stretching mode;
[[File:kjo3118_co2_spectrum.png|600px]]

===Charge distribution in CO2===
In a CO2 molecule, there is a difference in electronegativity on the atoms present: Carbon has electronegativity of 2.55<ref>Mark Winter, U. (2019). WebElements Periodic Table » Carbon » electronegativity. [online] Webelements.com. Available at: https://www.webelements.com/carbon/electronegativity.html [Accessed 8 Mar. 2019].</ref> and oxygen has an elecrtonegativity of 3.44<ref>Mark Winter, U. (2019). WebElements Periodic Table » Oxygen » electronegativity. [online] Webelements.com. Available at: https://www.webelements.com/oxygen/electronegativity.html [Accessed 8 Mar. 2019].</ref>. Therfore the C=O bonds are polarised, with a slight negative charge on the oxygen atoms, and a partial positive charge on each carbon atom:

[[File:kjo3118_co2_charge.png|600px]]

===CO2 molecular orbitals===
*The first molecular orbital chosen is a mixture of many orbitals in varying degrees: 1s,2s and 3s from carbon and 2s, 2p, 3s, and 3p from oxygen. [[File:kjo_mod_5.png|100px]] This orbital is full and it is one of the orbitals relatively close to the HOMO in energy. Since these overals are all 'head on' this orbital contributes to the σ bonding of the molecule.

[[File:kjo_co2mo_7.png|600px]]

*The next MO chosen is again a mixture of many orbitals: 2p, 3p on carbon and 2s, 2p, 3s, 3p on oxygen: [[File:kjo_mod_6.png|100px]]. This orbital also contributes to the σ bonding.

[[File:kjo_co2mo_6.png|600px]]

*The next two chosen MOs are a product of 2p and 3p, to a smaller degree than 2p, orbitals from both oxygens and the carbon interacting constructively, in phase, resulting in a bonding π orbital, which is full. this comtributes to the 'double' bonds between carbon and ocygen. It can be shown on this simple diagram:[[File:kjo_mod_1.png|100px]] Note that the orbital visualisations look almost exactly the same - they only differ in orientation, they are orthogonal to each other (they are also degenerate):

[[File:kjo_co2mo_5.png|600px]]

[[File:kjo_co2mo_4.png|600px]]

*The next pair or MOs chosen arises when the 2p and 3p, to a smaller degree than 2p,orbitals on oxygen atoms interfere out of phase: [[File:kjo_mod_4.png|100px]]. These two orbitals are the highest occupied molecular orbitals (HOMO). As with the previous case, these are degenerate, only differing in the orientation.

[[File:kjo_co2mo_3.png|600px]]

[[File:kjo_co2mo_2.png|600px]]

* The last pair of MOs chosen are a result of 2p and 3p orbitals on all 3 atoms interfering out of phase:[[file:kjo_mod_2.png|100px]]. This creates the lowest unoccupied molecular orbital (LUMO), which in this case is a π* antibonding orbital. Again these two orbitals are are degenerate and only differ in the orientation.

[[File:kjo_co2mo_8.png|600px]]

[[File:kjo_co2mo_1.png|600px]]

===CO2 HOMO-LUMO transition energy gap===
From my molecular orbital model i found that the HOMO has an energy of -0.36997a.u. and the LUMO has an energy of 0.02992. This gives the energy gap between HOMO and LUMO to be 0.39989a.u., which is 10.9 eV.

The HOMO-LUMO transition can be observed experimentally by spectroscopy. According to an absorption or emission peak wavenumber, the energy gap between the orbitals can be estimated. This value is found to be 9.9eV<ref>Zhan, C., Nichols, J. and Dixon, D. (2003). Ionization Potential, Electron Affinity, Electronegativity, Hardness, and Electron Excitation Energy: Molecular Properties from Density Functional Theory Orbital Energies. The Journal of Physical Chemistry A, 107(20), pp.4184-4195.</ref>.

Our compiled value is not exact, but does provide an estimate of the size of the HOMO-LUMO energy gap in a carbon dioxide molecule (the difference is only 10.1%).

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