ChemWiki - User contributions [en]

Rrp17

2019-05-24T13:40:10Z

Rrp17: /* The approximate position of the transition state */

= Molecular Reaction Dynamics =

= EXERCISE 1: H + H2 system =

If H collides with a H2 molecule, there system can be defined as an AB and C.
==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V2/∂q1q12>0and the second derivative ∂V2/∂q2q22<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction. The transition state is a saddle point, it is where the minimum , q1 and max ,q2 intersect.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077 Å.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where we displayed the transition state position by r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that with the last two contour plots, the products do not form-another variable affects the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains how the rate of a chemical reaction by which the reaction passes through a transition state and it has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|center|Surface plot of the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the saddle point (where there is no momentum) was determined. AB=1.908 Å , BC=0.745 Å. The use of an internuclear distance vs time also highlighted this.

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|Finding the activation energy for the H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|Finding the activation energy for the H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|Energy vs Time |250px]]
[[File:rrpmvt.PNG|thumb|center|Momentum vs Time |250px]]

====Distribution of energy between different modes====

Polanyi's rules1 explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

{| class="wikitable" border="1"
|-
! Incomplete !! Complete !!
|-
| [[File:rrph2com.PNG|thumb| High translational energy leads to a complete reaction|250px]] || [[File:rrph2non.PNG|thumb|Low translational energy does not lead to a complete reaction|250px]] ||
|}

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

{| class="wikitable" border="1"
|-
! Complete !! inComplete !!
|-
| [[File:rrpyes.PNG|thumb|High vibrational energy, reaction complete|250px]] || [[File:rrpno.PNG|thumb|Low vibrational energy, reaction incomplete|250px]] ||
|}

=References=

[1]Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction, Zhaojun Zhang, Yong Zhou,† and Dong H. Zhang, dx.doi.org/10.1021/jz301649w | J. Phys. Chem. Lett. 2012, 3, 3416−3419

Rrp17

2019-05-24T13:39:20Z

Rrp17: /* References */

= Molecular Reaction Dynamics =

= EXERCISE 1: H + H2 system =

If H collides with a H2 molecule, there system can be defined as an AB and C.
==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V2/∂q1q12>0and the second derivative ∂V2/∂q2q22<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction. The transition state is a saddle point, it is where the minimum , q1 and max ,q2 intersect.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077 Å.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where we displayed the transition state position by r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that with the last two contour plots, the products do not form-another variable affects the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains how the rate of a chemical reaction by which the reaction passes through a transition state and it has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|center|Surface plot of the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the saddle point (where there is no momentum) was determined. AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|Finding the activation energy for the H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|Finding the activation energy for the H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|Energy vs Time |250px]]
[[File:rrpmvt.PNG|thumb|center|Momentum vs Time |250px]]

====Distribution of energy between different modes====

Polanyi's rules1 explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

{| class="wikitable" border="1"
|-
! Incomplete !! Complete !!
|-
| [[File:rrph2com.PNG|thumb| High translational energy leads to a complete reaction|250px]] || [[File:rrph2non.PNG|thumb|Low translational energy does not lead to a complete reaction|250px]] ||
|}

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

{| class="wikitable" border="1"
|-
! Complete !! inComplete !!
|-
| [[File:rrpyes.PNG|thumb|High vibrational energy, reaction complete|250px]] || [[File:rrpno.PNG|thumb|Low vibrational energy, reaction incomplete|250px]] ||
|}

=References=

[1]Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction, Zhaojun Zhang, Yong Zhou,† and Dong H. Zhang, dx.doi.org/10.1021/jz301649w | J. Phys. Chem. Lett. 2012, 3, 3416−3419

Rrp17

2019-05-24T13:39:13Z

Rrp17: /* Distribution of energy between different modes */

= Molecular Reaction Dynamics =

= EXERCISE 1: H + H2 system =

If H collides with a H2 molecule, there system can be defined as an AB and C.
==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V2/∂q1q12>0and the second derivative ∂V2/∂q2q22<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction. The transition state is a saddle point, it is where the minimum , q1 and max ,q2 intersect.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077 Å.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where we displayed the transition state position by r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that with the last two contour plots, the products do not form-another variable affects the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains how the rate of a chemical reaction by which the reaction passes through a transition state and it has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|center|Surface plot of the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the saddle point (where there is no momentum) was determined. AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|Finding the activation energy for the H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|Finding the activation energy for the H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|Energy vs Time |250px]]
[[File:rrpmvt.PNG|thumb|center|Momentum vs Time |250px]]

====Distribution of energy between different modes====

Polanyi's rules1 explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

{| class="wikitable" border="1"
|-
! Incomplete !! Complete !!
|-
| [[File:rrph2com.PNG|thumb| High translational energy leads to a complete reaction|250px]] || [[File:rrph2non.PNG|thumb|Low translational energy does not lead to a complete reaction|250px]] ||
|}

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

{| class="wikitable" border="1"
|-
! Complete !! inComplete !!
|-
| [[File:rrpyes.PNG|thumb|High vibrational energy, reaction complete|250px]] || [[File:rrpno.PNG|thumb|Low vibrational energy, reaction incomplete|250px]] ||
|}

=References=

Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction, Zhaojun Zhang, Yong Zhou,† and Dong H. Zhang, dx.doi.org/10.1021/jz301649w | J. Phys. Chem. Lett. 2012, 3, 3416−3419

Rrp17

2019-05-24T13:37:10Z

Rrp17: /* EXERCISE 1: H + H2 system */

= Molecular Reaction Dynamics =

= EXERCISE 1: H + H2 system =

If H collides with a H2 molecule, there system can be defined as an AB and C.
==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V2/∂q1q12>0and the second derivative ∂V2/∂q2q22<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction. The transition state is a saddle point, it is where the minimum , q1 and max ,q2 intersect.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077 Å.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where we displayed the transition state position by r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that with the last two contour plots, the products do not form-another variable affects the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains how the rate of a chemical reaction by which the reaction passes through a transition state and it has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|center|Surface plot of the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the saddle point (where there is no momentum) was determined. AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|Finding the activation energy for the H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|Finding the activation energy for the H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|Energy vs Time |250px]]
[[File:rrpmvt.PNG|thumb|center|Momentum vs Time |250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

{| class="wikitable" border="1"
|-
! Incomplete !! Complete !!
|-
| [[File:rrph2com.PNG|thumb| High translational energy leads to a complete reaction|250px]] || [[File:rrph2non.PNG|thumb|Low translational energy does not lead to a complete reaction|250px]] ||
|}

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

{| class="wikitable" border="1"
|-
! Complete !! inComplete !!
|-
| [[File:rrpyes.PNG|thumb|High vibrational energy, reaction complete|250px]] || [[File:rrpno.PNG|thumb|Low vibrational energy, reaction incomplete|250px]] ||
|}

Rrp17

2019-05-24T13:34:59Z

Rrp17: /* Transition state position */

= Molecular Reaction Dynamics =

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V2/∂q1q12>0and the second derivative ∂V2/∂q2q22<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction. The transition state is a saddle point, it is where the minimum , q1 and max ,q2 intersect.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077 Å.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where we displayed the transition state position by r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that with the last two contour plots, the products do not form-another variable affects the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains how the rate of a chemical reaction by which the reaction passes through a transition state and it has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|center|Surface plot of the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the saddle point (where there is no momentum) was determined. AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|Finding the activation energy for the H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|Finding the activation energy for the H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|Energy vs Time |250px]]
[[File:rrpmvt.PNG|thumb|center|Momentum vs Time |250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

{| class="wikitable" border="1"
|-
! Incomplete !! Complete !!
|-
| [[File:rrph2com.PNG|thumb| High translational energy leads to a complete reaction|250px]] || [[File:rrph2non.PNG|thumb|Low translational energy does not lead to a complete reaction|250px]] ||
|}

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

{| class="wikitable" border="1"
|-
! Complete !! inComplete !!
|-
| [[File:rrpyes.PNG|thumb|High vibrational energy, reaction complete|250px]] || [[File:rrpno.PNG|thumb|Low vibrational energy, reaction incomplete|250px]] ||
|}

Rrp17

2019-05-24T13:34:43Z

Rrp17: /* The approximate position of the transition state */

= Molecular Reaction Dynamics =

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V2/∂q1q12>0and the second derivative ∂V2/∂q2q22<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction. The transition state is a saddle point, it is where the minimum , q1 and max ,q2 intersect.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where we displayed the transition state position by r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that with the last two contour plots, the products do not form-another variable affects the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains how the rate of a chemical reaction by which the reaction passes through a transition state and it has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|center|Surface plot of the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the saddle point (where there is no momentum) was determined. AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|Finding the activation energy for the H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|Finding the activation energy for the H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|Energy vs Time |250px]]
[[File:rrpmvt.PNG|thumb|center|Momentum vs Time |250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

{| class="wikitable" border="1"
|-
! Incomplete !! Complete !!
|-
| [[File:rrph2com.PNG|thumb| High translational energy leads to a complete reaction|250px]] || [[File:rrph2non.PNG|thumb|Low translational energy does not lead to a complete reaction|250px]] ||
|}

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

{| class="wikitable" border="1"
|-
! Complete !! inComplete !!
|-
| [[File:rrpyes.PNG|thumb|High vibrational energy, reaction complete|250px]] || [[File:rrpno.PNG|thumb|Low vibrational energy, reaction incomplete|250px]] ||
|}

Rrp17

2019-05-24T13:32:47Z

Rrp17: /* Mechanism for the release of the reaction energy */

= Molecular Reaction Dynamics =

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V2/∂q1q12>0and the second derivative ∂V2/∂q2q22<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction. The transition state is a saddle point, it is where the minimum , q1 and max ,q2 intersect.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where we displayed the transition state position by r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that with the last two contour plots, the products do not form-another variable affects the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains how the rate of a chemical reaction by which the reaction passes through a transition state and it has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|center|Surface plot of the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|Finding the activation energy for the H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|Finding the activation energy for the H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|Energy vs Time |250px]]
[[File:rrpmvt.PNG|thumb|center|Momentum vs Time |250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

{| class="wikitable" border="1"
|-
! Incomplete !! Complete !!
|-
| [[File:rrph2com.PNG|thumb| High translational energy leads to a complete reaction|250px]] || [[File:rrph2non.PNG|thumb|Low translational energy does not lead to a complete reaction|250px]] ||
|}

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

{| class="wikitable" border="1"
|-
! Complete !! inComplete !!
|-
| [[File:rrpyes.PNG|thumb|High vibrational energy, reaction complete|250px]] || [[File:rrpno.PNG|thumb|Low vibrational energy, reaction incomplete|250px]] ||
|}

Rrp17

2019-05-24T13:32:36Z

Rrp17: /* Mechanism for the release of the reaction energy */

= Molecular Reaction Dynamics =

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V2/∂q1q12>0and the second derivative ∂V2/∂q2q22<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction. The transition state is a saddle point, it is where the minimum , q1 and max ,q2 intersect.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where we displayed the transition state position by r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that with the last two contour plots, the products do not form-another variable affects the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains how the rate of a chemical reaction by which the reaction passes through a transition state and it has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|center|Surface plot of the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|Finding the activation energy for the H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|Finding the activation energy for the H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|Energy vs Time250px]]
[[File:rrpmvt.PNG|thumb|center|Momentum vs Time 250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

{| class="wikitable" border="1"
|-
! Incomplete !! Complete !!
|-
| [[File:rrph2com.PNG|thumb| High translational energy leads to a complete reaction|250px]] || [[File:rrph2non.PNG|thumb|Low translational energy does not lead to a complete reaction|250px]] ||
|}

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

{| class="wikitable" border="1"
|-
! Complete !! inComplete !!
|-
| [[File:rrpyes.PNG|thumb|High vibrational energy, reaction complete|250px]] || [[File:rrpno.PNG|thumb|Low vibrational energy, reaction incomplete|250px]] ||
|}

Rrp17

2019-05-24T13:32:05Z

Rrp17: /* Activation energy */

= Molecular Reaction Dynamics =

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V2/∂q1q12>0and the second derivative ∂V2/∂q2q22<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction. The transition state is a saddle point, it is where the minimum , q1 and max ,q2 intersect.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where we displayed the transition state position by r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that with the last two contour plots, the products do not form-another variable affects the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains how the rate of a chemical reaction by which the reaction passes through a transition state and it has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|center|Surface plot of the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|Finding the activation energy for the H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|Finding the activation energy for the H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

{| class="wikitable" border="1"
|-
! Incomplete !! Complete !!
|-
| [[File:rrph2com.PNG|thumb| High translational energy leads to a complete reaction|250px]] || [[File:rrph2non.PNG|thumb|Low translational energy does not lead to a complete reaction|250px]] ||
|}

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

{| class="wikitable" border="1"
|-
! Complete !! inComplete !!
|-
| [[File:rrpyes.PNG|thumb|High vibrational energy, reaction complete|250px]] || [[File:rrpno.PNG|thumb|Low vibrational energy, reaction incomplete|250px]] ||
|}

Rrp17

2019-05-24T13:29:22Z

Rrp17: /* Assumptions of Transition State Theory */

= Molecular Reaction Dynamics =

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V2/∂q1q12>0and the second derivative ∂V2/∂q2q22<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction. The transition state is a saddle point, it is where the minimum , q1 and max ,q2 intersect.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where we displayed the transition state position by r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that with the last two contour plots, the products do not form-another variable affects the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains how the rate of a chemical reaction by which the reaction passes through a transition state and it has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|center|Surface plot of the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

{| class="wikitable" border="1"
|-
! Incomplete !! Complete !!
|-
| [[File:rrph2com.PNG|thumb| High translational energy leads to a complete reaction|250px]] || [[File:rrph2non.PNG|thumb|Low translational energy does not lead to a complete reaction|250px]] ||
|}

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

{| class="wikitable" border="1"
|-
! Complete !! inComplete !!
|-
| [[File:rrpyes.PNG|thumb|High vibrational energy, reaction complete|250px]] || [[File:rrpno.PNG|thumb|Low vibrational energy, reaction incomplete|250px]] ||
|}

Rrp17

2019-05-24T13:27:52Z

Rrp17: /* How is the transition state defined? */

= Molecular Reaction Dynamics =

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V2/∂q1q12>0and the second derivative ∂V2/∂q2q22<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction. The transition state is a saddle point, it is where the minimum , q1 and max ,q2 intersect.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where we displayed the transition state position by r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that with the last two contour plots, the products do not form-another variable affects the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains how the rate of a chemical reaction by which the reaction passes through a transition state and it has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|center|Surface plot of the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

{| class="wikitable" border="1"
|-
! Incomplete !! Complete !!
|-
| [[File:rrph2com.PNG|thumb| High translational energy leads to a complete reaction|250px]] || [[File:rrph2non.PNG|thumb|Low translational energy does not lead to a complete reaction|250px]] ||
|}

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

{| class="wikitable" border="1"
|-
! Complete !! inComplete !!
|-
| [[File:rrpyes.PNG|thumb|High vibrational energy, reaction complete|250px]] || [[File:rrpno.PNG|thumb|Low vibrational energy, reaction incomplete|250px]] ||
|}

Rrp17

2019-05-24T13:27:40Z

Rrp17: /* How is the transition state defined? */

= Molecular Reaction Dynamics =

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V2/∂q1q12<0and the second derivative ∂V2/∂q2q22<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction. The transition state is a saddle point, it is where the minimum , q1 and max ,q2 intersect.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where we displayed the transition state position by r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that with the last two contour plots, the products do not form-another variable affects the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains how the rate of a chemical reaction by which the reaction passes through a transition state and it has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|center|Surface plot of the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

{| class="wikitable" border="1"
|-
! Incomplete !! Complete !!
|-
| [[File:rrph2com.PNG|thumb| High translational energy leads to a complete reaction|250px]] || [[File:rrph2non.PNG|thumb|Low translational energy does not lead to a complete reaction|250px]] ||
|}

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

{| class="wikitable" border="1"
|-
! Complete !! inComplete !!
|-
| [[File:rrpyes.PNG|thumb|High vibrational energy, reaction complete|250px]] || [[File:rrpno.PNG|thumb|Low vibrational energy, reaction incomplete|250px]] ||
|}

Rrp17

2019-05-24T13:24:09Z

Rrp17: /* How is the transition state defined? */

= Molecular Reaction Dynamics =

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction. The transition state is a saddle point, it is where the minimum , q1 and max ,q2 intersect.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where we displayed the transition state position by r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that with the last two contour plots, the products do not form-another variable affects the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains how the rate of a chemical reaction by which the reaction passes through a transition state and it has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|center|Surface plot of the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

{| class="wikitable" border="1"
|-
! Incomplete !! Complete !!
|-
| [[File:rrph2com.PNG|thumb| High translational energy leads to a complete reaction|250px]] || [[File:rrph2non.PNG|thumb|Low translational energy does not lead to a complete reaction|250px]] ||
|}

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

{| class="wikitable" border="1"
|-
! Complete !! inComplete !!
|-
| [[File:rrpyes.PNG|thumb|High vibrational energy, reaction complete|250px]] || [[File:rrpno.PNG|thumb|Low vibrational energy, reaction incomplete|250px]] ||
|}

Rrp17

2019-05-24T13:21:18Z

Rrp17: /* Distribution of energy between different modes */

= Molecular Reaction Dynamics =

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where we displayed the transition state position by r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that with the last two contour plots, the products do not form-another variable affects the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains how the rate of a chemical reaction by which the reaction passes through a transition state and it has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|center|Surface plot of the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

{| class="wikitable" border="1"
|-
! Incomplete !! Complete !!
|-
| [[File:rrph2com.PNG|thumb| High translational energy leads to a complete reaction|250px]] || [[File:rrph2non.PNG|thumb|Low translational energy does not lead to a complete reaction|250px]] ||
|}

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

{| class="wikitable" border="1"
|-
! Complete !! inComplete !!
|-
| [[File:rrpyes.PNG|thumb|High vibrational energy, reaction complete|250px]] || [[File:rrpno.PNG|thumb|Low vibrational energy, reaction incomplete|250px]] ||
|}

Rrp17

2019-05-24T13:20:07Z

Rrp17: /* Distribution of energy between different modes */

= Molecular Reaction Dynamics =

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where we displayed the transition state position by r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that with the last two contour plots, the products do not form-another variable affects the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains how the rate of a chemical reaction by which the reaction passes through a transition state and it has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|center|Surface plot of the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

{| class="wikitable" border="1"
|-
! Incomplete !! Complete !!
|-
| [[File:rrph2com.PNG|thumb| High translational energy leads to a complete reaction|250px]] ||
[[File:rrph2non.PNG|thumb|Low translational energy does not lead to a complete reaction|250px]]

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

{| class="wikitable" border="1"
|-
! Complete !! inComplete !!
|-
| [[File:rrpyes.PNG|thumb|High vibrational energy, reaction complete|250px]] ||
[[File:rrpno.PNG|thumb|Low vibrational energy, reaction incomplete|250px]]

Rrp17

2019-05-24T13:18:46Z

Rrp17: /* Distribution of energy between different modes */

= Molecular Reaction Dynamics =

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where we displayed the transition state position by r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that with the last two contour plots, the products do not form-another variable affects the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains how the rate of a chemical reaction by which the reaction passes through a transition state and it has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|center|Surface plot of the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

[[File:rrph2non.PNG|thumb|Low translational energy does not lead to a complete reaction|250px]]

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.
[[File:rrpyes.PNG|thumb|High vibrational energy, reaction complete|250px]]
[[File:rrpno.PNG|thumb|Low vibrational energy, reaction incomplete|250px]]

{| class="wikitable" border="1"
|-
! Incomplete !! Complete !!
|-
| [[File:rrph2com.PNG|thumb| High translational energy leads to a complete reaction|250px]] ||
[[File:rrph2non.PNG|thumb|Low translational energy does not lead to a complete reaction|250px]]

Rrp17

2019-05-24T13:17:31Z

Rrp17: /* Assumptions of Transition State Theory */

= Molecular Reaction Dynamics =

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where we displayed the transition state position by r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that with the last two contour plots, the products do not form-another variable affects the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains how the rate of a chemical reaction by which the reaction passes through a transition state and it has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|center|Surface plot of the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

[[File:rrph2com.PNG|thumb| High translational energy leads to a complete reaction|250px]]
[[File:rrph2non.PNG|thumb|Low translational energy does not lead to a complete reaction|250px]]

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.
[[File:rrpyes.PNG|thumb|High vibrational energy, reaction complete|250px]]
[[File:rrpno.PNG|thumb|Low vibrational energy, reaction incomplete|250px]]

Rrp17

2019-05-24T13:14:52Z

Rrp17: /* Molecular Reaction Dynamics */

= Molecular Reaction Dynamics =

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where we displayed the transition state position by r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that with the last two contour plots, the products do not form-another variable affects the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|center|Surface plot of the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

[[File:rrph2com.PNG|thumb| High translational energy leads to a complete reaction|250px]]
[[File:rrph2non.PNG|thumb|Low translational energy does not lead to a complete reaction|250px]]

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.
[[File:rrpyes.PNG|thumb|High vibrational energy, reaction complete|250px]]
[[File:rrpno.PNG|thumb|Low vibrational energy, reaction incomplete|250px]]

Rrp17

2019-05-24T12:37:06Z

Rrp17: /* Exercise 2: F-H-H system */

= Molecular Reaction Dynamics =

This question concerns an atom-diatom collision. By following the trajectory between

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where we displayed the transition state position by r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that with the last two contour plots, the products do not form-another variable affects the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|center|Surface plot of the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

[[File:rrph2com.PNG|thumb| High translational energy leads to a complete reaction|250px]]
[[File:rrph2non.PNG|thumb|Low translational energy does not lead to a complete reaction|250px]]

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.
[[File:rrpyes.PNG|thumb|High vibrational energy, reaction complete|250px]]
[[File:rrpno.PNG|thumb|Low vibrational energy, reaction incomplete|250px]]

Rrp17

2019-05-24T12:36:51Z

Rrp17: /* Exercise 2: F-H-H system */

= Molecular Reaction Dynamics =

This question concerns an atom-diatom collision. By following the trajectory between

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where we displayed the transition state position by r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that with the last two contour plots, the products do not form-another variable affects the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|center|Surface plotog the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

[[File:rrph2com.PNG|thumb| High translational energy leads to a complete reaction|250px]]
[[File:rrph2non.PNG|thumb|Low translational energy does not lead to a complete reaction|250px]]

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.
[[File:rrpyes.PNG|thumb|High vibrational energy, reaction complete|250px]]
[[File:rrpno.PNG|thumb|Low vibrational energy, reaction incomplete|250px]]

Rrp17

2019-05-24T12:36:20Z

Rrp17: /* Reactive and nonreactive trajectories */

= Molecular Reaction Dynamics =

This question concerns an atom-diatom collision. By following the trajectory between

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where we displayed the transition state position by r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that with the last two contour plots, the products do not form-another variable affects the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|Surface plotog the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

[[File:rrph2com.PNG|thumb| High translational energy leads to a complete reaction|250px]]
[[File:rrph2non.PNG|thumb|Low translational energy does not lead to a complete reaction|250px]]

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.
[[File:rrpyes.PNG|thumb|High vibrational energy, reaction complete|250px]]
[[File:rrpno.PNG|thumb|Low vibrational energy, reaction incomplete|250px]]

Rrp17

2019-05-24T12:35:31Z

Rrp17: /* Minimum energy path and trajectory */

= Molecular Reaction Dynamics =

This question concerns an atom-diatom collision. By following the trajectory between

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where we displayed the transition state position by r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that in the last two contour plots, the products did not form- something else must affect the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|Surface plotog the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

[[File:rrph2com.PNG|thumb| High translational energy leads to a complete reaction|250px]]
[[File:rrph2non.PNG|thumb|Low translational energy does not lead to a complete reaction|250px]]

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.
[[File:rrpyes.PNG|thumb|High vibrational energy, reaction complete|250px]]
[[File:rrpno.PNG|thumb|Low vibrational energy, reaction incomplete|250px]]

Rrp17

2019-05-23T16:15:47Z

Rrp17: /* Distribution of energy between different modes */

= Molecular Reaction Dynamics =

This question concerns an atom-diatom collision. By following the trajectory between

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that in the last two contour plots, the products did not form- something else must affect the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|Surface plotog the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

[[File:rrph2com.PNG|thumb| High translational energy leads to a complete reaction|250px]]
[[File:rrph2non.PNG|thumb|Low translational energy does not lead to a complete reaction|250px]]

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.
[[File:rrpyes.PNG|thumb|High vibrational energy, reaction complete|250px]]
[[File:rrpno.PNG|thumb|Low vibrational energy, reaction incomplete|250px]]

Rrp17

2019-05-23T16:15:32Z

Rrp17: /* Distribution of energy between different modes */

= Molecular Reaction Dynamics =

This question concerns an atom-diatom collision. By following the trajectory between

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that in the last two contour plots, the products did not form- something else must affect the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|Surface plotog the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

[[File:rrph2com.PNG| High translational energy leads to a complete reaction|250px]]
[[File:rrph2non.PNG|Low translational energy does not lead to a complete reaction|250px]]

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.
[[File:rrpyes.PNG|thumb|High vibrational energy, reaction complete|250px]]
[[File:rrpno.PNG|thumb|Low vibrational energy, reaction incomplete|250px]]

File:Rrpno.PNG

2019-05-23T16:15:02Z

Rrp17:

Rrp17

2019-05-23T16:14:44Z

Rrp17: /* Distribution of energy between different modes */

= Molecular Reaction Dynamics =

This question concerns an atom-diatom collision. By following the trajectory between

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that in the last two contour plots, the products did not form- something else must affect the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|Surface plotog the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

[[File:rrph2com.PNG| High translational energy leads to a complete reaction|250px]]
[[File:rrph2non.PNG|Low translational energy does not lead to a complete reaction|250px]]

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.
[[File:rrpyes.PNG|High vibrational energy, reaction complete|250px]]
[[File:rrpno.PNG|Low vibrational energy, reaction incomplete|250px]]

Rrp17

2019-05-23T16:12:58Z

Rrp17: /* Distribution of energy between different modes */

= Molecular Reaction Dynamics =

This question concerns an atom-diatom collision. By following the trajectory between

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that in the last two contour plots, the products did not form- something else must affect the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|Surface plotog the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

[[File:rrph2com.PNG| High translational energy leads to a complete reaction|250px]]
[[File:rrph2non.PNG|Low translational energy does not lead to a complete reaction|250px]]

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.
[[File:rrpyes.PNG]]

File:Rrpyes.PNG

2019-05-23T16:12:40Z

Rrp17:

File:Rrph2non.PNG

2019-05-23T16:02:17Z

Rrp17:

File:Rrph2com.PNG

2019-05-23T16:00:48Z

Rrp17:

Rrp17

2019-05-23T15:46:27Z

Rrp17: /* The approximate position of the transition state */

= Molecular Reaction Dynamics =

This question concerns an atom-diatom collision. By following the trajectory between

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that in the last two contour plots, the products did not form- something else must affect the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|Surface plotog the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|Contour plot of trajectory| 250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen that when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

Rrp17

2019-05-23T15:45:17Z

Rrp17: /* Exercise 2: F-H-H system */

= Molecular Reaction Dynamics =

This question concerns an atom-diatom collision. By following the trajectory between

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that in the last two contour plots, the products did not form- something else must affect the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|Surface plotog the F-H-H system with AB=HF and BC=H2|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen that when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

Rrp17

2019-05-23T15:43:36Z

Rrp17: /* Distribution of energy between different modes */

= Molecular Reaction Dynamics =

This question concerns an atom-diatom collision. By following the trajectory between

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that in the last two contour plots, the products did not form- something else must affect the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The rHH momentum corresponds to the vibrational energy and the rHF corresponds to the translational energy.

-For the H2 + F reaction, it can be seen that when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

Rrp17

2019-05-23T15:25:53Z

Rrp17: /* The approximate position of the transition state */

= Molecular Reaction Dynamics =

This question concerns an atom-diatom collision. By following the trajectory between

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that in the last two contour plots, the products did not form- something else must affect the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|250px]] || [[File:rrpzm.PNG|thumb| Trajectory zoomed in]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa.

-For the H2 + F reaction, the rHH momentum correspond to the vibrational energy and the rHF corresponds to the translational energy.
This can be shown with the contour plos below. It can be seen that when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.,

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

Rrp17

2019-05-23T15:25:27Z

Rrp17: /* The approximate position of the transition state */

= Molecular Reaction Dynamics =

This question concerns an atom-diatom collision. By following the trajectory between

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that in the last two contour plots, the products did not form- something else must affect the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpts.PNG|thumb|center|250px]] || [[File:rrpzm.PNG|thumb]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa.

-For the H2 + F reaction, the rHH momentum correspond to the vibrational energy and the rHF corresponds to the translational energy.
This can be shown with the contour plos below. It can be seen that when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.,

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

Rrp17

2019-05-23T15:25:08Z

Rrp17: /* The approximate position of the transition state */

= Molecular Reaction Dynamics =

This question concerns an atom-diatom collision. By following the trajectory between

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that in the last two contour plots, the products did not form- something else must affect the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| File:rrpts.PNG|thumb|center|250px]] || [[File:rrpzm.PNG|thumb]]
|}
[[File:rrpzm.PNG|thumb]]

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa.

-For the H2 + F reaction, the rHH momentum correspond to the vibrational energy and the rHF corresponds to the translational energy.
This can be shown with the contour plos below. It can be seen that when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.,

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

Rrp17

2019-05-23T15:24:47Z

Rrp17: /* The approximate position of the transition state */

= Molecular Reaction Dynamics =

This question concerns an atom-diatom collision. By following the trajectory between

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that in the last two contour plots, the products did not form- something else must affect the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| File:rrpts.PNG|thumb|center|250px]] || [[File:rrpzm.PNG|thumb]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa.

-For the H2 + F reaction, the rHH momentum correspond to the vibrational energy and the rHF corresponds to the translational energy.
This can be shown with the contour plos below. It can be seen that when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.,

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

Rrp17

2019-05-23T15:24:34Z

Rrp17: /* The approximate position of the transition state */

= Molecular Reaction Dynamics =

This question concerns an atom-diatom collision. By following the trajectory between

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that in the last two contour plots, the products did not form- something else must affect the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| File:rrpts.PNG|thumb|center|250px]] || [[File:rrpzm.PNG|250px]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa.

-For the H2 + F reaction, the rHH momentum correspond to the vibrational energy and the rHF corresponds to the translational energy.
This can be shown with the contour plos below. It can be seen that when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.,

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

Rrp17

2019-05-23T15:24:13Z

Rrp17: /* The approximate position of the transition state */

= Molecular Reaction Dynamics =

This question concerns an atom-diatom collision. By following the trajectory between

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that in the last two contour plots, the products did not form- something else must affect the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| File:rrpts.PNG|thumb|center|250px]] || [[File:rrpzm.PNG|thumb|center|250px]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa.

-For the H2 + F reaction, the rHH momentum correspond to the vibrational energy and the rHF corresponds to the translational energy.
This can be shown with the contour plos below. It can be seen that when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.,

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

File:Rrpts.PNG

2019-05-23T15:23:37Z

Rrp17: Rrp17 uploaded a new version of File:Rrpts.PNG

Rrp17

2019-05-23T15:21:48Z

Rrp17: /* Activation energy */

= Molecular Reaction Dynamics =

This question concerns an atom-diatom collision. By following the trajectory between

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that in the last two contour plots, the products did not form- something else must affect the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpzm.PNG|thumb|center|250px]] || File:rrpts.PNG|thumb|center|250px]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrphfa.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa.

-For the H2 + F reaction, the rHH momentum correspond to the vibrational energy and the rHF corresponds to the translational energy.
This can be shown with the contour plos below. It can be seen that when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.,

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

File:Rrphfa.PNG

2019-05-23T15:21:38Z

Rrp17:

Rrp17

2019-05-23T15:21:01Z

Rrp17: /* Activation energy */

= Molecular Reaction Dynamics =

This question concerns an atom-diatom collision. By following the trajectory between

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that in the last two contour plots, the products did not form- something else must affect the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpzm.PNG|thumb|center|250px]] || File:rrpts.PNG|thumb|center|250px]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]
[[File:rrp.PNG|thumb|center|The H + HF reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa.

-For the H2 + F reaction, the rHH momentum correspond to the vibrational energy and the rHF corresponds to the translational energy.
This can be shown with the contour plos below. It can be seen that when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.,

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

Rrp17

2019-05-23T15:19:54Z

Rrp17: /* Activation energy */

= Molecular Reaction Dynamics =

This question concerns an atom-diatom collision. By following the trajectory between

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that in the last two contour plots, the products did not form- something else must affect the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpzm.PNG|thumb|center|250px]] || File:rrpts.PNG|thumb|center|250px]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa.

-For the H2 + F reaction, the rHH momentum correspond to the vibrational energy and the rHF corresponds to the translational energy.
This can be shown with the contour plos below. It can be seen that when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.,

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

Rrp17

2019-05-23T15:19:39Z

Rrp17: /* Activation energy */

= Molecular Reaction Dynamics =

This question concerns an atom-diatom collision. By following the trajectory between

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that in the last two contour plots, the products did not form- something else must affect the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpzm.PNG|thumb|center|250px]] || File:rrpts.PNG|thumb|center|250px]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|The H2 + F reaction|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa.

-For the H2 + F reaction, the rHH momentum correspond to the vibrational energy and the rHF corresponds to the translational energy.
This can be shown with the contour plos below. It can be seen that when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.,

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

Rrp17

2019-05-23T15:19:00Z

Rrp17: /* Mechanism for the release of the reaction energy */

= Molecular Reaction Dynamics =

This question concerns an atom-diatom collision. By following the trajectory between

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that in the last two contour plots, the products did not form- something else must affect the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpzm.PNG|thumb|center|250px]] || File:rrpts.PNG|thumb|center|250px]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]
[[File:rrpmvt.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa.

-For the H2 + F reaction, the rHH momentum correspond to the vibrational energy and the rHF corresponds to the translational energy.
This can be shown with the contour plos below. It can be seen that when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.,

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

File:Rrpmvt.PNG

2019-05-23T15:18:47Z

Rrp17:

Rrp17

2019-05-23T15:16:35Z

Rrp17: /* Mechanism for the release of the reaction energy */

= Molecular Reaction Dynamics =

This question concerns an atom-diatom collision. By following the trajectory between

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that in the last two contour plots, the products did not form- something else must affect the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpzm.PNG|thumb|center|250px]] || File:rrpts.PNG|thumb|center|250px]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HF momentum of -2.2
HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa.

-For the H2 + F reaction, the rHH momentum correspond to the vibrational energy and the rHF corresponds to the translational energy.
This can be shown with the contour plos below. It can be seen that when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.,

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

Rrp17

2019-05-23T15:02:31Z

Rrp17: /* Activation energy */

= Molecular Reaction Dynamics =

This question concerns an atom-diatom collision. By following the trajectory between

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that in the last two contour plots, the products did not form- something else must affect the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpzm.PNG|thumb|center|250px]] || File:rrpts.PNG|thumb|center|250px]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrphf.PNG|thumb|center|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with bomb calorimetry.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HH momentum of -1.9
HF momentum of -2.2

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa.

-For the H2 + F reaction, the rHH momentum correspond to the vibrational energy and the rHF corresponds to the translational energy.
This can be shown with the contour plos below. It can be seen that when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.,

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

Rrp17

2019-05-23T15:00:25Z

Rrp17: /* Exercise 2: F-H-H system */

= Molecular Reaction Dynamics =

This question concerns an atom-diatom collision. By following the trajectory between

= EXERCISE 1: H + H2 system =

==== How is the transition state defined? ====

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(ri)/∂ri=0, the gradient of the potential energy is zero and the distance ri = rAB = rBC .

[[File:rrps_plot.PNG|thumb|center|Plot 1:The surface plot of the reaction between H1-H2 and H3. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.|250px]]

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V/∂rBC=0 and the second derivative ∂V2/∂qBC2<0
This would give you a maximum however, a tangent perpendicular to the qBC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction.

==== Transition state position ====

H + H2 is symmetric and hence, the transition state must have r1 = r2, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (rts) was found to be 0.9077.
This was determined by choosing a random value of r1 and changing the inter-nuclear distance until the force equalled zero.

[[File:rrp17_i_t.PNG|thumb|center|Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.|200px ]]

====Minimum energy path and trajectory====

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where r1=rts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

{| class="wikitable" border="1"
|-
! MEP !! Dynamic !!
|-
| [[File:Rrpep.PNG|thumb|Plot 3 The MEP surface plot.|250px ]] || [[File:Rrpyn.PNG|thumb|Plot 4 The trajectory surface plot|250px]]
|}

====Reactive and nonreactive trajectories====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy (Kcal/mol) !! Reactive? !! Trajectory contour plot !! Dynamic description
|-
| -1.25 || -2.5 || -99.018 || yes || [[File:rrpc1.PNG|thumb]] ||The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
|-
| -1.5 || -2 || -100.456 || no ||[[File:RrpC2.PNG|thumb]] ||The trajectory does not pass through the activation barrier as there is not enough energy
|-
| -1.5 || -2.5 || -98.956 || yes || [[File:rrpc3.PNG|thumb]] ||The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
|-
| -2.5 || -5 || -84.956 || no || [[File:rrpc4.PNG|thumb]] || The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
|-
| -2.5 || -5.2 || -83.416 || yes || [[File:rrpc5.PNG|thumb]] || the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.
|}

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that in the last two contour plots, the products did not form- something else must affect the outcome of the reaction other than activation energy!

====Assumptions of Transition State Theory====

The Transition State Theory explains the rate of an elementary chemical reaction which the reaction passing through a transition state/complex and has three assumptions:

-The energy of the particles follow a Boltzmann distribution
-The reactants and transition state structure is in equilibrium
-The transition state structure does not revert back to the reactants once the reactants form the transition state

=Exercise 2: F-H-H system=

Using the energy surface, the F + H2 is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

[[File:rrpf.PNG|thumb|250px]]

====The approximate position of the transition state====

Using a MEP contour plot, the point at which the trajectory stays in the transition state the transition state was found to be AB=1.908 Å , BC=0.745 Å

{| class="wikitable" border="1"
|-
! Contour Plot !! Zoom of Contour !!
|-
| [[File:rrpzm.PNG|thumb|center|250px]] || File:rrpts.PNG|thumb|center|250px]]
|}

====Activation energy====

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H2 + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

[[File:rrph2.PNG|thumb|center|250px]]
[[File:rrphf.PNG|thumb|center|250px]]

====Mechanism for the release of the reaction energy====

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with bomb calorimetry.

The initial conditions that result in a reactive trajectory:
HH distance of 0.74
HF distance of 2.3
HH momentum of -1.9
HF momentum of -2.2

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

[[File:rrpcons.PNG|thumb|center|250px]]

====Distribution of energy between different modes====

Polanyi's rules explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa.

-For the H2 + F reaction, the rHH momentum correspond to the vibrational energy and the rHF corresponds to the translational energy.
This can be shown with the contour plos below. It can be seen that when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.,

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.