ChemWiki - User contributions [en]

Talk:MRD:LaraMetcalf

2017-05-31T16:42:58Z

Je714:

Good report, with extra retrieval of information, which is nice! See my inline comments for suggestions.

[[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:42, 31 May 2017 (BST)

Talk:MRD:LaraMetcalf

2017-05-31T16:42:35Z

Je714: Created page with "Good report. See my inline comments for suggestions."

Good report. See my inline comments for suggestions.

MRD:LaraMetcalf

2017-05-31T16:42:13Z

Je714: /* Polanyi's empirical rules */

= Molecular Reaction Dynamics Lab =
==EXERCISE 1: H + H2 system==

A potential energy surface (PES) is a mathematical function which describes the electronic potential energy of a system as a function of the relative atomic positions in space and is widely used as a tool to carry out theoretical studies on molecular reaction dynamics <ref name="a" />.

===Dynamics from the transition state region===

The total gradient of the potential energy surface at a minimum and transition structure is equal to zero. The distinction between the two structures comes from the sign corresponding to their rate of change of gradient. The potential energy second derivative is positive for a minima and negative for a transition state <ref name="ref1" />. {{fontcolor1|gray| Which one? Along which direction? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:36, 31 May 2017 (BST)}}
The transition structure linking the two minima is a maximum along the minima energy reaction pathway and thus is called a saddle point <ref name="ref2" />.
This can be expressed mathematically by Taylor's Theorem <ref name="ref3" />:
:If fxxfyy − fx2y < 0 at (a, b) then (a, b) is a saddle point.
:If fxx >0 and fyy > 0 at (a,b) then (a,b) is a minimum point.

{{fontcolor1|gray| At the TS one of the partial 2nd derivatives will be >0 along one of the variables, and <0 along all the rest. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:36, 31 May 2017 (BST)}}

===Trajectories from r1 = r2: locating the transition state===

The best estimate of the transition state position rts is found to be equal to 0.90775 Å. The detection of the transition structure can be observed in an internuclear distances vs time plot. The transition state corresponds to where the gradient of the potential energy surface is zero, so it is expected that at rts the H atoms will be stationary - their internuclear separations will remain constant over time (fig 1). At a distance of 0.90775 Å the trajectory disappears (fig 2) confirming this to be a good approximation of the transition state position. Varying the interatomic distance away from the transition state position results in a trajectory that rolls back toward the reactants/products and thus a changing internuclear distance over time is observed (fig 3). {{fontcolor1|gray| Really? That's not what figure 3 is showing. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:37, 31 May 2017 (BST)}}

[[File:lem215_fig1.png|thumb|center|Figure 1 - internuclear distance vs time plot at the transition state position.]]
[[File:lem215_fig2.png|thumb|center|Figure 2 - disappearance of the trajectory at the transition state position.]]
[[File:lem215_fig3.png|thumb|center|Figure 3 - internuclear distance vs time plot at r=1.5 A.]]
[[File:lem215_fig4.png|thumb|center|Figure 4 - surface plot displaying the trajectory at r=1.5 A.]]

===Minimum Energy Path (MEP) and Dynamic Calculations===

Both calculations produce trajectories that rolls from the reactants H1-H2 H3 in the direction of increasing r1 towards H1 H2-H3 (r1 = A-B) Upon energy minimisation the trajectory converges to the MEP <ref name="ref4" />.
The MEP calculation generates the minimum energy trajectory thus is restricted to only one degree of freedom, the interatomic positions r1 and r2. The exchange between kinetic and potential energy (due to vibrations) does not occur as a result of the minimisation, and so the generated trajectory does not have an oscillative nature - the kinetic energy is fixed at zero.
On the contrary the dynamics calculation does not involve such limitations allowing it to have more degrees of freedom generating an oscillating trajectory, signifying influence of a non-zero kinetic energy.

[[File:lem215_mep.png|thumb|center|MEP Calculation Trajectory.]]
[[File:lem215_dynamics.png|thumb|center|Dynamic Trajectory Calculation.]]

At large t (t): (3 decimal places)
:r1(1.485)= 5.247 Å
:r2(1.485) = 0.739 Å
:p1(1.45) = 2.481
:p2(1.45) = 1.260

Reverting the initial conditions so that r1 = rts + 0.01 and r2 = rts then the reaction would have a trajectory that rolls from H1 H2-H3 towards H1-H2 H3 in the direction of increasing r2.

If the initial positions and final positions were swapped then the corresponding values for final momenta will have the opposite sign.

===Reactive and unreactive trajectories===

A series of initial condition were tested to determine whether or not they produce a reactive trajectory that starts from the reactants HA-HB and dissociated atom HC to the molecular product HB-HC with dissociated atom HA. Initial momenta p1 of atom HC approaching HB before the transition state and final momenta p2 of atom HA moving away from the molecular product HB-HC

{| class="wikitable" border="1" style="margin-left: auto; margin-right: auto; border: none; width: 60%; height: 200px"
|+ Initial Conditions
! p1 !! p2 !! Reactivity
|- style="text-align: center;"
| -1.25 || -2.5 || Reactive
|- style="text-align: center;"
| -1.5 || -2.0 || Unreactive
|- style="text-align: center;"
| -1.5 || -2.5 || Reactive
|- style="text-align: center;"
| -2.5 || -5 || Unreactive
|- style="text-align: center;"
| -2.5 || -5.2 || Reactive
|}

[[File:lem215_fig5.png|thumb|center|Figure 5 - initial conditions p1=-1.25, p2=-2.5.]]
The first set of conditions generates a linear trajectory path from HA-HB HC to the transition structure at approximately 0.908 Å. Before the collision r1 remains constant and r2 steadily decreases as H3 approaches H2. There is sufficient energy to overcome the activation energy barrier allowing the trajectory to roll towards the products. After the transition structure r1 steadily increases as HA moves away from HB from the molecule formed HB-HC, the oscillating trajectory signifies the vibrational energy of the HB-HC bond.

[[File:lem215_fig6.png|thumb|center|Figure 6 - initial conditions p1=-1.5, p2=-2.0.]]
Increasing p1 and p2 (initial momenta), relative to the first set of conditions, results in an unreactive trajectory. Before the transition structure r1 has a slight oscillation (vibrational energy of A-B) and r2 steadily decreases as HC approaches HB. At r1=0.810 Å and r2=1.120 Å the trajectory bounces back toward the reactants. This suggests amount of energy available to the system is not sufficient to overcome the activation energy barrier - the reaction is unsuccessful. It is evident that the vibrational energy of HA-HB is retained upon the return of the system to the reactants.

[[File:lem215_fig7.png|thumb|center|Figure 7 - initial conditions p1=-1.5, p2=-2.5.]]
Increasing p2 from -2.0 to -2.5 provides the energy required to overcome the barrier thus allowing the reaction to proceed. The trajectory rolls from the reactants to the products passing through the transition structure at 0.893 Å. Again the vibrational energy of the HA-HB then HB-HC bonds are observed before and after passing through the transition structure respectively.

[[File:lem215_fig8.png|thumb|center|Figure 8 - initial conditions p1=-2.5, p2=-5.0.]]
With the fourth set of initial momenta conditions the trajectory rolls from the reactants and passes through the transition structure illustrating an initial formation of the HB-HC bond. Soon after crossing the transition region the trajectory returns to the reactants - this is an example barrier re-crossing. The trajectory bounces several times within a small region about transition state position before returning back to the reactants.

[[File:lem215_fig9.png|thumb|center|Figure 9 - initial conditions p1=-2.5, p2=-5.2.]]
The trajectory approaches the transition structure along the direction of decreasing r2 at a constant r1. Again the trajectory displays barrier re-crossing before eventually bouncing across the barrier toward the product HB-HC and dissociated atom HA.

The final two sets of initial momenta have relatively large p2 values. Both systems have significant translational kinetic energy as H atom A moves away from the formed molecule HB-HC. At high kinetic translational energies of atoms/molecules it becomes increasingly difficult to justify the differences in the reactivity of the generated trajectories. This suggests the application of Transition State Theory is unsuitable for dynamic molecular systems of this type - this method is limiting.

====Transition State Theory====

The transition state is the known as the 'critical configuration' that corresponds to the highest point on a minimum energy reaction pathway on a potential energy surface <ref name="ref5" />. If the system is able to reach the critical geometry then a successful reaction is highly probable. The transition structure is a stationary point thus is identified as the saddle point, with associated reaction coordinates, on a potential energy surface.
The transition state theory is a cumulative a series of assumptions used to approximate the equilibrium rate constants for thermally driven molecular reactions <ref name="t" />. The TST stemmed from thermodynamic, kinetic-theory and statistical mechanic treatments <ref name="t" />.

=====Development of the Transition State Theory=====

The thermodynamic treatment stems from the Van't Hoff equation <ref name="t" /> which relates the change in the equilibrium constant (keq) to a given change in temperature (T) for a given enthalpy change (delta H):
:<math>\ln K_\mathrm{eq} = - \frac{{\Delta H^\ominus}}{RT}+ \frac{{\Delta S^\ominus }}{R}.</math>.
The kinetic-theory <ref name="t" /> explains the temperature dependence of reaction rates as a function of collision frequency and the statistical mechanical treatment focuses on equilibrium reaction rates in terms of the motion of molecules and statistical distribution of molecular speeds <ref name="t" /> - initially modelled by Maxwell and Boltzmann.

=====Main Features and Assumptions of the Transition State Theory <ref name="t"/> =====

:(1) Rates can be calculated by focusing on the "activated complex" - what happens before the transition state is reached are not important if one rate dominates the overall rate of reaction.
:(2) The activated complex is formed in 'quasi-equilibrium' i.e. the complex is in equilibrium with the reactants and products passing through - the concentrations of reactants and products can be determined by changing the concentration of one species e.g. removing the products.
:(3) Motion of the system is treated as 'free translational motion' - expressed by kinetic-theory.

=====Using the Transition State Theory to Determine Reactivity=====

The TST ignores the dynamical recrossings that can lead back to the reactants (also known as tunneling) so this calculation tends to overestimate the rates within the classical mechanic framework <ref name ="t"/>. For a successful reaction the molecular system must have sufficient energy to overcome the activation energy barrier and thus generate a reactive trajectory. The Arrhenius equation relates the activate energy to the reaction rate for a particular reaction scheme, k=Ae(-Ea/RT) <ref name ="ar"/>. The relationship shows how a small activation energy barrier will correspond to a larger reaction rate constant - the reaction will go faster. It is likely that the experimental activation energy barrier will be larger than the calculated value thus TST will likely overestimate the rate relative the the experimental rate for a given reaction.

{{fontcolor1|gray| Good job looking up information. Just a small point -- the statistical mechanical treatment does not really hold here, since we are not looking at an ensemble of particles, just at a simple triatomic collision in isolation. The biggest point to be made here, was the recrossing limitation, and how that overestimates the rates when compared to experiment. Because that's the most relevant limitation that you observe in this lab. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:40, 31 May 2017 (BST)}}

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

===Reaction Energetics===

The F + H2 reaction is exothermic. An exothermic reaction is defined as the reaction path from a higher (reactants) to lower total potential energy (products) <ref name="ref6"/>. The trajectory shows the approach of F to H2 from infinite separation (as it is dissociated): as A-B separation decreases the potential energy decreases to a minimum indicating the H-F bond formation and H-H bond dissociation is an exothermic process.
[[File:lem215_1.png|thumb|center|Figure 10 - Surface Plot Trajectory from F + H2 to F-H + H.]]

By definition the reverse of an exothermic reaction is endothermic <ref name="ref6"/>, this is indicated by the trajectory generate by the H-F + H reaction. H-F bond breaking and H-H bond forming is an endothermic and enthalpically unfavourable process. The reactants have a lower total potential energy than the products.
[[File:lem215_2.png|thumb|center|Figure 11 - Surface Plot Trajectory from F-H + H to F + H2]]

The H-F bond formation has a more negative (exothermic) reaction enthalpy indicating H-F to have a stronger bond than H2.

===Location of Transition State and Determination of the Activation Energies===

Rates of reactions are controlled by the free energy of the transition state and Hammond postulated that the structure of a given transition state will more closely resemble that of the reactants or products depending on which is closer in energy <ref name="h"/>. He concluded that for an exothermic reaction the transition state will be more structurally similar to the reactants i.e. the interconversion between the reactants and transition state will involve only a small reorganization of molecular structure. And conversely for an endothermic reaction the transition state will be more structurally related to the products.
The transition structure is expected to be closer in energy and structure to the reactants - longer A-B and shorter B-C distance so point of intersection further right in the saddle region was selected to determine the location (see figure 12) and energy of the transition state.

Approximate location of the transition state (3 dp):
:A-B (F-H) = 1.713 Å
:B-C (H-H) = 0.748 Å

[[File:lem215_Hpost.png|thumb|center|Figure 12 - Locating the transition state for a F-H-H system.]]

Activation energies (2 dp):

[[File:lem215_Hpost1.png|thumb|center|Figure 13 - Determination of the absolute energy for products (F-H + H).]]

:Energy of the reactants (F + H2) = -103.80 kcal/mol
:Energy of the products (F-H + H) = -134.00 kcal/mol
:Energy of the transition structure = -96.47 kcal/mol
:Activation energy for formation of HF + H = +7.33 kcal/mol (=-96.47--103.80)
:Activation energy for formation of F + H2 = +37.53 kcal/mol (=-96.47--134.00)

{{fontcolor1|gray| How did you get the Eas? They're not very accurate [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:41, 31 May 2017 (BST)}}

===Reaction Dynamics===
====Mechanism of Release of Reactant Energy====

[[File:lem215_rt.png|thumb|center|Figure 14 - Reactive Trajectory for F + H2.]]

The amplitude of the oscillating trajectory reflects the magnitude of the defined initial momenta values. The reactants produce an oscillation of smaller amplitude than the products as p1 (H-H) < p2 (F-H).
The trajectories before and after the transition state however differ by the period of the generated oscillations - this indicates differences in the translational kinetic energy in the reactants and products. It has been discussed previously that the molecular product H-F has a stronger bond than H-H so it is expected to ‘see’ a greater translational kinetic energy in the products than the reactants. This is displayed by the trajectory, the oscillation period for the product H-F (A-B) is larger than that of the reactant H-H (B-C).
As a result of the conservation of energy, the total energy of the system must remain constant <ref name="ref7"/>. Therefore the product state should have a lower potential energy than the reactants to balance the total energy (kinetic and potential) of the system so that it remains constant, suggesting the reaction to be exothermic - earlier concluded to be the case as for the reaction for F + H2. This could be confirmed by different spectroscopic methods such as calorimetry and IR/Raman spectroscopy.

:Heat-flow calorimetry is commonly used to thermally monitor reactions and thus determine whether a reaction is exothermic or endothermic. The heat transfer under e.g. constant pressure is monitored as the reaction proceeds and consequently the enthalpy change can be determined. Thermodynamically ΔH=qp at constant pressure (where qp is the heat flow at constant pressure), otherwise known as Hess's Law <ref name="ref8"/>. Exothermic reactions release heat thus if exothermic the calculated value for the enthalpy change will be negative in sign (visa versa for endothermic reactions).

:IR/Raman spectroscopy is used to determine the vibrational states of molecules/matter. If the reactants possess less translational kinetic energy and thus vibrational energy then the IR spectra for the reactants should display lower intensity peaks relative to the peaks in the IR spectra of the products.

{{fontcolor1|gray| Good experimental suggestions. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:41, 31 May 2017 (BST)}}

====Polanyi's empirical rules====

Polanyi's empirical rules focus on the importance of the transition state location in enhancing reactivity for activated reactions <ref name="ref9"/>. This model has enabled the prediction of the relative efficacy of vibrational and translational excitations in promoting different molecular dynamic reactions.
:Polanyi's rules <ref name="ref9"/> state that vibrational energy is more efficient at promoting late transition states (endothermic reactions according to Hammond Postulate <ref name="h"/>) than translational energy. And conversely translational kinetic energy is more efficient at promoting 'early' transition states (exothermic reactions).

The reaction from H + H-F provides an illustration of these rules. This reaction is endothermic, thus is an example of a late-barrier reaction - the transition state is closer to the products in energy and structure according to Hammond's Postulate <ref name="h"/>. According to the Polanyi's Empirical Rules the reactivity should be enhanced more efficiently by high vibrational energy (and low kinetic energy). When initial conditions were chosen so that the reactant H-H had high vibrational energy (high pHH) and low translational kinetic energy (small pHF) a kinetically driven trajectory was generated.

Not all systems obey Polanyi's Rules as a result of their empirical nature (experimentally determined). As the rules are experimentally determined they effectively relate to the average behaviour of a reacting system <ref name="ref10"/>. The reaction H2 + F is an example where the calculated trajectory deviates from Polanyi's predictions. This reaction is exothermic so according to Polanyi's rules it is expected that high translational kinetic energy would promote the early transition state and thus enhance the reactivity. However for the reactive trajectory to reach completion a higher initial momenta for H-H (pHH) than initial translational kinetic energy (pHF) was required. This may be due to the errors in the computational calculation of the reaction trajectory thus does not accurately portray was would be observed experimentally.

{{fontcolor1|gray| Missing some examples here with your trajectories. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:42, 31 May 2017 (BST)}}
References:
<references>
<ref name="a">Laganà, Antonio, and Antonio Riganelli. Reaction And Molecular Dynamics. 1st ed. Berlin, Heidelberg: Springer Berlin Heidelberg, 2000. Print..</ref>
<ref name="ref1">Collett, Charles T, and Christopher D Robson. Handbook Of Computational Chemistry Research. 1st ed. New York: Nova Science Publishers, 2010.</ref>
<ref name="ref2">Lewars, Errol G. "The Concept Of The Potential Energy Surface". Computational Chemistry (2010): 9-43. Web. 11 May 2017.</ref>
<ref name="ref3">Jeffrey, Alan. Essentials Of Engineering Mathematics. 1st ed. Boca Raton: Chapman & Hall/CRC Press, 2004. Print.</ref>
<ref name="ref4">Koistinen, O-P. et al. "Minimum Energy Path Calculations With Gaussian Process Regression". Nanosystems: Physics, Chemistry, Mathematics (2016): 925-935. Web. 20 May</ref>
<ref name="t">Laidler, Keith J., and M. Christine King. "Development Of Transition-State Theory". The Journal of Physical Chemistry 87.15 (1983): 2657-2664. Web. 26 May 2017.</ref>
<ref name="ref5">Fueno, Takayuki. Transition State: A Theoretical Approach. 1st ed. Tokyo: Gordon and Breach Science Publishers, 1999. Print.</ref>
<ref name="ar">Kotz, John C, and Paul Treichel. Chemistry & Chemical Reactivity. 1st ed. Fort Worth: Saunders College Pub., 1996. Print.</ref>
<ref name="ref6">Petrucci, Ralph H. General Chemistry. 1st ed. Toronto, Ont.: Pearson Canada, 2011. Print.</ref>
<ref name="h">Carey, Francis A, and Richard J Sundberg. Advanced Organic Chemistry. 1st ed. Norwell: Springer, 2007. Print.</ref>
<ref name="ref7">Khine, Myint Swe, and Issa M Saleh. Models And Modeling. 1st ed. Dordrecht: Springer Science+Business Media B.V, 2011</ref>
<ref name="ref8">Mikulecky, Peter, Michelle Rose Gilman, and Kate Brutlag. AP Chemistry For Dummies. 1st ed. Hoboken, N.J.: Wiley Pub., 2009. Print.</ref>
<ref name="ref9">J. Clayden, N. Greeves and S. G. Warren, Organic Chemistry, Oxford University Press, Oxford, 2012</ref>
<ref name="ref10">Zhang, Zhaojun et al. "Theoretical Study Of The Validity Of The Polanyi Rules For The Late-Barrier Cl + Chd3reaction". The Journal of Physical Chemistry Letters 3.23 (2012): 3416-3419. Web. 26 May 2017.</ref>
</ref>
</references>

MRD:LaraMetcalf

2017-05-31T16:41:40Z

Je714: /* Mechanism of Release of Reactant Energy */

= Molecular Reaction Dynamics Lab =
==EXERCISE 1: H + H2 system==

A potential energy surface (PES) is a mathematical function which describes the electronic potential energy of a system as a function of the relative atomic positions in space and is widely used as a tool to carry out theoretical studies on molecular reaction dynamics <ref name="a" />.

===Dynamics from the transition state region===

The total gradient of the potential energy surface at a minimum and transition structure is equal to zero. The distinction between the two structures comes from the sign corresponding to their rate of change of gradient. The potential energy second derivative is positive for a minima and negative for a transition state <ref name="ref1" />. {{fontcolor1|gray| Which one? Along which direction? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:36, 31 May 2017 (BST)}}
The transition structure linking the two minima is a maximum along the minima energy reaction pathway and thus is called a saddle point <ref name="ref2" />.
This can be expressed mathematically by Taylor's Theorem <ref name="ref3" />:
:If fxxfyy − fx2y < 0 at (a, b) then (a, b) is a saddle point.
:If fxx >0 and fyy > 0 at (a,b) then (a,b) is a minimum point.

{{fontcolor1|gray| At the TS one of the partial 2nd derivatives will be >0 along one of the variables, and <0 along all the rest. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:36, 31 May 2017 (BST)}}

===Trajectories from r1 = r2: locating the transition state===

The best estimate of the transition state position rts is found to be equal to 0.90775 Å. The detection of the transition structure can be observed in an internuclear distances vs time plot. The transition state corresponds to where the gradient of the potential energy surface is zero, so it is expected that at rts the H atoms will be stationary - their internuclear separations will remain constant over time (fig 1). At a distance of 0.90775 Å the trajectory disappears (fig 2) confirming this to be a good approximation of the transition state position. Varying the interatomic distance away from the transition state position results in a trajectory that rolls back toward the reactants/products and thus a changing internuclear distance over time is observed (fig 3). {{fontcolor1|gray| Really? That's not what figure 3 is showing. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:37, 31 May 2017 (BST)}}

[[File:lem215_fig1.png|thumb|center|Figure 1 - internuclear distance vs time plot at the transition state position.]]
[[File:lem215_fig2.png|thumb|center|Figure 2 - disappearance of the trajectory at the transition state position.]]
[[File:lem215_fig3.png|thumb|center|Figure 3 - internuclear distance vs time plot at r=1.5 A.]]
[[File:lem215_fig4.png|thumb|center|Figure 4 - surface plot displaying the trajectory at r=1.5 A.]]

===Minimum Energy Path (MEP) and Dynamic Calculations===

Both calculations produce trajectories that rolls from the reactants H1-H2 H3 in the direction of increasing r1 towards H1 H2-H3 (r1 = A-B) Upon energy minimisation the trajectory converges to the MEP <ref name="ref4" />.
The MEP calculation generates the minimum energy trajectory thus is restricted to only one degree of freedom, the interatomic positions r1 and r2. The exchange between kinetic and potential energy (due to vibrations) does not occur as a result of the minimisation, and so the generated trajectory does not have an oscillative nature - the kinetic energy is fixed at zero.
On the contrary the dynamics calculation does not involve such limitations allowing it to have more degrees of freedom generating an oscillating trajectory, signifying influence of a non-zero kinetic energy.

[[File:lem215_mep.png|thumb|center|MEP Calculation Trajectory.]]
[[File:lem215_dynamics.png|thumb|center|Dynamic Trajectory Calculation.]]

At large t (t): (3 decimal places)
:r1(1.485)= 5.247 Å
:r2(1.485) = 0.739 Å
:p1(1.45) = 2.481
:p2(1.45) = 1.260

Reverting the initial conditions so that r1 = rts + 0.01 and r2 = rts then the reaction would have a trajectory that rolls from H1 H2-H3 towards H1-H2 H3 in the direction of increasing r2.

If the initial positions and final positions were swapped then the corresponding values for final momenta will have the opposite sign.

===Reactive and unreactive trajectories===

A series of initial condition were tested to determine whether or not they produce a reactive trajectory that starts from the reactants HA-HB and dissociated atom HC to the molecular product HB-HC with dissociated atom HA. Initial momenta p1 of atom HC approaching HB before the transition state and final momenta p2 of atom HA moving away from the molecular product HB-HC

{| class="wikitable" border="1" style="margin-left: auto; margin-right: auto; border: none; width: 60%; height: 200px"
|+ Initial Conditions
! p1 !! p2 !! Reactivity
|- style="text-align: center;"
| -1.25 || -2.5 || Reactive
|- style="text-align: center;"
| -1.5 || -2.0 || Unreactive
|- style="text-align: center;"
| -1.5 || -2.5 || Reactive
|- style="text-align: center;"
| -2.5 || -5 || Unreactive
|- style="text-align: center;"
| -2.5 || -5.2 || Reactive
|}

[[File:lem215_fig5.png|thumb|center|Figure 5 - initial conditions p1=-1.25, p2=-2.5.]]
The first set of conditions generates a linear trajectory path from HA-HB HC to the transition structure at approximately 0.908 Å. Before the collision r1 remains constant and r2 steadily decreases as H3 approaches H2. There is sufficient energy to overcome the activation energy barrier allowing the trajectory to roll towards the products. After the transition structure r1 steadily increases as HA moves away from HB from the molecule formed HB-HC, the oscillating trajectory signifies the vibrational energy of the HB-HC bond.

[[File:lem215_fig6.png|thumb|center|Figure 6 - initial conditions p1=-1.5, p2=-2.0.]]
Increasing p1 and p2 (initial momenta), relative to the first set of conditions, results in an unreactive trajectory. Before the transition structure r1 has a slight oscillation (vibrational energy of A-B) and r2 steadily decreases as HC approaches HB. At r1=0.810 Å and r2=1.120 Å the trajectory bounces back toward the reactants. This suggests amount of energy available to the system is not sufficient to overcome the activation energy barrier - the reaction is unsuccessful. It is evident that the vibrational energy of HA-HB is retained upon the return of the system to the reactants.

[[File:lem215_fig7.png|thumb|center|Figure 7 - initial conditions p1=-1.5, p2=-2.5.]]
Increasing p2 from -2.0 to -2.5 provides the energy required to overcome the barrier thus allowing the reaction to proceed. The trajectory rolls from the reactants to the products passing through the transition structure at 0.893 Å. Again the vibrational energy of the HA-HB then HB-HC bonds are observed before and after passing through the transition structure respectively.

[[File:lem215_fig8.png|thumb|center|Figure 8 - initial conditions p1=-2.5, p2=-5.0.]]
With the fourth set of initial momenta conditions the trajectory rolls from the reactants and passes through the transition structure illustrating an initial formation of the HB-HC bond. Soon after crossing the transition region the trajectory returns to the reactants - this is an example barrier re-crossing. The trajectory bounces several times within a small region about transition state position before returning back to the reactants.

[[File:lem215_fig9.png|thumb|center|Figure 9 - initial conditions p1=-2.5, p2=-5.2.]]
The trajectory approaches the transition structure along the direction of decreasing r2 at a constant r1. Again the trajectory displays barrier re-crossing before eventually bouncing across the barrier toward the product HB-HC and dissociated atom HA.

The final two sets of initial momenta have relatively large p2 values. Both systems have significant translational kinetic energy as H atom A moves away from the formed molecule HB-HC. At high kinetic translational energies of atoms/molecules it becomes increasingly difficult to justify the differences in the reactivity of the generated trajectories. This suggests the application of Transition State Theory is unsuitable for dynamic molecular systems of this type - this method is limiting.

====Transition State Theory====

The transition state is the known as the 'critical configuration' that corresponds to the highest point on a minimum energy reaction pathway on a potential energy surface <ref name="ref5" />. If the system is able to reach the critical geometry then a successful reaction is highly probable. The transition structure is a stationary point thus is identified as the saddle point, with associated reaction coordinates, on a potential energy surface.
The transition state theory is a cumulative a series of assumptions used to approximate the equilibrium rate constants for thermally driven molecular reactions <ref name="t" />. The TST stemmed from thermodynamic, kinetic-theory and statistical mechanic treatments <ref name="t" />.

=====Development of the Transition State Theory=====

The thermodynamic treatment stems from the Van't Hoff equation <ref name="t" /> which relates the change in the equilibrium constant (keq) to a given change in temperature (T) for a given enthalpy change (delta H):
:<math>\ln K_\mathrm{eq} = - \frac{{\Delta H^\ominus}}{RT}+ \frac{{\Delta S^\ominus }}{R}.</math>.
The kinetic-theory <ref name="t" /> explains the temperature dependence of reaction rates as a function of collision frequency and the statistical mechanical treatment focuses on equilibrium reaction rates in terms of the motion of molecules and statistical distribution of molecular speeds <ref name="t" /> - initially modelled by Maxwell and Boltzmann.

=====Main Features and Assumptions of the Transition State Theory <ref name="t"/> =====

:(1) Rates can be calculated by focusing on the "activated complex" - what happens before the transition state is reached are not important if one rate dominates the overall rate of reaction.
:(2) The activated complex is formed in 'quasi-equilibrium' i.e. the complex is in equilibrium with the reactants and products passing through - the concentrations of reactants and products can be determined by changing the concentration of one species e.g. removing the products.
:(3) Motion of the system is treated as 'free translational motion' - expressed by kinetic-theory.

=====Using the Transition State Theory to Determine Reactivity=====

The TST ignores the dynamical recrossings that can lead back to the reactants (also known as tunneling) so this calculation tends to overestimate the rates within the classical mechanic framework <ref name ="t"/>. For a successful reaction the molecular system must have sufficient energy to overcome the activation energy barrier and thus generate a reactive trajectory. The Arrhenius equation relates the activate energy to the reaction rate for a particular reaction scheme, k=Ae(-Ea/RT) <ref name ="ar"/>. The relationship shows how a small activation energy barrier will correspond to a larger reaction rate constant - the reaction will go faster. It is likely that the experimental activation energy barrier will be larger than the calculated value thus TST will likely overestimate the rate relative the the experimental rate for a given reaction.

{{fontcolor1|gray| Good job looking up information. Just a small point -- the statistical mechanical treatment does not really hold here, since we are not looking at an ensemble of particles, just at a simple triatomic collision in isolation. The biggest point to be made here, was the recrossing limitation, and how that overestimates the rates when compared to experiment. Because that's the most relevant limitation that you observe in this lab. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:40, 31 May 2017 (BST)}}

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

===Reaction Energetics===

The F + H2 reaction is exothermic. An exothermic reaction is defined as the reaction path from a higher (reactants) to lower total potential energy (products) <ref name="ref6"/>. The trajectory shows the approach of F to H2 from infinite separation (as it is dissociated): as A-B separation decreases the potential energy decreases to a minimum indicating the H-F bond formation and H-H bond dissociation is an exothermic process.
[[File:lem215_1.png|thumb|center|Figure 10 - Surface Plot Trajectory from F + H2 to F-H + H.]]

By definition the reverse of an exothermic reaction is endothermic <ref name="ref6"/>, this is indicated by the trajectory generate by the H-F + H reaction. H-F bond breaking and H-H bond forming is an endothermic and enthalpically unfavourable process. The reactants have a lower total potential energy than the products.
[[File:lem215_2.png|thumb|center|Figure 11 - Surface Plot Trajectory from F-H + H to F + H2]]

The H-F bond formation has a more negative (exothermic) reaction enthalpy indicating H-F to have a stronger bond than H2.

===Location of Transition State and Determination of the Activation Energies===

Rates of reactions are controlled by the free energy of the transition state and Hammond postulated that the structure of a given transition state will more closely resemble that of the reactants or products depending on which is closer in energy <ref name="h"/>. He concluded that for an exothermic reaction the transition state will be more structurally similar to the reactants i.e. the interconversion between the reactants and transition state will involve only a small reorganization of molecular structure. And conversely for an endothermic reaction the transition state will be more structurally related to the products.
The transition structure is expected to be closer in energy and structure to the reactants - longer A-B and shorter B-C distance so point of intersection further right in the saddle region was selected to determine the location (see figure 12) and energy of the transition state.

Approximate location of the transition state (3 dp):
:A-B (F-H) = 1.713 Å
:B-C (H-H) = 0.748 Å

[[File:lem215_Hpost.png|thumb|center|Figure 12 - Locating the transition state for a F-H-H system.]]

Activation energies (2 dp):

[[File:lem215_Hpost1.png|thumb|center|Figure 13 - Determination of the absolute energy for products (F-H + H).]]

:Energy of the reactants (F + H2) = -103.80 kcal/mol
:Energy of the products (F-H + H) = -134.00 kcal/mol
:Energy of the transition structure = -96.47 kcal/mol
:Activation energy for formation of HF + H = +7.33 kcal/mol (=-96.47--103.80)
:Activation energy for formation of F + H2 = +37.53 kcal/mol (=-96.47--134.00)

{{fontcolor1|gray| How did you get the Eas? They're not very accurate [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:41, 31 May 2017 (BST)}}

===Reaction Dynamics===
====Mechanism of Release of Reactant Energy====

[[File:lem215_rt.png|thumb|center|Figure 14 - Reactive Trajectory for F + H2.]]

The amplitude of the oscillating trajectory reflects the magnitude of the defined initial momenta values. The reactants produce an oscillation of smaller amplitude than the products as p1 (H-H) < p2 (F-H).
The trajectories before and after the transition state however differ by the period of the generated oscillations - this indicates differences in the translational kinetic energy in the reactants and products. It has been discussed previously that the molecular product H-F has a stronger bond than H-H so it is expected to ‘see’ a greater translational kinetic energy in the products than the reactants. This is displayed by the trajectory, the oscillation period for the product H-F (A-B) is larger than that of the reactant H-H (B-C).
As a result of the conservation of energy, the total energy of the system must remain constant <ref name="ref7"/>. Therefore the product state should have a lower potential energy than the reactants to balance the total energy (kinetic and potential) of the system so that it remains constant, suggesting the reaction to be exothermic - earlier concluded to be the case as for the reaction for F + H2. This could be confirmed by different spectroscopic methods such as calorimetry and IR/Raman spectroscopy.

:Heat-flow calorimetry is commonly used to thermally monitor reactions and thus determine whether a reaction is exothermic or endothermic. The heat transfer under e.g. constant pressure is monitored as the reaction proceeds and consequently the enthalpy change can be determined. Thermodynamically ΔH=qp at constant pressure (where qp is the heat flow at constant pressure), otherwise known as Hess's Law <ref name="ref8"/>. Exothermic reactions release heat thus if exothermic the calculated value for the enthalpy change will be negative in sign (visa versa for endothermic reactions).

:IR/Raman spectroscopy is used to determine the vibrational states of molecules/matter. If the reactants possess less translational kinetic energy and thus vibrational energy then the IR spectra for the reactants should display lower intensity peaks relative to the peaks in the IR spectra of the products.

{{fontcolor1|gray| Good experimental suggestions. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:41, 31 May 2017 (BST)}}

====Polanyi's empirical rules====

Polanyi's empirical rules focus on the importance of the transition state location in enhancing reactivity for activated reactions <ref name="ref9"/>. This model has enabled the prediction of the relative efficacy of vibrational and translational excitations in promoting different molecular dynamic reactions.
:Polanyi's rules <ref name="ref9"/> state that vibrational energy is more efficient at promoting late transition states (endothermic reactions according to Hammond Postulate <ref name="h"/>) than translational energy. And conversely translational kinetic energy is more efficient at promoting 'early' transition states (exothermic reactions).

The reaction from H + H-F provides an illustration of these rules. This reaction is endothermic, thus is an example of a late-barrier reaction - the transition state is closer to the products in energy and structure according to Hammond's Postulate <ref name="h"/>. According to the Polanyi's Empirical Rules the reactivity should be enhanced more efficiently by high vibrational energy (and low kinetic energy). When initial conditions were chosen so that the reactant H-H had high vibrational energy (high pHH) and low translational kinetic energy (small pHF) a kinetically driven trajectory was generated.

Not all systems obey Polanyi's Rules as a result of their empirical nature (experimentally determined). As the rules are experimentally determined they effectively relate to the average behaviour of a reacting system <ref name="ref10"/>. The reaction H2 + F is an example where the calculated trajectory deviates from Polanyi's predictions. This reaction is exothermic so according to Polanyi's rules it is expected that high translational kinetic energy would promote the early transition state and thus enhance the reactivity. However for the reactive trajectory to reach completion a higher initial momenta for H-H (pHH) than initial translational kinetic energy (pHF) was required. This may be due to the errors in the computational calculation of the reaction trajectory thus does not accurately portray was would be observed experimentally.

References:
<references>
<ref name="a">Laganà, Antonio, and Antonio Riganelli. Reaction And Molecular Dynamics. 1st ed. Berlin, Heidelberg: Springer Berlin Heidelberg, 2000. Print..</ref>
<ref name="ref1">Collett, Charles T, and Christopher D Robson. Handbook Of Computational Chemistry Research. 1st ed. New York: Nova Science Publishers, 2010.</ref>
<ref name="ref2">Lewars, Errol G. "The Concept Of The Potential Energy Surface". Computational Chemistry (2010): 9-43. Web. 11 May 2017.</ref>
<ref name="ref3">Jeffrey, Alan. Essentials Of Engineering Mathematics. 1st ed. Boca Raton: Chapman & Hall/CRC Press, 2004. Print.</ref>
<ref name="ref4">Koistinen, O-P. et al. "Minimum Energy Path Calculations With Gaussian Process Regression". Nanosystems: Physics, Chemistry, Mathematics (2016): 925-935. Web. 20 May</ref>
<ref name="t">Laidler, Keith J., and M. Christine King. "Development Of Transition-State Theory". The Journal of Physical Chemistry 87.15 (1983): 2657-2664. Web. 26 May 2017.</ref>
<ref name="ref5">Fueno, Takayuki. Transition State: A Theoretical Approach. 1st ed. Tokyo: Gordon and Breach Science Publishers, 1999. Print.</ref>
<ref name="ar">Kotz, John C, and Paul Treichel. Chemistry & Chemical Reactivity. 1st ed. Fort Worth: Saunders College Pub., 1996. Print.</ref>
<ref name="ref6">Petrucci, Ralph H. General Chemistry. 1st ed. Toronto, Ont.: Pearson Canada, 2011. Print.</ref>
<ref name="h">Carey, Francis A, and Richard J Sundberg. Advanced Organic Chemistry. 1st ed. Norwell: Springer, 2007. Print.</ref>
<ref name="ref7">Khine, Myint Swe, and Issa M Saleh. Models And Modeling. 1st ed. Dordrecht: Springer Science+Business Media B.V, 2011</ref>
<ref name="ref8">Mikulecky, Peter, Michelle Rose Gilman, and Kate Brutlag. AP Chemistry For Dummies. 1st ed. Hoboken, N.J.: Wiley Pub., 2009. Print.</ref>
<ref name="ref9">J. Clayden, N. Greeves and S. G. Warren, Organic Chemistry, Oxford University Press, Oxford, 2012</ref>
<ref name="ref10">Zhang, Zhaojun et al. "Theoretical Study Of The Validity Of The Polanyi Rules For The Late-Barrier Cl + Chd3reaction". The Journal of Physical Chemistry Letters 3.23 (2012): 3416-3419. Web. 26 May 2017.</ref>
</ref>
</references>

MRD:LaraMetcalf

2017-05-31T16:41:11Z

Je714: /* Location of Transition State and Determination of the Activation Energies */

= Molecular Reaction Dynamics Lab =
==EXERCISE 1: H + H2 system==

A potential energy surface (PES) is a mathematical function which describes the electronic potential energy of a system as a function of the relative atomic positions in space and is widely used as a tool to carry out theoretical studies on molecular reaction dynamics <ref name="a" />.

===Dynamics from the transition state region===

The total gradient of the potential energy surface at a minimum and transition structure is equal to zero. The distinction between the two structures comes from the sign corresponding to their rate of change of gradient. The potential energy second derivative is positive for a minima and negative for a transition state <ref name="ref1" />. {{fontcolor1|gray| Which one? Along which direction? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:36, 31 May 2017 (BST)}}
The transition structure linking the two minima is a maximum along the minima energy reaction pathway and thus is called a saddle point <ref name="ref2" />.
This can be expressed mathematically by Taylor's Theorem <ref name="ref3" />:
:If fxxfyy − fx2y < 0 at (a, b) then (a, b) is a saddle point.
:If fxx >0 and fyy > 0 at (a,b) then (a,b) is a minimum point.

{{fontcolor1|gray| At the TS one of the partial 2nd derivatives will be >0 along one of the variables, and <0 along all the rest. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:36, 31 May 2017 (BST)}}

===Trajectories from r1 = r2: locating the transition state===

The best estimate of the transition state position rts is found to be equal to 0.90775 Å. The detection of the transition structure can be observed in an internuclear distances vs time plot. The transition state corresponds to where the gradient of the potential energy surface is zero, so it is expected that at rts the H atoms will be stationary - their internuclear separations will remain constant over time (fig 1). At a distance of 0.90775 Å the trajectory disappears (fig 2) confirming this to be a good approximation of the transition state position. Varying the interatomic distance away from the transition state position results in a trajectory that rolls back toward the reactants/products and thus a changing internuclear distance over time is observed (fig 3). {{fontcolor1|gray| Really? That's not what figure 3 is showing. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:37, 31 May 2017 (BST)}}

[[File:lem215_fig1.png|thumb|center|Figure 1 - internuclear distance vs time plot at the transition state position.]]
[[File:lem215_fig2.png|thumb|center|Figure 2 - disappearance of the trajectory at the transition state position.]]
[[File:lem215_fig3.png|thumb|center|Figure 3 - internuclear distance vs time plot at r=1.5 A.]]
[[File:lem215_fig4.png|thumb|center|Figure 4 - surface plot displaying the trajectory at r=1.5 A.]]

===Minimum Energy Path (MEP) and Dynamic Calculations===

Both calculations produce trajectories that rolls from the reactants H1-H2 H3 in the direction of increasing r1 towards H1 H2-H3 (r1 = A-B) Upon energy minimisation the trajectory converges to the MEP <ref name="ref4" />.
The MEP calculation generates the minimum energy trajectory thus is restricted to only one degree of freedom, the interatomic positions r1 and r2. The exchange between kinetic and potential energy (due to vibrations) does not occur as a result of the minimisation, and so the generated trajectory does not have an oscillative nature - the kinetic energy is fixed at zero.
On the contrary the dynamics calculation does not involve such limitations allowing it to have more degrees of freedom generating an oscillating trajectory, signifying influence of a non-zero kinetic energy.

[[File:lem215_mep.png|thumb|center|MEP Calculation Trajectory.]]
[[File:lem215_dynamics.png|thumb|center|Dynamic Trajectory Calculation.]]

At large t (t): (3 decimal places)
:r1(1.485)= 5.247 Å
:r2(1.485) = 0.739 Å
:p1(1.45) = 2.481
:p2(1.45) = 1.260

Reverting the initial conditions so that r1 = rts + 0.01 and r2 = rts then the reaction would have a trajectory that rolls from H1 H2-H3 towards H1-H2 H3 in the direction of increasing r2.

If the initial positions and final positions were swapped then the corresponding values for final momenta will have the opposite sign.

===Reactive and unreactive trajectories===

A series of initial condition were tested to determine whether or not they produce a reactive trajectory that starts from the reactants HA-HB and dissociated atom HC to the molecular product HB-HC with dissociated atom HA. Initial momenta p1 of atom HC approaching HB before the transition state and final momenta p2 of atom HA moving away from the molecular product HB-HC

{| class="wikitable" border="1" style="margin-left: auto; margin-right: auto; border: none; width: 60%; height: 200px"
|+ Initial Conditions
! p1 !! p2 !! Reactivity
|- style="text-align: center;"
| -1.25 || -2.5 || Reactive
|- style="text-align: center;"
| -1.5 || -2.0 || Unreactive
|- style="text-align: center;"
| -1.5 || -2.5 || Reactive
|- style="text-align: center;"
| -2.5 || -5 || Unreactive
|- style="text-align: center;"
| -2.5 || -5.2 || Reactive
|}

[[File:lem215_fig5.png|thumb|center|Figure 5 - initial conditions p1=-1.25, p2=-2.5.]]
The first set of conditions generates a linear trajectory path from HA-HB HC to the transition structure at approximately 0.908 Å. Before the collision r1 remains constant and r2 steadily decreases as H3 approaches H2. There is sufficient energy to overcome the activation energy barrier allowing the trajectory to roll towards the products. After the transition structure r1 steadily increases as HA moves away from HB from the molecule formed HB-HC, the oscillating trajectory signifies the vibrational energy of the HB-HC bond.

[[File:lem215_fig6.png|thumb|center|Figure 6 - initial conditions p1=-1.5, p2=-2.0.]]
Increasing p1 and p2 (initial momenta), relative to the first set of conditions, results in an unreactive trajectory. Before the transition structure r1 has a slight oscillation (vibrational energy of A-B) and r2 steadily decreases as HC approaches HB. At r1=0.810 Å and r2=1.120 Å the trajectory bounces back toward the reactants. This suggests amount of energy available to the system is not sufficient to overcome the activation energy barrier - the reaction is unsuccessful. It is evident that the vibrational energy of HA-HB is retained upon the return of the system to the reactants.

[[File:lem215_fig7.png|thumb|center|Figure 7 - initial conditions p1=-1.5, p2=-2.5.]]
Increasing p2 from -2.0 to -2.5 provides the energy required to overcome the barrier thus allowing the reaction to proceed. The trajectory rolls from the reactants to the products passing through the transition structure at 0.893 Å. Again the vibrational energy of the HA-HB then HB-HC bonds are observed before and after passing through the transition structure respectively.

[[File:lem215_fig8.png|thumb|center|Figure 8 - initial conditions p1=-2.5, p2=-5.0.]]
With the fourth set of initial momenta conditions the trajectory rolls from the reactants and passes through the transition structure illustrating an initial formation of the HB-HC bond. Soon after crossing the transition region the trajectory returns to the reactants - this is an example barrier re-crossing. The trajectory bounces several times within a small region about transition state position before returning back to the reactants.

[[File:lem215_fig9.png|thumb|center|Figure 9 - initial conditions p1=-2.5, p2=-5.2.]]
The trajectory approaches the transition structure along the direction of decreasing r2 at a constant r1. Again the trajectory displays barrier re-crossing before eventually bouncing across the barrier toward the product HB-HC and dissociated atom HA.

The final two sets of initial momenta have relatively large p2 values. Both systems have significant translational kinetic energy as H atom A moves away from the formed molecule HB-HC. At high kinetic translational energies of atoms/molecules it becomes increasingly difficult to justify the differences in the reactivity of the generated trajectories. This suggests the application of Transition State Theory is unsuitable for dynamic molecular systems of this type - this method is limiting.

====Transition State Theory====

The transition state is the known as the 'critical configuration' that corresponds to the highest point on a minimum energy reaction pathway on a potential energy surface <ref name="ref5" />. If the system is able to reach the critical geometry then a successful reaction is highly probable. The transition structure is a stationary point thus is identified as the saddle point, with associated reaction coordinates, on a potential energy surface.
The transition state theory is a cumulative a series of assumptions used to approximate the equilibrium rate constants for thermally driven molecular reactions <ref name="t" />. The TST stemmed from thermodynamic, kinetic-theory and statistical mechanic treatments <ref name="t" />.

=====Development of the Transition State Theory=====

The thermodynamic treatment stems from the Van't Hoff equation <ref name="t" /> which relates the change in the equilibrium constant (keq) to a given change in temperature (T) for a given enthalpy change (delta H):
:<math>\ln K_\mathrm{eq} = - \frac{{\Delta H^\ominus}}{RT}+ \frac{{\Delta S^\ominus }}{R}.</math>.
The kinetic-theory <ref name="t" /> explains the temperature dependence of reaction rates as a function of collision frequency and the statistical mechanical treatment focuses on equilibrium reaction rates in terms of the motion of molecules and statistical distribution of molecular speeds <ref name="t" /> - initially modelled by Maxwell and Boltzmann.

=====Main Features and Assumptions of the Transition State Theory <ref name="t"/> =====

:(1) Rates can be calculated by focusing on the "activated complex" - what happens before the transition state is reached are not important if one rate dominates the overall rate of reaction.
:(2) The activated complex is formed in 'quasi-equilibrium' i.e. the complex is in equilibrium with the reactants and products passing through - the concentrations of reactants and products can be determined by changing the concentration of one species e.g. removing the products.
:(3) Motion of the system is treated as 'free translational motion' - expressed by kinetic-theory.

=====Using the Transition State Theory to Determine Reactivity=====

The TST ignores the dynamical recrossings that can lead back to the reactants (also known as tunneling) so this calculation tends to overestimate the rates within the classical mechanic framework <ref name ="t"/>. For a successful reaction the molecular system must have sufficient energy to overcome the activation energy barrier and thus generate a reactive trajectory. The Arrhenius equation relates the activate energy to the reaction rate for a particular reaction scheme, k=Ae(-Ea/RT) <ref name ="ar"/>. The relationship shows how a small activation energy barrier will correspond to a larger reaction rate constant - the reaction will go faster. It is likely that the experimental activation energy barrier will be larger than the calculated value thus TST will likely overestimate the rate relative the the experimental rate for a given reaction.

{{fontcolor1|gray| Good job looking up information. Just a small point -- the statistical mechanical treatment does not really hold here, since we are not looking at an ensemble of particles, just at a simple triatomic collision in isolation. The biggest point to be made here, was the recrossing limitation, and how that overestimates the rates when compared to experiment. Because that's the most relevant limitation that you observe in this lab. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:40, 31 May 2017 (BST)}}

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

===Reaction Energetics===

The F + H2 reaction is exothermic. An exothermic reaction is defined as the reaction path from a higher (reactants) to lower total potential energy (products) <ref name="ref6"/>. The trajectory shows the approach of F to H2 from infinite separation (as it is dissociated): as A-B separation decreases the potential energy decreases to a minimum indicating the H-F bond formation and H-H bond dissociation is an exothermic process.
[[File:lem215_1.png|thumb|center|Figure 10 - Surface Plot Trajectory from F + H2 to F-H + H.]]

By definition the reverse of an exothermic reaction is endothermic <ref name="ref6"/>, this is indicated by the trajectory generate by the H-F + H reaction. H-F bond breaking and H-H bond forming is an endothermic and enthalpically unfavourable process. The reactants have a lower total potential energy than the products.
[[File:lem215_2.png|thumb|center|Figure 11 - Surface Plot Trajectory from F-H + H to F + H2]]

The H-F bond formation has a more negative (exothermic) reaction enthalpy indicating H-F to have a stronger bond than H2.

===Location of Transition State and Determination of the Activation Energies===

Rates of reactions are controlled by the free energy of the transition state and Hammond postulated that the structure of a given transition state will more closely resemble that of the reactants or products depending on which is closer in energy <ref name="h"/>. He concluded that for an exothermic reaction the transition state will be more structurally similar to the reactants i.e. the interconversion between the reactants and transition state will involve only a small reorganization of molecular structure. And conversely for an endothermic reaction the transition state will be more structurally related to the products.
The transition structure is expected to be closer in energy and structure to the reactants - longer A-B and shorter B-C distance so point of intersection further right in the saddle region was selected to determine the location (see figure 12) and energy of the transition state.

Approximate location of the transition state (3 dp):
:A-B (F-H) = 1.713 Å
:B-C (H-H) = 0.748 Å

[[File:lem215_Hpost.png|thumb|center|Figure 12 - Locating the transition state for a F-H-H system.]]

Activation energies (2 dp):

[[File:lem215_Hpost1.png|thumb|center|Figure 13 - Determination of the absolute energy for products (F-H + H).]]

:Energy of the reactants (F + H2) = -103.80 kcal/mol
:Energy of the products (F-H + H) = -134.00 kcal/mol
:Energy of the transition structure = -96.47 kcal/mol
:Activation energy for formation of HF + H = +7.33 kcal/mol (=-96.47--103.80)
:Activation energy for formation of F + H2 = +37.53 kcal/mol (=-96.47--134.00)

{{fontcolor1|gray| How did you get the Eas? They're not very accurate [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:41, 31 May 2017 (BST)}}

===Reaction Dynamics===
====Mechanism of Release of Reactant Energy====

[[File:lem215_rt.png|thumb|center|Figure 14 - Reactive Trajectory for F + H2.]]

The amplitude of the oscillating trajectory reflects the magnitude of the defined initial momenta values. The reactants produce an oscillation of smaller amplitude than the products as p1 (H-H) < p2 (F-H).
The trajectories before and after the transition state however differ by the period of the generated oscillations - this indicates differences in the translational kinetic energy in the reactants and products. It has been discussed previously that the molecular product H-F has a stronger bond than H-H so it is expected to ‘see’ a greater translational kinetic energy in the products than the reactants. This is displayed by the trajectory, the oscillation period for the product H-F (A-B) is larger than that of the reactant H-H (B-C).
As a result of the conservation of energy, the total energy of the system must remain constant <ref name="ref7"/>. Therefore the product state should have a lower potential energy than the reactants to balance the total energy (kinetic and potential) of the system so that it remains constant, suggesting the reaction to be exothermic - earlier concluded to be the case as for the reaction for F + H2. This could be confirmed by different spectroscopic methods such as calorimetry and IR/Raman spectroscopy.

:Heat-flow calorimetry is commonly used to thermally monitor reactions and thus determine whether a reaction is exothermic or endothermic. The heat transfer under e.g. constant pressure is monitored as the reaction proceeds and consequently the enthalpy change can be determined. Thermodynamically ΔH=qp at constant pressure (where qp is the heat flow at constant pressure), otherwise known as Hess's Law <ref name="ref8"/>. Exothermic reactions release heat thus if exothermic the calculated value for the enthalpy change will be negative in sign (visa versa for endothermic reactions).

:IR/Raman spectroscopy is used to determine the vibrational states of molecules/matter. If the reactants possess less translational kinetic energy and thus vibrational energy then the IR spectra for the reactants should display lower intensity peaks relative to the peaks in the IR spectra of the products.

====Polanyi's empirical rules====

Polanyi's empirical rules focus on the importance of the transition state location in enhancing reactivity for activated reactions <ref name="ref9"/>. This model has enabled the prediction of the relative efficacy of vibrational and translational excitations in promoting different molecular dynamic reactions.
:Polanyi's rules <ref name="ref9"/> state that vibrational energy is more efficient at promoting late transition states (endothermic reactions according to Hammond Postulate <ref name="h"/>) than translational energy. And conversely translational kinetic energy is more efficient at promoting 'early' transition states (exothermic reactions).

The reaction from H + H-F provides an illustration of these rules. This reaction is endothermic, thus is an example of a late-barrier reaction - the transition state is closer to the products in energy and structure according to Hammond's Postulate <ref name="h"/>. According to the Polanyi's Empirical Rules the reactivity should be enhanced more efficiently by high vibrational energy (and low kinetic energy). When initial conditions were chosen so that the reactant H-H had high vibrational energy (high pHH) and low translational kinetic energy (small pHF) a kinetically driven trajectory was generated.

Not all systems obey Polanyi's Rules as a result of their empirical nature (experimentally determined). As the rules are experimentally determined they effectively relate to the average behaviour of a reacting system <ref name="ref10"/>. The reaction H2 + F is an example where the calculated trajectory deviates from Polanyi's predictions. This reaction is exothermic so according to Polanyi's rules it is expected that high translational kinetic energy would promote the early transition state and thus enhance the reactivity. However for the reactive trajectory to reach completion a higher initial momenta for H-H (pHH) than initial translational kinetic energy (pHF) was required. This may be due to the errors in the computational calculation of the reaction trajectory thus does not accurately portray was would be observed experimentally.

References:
<references>
<ref name="a">Laganà, Antonio, and Antonio Riganelli. Reaction And Molecular Dynamics. 1st ed. Berlin, Heidelberg: Springer Berlin Heidelberg, 2000. Print..</ref>
<ref name="ref1">Collett, Charles T, and Christopher D Robson. Handbook Of Computational Chemistry Research. 1st ed. New York: Nova Science Publishers, 2010.</ref>
<ref name="ref2">Lewars, Errol G. "The Concept Of The Potential Energy Surface". Computational Chemistry (2010): 9-43. Web. 11 May 2017.</ref>
<ref name="ref3">Jeffrey, Alan. Essentials Of Engineering Mathematics. 1st ed. Boca Raton: Chapman & Hall/CRC Press, 2004. Print.</ref>
<ref name="ref4">Koistinen, O-P. et al. "Minimum Energy Path Calculations With Gaussian Process Regression". Nanosystems: Physics, Chemistry, Mathematics (2016): 925-935. Web. 20 May</ref>
<ref name="t">Laidler, Keith J., and M. Christine King. "Development Of Transition-State Theory". The Journal of Physical Chemistry 87.15 (1983): 2657-2664. Web. 26 May 2017.</ref>
<ref name="ref5">Fueno, Takayuki. Transition State: A Theoretical Approach. 1st ed. Tokyo: Gordon and Breach Science Publishers, 1999. Print.</ref>
<ref name="ar">Kotz, John C, and Paul Treichel. Chemistry & Chemical Reactivity. 1st ed. Fort Worth: Saunders College Pub., 1996. Print.</ref>
<ref name="ref6">Petrucci, Ralph H. General Chemistry. 1st ed. Toronto, Ont.: Pearson Canada, 2011. Print.</ref>
<ref name="h">Carey, Francis A, and Richard J Sundberg. Advanced Organic Chemistry. 1st ed. Norwell: Springer, 2007. Print.</ref>
<ref name="ref7">Khine, Myint Swe, and Issa M Saleh. Models And Modeling. 1st ed. Dordrecht: Springer Science+Business Media B.V, 2011</ref>
<ref name="ref8">Mikulecky, Peter, Michelle Rose Gilman, and Kate Brutlag. AP Chemistry For Dummies. 1st ed. Hoboken, N.J.: Wiley Pub., 2009. Print.</ref>
<ref name="ref9">J. Clayden, N. Greeves and S. G. Warren, Organic Chemistry, Oxford University Press, Oxford, 2012</ref>
<ref name="ref10">Zhang, Zhaojun et al. "Theoretical Study Of The Validity Of The Polanyi Rules For The Late-Barrier Cl + Chd3reaction". The Journal of Physical Chemistry Letters 3.23 (2012): 3416-3419. Web. 26 May 2017.</ref>
</ref>
</references>

MRD:LaraMetcalf

2017-05-31T16:40:06Z

Je714: /* Using the Transition State Theory to Determine Reactivity */

= Molecular Reaction Dynamics Lab =
==EXERCISE 1: H + H2 system==

A potential energy surface (PES) is a mathematical function which describes the electronic potential energy of a system as a function of the relative atomic positions in space and is widely used as a tool to carry out theoretical studies on molecular reaction dynamics <ref name="a" />.

===Dynamics from the transition state region===

The total gradient of the potential energy surface at a minimum and transition structure is equal to zero. The distinction between the two structures comes from the sign corresponding to their rate of change of gradient. The potential energy second derivative is positive for a minima and negative for a transition state <ref name="ref1" />. {{fontcolor1|gray| Which one? Along which direction? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:36, 31 May 2017 (BST)}}
The transition structure linking the two minima is a maximum along the minima energy reaction pathway and thus is called a saddle point <ref name="ref2" />.
This can be expressed mathematically by Taylor's Theorem <ref name="ref3" />:
:If fxxfyy − fx2y < 0 at (a, b) then (a, b) is a saddle point.
:If fxx >0 and fyy > 0 at (a,b) then (a,b) is a minimum point.

{{fontcolor1|gray| At the TS one of the partial 2nd derivatives will be >0 along one of the variables, and <0 along all the rest. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:36, 31 May 2017 (BST)}}

===Trajectories from r1 = r2: locating the transition state===

The best estimate of the transition state position rts is found to be equal to 0.90775 Å. The detection of the transition structure can be observed in an internuclear distances vs time plot. The transition state corresponds to where the gradient of the potential energy surface is zero, so it is expected that at rts the H atoms will be stationary - their internuclear separations will remain constant over time (fig 1). At a distance of 0.90775 Å the trajectory disappears (fig 2) confirming this to be a good approximation of the transition state position. Varying the interatomic distance away from the transition state position results in a trajectory that rolls back toward the reactants/products and thus a changing internuclear distance over time is observed (fig 3). {{fontcolor1|gray| Really? That's not what figure 3 is showing. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:37, 31 May 2017 (BST)}}

[[File:lem215_fig1.png|thumb|center|Figure 1 - internuclear distance vs time plot at the transition state position.]]
[[File:lem215_fig2.png|thumb|center|Figure 2 - disappearance of the trajectory at the transition state position.]]
[[File:lem215_fig3.png|thumb|center|Figure 3 - internuclear distance vs time plot at r=1.5 A.]]
[[File:lem215_fig4.png|thumb|center|Figure 4 - surface plot displaying the trajectory at r=1.5 A.]]

===Minimum Energy Path (MEP) and Dynamic Calculations===

Both calculations produce trajectories that rolls from the reactants H1-H2 H3 in the direction of increasing r1 towards H1 H2-H3 (r1 = A-B) Upon energy minimisation the trajectory converges to the MEP <ref name="ref4" />.
The MEP calculation generates the minimum energy trajectory thus is restricted to only one degree of freedom, the interatomic positions r1 and r2. The exchange between kinetic and potential energy (due to vibrations) does not occur as a result of the minimisation, and so the generated trajectory does not have an oscillative nature - the kinetic energy is fixed at zero.
On the contrary the dynamics calculation does not involve such limitations allowing it to have more degrees of freedom generating an oscillating trajectory, signifying influence of a non-zero kinetic energy.

[[File:lem215_mep.png|thumb|center|MEP Calculation Trajectory.]]
[[File:lem215_dynamics.png|thumb|center|Dynamic Trajectory Calculation.]]

At large t (t): (3 decimal places)
:r1(1.485)= 5.247 Å
:r2(1.485) = 0.739 Å
:p1(1.45) = 2.481
:p2(1.45) = 1.260

Reverting the initial conditions so that r1 = rts + 0.01 and r2 = rts then the reaction would have a trajectory that rolls from H1 H2-H3 towards H1-H2 H3 in the direction of increasing r2.

If the initial positions and final positions were swapped then the corresponding values for final momenta will have the opposite sign.

===Reactive and unreactive trajectories===

A series of initial condition were tested to determine whether or not they produce a reactive trajectory that starts from the reactants HA-HB and dissociated atom HC to the molecular product HB-HC with dissociated atom HA. Initial momenta p1 of atom HC approaching HB before the transition state and final momenta p2 of atom HA moving away from the molecular product HB-HC

{| class="wikitable" border="1" style="margin-left: auto; margin-right: auto; border: none; width: 60%; height: 200px"
|+ Initial Conditions
! p1 !! p2 !! Reactivity
|- style="text-align: center;"
| -1.25 || -2.5 || Reactive
|- style="text-align: center;"
| -1.5 || -2.0 || Unreactive
|- style="text-align: center;"
| -1.5 || -2.5 || Reactive
|- style="text-align: center;"
| -2.5 || -5 || Unreactive
|- style="text-align: center;"
| -2.5 || -5.2 || Reactive
|}

[[File:lem215_fig5.png|thumb|center|Figure 5 - initial conditions p1=-1.25, p2=-2.5.]]
The first set of conditions generates a linear trajectory path from HA-HB HC to the transition structure at approximately 0.908 Å. Before the collision r1 remains constant and r2 steadily decreases as H3 approaches H2. There is sufficient energy to overcome the activation energy barrier allowing the trajectory to roll towards the products. After the transition structure r1 steadily increases as HA moves away from HB from the molecule formed HB-HC, the oscillating trajectory signifies the vibrational energy of the HB-HC bond.

[[File:lem215_fig6.png|thumb|center|Figure 6 - initial conditions p1=-1.5, p2=-2.0.]]
Increasing p1 and p2 (initial momenta), relative to the first set of conditions, results in an unreactive trajectory. Before the transition structure r1 has a slight oscillation (vibrational energy of A-B) and r2 steadily decreases as HC approaches HB. At r1=0.810 Å and r2=1.120 Å the trajectory bounces back toward the reactants. This suggests amount of energy available to the system is not sufficient to overcome the activation energy barrier - the reaction is unsuccessful. It is evident that the vibrational energy of HA-HB is retained upon the return of the system to the reactants.

[[File:lem215_fig7.png|thumb|center|Figure 7 - initial conditions p1=-1.5, p2=-2.5.]]
Increasing p2 from -2.0 to -2.5 provides the energy required to overcome the barrier thus allowing the reaction to proceed. The trajectory rolls from the reactants to the products passing through the transition structure at 0.893 Å. Again the vibrational energy of the HA-HB then HB-HC bonds are observed before and after passing through the transition structure respectively.

[[File:lem215_fig8.png|thumb|center|Figure 8 - initial conditions p1=-2.5, p2=-5.0.]]
With the fourth set of initial momenta conditions the trajectory rolls from the reactants and passes through the transition structure illustrating an initial formation of the HB-HC bond. Soon after crossing the transition region the trajectory returns to the reactants - this is an example barrier re-crossing. The trajectory bounces several times within a small region about transition state position before returning back to the reactants.

[[File:lem215_fig9.png|thumb|center|Figure 9 - initial conditions p1=-2.5, p2=-5.2.]]
The trajectory approaches the transition structure along the direction of decreasing r2 at a constant r1. Again the trajectory displays barrier re-crossing before eventually bouncing across the barrier toward the product HB-HC and dissociated atom HA.

The final two sets of initial momenta have relatively large p2 values. Both systems have significant translational kinetic energy as H atom A moves away from the formed molecule HB-HC. At high kinetic translational energies of atoms/molecules it becomes increasingly difficult to justify the differences in the reactivity of the generated trajectories. This suggests the application of Transition State Theory is unsuitable for dynamic molecular systems of this type - this method is limiting.

====Transition State Theory====

The transition state is the known as the 'critical configuration' that corresponds to the highest point on a minimum energy reaction pathway on a potential energy surface <ref name="ref5" />. If the system is able to reach the critical geometry then a successful reaction is highly probable. The transition structure is a stationary point thus is identified as the saddle point, with associated reaction coordinates, on a potential energy surface.
The transition state theory is a cumulative a series of assumptions used to approximate the equilibrium rate constants for thermally driven molecular reactions <ref name="t" />. The TST stemmed from thermodynamic, kinetic-theory and statistical mechanic treatments <ref name="t" />.

=====Development of the Transition State Theory=====

The thermodynamic treatment stems from the Van't Hoff equation <ref name="t" /> which relates the change in the equilibrium constant (keq) to a given change in temperature (T) for a given enthalpy change (delta H):
:<math>\ln K_\mathrm{eq} = - \frac{{\Delta H^\ominus}}{RT}+ \frac{{\Delta S^\ominus }}{R}.</math>.
The kinetic-theory <ref name="t" /> explains the temperature dependence of reaction rates as a function of collision frequency and the statistical mechanical treatment focuses on equilibrium reaction rates in terms of the motion of molecules and statistical distribution of molecular speeds <ref name="t" /> - initially modelled by Maxwell and Boltzmann.

=====Main Features and Assumptions of the Transition State Theory <ref name="t"/> =====

:(1) Rates can be calculated by focusing on the "activated complex" - what happens before the transition state is reached are not important if one rate dominates the overall rate of reaction.
:(2) The activated complex is formed in 'quasi-equilibrium' i.e. the complex is in equilibrium with the reactants and products passing through - the concentrations of reactants and products can be determined by changing the concentration of one species e.g. removing the products.
:(3) Motion of the system is treated as 'free translational motion' - expressed by kinetic-theory.

=====Using the Transition State Theory to Determine Reactivity=====

The TST ignores the dynamical recrossings that can lead back to the reactants (also known as tunneling) so this calculation tends to overestimate the rates within the classical mechanic framework <ref name ="t"/>. For a successful reaction the molecular system must have sufficient energy to overcome the activation energy barrier and thus generate a reactive trajectory. The Arrhenius equation relates the activate energy to the reaction rate for a particular reaction scheme, k=Ae(-Ea/RT) <ref name ="ar"/>. The relationship shows how a small activation energy barrier will correspond to a larger reaction rate constant - the reaction will go faster. It is likely that the experimental activation energy barrier will be larger than the calculated value thus TST will likely overestimate the rate relative the the experimental rate for a given reaction.

{{fontcolor1|gray| Good job looking up information. Just a small point -- the statistical mechanical treatment does not really hold here, since we are not looking at an ensemble of particles, just at a simple triatomic collision in isolation. The biggest point to be made here, was the recrossing limitation, and how that overestimates the rates when compared to experiment. Because that's the most relevant limitation that you observe in this lab. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:40, 31 May 2017 (BST)}}

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

===Reaction Energetics===

The F + H2 reaction is exothermic. An exothermic reaction is defined as the reaction path from a higher (reactants) to lower total potential energy (products) <ref name="ref6"/>. The trajectory shows the approach of F to H2 from infinite separation (as it is dissociated): as A-B separation decreases the potential energy decreases to a minimum indicating the H-F bond formation and H-H bond dissociation is an exothermic process.
[[File:lem215_1.png|thumb|center|Figure 10 - Surface Plot Trajectory from F + H2 to F-H + H.]]

By definition the reverse of an exothermic reaction is endothermic <ref name="ref6"/>, this is indicated by the trajectory generate by the H-F + H reaction. H-F bond breaking and H-H bond forming is an endothermic and enthalpically unfavourable process. The reactants have a lower total potential energy than the products.
[[File:lem215_2.png|thumb|center|Figure 11 - Surface Plot Trajectory from F-H + H to F + H2]]

The H-F bond formation has a more negative (exothermic) reaction enthalpy indicating H-F to have a stronger bond than H2.

===Location of Transition State and Determination of the Activation Energies===

Rates of reactions are controlled by the free energy of the transition state and Hammond postulated that the structure of a given transition state will more closely resemble that of the reactants or products depending on which is closer in energy <ref name="h"/>. He concluded that for an exothermic reaction the transition state will be more structurally similar to the reactants i.e. the interconversion between the reactants and transition state will involve only a small reorganization of molecular structure. And conversely for an endothermic reaction the transition state will be more structurally related to the products.
The transition structure is expected to be closer in energy and structure to the reactants - longer A-B and shorter B-C distance so point of intersection further right in the saddle region was selected to determine the location (see figure 12) and energy of the transition state.

Approximate location of the transition state (3 dp):
:A-B (F-H) = 1.713 Å
:B-C (H-H) = 0.748 Å

[[File:lem215_Hpost.png|thumb|center|Figure 12 - Locating the transition state for a F-H-H system.]]

Activation energies (2 dp):

[[File:lem215_Hpost1.png|thumb|center|Figure 13 - Determination of the absolute energy for products (F-H + H).]]

:Energy of the reactants (F + H2) = -103.80 kcal/mol
:Energy of the products (F-H + H) = -134.00 kcal/mol
:Energy of the transition structure = -96.47 kcal/mol
:Activation energy for formation of HF + H = +7.33 kcal/mol (=-96.47--103.80)
:Activation energy for formation of F + H2 = +37.53 kcal/mol (=-96.47--134.00)

===Reaction Dynamics===
====Mechanism of Release of Reactant Energy====

[[File:lem215_rt.png|thumb|center|Figure 14 - Reactive Trajectory for F + H2.]]

The amplitude of the oscillating trajectory reflects the magnitude of the defined initial momenta values. The reactants produce an oscillation of smaller amplitude than the products as p1 (H-H) < p2 (F-H).
The trajectories before and after the transition state however differ by the period of the generated oscillations - this indicates differences in the translational kinetic energy in the reactants and products. It has been discussed previously that the molecular product H-F has a stronger bond than H-H so it is expected to ‘see’ a greater translational kinetic energy in the products than the reactants. This is displayed by the trajectory, the oscillation period for the product H-F (A-B) is larger than that of the reactant H-H (B-C).
As a result of the conservation of energy, the total energy of the system must remain constant <ref name="ref7"/>. Therefore the product state should have a lower potential energy than the reactants to balance the total energy (kinetic and potential) of the system so that it remains constant, suggesting the reaction to be exothermic - earlier concluded to be the case as for the reaction for F + H2. This could be confirmed by different spectroscopic methods such as calorimetry and IR/Raman spectroscopy.

:Heat-flow calorimetry is commonly used to thermally monitor reactions and thus determine whether a reaction is exothermic or endothermic. The heat transfer under e.g. constant pressure is monitored as the reaction proceeds and consequently the enthalpy change can be determined. Thermodynamically ΔH=qp at constant pressure (where qp is the heat flow at constant pressure), otherwise known as Hess's Law <ref name="ref8"/>. Exothermic reactions release heat thus if exothermic the calculated value for the enthalpy change will be negative in sign (visa versa for endothermic reactions).

:IR/Raman spectroscopy is used to determine the vibrational states of molecules/matter. If the reactants possess less translational kinetic energy and thus vibrational energy then the IR spectra for the reactants should display lower intensity peaks relative to the peaks in the IR spectra of the products.

====Polanyi's empirical rules====

Polanyi's empirical rules focus on the importance of the transition state location in enhancing reactivity for activated reactions <ref name="ref9"/>. This model has enabled the prediction of the relative efficacy of vibrational and translational excitations in promoting different molecular dynamic reactions.
:Polanyi's rules <ref name="ref9"/> state that vibrational energy is more efficient at promoting late transition states (endothermic reactions according to Hammond Postulate <ref name="h"/>) than translational energy. And conversely translational kinetic energy is more efficient at promoting 'early' transition states (exothermic reactions).

The reaction from H + H-F provides an illustration of these rules. This reaction is endothermic, thus is an example of a late-barrier reaction - the transition state is closer to the products in energy and structure according to Hammond's Postulate <ref name="h"/>. According to the Polanyi's Empirical Rules the reactivity should be enhanced more efficiently by high vibrational energy (and low kinetic energy). When initial conditions were chosen so that the reactant H-H had high vibrational energy (high pHH) and low translational kinetic energy (small pHF) a kinetically driven trajectory was generated.

Not all systems obey Polanyi's Rules as a result of their empirical nature (experimentally determined). As the rules are experimentally determined they effectively relate to the average behaviour of a reacting system <ref name="ref10"/>. The reaction H2 + F is an example where the calculated trajectory deviates from Polanyi's predictions. This reaction is exothermic so according to Polanyi's rules it is expected that high translational kinetic energy would promote the early transition state and thus enhance the reactivity. However for the reactive trajectory to reach completion a higher initial momenta for H-H (pHH) than initial translational kinetic energy (pHF) was required. This may be due to the errors in the computational calculation of the reaction trajectory thus does not accurately portray was would be observed experimentally.

References:
<references>
<ref name="a">Laganà, Antonio, and Antonio Riganelli. Reaction And Molecular Dynamics. 1st ed. Berlin, Heidelberg: Springer Berlin Heidelberg, 2000. Print..</ref>
<ref name="ref1">Collett, Charles T, and Christopher D Robson. Handbook Of Computational Chemistry Research. 1st ed. New York: Nova Science Publishers, 2010.</ref>
<ref name="ref2">Lewars, Errol G. "The Concept Of The Potential Energy Surface". Computational Chemistry (2010): 9-43. Web. 11 May 2017.</ref>
<ref name="ref3">Jeffrey, Alan. Essentials Of Engineering Mathematics. 1st ed. Boca Raton: Chapman & Hall/CRC Press, 2004. Print.</ref>
<ref name="ref4">Koistinen, O-P. et al. "Minimum Energy Path Calculations With Gaussian Process Regression". Nanosystems: Physics, Chemistry, Mathematics (2016): 925-935. Web. 20 May</ref>
<ref name="t">Laidler, Keith J., and M. Christine King. "Development Of Transition-State Theory". The Journal of Physical Chemistry 87.15 (1983): 2657-2664. Web. 26 May 2017.</ref>
<ref name="ref5">Fueno, Takayuki. Transition State: A Theoretical Approach. 1st ed. Tokyo: Gordon and Breach Science Publishers, 1999. Print.</ref>
<ref name="ar">Kotz, John C, and Paul Treichel. Chemistry & Chemical Reactivity. 1st ed. Fort Worth: Saunders College Pub., 1996. Print.</ref>
<ref name="ref6">Petrucci, Ralph H. General Chemistry. 1st ed. Toronto, Ont.: Pearson Canada, 2011. Print.</ref>
<ref name="h">Carey, Francis A, and Richard J Sundberg. Advanced Organic Chemistry. 1st ed. Norwell: Springer, 2007. Print.</ref>
<ref name="ref7">Khine, Myint Swe, and Issa M Saleh. Models And Modeling. 1st ed. Dordrecht: Springer Science+Business Media B.V, 2011</ref>
<ref name="ref8">Mikulecky, Peter, Michelle Rose Gilman, and Kate Brutlag. AP Chemistry For Dummies. 1st ed. Hoboken, N.J.: Wiley Pub., 2009. Print.</ref>
<ref name="ref9">J. Clayden, N. Greeves and S. G. Warren, Organic Chemistry, Oxford University Press, Oxford, 2012</ref>
<ref name="ref10">Zhang, Zhaojun et al. "Theoretical Study Of The Validity Of The Polanyi Rules For The Late-Barrier Cl + Chd3reaction". The Journal of Physical Chemistry Letters 3.23 (2012): 3416-3419. Web. 26 May 2017.</ref>
</ref>
</references>

MRD:LaraMetcalf

2017-05-31T16:37:36Z

Je714: /* Trajectories from r1 = r2: locating the transition state */

= Molecular Reaction Dynamics Lab =
==EXERCISE 1: H + H2 system==

A potential energy surface (PES) is a mathematical function which describes the electronic potential energy of a system as a function of the relative atomic positions in space and is widely used as a tool to carry out theoretical studies on molecular reaction dynamics <ref name="a" />.

===Dynamics from the transition state region===

The total gradient of the potential energy surface at a minimum and transition structure is equal to zero. The distinction between the two structures comes from the sign corresponding to their rate of change of gradient. The potential energy second derivative is positive for a minima and negative for a transition state <ref name="ref1" />. {{fontcolor1|gray| Which one? Along which direction? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:36, 31 May 2017 (BST)}}
The transition structure linking the two minima is a maximum along the minima energy reaction pathway and thus is called a saddle point <ref name="ref2" />.
This can be expressed mathematically by Taylor's Theorem <ref name="ref3" />:
:If fxxfyy − fx2y < 0 at (a, b) then (a, b) is a saddle point.
:If fxx >0 and fyy > 0 at (a,b) then (a,b) is a minimum point.

{{fontcolor1|gray| At the TS one of the partial 2nd derivatives will be >0 along one of the variables, and <0 along all the rest. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:36, 31 May 2017 (BST)}}

===Trajectories from r1 = r2: locating the transition state===

The best estimate of the transition state position rts is found to be equal to 0.90775 Å. The detection of the transition structure can be observed in an internuclear distances vs time plot. The transition state corresponds to where the gradient of the potential energy surface is zero, so it is expected that at rts the H atoms will be stationary - their internuclear separations will remain constant over time (fig 1). At a distance of 0.90775 Å the trajectory disappears (fig 2) confirming this to be a good approximation of the transition state position. Varying the interatomic distance away from the transition state position results in a trajectory that rolls back toward the reactants/products and thus a changing internuclear distance over time is observed (fig 3). {{fontcolor1|gray| Really? That's not what figure 3 is showing. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:37, 31 May 2017 (BST)}}

[[File:lem215_fig1.png|thumb|center|Figure 1 - internuclear distance vs time plot at the transition state position.]]
[[File:lem215_fig2.png|thumb|center|Figure 2 - disappearance of the trajectory at the transition state position.]]
[[File:lem215_fig3.png|thumb|center|Figure 3 - internuclear distance vs time plot at r=1.5 A.]]
[[File:lem215_fig4.png|thumb|center|Figure 4 - surface plot displaying the trajectory at r=1.5 A.]]

===Minimum Energy Path (MEP) and Dynamic Calculations===

Both calculations produce trajectories that rolls from the reactants H1-H2 H3 in the direction of increasing r1 towards H1 H2-H3 (r1 = A-B) Upon energy minimisation the trajectory converges to the MEP <ref name="ref4" />.
The MEP calculation generates the minimum energy trajectory thus is restricted to only one degree of freedom, the interatomic positions r1 and r2. The exchange between kinetic and potential energy (due to vibrations) does not occur as a result of the minimisation, and so the generated trajectory does not have an oscillative nature - the kinetic energy is fixed at zero.
On the contrary the dynamics calculation does not involve such limitations allowing it to have more degrees of freedom generating an oscillating trajectory, signifying influence of a non-zero kinetic energy.

[[File:lem215_mep.png|thumb|center|MEP Calculation Trajectory.]]
[[File:lem215_dynamics.png|thumb|center|Dynamic Trajectory Calculation.]]

At large t (t): (3 decimal places)
:r1(1.485)= 5.247 Å
:r2(1.485) = 0.739 Å
:p1(1.45) = 2.481
:p2(1.45) = 1.260

Reverting the initial conditions so that r1 = rts + 0.01 and r2 = rts then the reaction would have a trajectory that rolls from H1 H2-H3 towards H1-H2 H3 in the direction of increasing r2.

If the initial positions and final positions were swapped then the corresponding values for final momenta will have the opposite sign.

===Reactive and unreactive trajectories===

A series of initial condition were tested to determine whether or not they produce a reactive trajectory that starts from the reactants HA-HB and dissociated atom HC to the molecular product HB-HC with dissociated atom HA. Initial momenta p1 of atom HC approaching HB before the transition state and final momenta p2 of atom HA moving away from the molecular product HB-HC

{| class="wikitable" border="1" style="margin-left: auto; margin-right: auto; border: none; width: 60%; height: 200px"
|+ Initial Conditions
! p1 !! p2 !! Reactivity
|- style="text-align: center;"
| -1.25 || -2.5 || Reactive
|- style="text-align: center;"
| -1.5 || -2.0 || Unreactive
|- style="text-align: center;"
| -1.5 || -2.5 || Reactive
|- style="text-align: center;"
| -2.5 || -5 || Unreactive
|- style="text-align: center;"
| -2.5 || -5.2 || Reactive
|}

[[File:lem215_fig5.png|thumb|center|Figure 5 - initial conditions p1=-1.25, p2=-2.5.]]
The first set of conditions generates a linear trajectory path from HA-HB HC to the transition structure at approximately 0.908 Å. Before the collision r1 remains constant and r2 steadily decreases as H3 approaches H2. There is sufficient energy to overcome the activation energy barrier allowing the trajectory to roll towards the products. After the transition structure r1 steadily increases as HA moves away from HB from the molecule formed HB-HC, the oscillating trajectory signifies the vibrational energy of the HB-HC bond.

[[File:lem215_fig6.png|thumb|center|Figure 6 - initial conditions p1=-1.5, p2=-2.0.]]
Increasing p1 and p2 (initial momenta), relative to the first set of conditions, results in an unreactive trajectory. Before the transition structure r1 has a slight oscillation (vibrational energy of A-B) and r2 steadily decreases as HC approaches HB. At r1=0.810 Å and r2=1.120 Å the trajectory bounces back toward the reactants. This suggests amount of energy available to the system is not sufficient to overcome the activation energy barrier - the reaction is unsuccessful. It is evident that the vibrational energy of HA-HB is retained upon the return of the system to the reactants.

[[File:lem215_fig7.png|thumb|center|Figure 7 - initial conditions p1=-1.5, p2=-2.5.]]
Increasing p2 from -2.0 to -2.5 provides the energy required to overcome the barrier thus allowing the reaction to proceed. The trajectory rolls from the reactants to the products passing through the transition structure at 0.893 Å. Again the vibrational energy of the HA-HB then HB-HC bonds are observed before and after passing through the transition structure respectively.

[[File:lem215_fig8.png|thumb|center|Figure 8 - initial conditions p1=-2.5, p2=-5.0.]]
With the fourth set of initial momenta conditions the trajectory rolls from the reactants and passes through the transition structure illustrating an initial formation of the HB-HC bond. Soon after crossing the transition region the trajectory returns to the reactants - this is an example barrier re-crossing. The trajectory bounces several times within a small region about transition state position before returning back to the reactants.

[[File:lem215_fig9.png|thumb|center|Figure 9 - initial conditions p1=-2.5, p2=-5.2.]]
The trajectory approaches the transition structure along the direction of decreasing r2 at a constant r1. Again the trajectory displays barrier re-crossing before eventually bouncing across the barrier toward the product HB-HC and dissociated atom HA.

The final two sets of initial momenta have relatively large p2 values. Both systems have significant translational kinetic energy as H atom A moves away from the formed molecule HB-HC. At high kinetic translational energies of atoms/molecules it becomes increasingly difficult to justify the differences in the reactivity of the generated trajectories. This suggests the application of Transition State Theory is unsuitable for dynamic molecular systems of this type - this method is limiting.

====Transition State Theory====

The transition state is the known as the 'critical configuration' that corresponds to the highest point on a minimum energy reaction pathway on a potential energy surface <ref name="ref5" />. If the system is able to reach the critical geometry then a successful reaction is highly probable. The transition structure is a stationary point thus is identified as the saddle point, with associated reaction coordinates, on a potential energy surface.
The transition state theory is a cumulative a series of assumptions used to approximate the equilibrium rate constants for thermally driven molecular reactions <ref name="t" />. The TST stemmed from thermodynamic, kinetic-theory and statistical mechanic treatments <ref name="t" />.

=====Development of the Transition State Theory=====

The thermodynamic treatment stems from the Van't Hoff equation <ref name="t" /> which relates the change in the equilibrium constant (keq) to a given change in temperature (T) for a given enthalpy change (delta H):
:<math>\ln K_\mathrm{eq} = - \frac{{\Delta H^\ominus}}{RT}+ \frac{{\Delta S^\ominus }}{R}.</math>.
The kinetic-theory <ref name="t" /> explains the temperature dependence of reaction rates as a function of collision frequency and the statistical mechanical treatment focuses on equilibrium reaction rates in terms of the motion of molecules and statistical distribution of molecular speeds <ref name="t" /> - initially modelled by Maxwell and Boltzmann.

=====Main Features and Assumptions of the Transition State Theory <ref name="t"/> =====

:(1) Rates can be calculated by focusing on the "activated complex" - what happens before the transition state is reached are not important if one rate dominates the overall rate of reaction.
:(2) The activated complex is formed in 'quasi-equilibrium' i.e. the complex is in equilibrium with the reactants and products passing through - the concentrations of reactants and products can be determined by changing the concentration of one species e.g. removing the products.
:(3) Motion of the system is treated as 'free translational motion' - expressed by kinetic-theory.

=====Using the Transition State Theory to Determine Reactivity=====

The TST ignores the dynamical recrossings that can lead back to the reactants (also known as tunneling) so this calculation tends to overestimate the rates within the classical mechanic framework <ref name ="t"/>. For a successful reaction the molecular system must have sufficient energy to overcome the activation energy barrier and thus generate a reactive trajectory. The Arrhenius equation relates the activate energy to the reaction rate for a particular reaction scheme, k=Ae(-Ea/RT) <ref name ="ar"/>. The relationship shows how a small activation energy barrier will correspond to a larger reaction rate constant - the reaction will go faster. It is likely that the experimental activation energy barrier will be larger than the calculated value thus TST will likely overestimate the rate relative the the experimental rate for a given reaction.

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

===Reaction Energetics===

The F + H2 reaction is exothermic. An exothermic reaction is defined as the reaction path from a higher (reactants) to lower total potential energy (products) <ref name="ref6"/>. The trajectory shows the approach of F to H2 from infinite separation (as it is dissociated): as A-B separation decreases the potential energy decreases to a minimum indicating the H-F bond formation and H-H bond dissociation is an exothermic process.
[[File:lem215_1.png|thumb|center|Figure 10 - Surface Plot Trajectory from F + H2 to F-H + H.]]

By definition the reverse of an exothermic reaction is endothermic <ref name="ref6"/>, this is indicated by the trajectory generate by the H-F + H reaction. H-F bond breaking and H-H bond forming is an endothermic and enthalpically unfavourable process. The reactants have a lower total potential energy than the products.
[[File:lem215_2.png|thumb|center|Figure 11 - Surface Plot Trajectory from F-H + H to F + H2]]

The H-F bond formation has a more negative (exothermic) reaction enthalpy indicating H-F to have a stronger bond than H2.

===Location of Transition State and Determination of the Activation Energies===

Rates of reactions are controlled by the free energy of the transition state and Hammond postulated that the structure of a given transition state will more closely resemble that of the reactants or products depending on which is closer in energy <ref name="h"/>. He concluded that for an exothermic reaction the transition state will be more structurally similar to the reactants i.e. the interconversion between the reactants and transition state will involve only a small reorganization of molecular structure. And conversely for an endothermic reaction the transition state will be more structurally related to the products.
The transition structure is expected to be closer in energy and structure to the reactants - longer A-B and shorter B-C distance so point of intersection further right in the saddle region was selected to determine the location (see figure 12) and energy of the transition state.

Approximate location of the transition state (3 dp):
:A-B (F-H) = 1.713 Å
:B-C (H-H) = 0.748 Å

[[File:lem215_Hpost.png|thumb|center|Figure 12 - Locating the transition state for a F-H-H system.]]

Activation energies (2 dp):

[[File:lem215_Hpost1.png|thumb|center|Figure 13 - Determination of the absolute energy for products (F-H + H).]]

:Energy of the reactants (F + H2) = -103.80 kcal/mol
:Energy of the products (F-H + H) = -134.00 kcal/mol
:Energy of the transition structure = -96.47 kcal/mol
:Activation energy for formation of HF + H = +7.33 kcal/mol (=-96.47--103.80)
:Activation energy for formation of F + H2 = +37.53 kcal/mol (=-96.47--134.00)

===Reaction Dynamics===
====Mechanism of Release of Reactant Energy====

[[File:lem215_rt.png|thumb|center|Figure 14 - Reactive Trajectory for F + H2.]]

The amplitude of the oscillating trajectory reflects the magnitude of the defined initial momenta values. The reactants produce an oscillation of smaller amplitude than the products as p1 (H-H) < p2 (F-H).
The trajectories before and after the transition state however differ by the period of the generated oscillations - this indicates differences in the translational kinetic energy in the reactants and products. It has been discussed previously that the molecular product H-F has a stronger bond than H-H so it is expected to ‘see’ a greater translational kinetic energy in the products than the reactants. This is displayed by the trajectory, the oscillation period for the product H-F (A-B) is larger than that of the reactant H-H (B-C).
As a result of the conservation of energy, the total energy of the system must remain constant <ref name="ref7"/>. Therefore the product state should have a lower potential energy than the reactants to balance the total energy (kinetic and potential) of the system so that it remains constant, suggesting the reaction to be exothermic - earlier concluded to be the case as for the reaction for F + H2. This could be confirmed by different spectroscopic methods such as calorimetry and IR/Raman spectroscopy.

:Heat-flow calorimetry is commonly used to thermally monitor reactions and thus determine whether a reaction is exothermic or endothermic. The heat transfer under e.g. constant pressure is monitored as the reaction proceeds and consequently the enthalpy change can be determined. Thermodynamically ΔH=qp at constant pressure (where qp is the heat flow at constant pressure), otherwise known as Hess's Law <ref name="ref8"/>. Exothermic reactions release heat thus if exothermic the calculated value for the enthalpy change will be negative in sign (visa versa for endothermic reactions).

:IR/Raman spectroscopy is used to determine the vibrational states of molecules/matter. If the reactants possess less translational kinetic energy and thus vibrational energy then the IR spectra for the reactants should display lower intensity peaks relative to the peaks in the IR spectra of the products.

====Polanyi's empirical rules====

Polanyi's empirical rules focus on the importance of the transition state location in enhancing reactivity for activated reactions <ref name="ref9"/>. This model has enabled the prediction of the relative efficacy of vibrational and translational excitations in promoting different molecular dynamic reactions.
:Polanyi's rules <ref name="ref9"/> state that vibrational energy is more efficient at promoting late transition states (endothermic reactions according to Hammond Postulate <ref name="h"/>) than translational energy. And conversely translational kinetic energy is more efficient at promoting 'early' transition states (exothermic reactions).

The reaction from H + H-F provides an illustration of these rules. This reaction is endothermic, thus is an example of a late-barrier reaction - the transition state is closer to the products in energy and structure according to Hammond's Postulate <ref name="h"/>. According to the Polanyi's Empirical Rules the reactivity should be enhanced more efficiently by high vibrational energy (and low kinetic energy). When initial conditions were chosen so that the reactant H-H had high vibrational energy (high pHH) and low translational kinetic energy (small pHF) a kinetically driven trajectory was generated.

Not all systems obey Polanyi's Rules as a result of their empirical nature (experimentally determined). As the rules are experimentally determined they effectively relate to the average behaviour of a reacting system <ref name="ref10"/>. The reaction H2 + F is an example where the calculated trajectory deviates from Polanyi's predictions. This reaction is exothermic so according to Polanyi's rules it is expected that high translational kinetic energy would promote the early transition state and thus enhance the reactivity. However for the reactive trajectory to reach completion a higher initial momenta for H-H (pHH) than initial translational kinetic energy (pHF) was required. This may be due to the errors in the computational calculation of the reaction trajectory thus does not accurately portray was would be observed experimentally.

References:
<references>
<ref name="a">Laganà, Antonio, and Antonio Riganelli. Reaction And Molecular Dynamics. 1st ed. Berlin, Heidelberg: Springer Berlin Heidelberg, 2000. Print..</ref>
<ref name="ref1">Collett, Charles T, and Christopher D Robson. Handbook Of Computational Chemistry Research. 1st ed. New York: Nova Science Publishers, 2010.</ref>
<ref name="ref2">Lewars, Errol G. "The Concept Of The Potential Energy Surface". Computational Chemistry (2010): 9-43. Web. 11 May 2017.</ref>
<ref name="ref3">Jeffrey, Alan. Essentials Of Engineering Mathematics. 1st ed. Boca Raton: Chapman & Hall/CRC Press, 2004. Print.</ref>
<ref name="ref4">Koistinen, O-P. et al. "Minimum Energy Path Calculations With Gaussian Process Regression". Nanosystems: Physics, Chemistry, Mathematics (2016): 925-935. Web. 20 May</ref>
<ref name="t">Laidler, Keith J., and M. Christine King. "Development Of Transition-State Theory". The Journal of Physical Chemistry 87.15 (1983): 2657-2664. Web. 26 May 2017.</ref>
<ref name="ref5">Fueno, Takayuki. Transition State: A Theoretical Approach. 1st ed. Tokyo: Gordon and Breach Science Publishers, 1999. Print.</ref>
<ref name="ar">Kotz, John C, and Paul Treichel. Chemistry & Chemical Reactivity. 1st ed. Fort Worth: Saunders College Pub., 1996. Print.</ref>
<ref name="ref6">Petrucci, Ralph H. General Chemistry. 1st ed. Toronto, Ont.: Pearson Canada, 2011. Print.</ref>
<ref name="h">Carey, Francis A, and Richard J Sundberg. Advanced Organic Chemistry. 1st ed. Norwell: Springer, 2007. Print.</ref>
<ref name="ref7">Khine, Myint Swe, and Issa M Saleh. Models And Modeling. 1st ed. Dordrecht: Springer Science+Business Media B.V, 2011</ref>
<ref name="ref8">Mikulecky, Peter, Michelle Rose Gilman, and Kate Brutlag. AP Chemistry For Dummies. 1st ed. Hoboken, N.J.: Wiley Pub., 2009. Print.</ref>
<ref name="ref9">J. Clayden, N. Greeves and S. G. Warren, Organic Chemistry, Oxford University Press, Oxford, 2012</ref>
<ref name="ref10">Zhang, Zhaojun et al. "Theoretical Study Of The Validity Of The Polanyi Rules For The Late-Barrier Cl + Chd3reaction". The Journal of Physical Chemistry Letters 3.23 (2012): 3416-3419. Web. 26 May 2017.</ref>
</ref>
</references>

MRD:LaraMetcalf

2017-05-31T16:36:38Z

Je714: /* Dynamics from the transition state region */

= Molecular Reaction Dynamics Lab =
==EXERCISE 1: H + H2 system==

A potential energy surface (PES) is a mathematical function which describes the electronic potential energy of a system as a function of the relative atomic positions in space and is widely used as a tool to carry out theoretical studies on molecular reaction dynamics <ref name="a" />.

===Dynamics from the transition state region===

The total gradient of the potential energy surface at a minimum and transition structure is equal to zero. The distinction between the two structures comes from the sign corresponding to their rate of change of gradient. The potential energy second derivative is positive for a minima and negative for a transition state <ref name="ref1" />. {{fontcolor1|gray| Which one? Along which direction? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:36, 31 May 2017 (BST)}}
The transition structure linking the two minima is a maximum along the minima energy reaction pathway and thus is called a saddle point <ref name="ref2" />.
This can be expressed mathematically by Taylor's Theorem <ref name="ref3" />:
:If fxxfyy − fx2y < 0 at (a, b) then (a, b) is a saddle point.
:If fxx >0 and fyy > 0 at (a,b) then (a,b) is a minimum point.

{{fontcolor1|gray| At the TS one of the partial 2nd derivatives will be >0 along one of the variables, and <0 along all the rest. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:36, 31 May 2017 (BST)}}

===Trajectories from r1 = r2: locating the transition state===

The best estimate of the transition state position rts is found to be equal to 0.90775 Å. The detection of the transition structure can be observed in an internuclear distances vs time plot. The transition state corresponds to where the gradient of the potential energy surface is zero, so it is expected that at rts the H atoms will be stationary - their internuclear separations will remain constant over time (fig 1). At a distance of 0.90775 Å the trajectory disappears (fig 2) confirming this to be a good approximation of the transition state position. Varying the interatomic distance away from the transition state position results in a trajectory that rolls back toward the reactants/products and thus a changing internuclear distance over time is observed (fig 3).

[[File:lem215_fig1.png|thumb|center|Figure 1 - internuclear distance vs time plot at the transition state position.]]
[[File:lem215_fig2.png|thumb|center|Figure 2 - disappearance of the trajectory at the transition state position.]]
[[File:lem215_fig3.png|thumb|center|Figure 3 - internuclear distance vs time plot at r=1.5 A.]]
[[File:lem215_fig4.png|thumb|center|Figure 4 - surface plot displaying the trajectory at r=1.5 A.]]

===Minimum Energy Path (MEP) and Dynamic Calculations===

Both calculations produce trajectories that rolls from the reactants H1-H2 H3 in the direction of increasing r1 towards H1 H2-H3 (r1 = A-B) Upon energy minimisation the trajectory converges to the MEP <ref name="ref4" />.
The MEP calculation generates the minimum energy trajectory thus is restricted to only one degree of freedom, the interatomic positions r1 and r2. The exchange between kinetic and potential energy (due to vibrations) does not occur as a result of the minimisation, and so the generated trajectory does not have an oscillative nature - the kinetic energy is fixed at zero.
On the contrary the dynamics calculation does not involve such limitations allowing it to have more degrees of freedom generating an oscillating trajectory, signifying influence of a non-zero kinetic energy.

[[File:lem215_mep.png|thumb|center|MEP Calculation Trajectory.]]
[[File:lem215_dynamics.png|thumb|center|Dynamic Trajectory Calculation.]]

At large t (t): (3 decimal places)
:r1(1.485)= 5.247 Å
:r2(1.485) = 0.739 Å
:p1(1.45) = 2.481
:p2(1.45) = 1.260

Reverting the initial conditions so that r1 = rts + 0.01 and r2 = rts then the reaction would have a trajectory that rolls from H1 H2-H3 towards H1-H2 H3 in the direction of increasing r2.

If the initial positions and final positions were swapped then the corresponding values for final momenta will have the opposite sign.

===Reactive and unreactive trajectories===

A series of initial condition were tested to determine whether or not they produce a reactive trajectory that starts from the reactants HA-HB and dissociated atom HC to the molecular product HB-HC with dissociated atom HA. Initial momenta p1 of atom HC approaching HB before the transition state and final momenta p2 of atom HA moving away from the molecular product HB-HC

{| class="wikitable" border="1" style="margin-left: auto; margin-right: auto; border: none; width: 60%; height: 200px"
|+ Initial Conditions
! p1 !! p2 !! Reactivity
|- style="text-align: center;"
| -1.25 || -2.5 || Reactive
|- style="text-align: center;"
| -1.5 || -2.0 || Unreactive
|- style="text-align: center;"
| -1.5 || -2.5 || Reactive
|- style="text-align: center;"
| -2.5 || -5 || Unreactive
|- style="text-align: center;"
| -2.5 || -5.2 || Reactive
|}

[[File:lem215_fig5.png|thumb|center|Figure 5 - initial conditions p1=-1.25, p2=-2.5.]]
The first set of conditions generates a linear trajectory path from HA-HB HC to the transition structure at approximately 0.908 Å. Before the collision r1 remains constant and r2 steadily decreases as H3 approaches H2. There is sufficient energy to overcome the activation energy barrier allowing the trajectory to roll towards the products. After the transition structure r1 steadily increases as HA moves away from HB from the molecule formed HB-HC, the oscillating trajectory signifies the vibrational energy of the HB-HC bond.

[[File:lem215_fig6.png|thumb|center|Figure 6 - initial conditions p1=-1.5, p2=-2.0.]]
Increasing p1 and p2 (initial momenta), relative to the first set of conditions, results in an unreactive trajectory. Before the transition structure r1 has a slight oscillation (vibrational energy of A-B) and r2 steadily decreases as HC approaches HB. At r1=0.810 Å and r2=1.120 Å the trajectory bounces back toward the reactants. This suggests amount of energy available to the system is not sufficient to overcome the activation energy barrier - the reaction is unsuccessful. It is evident that the vibrational energy of HA-HB is retained upon the return of the system to the reactants.

[[File:lem215_fig7.png|thumb|center|Figure 7 - initial conditions p1=-1.5, p2=-2.5.]]
Increasing p2 from -2.0 to -2.5 provides the energy required to overcome the barrier thus allowing the reaction to proceed. The trajectory rolls from the reactants to the products passing through the transition structure at 0.893 Å. Again the vibrational energy of the HA-HB then HB-HC bonds are observed before and after passing through the transition structure respectively.

[[File:lem215_fig8.png|thumb|center|Figure 8 - initial conditions p1=-2.5, p2=-5.0.]]
With the fourth set of initial momenta conditions the trajectory rolls from the reactants and passes through the transition structure illustrating an initial formation of the HB-HC bond. Soon after crossing the transition region the trajectory returns to the reactants - this is an example barrier re-crossing. The trajectory bounces several times within a small region about transition state position before returning back to the reactants.

[[File:lem215_fig9.png|thumb|center|Figure 9 - initial conditions p1=-2.5, p2=-5.2.]]
The trajectory approaches the transition structure along the direction of decreasing r2 at a constant r1. Again the trajectory displays barrier re-crossing before eventually bouncing across the barrier toward the product HB-HC and dissociated atom HA.

The final two sets of initial momenta have relatively large p2 values. Both systems have significant translational kinetic energy as H atom A moves away from the formed molecule HB-HC. At high kinetic translational energies of atoms/molecules it becomes increasingly difficult to justify the differences in the reactivity of the generated trajectories. This suggests the application of Transition State Theory is unsuitable for dynamic molecular systems of this type - this method is limiting.

====Transition State Theory====

The transition state is the known as the 'critical configuration' that corresponds to the highest point on a minimum energy reaction pathway on a potential energy surface <ref name="ref5" />. If the system is able to reach the critical geometry then a successful reaction is highly probable. The transition structure is a stationary point thus is identified as the saddle point, with associated reaction coordinates, on a potential energy surface.
The transition state theory is a cumulative a series of assumptions used to approximate the equilibrium rate constants for thermally driven molecular reactions <ref name="t" />. The TST stemmed from thermodynamic, kinetic-theory and statistical mechanic treatments <ref name="t" />.

=====Development of the Transition State Theory=====

The thermodynamic treatment stems from the Van't Hoff equation <ref name="t" /> which relates the change in the equilibrium constant (keq) to a given change in temperature (T) for a given enthalpy change (delta H):
:<math>\ln K_\mathrm{eq} = - \frac{{\Delta H^\ominus}}{RT}+ \frac{{\Delta S^\ominus }}{R}.</math>.
The kinetic-theory <ref name="t" /> explains the temperature dependence of reaction rates as a function of collision frequency and the statistical mechanical treatment focuses on equilibrium reaction rates in terms of the motion of molecules and statistical distribution of molecular speeds <ref name="t" /> - initially modelled by Maxwell and Boltzmann.

=====Main Features and Assumptions of the Transition State Theory <ref name="t"/> =====

:(1) Rates can be calculated by focusing on the "activated complex" - what happens before the transition state is reached are not important if one rate dominates the overall rate of reaction.
:(2) The activated complex is formed in 'quasi-equilibrium' i.e. the complex is in equilibrium with the reactants and products passing through - the concentrations of reactants and products can be determined by changing the concentration of one species e.g. removing the products.
:(3) Motion of the system is treated as 'free translational motion' - expressed by kinetic-theory.

=====Using the Transition State Theory to Determine Reactivity=====

The TST ignores the dynamical recrossings that can lead back to the reactants (also known as tunneling) so this calculation tends to overestimate the rates within the classical mechanic framework <ref name ="t"/>. For a successful reaction the molecular system must have sufficient energy to overcome the activation energy barrier and thus generate a reactive trajectory. The Arrhenius equation relates the activate energy to the reaction rate for a particular reaction scheme, k=Ae(-Ea/RT) <ref name ="ar"/>. The relationship shows how a small activation energy barrier will correspond to a larger reaction rate constant - the reaction will go faster. It is likely that the experimental activation energy barrier will be larger than the calculated value thus TST will likely overestimate the rate relative the the experimental rate for a given reaction.

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

===Reaction Energetics===

The F + H2 reaction is exothermic. An exothermic reaction is defined as the reaction path from a higher (reactants) to lower total potential energy (products) <ref name="ref6"/>. The trajectory shows the approach of F to H2 from infinite separation (as it is dissociated): as A-B separation decreases the potential energy decreases to a minimum indicating the H-F bond formation and H-H bond dissociation is an exothermic process.
[[File:lem215_1.png|thumb|center|Figure 10 - Surface Plot Trajectory from F + H2 to F-H + H.]]

By definition the reverse of an exothermic reaction is endothermic <ref name="ref6"/>, this is indicated by the trajectory generate by the H-F + H reaction. H-F bond breaking and H-H bond forming is an endothermic and enthalpically unfavourable process. The reactants have a lower total potential energy than the products.
[[File:lem215_2.png|thumb|center|Figure 11 - Surface Plot Trajectory from F-H + H to F + H2]]

The H-F bond formation has a more negative (exothermic) reaction enthalpy indicating H-F to have a stronger bond than H2.

===Location of Transition State and Determination of the Activation Energies===

Rates of reactions are controlled by the free energy of the transition state and Hammond postulated that the structure of a given transition state will more closely resemble that of the reactants or products depending on which is closer in energy <ref name="h"/>. He concluded that for an exothermic reaction the transition state will be more structurally similar to the reactants i.e. the interconversion between the reactants and transition state will involve only a small reorganization of molecular structure. And conversely for an endothermic reaction the transition state will be more structurally related to the products.
The transition structure is expected to be closer in energy and structure to the reactants - longer A-B and shorter B-C distance so point of intersection further right in the saddle region was selected to determine the location (see figure 12) and energy of the transition state.

Approximate location of the transition state (3 dp):
:A-B (F-H) = 1.713 Å
:B-C (H-H) = 0.748 Å

[[File:lem215_Hpost.png|thumb|center|Figure 12 - Locating the transition state for a F-H-H system.]]

Activation energies (2 dp):

[[File:lem215_Hpost1.png|thumb|center|Figure 13 - Determination of the absolute energy for products (F-H + H).]]

:Energy of the reactants (F + H2) = -103.80 kcal/mol
:Energy of the products (F-H + H) = -134.00 kcal/mol
:Energy of the transition structure = -96.47 kcal/mol
:Activation energy for formation of HF + H = +7.33 kcal/mol (=-96.47--103.80)
:Activation energy for formation of F + H2 = +37.53 kcal/mol (=-96.47--134.00)

===Reaction Dynamics===
====Mechanism of Release of Reactant Energy====

[[File:lem215_rt.png|thumb|center|Figure 14 - Reactive Trajectory for F + H2.]]

The amplitude of the oscillating trajectory reflects the magnitude of the defined initial momenta values. The reactants produce an oscillation of smaller amplitude than the products as p1 (H-H) < p2 (F-H).
The trajectories before and after the transition state however differ by the period of the generated oscillations - this indicates differences in the translational kinetic energy in the reactants and products. It has been discussed previously that the molecular product H-F has a stronger bond than H-H so it is expected to ‘see’ a greater translational kinetic energy in the products than the reactants. This is displayed by the trajectory, the oscillation period for the product H-F (A-B) is larger than that of the reactant H-H (B-C).
As a result of the conservation of energy, the total energy of the system must remain constant <ref name="ref7"/>. Therefore the product state should have a lower potential energy than the reactants to balance the total energy (kinetic and potential) of the system so that it remains constant, suggesting the reaction to be exothermic - earlier concluded to be the case as for the reaction for F + H2. This could be confirmed by different spectroscopic methods such as calorimetry and IR/Raman spectroscopy.

:Heat-flow calorimetry is commonly used to thermally monitor reactions and thus determine whether a reaction is exothermic or endothermic. The heat transfer under e.g. constant pressure is monitored as the reaction proceeds and consequently the enthalpy change can be determined. Thermodynamically ΔH=qp at constant pressure (where qp is the heat flow at constant pressure), otherwise known as Hess's Law <ref name="ref8"/>. Exothermic reactions release heat thus if exothermic the calculated value for the enthalpy change will be negative in sign (visa versa for endothermic reactions).

:IR/Raman spectroscopy is used to determine the vibrational states of molecules/matter. If the reactants possess less translational kinetic energy and thus vibrational energy then the IR spectra for the reactants should display lower intensity peaks relative to the peaks in the IR spectra of the products.

====Polanyi's empirical rules====

Polanyi's empirical rules focus on the importance of the transition state location in enhancing reactivity for activated reactions <ref name="ref9"/>. This model has enabled the prediction of the relative efficacy of vibrational and translational excitations in promoting different molecular dynamic reactions.
:Polanyi's rules <ref name="ref9"/> state that vibrational energy is more efficient at promoting late transition states (endothermic reactions according to Hammond Postulate <ref name="h"/>) than translational energy. And conversely translational kinetic energy is more efficient at promoting 'early' transition states (exothermic reactions).

The reaction from H + H-F provides an illustration of these rules. This reaction is endothermic, thus is an example of a late-barrier reaction - the transition state is closer to the products in energy and structure according to Hammond's Postulate <ref name="h"/>. According to the Polanyi's Empirical Rules the reactivity should be enhanced more efficiently by high vibrational energy (and low kinetic energy). When initial conditions were chosen so that the reactant H-H had high vibrational energy (high pHH) and low translational kinetic energy (small pHF) a kinetically driven trajectory was generated.

Not all systems obey Polanyi's Rules as a result of their empirical nature (experimentally determined). As the rules are experimentally determined they effectively relate to the average behaviour of a reacting system <ref name="ref10"/>. The reaction H2 + F is an example where the calculated trajectory deviates from Polanyi's predictions. This reaction is exothermic so according to Polanyi's rules it is expected that high translational kinetic energy would promote the early transition state and thus enhance the reactivity. However for the reactive trajectory to reach completion a higher initial momenta for H-H (pHH) than initial translational kinetic energy (pHF) was required. This may be due to the errors in the computational calculation of the reaction trajectory thus does not accurately portray was would be observed experimentally.

References:
<references>
<ref name="a">Laganà, Antonio, and Antonio Riganelli. Reaction And Molecular Dynamics. 1st ed. Berlin, Heidelberg: Springer Berlin Heidelberg, 2000. Print..</ref>
<ref name="ref1">Collett, Charles T, and Christopher D Robson. Handbook Of Computational Chemistry Research. 1st ed. New York: Nova Science Publishers, 2010.</ref>
<ref name="ref2">Lewars, Errol G. "The Concept Of The Potential Energy Surface". Computational Chemistry (2010): 9-43. Web. 11 May 2017.</ref>
<ref name="ref3">Jeffrey, Alan. Essentials Of Engineering Mathematics. 1st ed. Boca Raton: Chapman & Hall/CRC Press, 2004. Print.</ref>
<ref name="ref4">Koistinen, O-P. et al. "Minimum Energy Path Calculations With Gaussian Process Regression". Nanosystems: Physics, Chemistry, Mathematics (2016): 925-935. Web. 20 May</ref>
<ref name="t">Laidler, Keith J., and M. Christine King. "Development Of Transition-State Theory". The Journal of Physical Chemistry 87.15 (1983): 2657-2664. Web. 26 May 2017.</ref>
<ref name="ref5">Fueno, Takayuki. Transition State: A Theoretical Approach. 1st ed. Tokyo: Gordon and Breach Science Publishers, 1999. Print.</ref>
<ref name="ar">Kotz, John C, and Paul Treichel. Chemistry & Chemical Reactivity. 1st ed. Fort Worth: Saunders College Pub., 1996. Print.</ref>
<ref name="ref6">Petrucci, Ralph H. General Chemistry. 1st ed. Toronto, Ont.: Pearson Canada, 2011. Print.</ref>
<ref name="h">Carey, Francis A, and Richard J Sundberg. Advanced Organic Chemistry. 1st ed. Norwell: Springer, 2007. Print.</ref>
<ref name="ref7">Khine, Myint Swe, and Issa M Saleh. Models And Modeling. 1st ed. Dordrecht: Springer Science+Business Media B.V, 2011</ref>
<ref name="ref8">Mikulecky, Peter, Michelle Rose Gilman, and Kate Brutlag. AP Chemistry For Dummies. 1st ed. Hoboken, N.J.: Wiley Pub., 2009. Print.</ref>
<ref name="ref9">J. Clayden, N. Greeves and S. G. Warren, Organic Chemistry, Oxford University Press, Oxford, 2012</ref>
<ref name="ref10">Zhang, Zhaojun et al. "Theoretical Study Of The Validity Of The Polanyi Rules For The Late-Barrier Cl + Chd3reaction". The Journal of Physical Chemistry Letters 3.23 (2012): 3416-3419. Web. 26 May 2017.</ref>
</ref>
</references>

Talk:MRD:kbg1501068978

2017-05-31T16:34:28Z

Je714: Created page with "Good effort, some of the answers are a bit scant, though. Check my inline comments for suggestions and corrections."

Good effort, some of the answers are a bit scant, though. Check my inline comments for suggestions and corrections.

MRD:kbg1501068978

2017-05-31T16:33:56Z

Je714: /* Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. */

= Molecular Reaction Dynamics: Applications to Triatomic systems =

==Part 1: H + H2 system==

=== '''What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.''' ===
The total gradient of the potential energy surface is zero at a minimum and at a transition structure.

When the second derivative of a stationary point is positive, the point is the minima. This is the point which has the smallest value compared to the the points nearby, showing a "valley.

For the transition structure, there is a saddle point in the potential energy surface. In terms of curvature, the gradient goes from (+) to 0 then (+) or from (-) to 0 then (-) again.

{{fontcolor1|gray| In which directions? Use the sign of second partial derivatives to be more precise with your definition. http://mathworld.wolfram.com/SecondDerivativeTest.html [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:28, 31 May 2017 (BST)}}

=== '''Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.''' ===
At transition state, r=0.908A. This is the radius that gives least oscillation when the momentum of both particles are zero. Going lower or higher than this value will cause the system to oscillate. The non-oscillating system shows that r=0.908A is the stationary point, where it is at the unstable equilibrium, slight increase or decrease in radius causes the system to fall to the stable equilibrium.
{{fontcolor1|gray| Why do you think you have 'least oscillation' at the TS? Think F = -dV/dr [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:29, 31 May 2017 (BST)}}

[[File:RTSkbg15.png|600px|thumb|Radius at which least oscillation is obtained.|none]]

=== '''Comment on how the ''mep'' and the trajectory you just calculated differ.''' ===
Minimum energy path (mep) gives trajectory which follows the floor of the valley, non-oscillating whereas dynamic gives oscillating trajectory.

In mep, the velocity is always reset to 0. Therefore, the atom will accelerate and move in the direction of steepest descent at each time step. Hence, it will follow the floor of the valley, minima of the potential surface.

In dynamic trajectory calculation, at each time step, the atom has an initial velocity. Therefore, its motion is the coupling of the initial velocity and the acceleration due to the direction of steepest descent, which results in the oscillating path.

{{fontcolor1|gray| Good [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:29, 31 May 2017 (BST)}}

=== Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory. ===
{| class="wikitable"
!Condition
!p1
!p2
!Trajectory
!Reactivity
!Comments
|-
|1
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|[[File:Condition1-kbg15.png|300px]]
|Reactive
|Atoms have sufficient energy to pass through the transition state.
New B-C bond is formed and oscillates due to vibrations.
|-
|2
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|[[File:Condition2-kbg15.png|300px]]
|Unreactive
|Atoms approach each other but energy is not high enough to pass through transition state.
|-
|3
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|[[File:Condition3-kbg15.png|300px]]
|Reactive
|Atoms have sufficient energy to pass through the transition state.
New B-C bond is formed and oscillates due to vibrations.

Higher magnitude of vibration of A-B due to higher kinetic energy.

(Higher momentum, higher KE, KE=p2/2m)
|-
|4
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|[[File:Condition4-kbg15.png|300px]]
|Reactive but
returns to

initial state
|Atoms have sufficient energy to pass through the transition state but the products returns
to the initial state by recrossing the transition state.
|-
|5
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|[[File:Condition5-kbg15.png|300px]]
|Reactive
|Atoms have sufficient energy to pass through the transition state twice, forming products.
|}

=== '''State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''' ===

==== Quasi-equilibrium assumption ====
When the reactants and products are not in equilibrium with each other, the activated complexes are in quasi-equilibrium with the reactants.

The flux of activated complexes in the two directions are independent of each other.

Each intermediate is long-lived enough to reach a Boltzmann distribution of energies before continuing to the next step.

Unless atoms or molecules collide with enough energy to form the transition structure, then the reaction does not occur.

The reaction system will pass over the lowest energy saddle point on the potential energy surface.

By obtaining the ΔG≠ from the trajectory, the rate could be predicted using the Arrhenius equation. The value can hence be compared with the experimental values.

{{fontcolor1|gray| The 1st and 3rt points don't apply here -- we're simulating a triatomic collision in isolation, not an ensemble of particles. You need to talk about barrier recrossing, which is what you observe in some of the previous examples. You forgot to say if you expect the calculated reaction rate values to be higher or lower than the experimental ones. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:32, 31 May 2017 (BST)}}

== Part 2: F-H-H system ==

=== Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===
1) F + H2 -> HF + H , exothermic

2) H + HF -> H2 + F , endothermic

{{fontcolor1|gray| Need to relate this to the PES, if possible [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:32, 31 May 2017 (BST)}}

The exothermic reaction in 1 suggests that H-F bond is stronger than H-H bond. Since the formation of more stable or stronger bond in the product releases more energy than formation of weaker bond. This yields larger release in energy in 1 compared to 2. This is in agreement with literature BE (H-F) = 565 kJ mol-1 , BE (H-H) = 432 kJ mol-1.

=== Locate the approximate position of the transition state. ===
[[File:TS-FHH-kbg15.png|600px|thumb|Radius at which least oscillation is obtained.|none]]rHF = 1.81A

rHH = 0.745A

=== Report the activation energy for both reactions. ===
Energy of transition state = -103.3 kcal mol-1

Energy of HF and H = -133.9 kcal mol-1

Energy of H2 and F = -104 kcal mol-1

'''Thus,'''

Reaction 1: Ea = 0.7 kcal mol-1 (lit. 1.6 kcal mol-1)

Reaction 2: Ea = 30.6 kcal mol-1

ΔrH (Reaction 1) = -29.9 kcal mol-1 (lit. -31.9 kcal mol-1)

=== Precise estimate of final energy ===
{| class="wikitable"
!Potential Energy vs Time
!Energy / kcal mol-1
|-
||[[File:High scan 1-kbg15.png|600x600px]]
|<nowiki>-133.9</nowiki>
|-
||[[File:High scan 2- kbg15.png|600x600px]]
|<nowiki>-104</nowiki>
|}

=== Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally? ===
[[File:Reactive internuclear-kbg15.png|570x570px|thumb|Internuclear Momenta vs Time.]][[File:Reactive surface-kbg15.png|570x570px|thumb|Surface plot of a reactive trajectory.|none]]Initial conditions are rHF = 2.5 A, rHH = 0.73 A, pHF = -1.2, pHH = -1

From the internuclear momenta-time graph, H-H changes from small oscillatory vibration to a straight line, whereas H-F changes from straight line to large oscillatory vibration.

Therefore, due to the conservation of energy, the energy released is stored as vibrational energy in H-F. The system goes from small vibrational energy of the reactant to large vibrational energy of the product. The reaction is exorthermic, an increase in temperature of the reaction mixture indicates the reaction is successful.

Infrared chemiluminescence can be used to confirm the release of vibrational energy. It refers to the emission of infrared photons from vibrationally excited product molecules immediately after their formation. The intensities of infrared emission lines from vibrationally excited molecules are used to measure the populations of vibrational states of product molecules.

=== Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. ===
[[File:Low p(HH)-kbg15.png|570x570px|thumb|Low p(HH)]][[File:High p(HH)-kbg15.png|570x570px|thumb|High p(HH)|none]]

By using the equation KE=p2/2m, the kinetic energy of high p(HH) is much higher than low p(HH). This means that lower amount of input energy is needed for low p(HH).

Polanyi's rules state that vibrational energy is more efficient than translational energy in promoting reactions with late transition states. This would imply to reactions which are endothermic according to the Hammond's postulate. This is shown in the trajectories above, where low p(HH) requires less initial energy for the reaction to be reactive when the energy is stored in the vibrational motion of H-F. Thus, exothermic reactions are more energetically efficient when the reactant energy is stored in the translational mode, whereas endothermic reactions are more energetically efficient when the reactant energy is stored in vibrational mode.

{{fontcolor1|gray| Do you have examples of the opposite case? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:33, 31 May 2017 (BST)}}

MRD:kbg1501068978

2017-05-31T16:32:51Z

Je714: /* Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? */

MRD:kbg1501068978

2017-05-31T16:32:22Z

Je714: /* Quasi-equilibrium assumption */

= Molecular Reaction Dynamics: Applications to Triatomic systems =

==Part 1: H + H2 system==

=== '''What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.''' ===
The total gradient of the potential energy surface is zero at a minimum and at a transition structure.

When the second derivative of a stationary point is positive, the point is the minima. This is the point which has the smallest value compared to the the points nearby, showing a "valley.

For the transition structure, there is a saddle point in the potential energy surface. In terms of curvature, the gradient goes from (+) to 0 then (+) or from (-) to 0 then (-) again.

{{fontcolor1|gray| In which directions? Use the sign of second partial derivatives to be more precise with your definition. http://mathworld.wolfram.com/SecondDerivativeTest.html [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:28, 31 May 2017 (BST)}}

=== '''Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.''' ===
At transition state, r=0.908A. This is the radius that gives least oscillation when the momentum of both particles are zero. Going lower or higher than this value will cause the system to oscillate. The non-oscillating system shows that r=0.908A is the stationary point, where it is at the unstable equilibrium, slight increase or decrease in radius causes the system to fall to the stable equilibrium.
{{fontcolor1|gray| Why do you think you have 'least oscillation' at the TS? Think F = -dV/dr [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:29, 31 May 2017 (BST)}}

[[File:RTSkbg15.png|600px|thumb|Radius at which least oscillation is obtained.|none]]

=== '''Comment on how the ''mep'' and the trajectory you just calculated differ.''' ===
Minimum energy path (mep) gives trajectory which follows the floor of the valley, non-oscillating whereas dynamic gives oscillating trajectory.

In mep, the velocity is always reset to 0. Therefore, the atom will accelerate and move in the direction of steepest descent at each time step. Hence, it will follow the floor of the valley, minima of the potential surface.

In dynamic trajectory calculation, at each time step, the atom has an initial velocity. Therefore, its motion is the coupling of the initial velocity and the acceleration due to the direction of steepest descent, which results in the oscillating path.

{{fontcolor1|gray| Good [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:29, 31 May 2017 (BST)}}

=== Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory. ===
{| class="wikitable"
!Condition
!p1
!p2
!Trajectory
!Reactivity
!Comments
|-
|1
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|[[File:Condition1-kbg15.png|300px]]
|Reactive
|Atoms have sufficient energy to pass through the transition state.
New B-C bond is formed and oscillates due to vibrations.
|-
|2
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|[[File:Condition2-kbg15.png|300px]]
|Unreactive
|Atoms approach each other but energy is not high enough to pass through transition state.
|-
|3
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|[[File:Condition3-kbg15.png|300px]]
|Reactive
|Atoms have sufficient energy to pass through the transition state.
New B-C bond is formed and oscillates due to vibrations.

Higher magnitude of vibration of A-B due to higher kinetic energy.

(Higher momentum, higher KE, KE=p2/2m)
|-
|4
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|[[File:Condition4-kbg15.png|300px]]
|Reactive but
returns to

initial state
|Atoms have sufficient energy to pass through the transition state but the products returns
to the initial state by recrossing the transition state.
|-
|5
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|[[File:Condition5-kbg15.png|300px]]
|Reactive
|Atoms have sufficient energy to pass through the transition state twice, forming products.
|}

=== '''State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''' ===

==== Quasi-equilibrium assumption ====
When the reactants and products are not in equilibrium with each other, the activated complexes are in quasi-equilibrium with the reactants.

The flux of activated complexes in the two directions are independent of each other.

Each intermediate is long-lived enough to reach a Boltzmann distribution of energies before continuing to the next step.

Unless atoms or molecules collide with enough energy to form the transition structure, then the reaction does not occur.

The reaction system will pass over the lowest energy saddle point on the potential energy surface.

By obtaining the ΔG≠ from the trajectory, the rate could be predicted using the Arrhenius equation. The value can hence be compared with the experimental values.

{{fontcolor1|gray| The 1st and 3rt points don't apply here -- we're simulating a triatomic collision in isolation, not an ensemble of particles. You need to talk about barrier recrossing, which is what you observe in some of the previous examples. You forgot to say if you expect the calculated reaction rate values to be higher or lower than the experimental ones. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:32, 31 May 2017 (BST)}}

== Part 2: F-H-H system ==

=== Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===
1) F + H2 -> HF + H , exothermic

2) H + HF -> H2 + F , endothermic

The exothermic reaction in 1 suggests that H-F bond is stronger than H-H bond. Since the formation of more stable or stronger bond in the product releases more energy than formation of weaker bond. This yields larger release in energy in 1 compared to 2. This is in agreement with literature BE (H-F) = 565 kJ mol-1 , BE (H-H) = 432 kJ mol-1.

=== Locate the approximate position of the transition state. ===
[[File:TS-FHH-kbg15.png|600px|thumb|Radius at which least oscillation is obtained.|none]]rHF = 1.81A

rHH = 0.745A

=== Report the activation energy for both reactions. ===
Energy of transition state = -103.3 kcal mol-1

Energy of HF and H = -133.9 kcal mol-1

Energy of H2 and F = -104 kcal mol-1

'''Thus,'''

Reaction 1: Ea = 0.7 kcal mol-1 (lit. 1.6 kcal mol-1)

Reaction 2: Ea = 30.6 kcal mol-1

ΔrH (Reaction 1) = -29.9 kcal mol-1 (lit. -31.9 kcal mol-1)

=== Precise estimate of final energy ===
{| class="wikitable"
!Potential Energy vs Time
!Energy / kcal mol-1
|-
||[[File:High scan 1-kbg15.png|600x600px]]
|<nowiki>-133.9</nowiki>
|-
||[[File:High scan 2- kbg15.png|600x600px]]
|<nowiki>-104</nowiki>
|}

=== Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally? ===
[[File:Reactive internuclear-kbg15.png|570x570px|thumb|Internuclear Momenta vs Time.]][[File:Reactive surface-kbg15.png|570x570px|thumb|Surface plot of a reactive trajectory.|none]]Initial conditions are rHF = 2.5 A, rHH = 0.73 A, pHF = -1.2, pHH = -1

From the internuclear momenta-time graph, H-H changes from small oscillatory vibration to a straight line, whereas H-F changes from straight line to large oscillatory vibration.

Therefore, due to the conservation of energy, the energy released is stored as vibrational energy in H-F. The system goes from small vibrational energy of the reactant to large vibrational energy of the product. The reaction is exorthermic, an increase in temperature of the reaction mixture indicates the reaction is successful.

Infrared chemiluminescence can be used to confirm the release of vibrational energy. It refers to the emission of infrared photons from vibrationally excited product molecules immediately after their formation. The intensities of infrared emission lines from vibrationally excited molecules are used to measure the populations of vibrational states of product molecules.

=== Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. ===
[[File:Low p(HH)-kbg15.png|570x570px|thumb|Low p(HH)]][[File:High p(HH)-kbg15.png|570x570px|thumb|High p(HH)|none]]

By using the equation KE=p2/2m, the kinetic energy of high p(HH) is much higher than low p(HH). This means that lower amount of input energy is needed for low p(HH).

Polanyi's rules state that vibrational energy is more efficient than translational energy in promoting reactions with late transition states. This would imply to reactions which are endothermic according to the Hammond's postulate. This is shown in the trajectories above, where low p(HH) requires less initial energy for the reaction to be reactive when the energy is stored in the vibrational motion of H-F. Thus, exothermic reactions are more energetically efficient when the reactant energy is stored in the translational mode, whereas endothermic reactions are more energetically efficient when the reactant energy is stored in vibrational mode.

MRD:kbg1501068978

2017-05-31T16:29:56Z

Je714: /* Comment on how the mep and the trajectory you just calculated differ. */

= Molecular Reaction Dynamics: Applications to Triatomic systems =

==Part 1: H + H2 system==

=== '''What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.''' ===
The total gradient of the potential energy surface is zero at a minimum and at a transition structure.

When the second derivative of a stationary point is positive, the point is the minima. This is the point which has the smallest value compared to the the points nearby, showing a "valley.

For the transition structure, there is a saddle point in the potential energy surface. In terms of curvature, the gradient goes from (+) to 0 then (+) or from (-) to 0 then (-) again.

{{fontcolor1|gray| In which directions? Use the sign of second partial derivatives to be more precise with your definition. http://mathworld.wolfram.com/SecondDerivativeTest.html [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:28, 31 May 2017 (BST)}}

=== '''Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.''' ===
At transition state, r=0.908A. This is the radius that gives least oscillation when the momentum of both particles are zero. Going lower or higher than this value will cause the system to oscillate. The non-oscillating system shows that r=0.908A is the stationary point, where it is at the unstable equilibrium, slight increase or decrease in radius causes the system to fall to the stable equilibrium.
{{fontcolor1|gray| Why do you think you have 'least oscillation' at the TS? Think F = -dV/dr [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:29, 31 May 2017 (BST)}}

[[File:RTSkbg15.png|600px|thumb|Radius at which least oscillation is obtained.|none]]

=== '''Comment on how the ''mep'' and the trajectory you just calculated differ.''' ===
Minimum energy path (mep) gives trajectory which follows the floor of the valley, non-oscillating whereas dynamic gives oscillating trajectory.

In mep, the velocity is always reset to 0. Therefore, the atom will accelerate and move in the direction of steepest descent at each time step. Hence, it will follow the floor of the valley, minima of the potential surface.

In dynamic trajectory calculation, at each time step, the atom has an initial velocity. Therefore, its motion is the coupling of the initial velocity and the acceleration due to the direction of steepest descent, which results in the oscillating path.

{{fontcolor1|gray| Good [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:29, 31 May 2017 (BST)}}

=== Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory. ===
{| class="wikitable"
!Condition
!p1
!p2
!Trajectory
!Reactivity
!Comments
|-
|1
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|[[File:Condition1-kbg15.png|300px]]
|Reactive
|Atoms have sufficient energy to pass through the transition state.
New B-C bond is formed and oscillates due to vibrations.
|-
|2
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|[[File:Condition2-kbg15.png|300px]]
|Unreactive
|Atoms approach each other but energy is not high enough to pass through transition state.
|-
|3
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|[[File:Condition3-kbg15.png|300px]]
|Reactive
|Atoms have sufficient energy to pass through the transition state.
New B-C bond is formed and oscillates due to vibrations.

Higher magnitude of vibration of A-B due to higher kinetic energy.

(Higher momentum, higher KE, KE=p2/2m)
|-
|4
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|[[File:Condition4-kbg15.png|300px]]
|Reactive but
returns to

initial state
|Atoms have sufficient energy to pass through the transition state but the products returns
to the initial state by recrossing the transition state.
|-
|5
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|[[File:Condition5-kbg15.png|300px]]
|Reactive
|Atoms have sufficient energy to pass through the transition state twice, forming products.
|}

=== '''State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''' ===

==== Quasi-equilibrium assumption ====
When the reactants and products are not in equilibrium with each other, the activated complexes are in quasi-equilibrium with the reactants.

The flux of activated complexes in the two directions are independent of each other.

Each intermediate is long-lived enough to reach a Boltzmann distribution of energies before continuing to the next step.

Unless atoms or molecules collide with enough energy to form the transition structure, then the reaction does not occur.

The reaction system will pass over the lowest energy saddle point on the potential energy surface.

By obtaining the ΔG≠ from the trajectory, the rate could be predicted using the Arrhenius equation. The value can hence be compared with the experimental values.

== Part 2: F-H-H system ==

=== Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===
1) F + H2 -> HF + H , exothermic

2) H + HF -> H2 + F , endothermic

The exothermic reaction in 1 suggests that H-F bond is stronger than H-H bond. Since the formation of more stable or stronger bond in the product releases more energy than formation of weaker bond. This yields larger release in energy in 1 compared to 2. This is in agreement with literature BE (H-F) = 565 kJ mol-1 , BE (H-H) = 432 kJ mol-1.

=== Locate the approximate position of the transition state. ===
[[File:TS-FHH-kbg15.png|600px|thumb|Radius at which least oscillation is obtained.|none]]rHF = 1.81A

rHH = 0.745A

=== Report the activation energy for both reactions. ===
Energy of transition state = -103.3 kcal mol-1

Energy of HF and H = -133.9 kcal mol-1

Energy of H2 and F = -104 kcal mol-1

'''Thus,'''

Reaction 1: Ea = 0.7 kcal mol-1 (lit. 1.6 kcal mol-1)

Reaction 2: Ea = 30.6 kcal mol-1

ΔrH (Reaction 1) = -29.9 kcal mol-1 (lit. -31.9 kcal mol-1)

=== Precise estimate of final energy ===
{| class="wikitable"
!Potential Energy vs Time
!Energy / kcal mol-1
|-
||[[File:High scan 1-kbg15.png|600x600px]]
|<nowiki>-133.9</nowiki>
|-
||[[File:High scan 2- kbg15.png|600x600px]]
|<nowiki>-104</nowiki>
|}

=== Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally? ===
[[File:Reactive internuclear-kbg15.png|570x570px|thumb|Internuclear Momenta vs Time.]][[File:Reactive surface-kbg15.png|570x570px|thumb|Surface plot of a reactive trajectory.|none]]Initial conditions are rHF = 2.5 A, rHH = 0.73 A, pHF = -1.2, pHH = -1

From the internuclear momenta-time graph, H-H changes from small oscillatory vibration to a straight line, whereas H-F changes from straight line to large oscillatory vibration.

Therefore, due to the conservation of energy, the energy released is stored as vibrational energy in H-F. The system goes from small vibrational energy of the reactant to large vibrational energy of the product. The reaction is exorthermic, an increase in temperature of the reaction mixture indicates the reaction is successful.

Infrared chemiluminescence can be used to confirm the release of vibrational energy. It refers to the emission of infrared photons from vibrationally excited product molecules immediately after their formation. The intensities of infrared emission lines from vibrationally excited molecules are used to measure the populations of vibrational states of product molecules.

=== Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. ===
[[File:Low p(HH)-kbg15.png|570x570px|thumb|Low p(HH)]][[File:High p(HH)-kbg15.png|570x570px|thumb|High p(HH)|none]]

By using the equation KE=p2/2m, the kinetic energy of high p(HH) is much higher than low p(HH). This means that lower amount of input energy is needed for low p(HH).

Polanyi's rules state that vibrational energy is more efficient than translational energy in promoting reactions with late transition states. This would imply to reactions which are endothermic according to the Hammond's postulate. This is shown in the trajectories above, where low p(HH) requires less initial energy for the reaction to be reactive when the energy is stored in the vibrational motion of H-F. Thus, exothermic reactions are more energetically efficient when the reactant energy is stored in the translational mode, whereas endothermic reactions are more energetically efficient when the reactant energy is stored in vibrational mode.

MRD:kbg1501068978

2017-05-31T16:29:22Z

Je714: /* Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory. */

= Molecular Reaction Dynamics: Applications to Triatomic systems =

==Part 1: H + H2 system==

=== '''What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.''' ===
The total gradient of the potential energy surface is zero at a minimum and at a transition structure.

When the second derivative of a stationary point is positive, the point is the minima. This is the point which has the smallest value compared to the the points nearby, showing a "valley.

For the transition structure, there is a saddle point in the potential energy surface. In terms of curvature, the gradient goes from (+) to 0 then (+) or from (-) to 0 then (-) again.

{{fontcolor1|gray| In which directions? Use the sign of second partial derivatives to be more precise with your definition. http://mathworld.wolfram.com/SecondDerivativeTest.html [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:28, 31 May 2017 (BST)}}

=== '''Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.''' ===
At transition state, r=0.908A. This is the radius that gives least oscillation when the momentum of both particles are zero. Going lower or higher than this value will cause the system to oscillate. The non-oscillating system shows that r=0.908A is the stationary point, where it is at the unstable equilibrium, slight increase or decrease in radius causes the system to fall to the stable equilibrium.
{{fontcolor1|gray| Why do you think you have 'least oscillation' at the TS? Think F = -dV/dr [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:29, 31 May 2017 (BST)}}

[[File:RTSkbg15.png|600px|thumb|Radius at which least oscillation is obtained.|none]]

=== '''Comment on how the ''mep'' and the trajectory you just calculated differ.''' ===
Minimum energy path (mep) gives trajectory which follows the floor of the valley, non-oscillating whereas dynamic gives oscillating trajectory.

In mep, the velocity is always reset to 0. Therefore, the atom will accelerate and move in the direction of steepest descent at each time step. Hence, it will follow the floor of the valley, minima of the potential surface.

In dynamic trajectory calculation, at each time step, the atom has an initial velocity. Therefore, its motion is the coupling of the initial velocity and the acceleration due to the direction of steepest descent, which results in the oscillating path.

=== Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory. ===
{| class="wikitable"
!Condition
!p1
!p2
!Trajectory
!Reactivity
!Comments
|-
|1
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|[[File:Condition1-kbg15.png|300px]]
|Reactive
|Atoms have sufficient energy to pass through the transition state.
New B-C bond is formed and oscillates due to vibrations.
|-
|2
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|[[File:Condition2-kbg15.png|300px]]
|Unreactive
|Atoms approach each other but energy is not high enough to pass through transition state.
|-
|3
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|[[File:Condition3-kbg15.png|300px]]
|Reactive
|Atoms have sufficient energy to pass through the transition state.
New B-C bond is formed and oscillates due to vibrations.

Higher magnitude of vibration of A-B due to higher kinetic energy.

(Higher momentum, higher KE, KE=p2/2m)
|-
|4
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|[[File:Condition4-kbg15.png|300px]]
|Reactive but
returns to

initial state
|Atoms have sufficient energy to pass through the transition state but the products returns
to the initial state by recrossing the transition state.
|-
|5
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|[[File:Condition5-kbg15.png|300px]]
|Reactive
|Atoms have sufficient energy to pass through the transition state twice, forming products.
|}

=== '''State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''' ===

==== Quasi-equilibrium assumption ====
When the reactants and products are not in equilibrium with each other, the activated complexes are in quasi-equilibrium with the reactants.

The flux of activated complexes in the two directions are independent of each other.

Each intermediate is long-lived enough to reach a Boltzmann distribution of energies before continuing to the next step.

Unless atoms or molecules collide with enough energy to form the transition structure, then the reaction does not occur.

The reaction system will pass over the lowest energy saddle point on the potential energy surface.

By obtaining the ΔG≠ from the trajectory, the rate could be predicted using the Arrhenius equation. The value can hence be compared with the experimental values.

== Part 2: F-H-H system ==

=== Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===
1) F + H2 -> HF + H , exothermic

2) H + HF -> H2 + F , endothermic

The exothermic reaction in 1 suggests that H-F bond is stronger than H-H bond. Since the formation of more stable or stronger bond in the product releases more energy than formation of weaker bond. This yields larger release in energy in 1 compared to 2. This is in agreement with literature BE (H-F) = 565 kJ mol-1 , BE (H-H) = 432 kJ mol-1.

=== Locate the approximate position of the transition state. ===
[[File:TS-FHH-kbg15.png|600px|thumb|Radius at which least oscillation is obtained.|none]]rHF = 1.81A

rHH = 0.745A

=== Report the activation energy for both reactions. ===
Energy of transition state = -103.3 kcal mol-1

Energy of HF and H = -133.9 kcal mol-1

Energy of H2 and F = -104 kcal mol-1

'''Thus,'''

Reaction 1: Ea = 0.7 kcal mol-1 (lit. 1.6 kcal mol-1)

Reaction 2: Ea = 30.6 kcal mol-1

ΔrH (Reaction 1) = -29.9 kcal mol-1 (lit. -31.9 kcal mol-1)

=== Precise estimate of final energy ===
{| class="wikitable"
!Potential Energy vs Time
!Energy / kcal mol-1
|-
||[[File:High scan 1-kbg15.png|600x600px]]
|<nowiki>-133.9</nowiki>
|-
||[[File:High scan 2- kbg15.png|600x600px]]
|<nowiki>-104</nowiki>
|}

=== Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally? ===
[[File:Reactive internuclear-kbg15.png|570x570px|thumb|Internuclear Momenta vs Time.]][[File:Reactive surface-kbg15.png|570x570px|thumb|Surface plot of a reactive trajectory.|none]]Initial conditions are rHF = 2.5 A, rHH = 0.73 A, pHF = -1.2, pHH = -1

From the internuclear momenta-time graph, H-H changes from small oscillatory vibration to a straight line, whereas H-F changes from straight line to large oscillatory vibration.

Therefore, due to the conservation of energy, the energy released is stored as vibrational energy in H-F. The system goes from small vibrational energy of the reactant to large vibrational energy of the product. The reaction is exorthermic, an increase in temperature of the reaction mixture indicates the reaction is successful.

Infrared chemiluminescence can be used to confirm the release of vibrational energy. It refers to the emission of infrared photons from vibrationally excited product molecules immediately after their formation. The intensities of infrared emission lines from vibrationally excited molecules are used to measure the populations of vibrational states of product molecules.

=== Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. ===
[[File:Low p(HH)-kbg15.png|570x570px|thumb|Low p(HH)]][[File:High p(HH)-kbg15.png|570x570px|thumb|High p(HH)|none]]

By using the equation KE=p2/2m, the kinetic energy of high p(HH) is much higher than low p(HH). This means that lower amount of input energy is needed for low p(HH).

Polanyi's rules state that vibrational energy is more efficient than translational energy in promoting reactions with late transition states. This would imply to reactions which are endothermic according to the Hammond's postulate. This is shown in the trajectories above, where low p(HH) requires less initial energy for the reaction to be reactive when the energy is stored in the vibrational motion of H-F. Thus, exothermic reactions are more energetically efficient when the reactant energy is stored in the translational mode, whereas endothermic reactions are more energetically efficient when the reactant energy is stored in vibrational mode.

MRD:kbg1501068978

2017-05-31T16:29:02Z

= Molecular Reaction Dynamics: Applications to Triatomic systems =

==Part 1: H + H2 system==

=== '''What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.''' ===
The total gradient of the potential energy surface is zero at a minimum and at a transition structure.

When the second derivative of a stationary point is positive, the point is the minima. This is the point which has the smallest value compared to the the points nearby, showing a "valley.

For the transition structure, there is a saddle point in the potential energy surface. In terms of curvature, the gradient goes from (+) to 0 then (+) or from (-) to 0 then (-) again.

{{fontcolor1|gray| In which directions? Use the sign of second partial derivatives to be more precise with your definition. http://mathworld.wolfram.com/SecondDerivativeTest.html [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:28, 31 May 2017 (BST)}}

=== '''Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.''' ===
At transition state, r=0.908A. This is the radius that gives least oscillation when the momentum of both particles are zero. Going lower or higher than this value will cause the system to oscillate. The non-oscillating system shows that r=0.908A is the stationary point, where it is at the unstable equilibrium, slight increase or decrease in radius causes the system to fall to the stable equilibrium.

[[File:RTSkbg15.png|600px|thumb|Radius at which least oscillation is obtained.|none]]

{{fontcolor1|gray| Why do you think you have 'least oscillation' at the TS? Think F = -dV/dr [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:29, 31 May 2017 (BST)}}

=== '''Comment on how the ''mep'' and the trajectory you just calculated differ.''' ===
Minimum energy path (mep) gives trajectory which follows the floor of the valley, non-oscillating whereas dynamic gives oscillating trajectory.

In mep, the velocity is always reset to 0. Therefore, the atom will accelerate and move in the direction of steepest descent at each time step. Hence, it will follow the floor of the valley, minima of the potential surface.

In dynamic trajectory calculation, at each time step, the atom has an initial velocity. Therefore, its motion is the coupling of the initial velocity and the acceleration due to the direction of steepest descent, which results in the oscillating path.

=== Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory. ===
{| class="wikitable"
!Condition
!p1
!p2
!Trajectory
!Reactivity
!Comments
|-
|1
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|[[File:Condition1-kbg15.png|300px]]
|Reactive
|Atoms have sufficient energy to pass through the transition state.
New B-C bond is formed and oscillates due to vibrations.
|-
|2
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|[[File:Condition2-kbg15.png|300px]]
|Unreactive
|Atoms approach each other but energy is not high enough to pass through transition state.
|-
|3
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|[[File:Condition3-kbg15.png|300px]]
|Reactive
|Atoms have sufficient energy to pass through the transition state.
New B-C bond is formed and oscillates due to vibrations.

Higher magnitude of vibration of A-B due to higher kinetic energy.

(Higher momentum, higher KE, KE=p2/2m)
|-
|4
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|[[File:Condition4-kbg15.png|300px]]
|Reactive but
returns to

initial state
|Atoms have sufficient energy to pass through the transition state but the products returns
to the initial state by recrossing the transition state.
|-
|5
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|[[File:Condition5-kbg15.png|300px]]
|Reactive
|Atoms have sufficient energy to pass through the transition state twice, forming products.
|}

=== '''State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''' ===

==== Quasi-equilibrium assumption ====
When the reactants and products are not in equilibrium with each other, the activated complexes are in quasi-equilibrium with the reactants.

The flux of activated complexes in the two directions are independent of each other.

Each intermediate is long-lived enough to reach a Boltzmann distribution of energies before continuing to the next step.

Unless atoms or molecules collide with enough energy to form the transition structure, then the reaction does not occur.

The reaction system will pass over the lowest energy saddle point on the potential energy surface.

By obtaining the ΔG≠ from the trajectory, the rate could be predicted using the Arrhenius equation. The value can hence be compared with the experimental values.

== Part 2: F-H-H system ==

=== Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===
1) F + H2 -> HF + H , exothermic

2) H + HF -> H2 + F , endothermic

The exothermic reaction in 1 suggests that H-F bond is stronger than H-H bond. Since the formation of more stable or stronger bond in the product releases more energy than formation of weaker bond. This yields larger release in energy in 1 compared to 2. This is in agreement with literature BE (H-F) = 565 kJ mol-1 , BE (H-H) = 432 kJ mol-1.

=== Locate the approximate position of the transition state. ===
[[File:TS-FHH-kbg15.png|600px|thumb|Radius at which least oscillation is obtained.|none]]rHF = 1.81A

rHH = 0.745A

=== Report the activation energy for both reactions. ===
Energy of transition state = -103.3 kcal mol-1

Energy of HF and H = -133.9 kcal mol-1

Energy of H2 and F = -104 kcal mol-1

'''Thus,'''

Reaction 1: Ea = 0.7 kcal mol-1 (lit. 1.6 kcal mol-1)

Reaction 2: Ea = 30.6 kcal mol-1

ΔrH (Reaction 1) = -29.9 kcal mol-1 (lit. -31.9 kcal mol-1)

=== Precise estimate of final energy ===
{| class="wikitable"
!Potential Energy vs Time
!Energy / kcal mol-1
|-
||[[File:High scan 1-kbg15.png|600x600px]]
|<nowiki>-133.9</nowiki>
|-
||[[File:High scan 2- kbg15.png|600x600px]]
|<nowiki>-104</nowiki>
|}

=== Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally? ===
[[File:Reactive internuclear-kbg15.png|570x570px|thumb|Internuclear Momenta vs Time.]][[File:Reactive surface-kbg15.png|570x570px|thumb|Surface plot of a reactive trajectory.|none]]Initial conditions are rHF = 2.5 A, rHH = 0.73 A, pHF = -1.2, pHH = -1

From the internuclear momenta-time graph, H-H changes from small oscillatory vibration to a straight line, whereas H-F changes from straight line to large oscillatory vibration.

Therefore, due to the conservation of energy, the energy released is stored as vibrational energy in H-F. The system goes from small vibrational energy of the reactant to large vibrational energy of the product. The reaction is exorthermic, an increase in temperature of the reaction mixture indicates the reaction is successful.

Infrared chemiluminescence can be used to confirm the release of vibrational energy. It refers to the emission of infrared photons from vibrationally excited product molecules immediately after their formation. The intensities of infrared emission lines from vibrationally excited molecules are used to measure the populations of vibrational states of product molecules.

=== Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. ===
[[File:Low p(HH)-kbg15.png|570x570px|thumb|Low p(HH)]][[File:High p(HH)-kbg15.png|570x570px|thumb|High p(HH)|none]]

By using the equation KE=p2/2m, the kinetic energy of high p(HH) is much higher than low p(HH). This means that lower amount of input energy is needed for low p(HH).

Polanyi's rules state that vibrational energy is more efficient than translational energy in promoting reactions with late transition states. This would imply to reactions which are endothermic according to the Hammond's postulate. This is shown in the trajectories above, where low p(HH) requires less initial energy for the reaction to be reactive when the energy is stored in the vibrational motion of H-F. Thus, exothermic reactions are more energetically efficient when the reactant energy is stored in the translational mode, whereas endothermic reactions are more energetically efficient when the reactant energy is stored in vibrational mode.

MRD:kbg1501068978

2017-05-31T16:28:17Z

Je714: /* What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface. */

= Molecular Reaction Dynamics: Applications to Triatomic systems =

==Part 1: H + H2 system==

=== '''What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.''' ===
The total gradient of the potential energy surface is zero at a minimum and at a transition structure.

When the second derivative of a stationary point is positive, the point is the minima. This is the point which has the smallest value compared to the the points nearby, showing a "valley.

For the transition structure, there is a saddle point in the potential energy surface. In terms of curvature, the gradient goes from (+) to 0 then (+) or from (-) to 0 then (-) again.

{{fontcolor1|gray| In which directions? Use the sign of second partial derivatives to be more precise with your definition. http://mathworld.wolfram.com/SecondDerivativeTest.html [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:28, 31 May 2017 (BST)}}

=== '''Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.''' ===
At transition state, r=0.908A. This is the radius that gives least oscillation when the momentum of both particles are zero. Going lower or higher than this value will cause the system to oscillate. The non-oscillating system shows that r=0.908A is the stationary point, where it is at the unstable equilibrium, slight increase or decrease in radius causes the system to fall to the stable equilibrium.

[[File:RTSkbg15.png|600px|thumb|Radius at which least oscillation is obtained.|none]]

=== '''Comment on how the ''mep'' and the trajectory you just calculated differ.''' ===
Minimum energy path (mep) gives trajectory which follows the floor of the valley, non-oscillating whereas dynamic gives oscillating trajectory.

In mep, the velocity is always reset to 0. Therefore, the atom will accelerate and move in the direction of steepest descent at each time step. Hence, it will follow the floor of the valley, minima of the potential surface.

In dynamic trajectory calculation, at each time step, the atom has an initial velocity. Therefore, its motion is the coupling of the initial velocity and the acceleration due to the direction of steepest descent, which results in the oscillating path.

=== Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory. ===
{| class="wikitable"
!Condition
!p1
!p2
!Trajectory
!Reactivity
!Comments
|-
|1
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|[[File:Condition1-kbg15.png|300px]]
|Reactive
|Atoms have sufficient energy to pass through the transition state.
New B-C bond is formed and oscillates due to vibrations.
|-
|2
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|[[File:Condition2-kbg15.png|300px]]
|Unreactive
|Atoms approach each other but energy is not high enough to pass through transition state.
|-
|3
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|[[File:Condition3-kbg15.png|300px]]
|Reactive
|Atoms have sufficient energy to pass through the transition state.
New B-C bond is formed and oscillates due to vibrations.

Higher magnitude of vibration of A-B due to higher kinetic energy.

(Higher momentum, higher KE, KE=p2/2m)
|-
|4
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|[[File:Condition4-kbg15.png|300px]]
|Reactive but
returns to

initial state
|Atoms have sufficient energy to pass through the transition state but the products returns
to the initial state by recrossing the transition state.
|-
|5
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|[[File:Condition5-kbg15.png|300px]]
|Reactive
|Atoms have sufficient energy to pass through the transition state twice, forming products.
|}

=== '''State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''' ===

==== Quasi-equilibrium assumption ====
When the reactants and products are not in equilibrium with each other, the activated complexes are in quasi-equilibrium with the reactants.

The flux of activated complexes in the two directions are independent of each other.

Each intermediate is long-lived enough to reach a Boltzmann distribution of energies before continuing to the next step.

Unless atoms or molecules collide with enough energy to form the transition structure, then the reaction does not occur.

The reaction system will pass over the lowest energy saddle point on the potential energy surface.

By obtaining the ΔG≠ from the trajectory, the rate could be predicted using the Arrhenius equation. The value can hence be compared with the experimental values.

== Part 2: F-H-H system ==

=== Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===
1) F + H2 -> HF + H , exothermic

2) H + HF -> H2 + F , endothermic

The exothermic reaction in 1 suggests that H-F bond is stronger than H-H bond. Since the formation of more stable or stronger bond in the product releases more energy than formation of weaker bond. This yields larger release in energy in 1 compared to 2. This is in agreement with literature BE (H-F) = 565 kJ mol-1 , BE (H-H) = 432 kJ mol-1.

=== Locate the approximate position of the transition state. ===
[[File:TS-FHH-kbg15.png|600px|thumb|Radius at which least oscillation is obtained.|none]]rHF = 1.81A

rHH = 0.745A

=== Report the activation energy for both reactions. ===
Energy of transition state = -103.3 kcal mol-1

Energy of HF and H = -133.9 kcal mol-1

Energy of H2 and F = -104 kcal mol-1

'''Thus,'''

Reaction 1: Ea = 0.7 kcal mol-1 (lit. 1.6 kcal mol-1)

Reaction 2: Ea = 30.6 kcal mol-1

ΔrH (Reaction 1) = -29.9 kcal mol-1 (lit. -31.9 kcal mol-1)

=== Precise estimate of final energy ===
{| class="wikitable"
!Potential Energy vs Time
!Energy / kcal mol-1
|-
||[[File:High scan 1-kbg15.png|600x600px]]
|<nowiki>-133.9</nowiki>
|-
||[[File:High scan 2- kbg15.png|600x600px]]
|<nowiki>-104</nowiki>
|}

=== Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally? ===
[[File:Reactive internuclear-kbg15.png|570x570px|thumb|Internuclear Momenta vs Time.]][[File:Reactive surface-kbg15.png|570x570px|thumb|Surface plot of a reactive trajectory.|none]]Initial conditions are rHF = 2.5 A, rHH = 0.73 A, pHF = -1.2, pHH = -1

From the internuclear momenta-time graph, H-H changes from small oscillatory vibration to a straight line, whereas H-F changes from straight line to large oscillatory vibration.

Therefore, due to the conservation of energy, the energy released is stored as vibrational energy in H-F. The system goes from small vibrational energy of the reactant to large vibrational energy of the product. The reaction is exorthermic, an increase in temperature of the reaction mixture indicates the reaction is successful.

Infrared chemiluminescence can be used to confirm the release of vibrational energy. It refers to the emission of infrared photons from vibrationally excited product molecules immediately after their formation. The intensities of infrared emission lines from vibrationally excited molecules are used to measure the populations of vibrational states of product molecules.

=== Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. ===
[[File:Low p(HH)-kbg15.png|570x570px|thumb|Low p(HH)]][[File:High p(HH)-kbg15.png|570x570px|thumb|High p(HH)|none]]

By using the equation KE=p2/2m, the kinetic energy of high p(HH) is much higher than low p(HH). This means that lower amount of input energy is needed for low p(HH).

Polanyi's rules state that vibrational energy is more efficient than translational energy in promoting reactions with late transition states. This would imply to reactions which are endothermic according to the Hammond's postulate. This is shown in the trajectories above, where low p(HH) requires less initial energy for the reaction to be reactive when the energy is stored in the vibrational motion of H-F. Thus, exothermic reactions are more energetically efficient when the reactant energy is stored in the translational mode, whereas endothermic reactions are more energetically efficient when the reactant energy is stored in vibrational mode.

Talk:MRD:wjb115

2017-05-31T16:26:42Z

Je714: Created page with "Not bad. See my inline comments for improvements. The last answer is incomplete. ~~~~"

Not bad. See my inline comments for improvements. The last answer is incomplete.

[[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:26, 31 May 2017 (BST)

MRD:wjb115

2017-05-31T16:26:18Z

Je714: /* Exercise 2: F-H-H system */

=== Molecular Reaction Dynamics ===

====Exercise 1:H + H2 system====

'''''What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.'''''

The total gradient of the potential energy surface, at a minimum, is zero, where ∂V(ri)/∂ri=0.
At a transition structure ∂V(r1)/∂r1∂,V(r2)/∂r2=0, which means the potential energy is again 0. The transition state is found at the maximum on the minimum energy path, and as a result moving in one direction will give a maxima and the other direction a minima. Consequently, the transition structure can be considered a saddle point.
Minima can be used to find the reactants, products and intermediates.
In order to determine transition structures, start a trajectory near the transition state and observe whether it "rolls" towards the reactants or products.

{{fontcolor1|gray| The rolling analogy is not wrong but it's just that, an analogy. A more rigorous discussion is needed here, either discussing curvature with words or using second partial derivatives and their signs. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:24, 31 May 2017 (BST)}}

'''''Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.'''''

The transition state can be found on the maximum of the minimum energy path. By setting the momentum to zero and making r1=r2, it was possible to locate the transition state. Initially the bond lengths were set equal at just under 1.0Å, which was chosen using a plot of internuclear distance vs time. After trial and error with the distance, it was found that at 0.907Å the transition state is most likely to be found, since at this distance the trajectory oscillated and "never fell off". {{fontcolor1|gray| Never is a strong word. You can always run a sufficiently long trajectory and the TS geometry will be lost -- the TS is a mathematical construct which you can only get close to with arbitrary precision in practice. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:24, 31 May 2017 (BST)}}

[[File:Internucdist_vs_time.PNG]]
Figure. 1 Plot of Internuclear distance vs time

The above plot shows the internuclear distance vs time at a distance of 0.907Å, when the oscillations appeared to be at their lowest.
'''
''Comment on how the mep and the trajectory you just calculated differ.''
'''
r1=0.917 r2=0.907

[[File:mepwjb115.PNG|300px]] [[File:vierawjb115.PNG|300px]]

Both the mep and dynamic calculated trajectories follow the minimum energy path in the valley, however, the plotted lines differ in each. In the mep the line is curved, but in dynamic it appears to be oscillating. This is a result of vibrational energy being neglected in the mep {{fontcolor1|gray| how is it neglected? That's what we're asking [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:24, 31 May 2017 (BST)}}, whilst in dynamic these changing vibrational energies are considered, and are responsible for the the wave-like oscillation of the dynamically calculated trajectory.

'''''Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.'''''

The initial positions used were r1 = 0.74 and r2 = 2.0, trajectories were then run with following momenta:

{| class="wikitable" border="1"
|+ Momenta dependent reactions:
!Reaction Number!!P1 !! P2 !! Reaction?
|-
| 1||-1.25 || -2.5 || Yes
|-
| 2||-1.5 || -2.0 || No
|-
| 3||-1.5 || -2.5 || Yes
|-
| 4||-2.5 || -5.0 || No
|-
| 5||-2.5 || -5.2 || Yes
|}

Reaction 1.[[File:1reactivewjb115.PNG|300px]]

This trajectory passes through the transition state and then proceeds to the products, making it a reactive process.

Reaction 2.[[File:2reactivewjb115.PNG|300px]]

The trajectory is unreactive since when it reaches the transition state it "bounces" back towards the reactants instead of towards the products, as the energy barrier of the transition can not be overcome.

Reaction 3.[[File:3reactivewjb115.PNG|300px]]

Again, the trajectory passes through the transition state and onto the products, making it a reactive process.

Reaction 4.[[File:4reactivewjb115.PNG|300px]]

The trajectory shows barrier recrossing where the transition state is crossed twice, and the trajectory reverts back to the reactants. Consequently, this is an unreactive process.

Reaction 5.[[File:5reactivewjb115.PNG|300px]]

Slightly different to the previous reaction, in this example the transition state appears to be crossed twice but instead of returning to the reactants there is sufficient energy to pass the activation energy of the reaction, so it can proceed towards products. This is therefore a reactive process.

'''''State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''''

The transition state theory is used to explain the reaction rates of elementary reactions. The theory assumes that the reactants and transition state structures, of a reaction, are in quasi-equilibrium which differs from classical chemical equilibrium. The theory is also based on the assumption that nuclei behave in classical mechanical sense, whereby a reaction can only proceed provided both the colliding molecules have sufficient energy to form the structure. This classical approach ignores the possibility of quantum tunneling, which states that reactions can occur even if the colliding molecules do not have the required energy to pass the energy barrier.
At high temperatures the theory is also less useful as for some reactions it can fail. The theory is based on an idea that the reaction will pass through lowest energy saddle point on the potential energy surface, which at low temperatures is always true. However, as higher temperatures are reached molecules can populate vibrational modes. As a result of this, transition states may be far away from the lowest energy saddle point.
Lastly, an assumption is made that every intermediate has a long enough lifetime for it reach a boltzmann distribution, before moving on to the next step of the reaction.

{{fontcolor1|gray| The quasi-equilibrium comment is not applicable in this case -- it's a statistical mechanic treatment of TST. We're looking at a triatomic collision in isollation -- we're not simulating an ensemble of particles. Otherwise good! [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:24, 31 May 2017 (BST)}}

====Exercise 2: F-H-H system====

'''''Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''''

The reaction 'F + H2 ---> FH + H' has its trajectory shown below:

[[File:F+h2.PNG|500px]]

One can see from the graph that the potential energy of the products are below the potential energy of reactants. Consequently, this is an exothermic reaction as energy is being released.
The converse can be said for the 'FH + H ---> H2' reaction, since it is simply the prior reaction in the reverse direction. Subsequently, the reactants would therefore site at lower potential energy than the products, so it would be considered an endothermic reaction.

This implies that the H2 molecule has a lower bond enthalpy than HF, since for a reaction to be exothermic 'bonds broken-bonds formed<0'. For this to be the case in 'F + H2 ---> FH + H' the bond dissociation energies of H2 must be lower than those of F-H, otherwise a negative value of bond enthalpy change could not be obtained.

'''''Locate the approximate position of the transition state.'''''

In order to determine the position of the transition state the same method as previously was used, where the momenta p1 and p2 were set to 0. The values of r1 and r2 were found using hammond's postulate. The reaction 'F + H2 ---> FH + H' has an early transition state which means it will resemble the reactants more than the products. With this in mind the transition state could be located using bond distances of r1=1.799 and r2=0.74. r1 was found more accurately using trial and error, after hammond's postulate aided in where to start to looking. {{fontcolor1|gray| Good [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:26, 31 May 2017 (BST)}}

[[File:tstatewjb115.PNG|500px]]

The above graph shows bond distances at the transition state.

[[File:thierryhenrywjb115.PNG|500px]]

The above confirms a transition state as there is minimal oscillation.

'''''Report the activation energy for both reactions.'''''

Activation energy is the difference between the energy of the reactants and the energy of the transition state.

'F + H2 ---> FH + H': The activation energy for this reaction was found to be (-103.3)-(-103.4)= 0.1 kJ/mol

[[File:bergkampwjb115.PNG|350px]] [[File:seamanwjb115.PNG|350px]]

'FH + H ---> H2': Activation energy = -103.3- -124= 20.7 kJ/mol

[[File:laurenwjb115.PNG|350px]] [[File:seamanwjb115.PNG|350px]]

{{fontcolor1|gray| Values are a bit off. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:26, 31 May 2017 (BST)}}

'''''In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?'''''

In any isolated system the total energy is always conserved. In this reaction potential energy is converted into kinetic and some vibrational energy. Initially, energy is stored as potential energy within the molecules, then as it proceeds towards the tranisiton state the potential energy is converted into kinetic energy. So as the potential energy decreases the kinetic energy increases. If, there is a successful reaction between the molecules, the energy is converted back to potential energy, as a transition structure is a potential energy maximum. Then as the reaction proceeds to make the products this potential energy is converted back into kinetic energy.

IR could be used to determine how much energy is converted into vibrational energy rather than kinetic. {{fontcolor1|gray| How? Give a bit more details. What vibration would you monitor? Any other experimental technique that you can think of? Think about temperature release... [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:26, 31 May 2017 (BST)}}

'''''Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''''

The question of different modes of energy can be approached using Polyani's rules, in which he states that vibrational energy is more efficient in promoting a later transition state than kinetic energy. This would imply that in an exothermic reaction, where the transition state is early and resembles the reactants, having more kinetic than vibrational energy is preferable and would make this type of reaction more likely. The converse would be the case in an endothermic reaction, where the transition state is late, whereby vibrational energy would make this type of reaction more likely. The rules also states, however, that if the kinetic in an exothermic or vibrational in endothermic is too high then the reaction may proceed to the products, but then also reverse back to the reactants.

{{fontcolor1|gray| Missing some examples here [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:26, 31 May 2017 (BST)}}

====References====

1. Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.

2. Atkins' Physical Chemistry. Atkins, P. and Paula, J. 10th edition. 2014.

3. Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419

4.The reaction of F + H2→ HF + H. A case study in reaction dynamics, Faraday Discuss. Chem. Soc., 1977,62, 267-290, John C. Polanyi and Jerry L. Schreiber

MRD:wjb115

2017-05-31T16:24:07Z

Je714: /* Exercise 1:H + H2 system */

=== Molecular Reaction Dynamics ===

====Exercise 1:H + H2 system====

'''''What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.'''''

The total gradient of the potential energy surface, at a minimum, is zero, where ∂V(ri)/∂ri=0.
At a transition structure ∂V(r1)/∂r1∂,V(r2)/∂r2=0, which means the potential energy is again 0. The transition state is found at the maximum on the minimum energy path, and as a result moving in one direction will give a maxima and the other direction a minima. Consequently, the transition structure can be considered a saddle point.
Minima can be used to find the reactants, products and intermediates.
In order to determine transition structures, start a trajectory near the transition state and observe whether it "rolls" towards the reactants or products.

{{fontcolor1|gray| The rolling analogy is not wrong but it's just that, an analogy. A more rigorous discussion is needed here, either discussing curvature with words or using second partial derivatives and their signs. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:24, 31 May 2017 (BST)}}

'''''Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.'''''

The transition state can be found on the maximum of the minimum energy path. By setting the momentum to zero and making r1=r2, it was possible to locate the transition state. Initially the bond lengths were set equal at just under 1.0Å, which was chosen using a plot of internuclear distance vs time. After trial and error with the distance, it was found that at 0.907Å the transition state is most likely to be found, since at this distance the trajectory oscillated and "never fell off". {{fontcolor1|gray| Never is a strong word. You can always run a sufficiently long trajectory and the TS geometry will be lost -- the TS is a mathematical construct which you can only get close to with arbitrary precision in practice. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:24, 31 May 2017 (BST)}}

[[File:Internucdist_vs_time.PNG]]
Figure. 1 Plot of Internuclear distance vs time

The above plot shows the internuclear distance vs time at a distance of 0.907Å, when the oscillations appeared to be at their lowest.
'''
''Comment on how the mep and the trajectory you just calculated differ.''
'''
r1=0.917 r2=0.907

[[File:mepwjb115.PNG|300px]] [[File:vierawjb115.PNG|300px]]

Both the mep and dynamic calculated trajectories follow the minimum energy path in the valley, however, the plotted lines differ in each. In the mep the line is curved, but in dynamic it appears to be oscillating. This is a result of vibrational energy being neglected in the mep {{fontcolor1|gray| how is it neglected? That's what we're asking [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:24, 31 May 2017 (BST)}}, whilst in dynamic these changing vibrational energies are considered, and are responsible for the the wave-like oscillation of the dynamically calculated trajectory.

'''''Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.'''''

The initial positions used were r1 = 0.74 and r2 = 2.0, trajectories were then run with following momenta:

{| class="wikitable" border="1"
|+ Momenta dependent reactions:
!Reaction Number!!P1 !! P2 !! Reaction?
|-
| 1||-1.25 || -2.5 || Yes
|-
| 2||-1.5 || -2.0 || No
|-
| 3||-1.5 || -2.5 || Yes
|-
| 4||-2.5 || -5.0 || No
|-
| 5||-2.5 || -5.2 || Yes
|}

Reaction 1.[[File:1reactivewjb115.PNG|300px]]

This trajectory passes through the transition state and then proceeds to the products, making it a reactive process.

Reaction 2.[[File:2reactivewjb115.PNG|300px]]

The trajectory is unreactive since when it reaches the transition state it "bounces" back towards the reactants instead of towards the products, as the energy barrier of the transition can not be overcome.

Reaction 3.[[File:3reactivewjb115.PNG|300px]]

Again, the trajectory passes through the transition state and onto the products, making it a reactive process.

Reaction 4.[[File:4reactivewjb115.PNG|300px]]

The trajectory shows barrier recrossing where the transition state is crossed twice, and the trajectory reverts back to the reactants. Consequently, this is an unreactive process.

Reaction 5.[[File:5reactivewjb115.PNG|300px]]

Slightly different to the previous reaction, in this example the transition state appears to be crossed twice but instead of returning to the reactants there is sufficient energy to pass the activation energy of the reaction, so it can proceed towards products. This is therefore a reactive process.

'''''State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''''

The transition state theory is used to explain the reaction rates of elementary reactions. The theory assumes that the reactants and transition state structures, of a reaction, are in quasi-equilibrium which differs from classical chemical equilibrium. The theory is also based on the assumption that nuclei behave in classical mechanical sense, whereby a reaction can only proceed provided both the colliding molecules have sufficient energy to form the structure. This classical approach ignores the possibility of quantum tunneling, which states that reactions can occur even if the colliding molecules do not have the required energy to pass the energy barrier.
At high temperatures the theory is also less useful as for some reactions it can fail. The theory is based on an idea that the reaction will pass through lowest energy saddle point on the potential energy surface, which at low temperatures is always true. However, as higher temperatures are reached molecules can populate vibrational modes. As a result of this, transition states may be far away from the lowest energy saddle point.
Lastly, an assumption is made that every intermediate has a long enough lifetime for it reach a boltzmann distribution, before moving on to the next step of the reaction.

{{fontcolor1|gray| The quasi-equilibrium comment is not applicable in this case -- it's a statistical mechanic treatment of TST. We're looking at a triatomic collision in isollation -- we're not simulating an ensemble of particles. Otherwise good! [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:24, 31 May 2017 (BST)}}

====Exercise 2: F-H-H system====

'''''Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''''

The reaction 'F + H2 ---> FH + H' has its trajectory shown below:

[[File:F+h2.PNG|500px]]

One can see from the graph that the potential energy of the products are below the potential energy of reactants. Consequently, this is an exothermic reaction as energy is being released.
The converse can be said for the 'FH + H ---> H2' reaction, since it is simply the prior reaction in the reverse direction. Subsequently, the reactants would therefore site at lower potential energy than the products, so it would be considered an endothermic reaction.

This implies that the H2 molecule has a lower bond enthalpy than HF, since for a reaction to be exothermic 'bonds broken-bonds formed<0'. For this to be the case in 'F + H2 ---> FH + H' the bond dissociation energies of H2 must be lower than those of F-H, otherwise a negative value of bond enthalpy change could not be obtained.

'''''Locate the approximate position of the transition state.'''''

In order to determine the position of the transition state the same method as previously was used, where the momenta p1 and p2 were set to 0. The values of r1 and r2 were found using hammond's postulate. The reaction 'F + H2 ---> FH + H' has an early transition state which means it will resemble the reactants more than the products. With this in mind the transition state could be located using bond distances of r1=1.799 and r2=0.74. r1 was found more accurately using trial and error, after hammond's postulate aided in where to start to looking.

[[File:tstatewjb115.PNG|500px]]

The above graph shows bond distances at the transition state.

[[File:thierryhenrywjb115.PNG|500px]]

The above confirms a transition state as there is minimal oscillation.

'''''Report the activation energy for both reactions.'''''

Activation energy is the difference between the energy of the reactants and the energy of the transition state.

'F + H2 ---> FH + H': The activation energy for this reaction was found to be (-103.3)-(-103.4)= 0.1 kJ/mol

[[File:bergkampwjb115.PNG|350px]] [[File:seamanwjb115.PNG|350px]]

'FH + H ---> H2': Activation energy = -103.3- -124= 20.7 kJ/mol

[[File:laurenwjb115.PNG|350px]] [[File:seamanwjb115.PNG|350px]]

'''''In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?'''''

In any isolated system the total energy is always conserved. In this reaction potential energy is converted into kinetic and some vibrational energy. Initially, energy is stored as potential energy within the molecules, then as it proceeds towards the tranisiton state the potential energy is converted into kinetic energy. So as the potential energy decreases the kinetic energy increases. If, there is a successful reaction between the molecules, the energy is converted back to potential energy, as a transition structure is a potential energy maximum. Then as the reaction proceeds to make the products this potential energy is converted back into kinetic energy.

IR could be used to determine how much energy is converted into vibrational energy rather than kinetic.

'''''Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''''

The question of different modes of energy can be approached using Polyani's rules, in which he states that vibrational energy is more efficient in promoting a later transition state than kinetic energy. This would imply that in an exothermic reaction, where the transition state is early and resembles the reactants, having more kinetic than vibrational energy is preferable and would make this type of reaction more likely. The converse would be the case in an endothermic reaction, where the transition state is late, whereby vibrational energy would make this type of reaction more likely. The rules also states, however, that if the kinetic in an exothermic or vibrational in endothermic is too high then the reaction may proceed to the products, but then also reverse back to the reactants.

====References====

1. Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.

2. Atkins' Physical Chemistry. Atkins, P. and Paula, J. 10th edition. 2014.

3. Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419

4.The reaction of F + H2→ HF + H. A case study in reaction dynamics, Faraday Discuss. Chem. Soc., 1977,62, 267-290, John C. Polanyi and Jerry L. Schreiber

Talk:MRD:01023138jw7815

2017-05-31T16:19:34Z

Je714: Created page with "Good report. Your last sections are very detailed, congrats. ~~~~"

Good report. Your last sections are very detailed, congrats.

[[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:19, 31 May 2017 (BST)

MRD:01023138jw7815

2017-05-31T16:18:44Z

Je714: /* State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? */

== Exercise 1: H + H2 system ==

=== What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface. ===
In a Potential Energy Surface (PES), the first derivative refers to the gradient of a point on the surface; while the second derivative refers to the rate of change of gradient of a point on the surface, also known as the curvature. A positive curvature implies that the gradient changes from negative to zero to positive (minimum point), a negative curvature implies that the gradient changes from positive to zero to negative (maximum point), and zero curvature implies that the gradient changes from negative to zero to negative or positive to zero to positive (saddle point).

The ''gradient'' of the potential energy surface at a minimum and at a transitions structure are both zero.

The minima and transition structures can be differentiated by their ''curvatures'' because the minima has a positive curvature (minimum point), while the transition structure has zero curvature (saddle point).

{{fontcolor1|gray| Not entirely sure what you mean here. At a TS, one of the second partial derivates is >0 and all the others are <0 [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:15, 31 May 2017 (BST)}}

=== Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory. ===
[[File:01023138 bestguess.png|1080x720px]]

From the "Internuclear distance vs Time" plot, the best estimate was obtained when rts = 0.9075 Å. With reference to the above Figure, the value was obtained by varying the values of r1 = r2 until the internuclear distance vs time behaviour switches from "oscillating" to just a steady horizontal line. A steady horizontal line means that the internuclear distances are constant with time, and is expected because a transition state will remain in its position indefinitely when the momentum p1 and p2 is zero. {{fontcolor1|gray| Why? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:16, 31 May 2017 (BST)}}

=== Comment on how the ''mep'' and the dynamic trajectory calculated differ. ===
Using r1 = rts + 0.1 and r2 = rts;

In ''mep'', the trajectory follows the valley floor as a non-oscillating straight line because the velocity is reset to zero in each time step, and the trajectory will only move in the direction of the steepest slope (towards the product).

In dynamic, the trajectory follows the valley as an oscilllating line because it takes into accout the initial velocity, which will influence its subsequent motion in the next time step.

{{fontcolor1|gray| Good [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:16, 31 May 2017 (BST)}}

=== Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory. ===
In the following analyses, it is important to point out that the momentum we are adjusting is directly proportional to the kinetic energy required to overcome the activation barrier ( KE = p2/2m).
{| class="wikitable"
!Reaction
!p1
!p2
!Reactive/Unreactive
!Trajectory
!Description
|-
|1
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|Reactive
|[[File:01023138 Case1.png|720x480px]]
|As A approaches B-C, the AB momentum is high enough such that the kinetic energy is sufficient to overcome the activation barrier, thus the trajectory will pass through the transition state to form A-B, with a bond distance of around 0.70. There is greater oscillation once the transition state is passed, implying that the product A-B has more vibrational energy than the reactants.
|-
|2
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|Unreactive
| [[File:01023138 Case2.png|720x480px]]
|As A approaches B-C, the AB momentum is not high enough and there is insufficient kinetic energy to overcome the activation barrier for the trajectory to pass through the transition state to form A-B. The trajectory thus shows an initial decrease in A-B bond length until around 1.1 Å, but increases back to its initial value (to reactants) and is unreactive.
|-
|3
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|Reactive
| [[File:01023138 Case3.png|720x480px]]
|As A approaches B-C, the AB momentum is high enough such that kinetic energy is sufficient to overcome the activation barrier, thus the trajectory will pass through the transition state to form A-B, with a bond distance of around 0.70. Unlike Reaction 1, the product A-B has only slightly more vibrational energy than the reactants as indicated by the greater degree of oscillation once it passes through the transition barrier.
|-
|4
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|Unreactive
| [[File:01023138 Case4.png|720x480px]]
|As A approaches B-C, the AB momentum is high enough such that kinetic energy is sufficient to overcome the activation barrier, however, while the trajectory initially crosses the transition barrier towards the product, it eventually recrosses back towards to reactants, and the reactant is reformed with a much higher vibration energy.
|-
|5
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|Reactive
| [[File:01023138 Case5.png|720x480px]]
|As A approaches B-C, the AB momentum is higher enough such that the kinetic energy is sufficient to overcome the activation barrier. In this case, the trajectory passes through the transition barrier to the products, but recrosses to the reactants and then back to the product A-B, which has a bond length of around 0.70 Å and possess a much greater vibrational energy than the reactants.
|}

=== State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
The transition state theory is used to qualitatively understand how chemical reactions take place. The following are key assumptions made in this theory:
# The theory assumes that the atomic nuclei follow classical mechanics, where sufficient kinetic energy must be present to overcome the activation barrier. This is not true because in quantum mechanics, atoms with insufficient energy can still tunnel through the activation barrier without actually "overcoming" it.
# The theory assumes that barrier recrossing does not occur, as seen in reaction 4 and 5. This means that if a trajectory passes through the activation barrier, it must lead to products. However, from reaction 4, we see that this is not necessarily true - recrossing occured to reform the reactants. This highlights that even if a molecule possesses sufficient kinetic energy to overcome the activation barrier, there are other factors that must be considered for a successful reaction.
# The theory assumes that the intermediates have a sufficiently long lifetime to reach a Boltzmann distribution of energies before forming the products. In reality, intermediates are not long-lived and exist on a timescale of around 10-13 seconds, which is insufficient for the Boltzmann distribution to be reached. {{fontcolor1|gray| This is a statistical mechanics consideration that is not really applicable in our simulations. We're just looking at a triatomic collision in isolation -- no ensemble of particles are considered here. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:18, 31 May 2017 (BST)}}
# The theory assumes that the reaction trajectory passes through the transition state (saddle point on the PES) before reaching the product. This is not necessarily the case, and we see in reaction 5 that s possible for trajectories to result in product without having passes through the saddle point.
# The theory only assumes 3 possible states in the reaction - the reactants, transition complex, and the products, anything else is not considered.
Apart from these assumptions, it is important to highlight that the transition state theory may not apply at elevated temperatures because molecules begin to populate higher vibrational modes and result in complex motion which may lead to transition states far away from the lowest energy saddle point.

{{fontcolor1|gray| So, does TST under or over estimate reaction rates? You forgot to discuss that. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:18, 31 May 2017 (BST)}}

== Exercise 2: F + H2 system ==

=== Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===
{| class="wikitable"
!Bond
!Bond Strength<ref>Y. R. Luo, ''Comprehensive handbook of chemical bond energies. CRC press.'', 2007.</ref>/kJ mol-1
|-
|H-F
|565
|-
|H-H
|432
|}
For F + H2, the reaction is exothermic as the product energy is lower than the reactant energy as seen in the potential energy surface. This is in agreement with the prediction based on bond strengths, as the H-F bond being formed is stronger than the H-H bond being broken, resulting in a net lowering of energy.

For H + HF, the reaction is endothermic as the product energy is higher than the reactant energy as seen in the potential energy surface. Once again, this is to be expected as the H-H bond being formed is weaker than the H-F bond being broken, resulting in a net raising of energy.

=== Locate the approximate position of the transition state. ===
[[File:01023138 Fhfts.png|none|thumb|1080x720px]]

From the "Internuclear distance vs Time" plot, a constant internuclear distance is obtained (as shown above) when rHH = 0.744 Å and rHF = 1.811 Å.

=== Report the activation energy for both reactions. ===
F + H2 -> HF + H

For this illustration, the reactants are F + H2 and the products are HF + H.
{| class="wikitable"
!Direction of Reaction
!Potential Energy vs Time
!Energy/kJ mol-1
!
|-
|Transition State -> HF + H
|[[File:01023138 Forward.png|720x480px]]
|Transition State = -103.8

Products (HF + H) = -133.9
|
|-
|Transition State -> F + H2
|[[File:01023138 Backward.png|720x480px]]
|Transition State = -103.8

Reactants (F + H2) = -104.0
|
|}

==== Forward Reaction: F + H2 -> HF + H ====
Activation Energy = -103.8 - (-133.9) = 30.1 kcal/mol

==== Backward Reaction: H + HF -> F + H2 ====
Activation Energy = -103.8 - (-104.0) = 0.2 kcal/mol

=== In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally? ===

[[File:01023138_Initialconditions.png|1080x720px]]

In the forward exothermic reaction F + H2 -> H + HF, the reactants posses more potential energy than the products. Because of conservation of energy, during the course of the reaction, the potential energy and is converted to kinetic energy. This reaction energy released in the form of kinetic energy can be converted to vibrational energy and translational energy, and can be detected through infrared chemiluminscence or through temperature measurements.

==== Infrared Chemiluminescence ====
From the above trajectory, it can be seen that the product H-F has a greater vibrational energy than the reactant H-H represented by the greater degree of oscillations. The infrared emission of the vibrationally excited product can be measured using methods such as infrared chemiluminescence <ref>J. C. Polanyi, ''J. Quant. Spectrosc. Radiat. Transf.'', 1963, '''3''', 471–496.</ref>.

==== Temperature Measurements ====
If reaction energy is released as heat (i.e. thermal energy), this can also be measured experimentally by monitoring the increase in temperature as the reaction progresses.

=== Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. ===
For an ''efficient'' reaction, the reactants must not only have sufficient energy, but should have the right distribution of kinetic energy between the vibrational and translational components. Polanyi's empircal rules states for reactions with late transition states, vibrational energy is more effective than translational energy for an efficient reaction, while for reactions with early transition states, translational energy is more effective than vibrational energy for an efficient reaction.

According to Hammond's postulate, endothermic reactions have a late transition state, while exothermic reactions have an early transition state. Thus, his rules can be reframed to mean endothermic reactions prefer vibrational energy, while exothermic reactions prefer translational energy.

{| class="wikitable"
!Direction of Reaction
!Thermodynamics of Reaction
!High Vibrational Energy,
Low Translational Energy
!Low Vibrational Energy,
High Translational Energy
!Analysis using Polanyi's Rule
|-
|Forward (F + H2 -> HF + H)
|Exothermic
|[[File:01023138Exo - low translational, high vibrational.png|320x240px]]

pHF = -0.1 (Low translational momentum)

pHH = 0.8 (High vibrational momentum)
|[[File:01023138Exo - high translational, low vibrational.png|320x240px]]

pHF = -0.8 (High translational momentum)

pHH = 0.1 (Low vibrational momentum)
|In both cases, the total energy is the same. In the first case with low translational energy, the trajectory is unreactive. However, with high translational energy, the trajectory becomes reactive.

This is in line with Polanyi's rule, since translational energy is more important for an effective reaction when the process is exothermic.
|-
|Backaward (H + HF -> F + H2)
|Endothermic
|[[File:01023138Endo - high vibrational, low translational.png|320x240px]]

pHF = 6.5 (High vibrational momentum)

pHH = -1.5 (Low translational momentum)
|[[File:01023138Endo - low vibrational high translational.png|320x240px]]

pHF = 1.5 (Low vibrational momentum)

pHH = -6.5 (High translational momentum)
|In both cases, the total energy is the same. In the first case with high vibrational energy, the trajectory is reactive. As soon as the vibrational energy is lowered in the second case, the trajectory becomes unreactive.

This is in line with Polanyi's rule, since vibrational energy is more important for an effective reaction when the process is endothermic.
|}

== References ==
<references />

MRD:01023138jw7815

2017-05-31T16:16:37Z

Je714: /* Comment on how the mep and the dynamic trajectory calculated differ. */

== Exercise 1: H + H2 system ==

=== What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface. ===
In a Potential Energy Surface (PES), the first derivative refers to the gradient of a point on the surface; while the second derivative refers to the rate of change of gradient of a point on the surface, also known as the curvature. A positive curvature implies that the gradient changes from negative to zero to positive (minimum point), a negative curvature implies that the gradient changes from positive to zero to negative (maximum point), and zero curvature implies that the gradient changes from negative to zero to negative or positive to zero to positive (saddle point).

The ''gradient'' of the potential energy surface at a minimum and at a transitions structure are both zero.

The minima and transition structures can be differentiated by their ''curvatures'' because the minima has a positive curvature (minimum point), while the transition structure has zero curvature (saddle point).

{{fontcolor1|gray| Not entirely sure what you mean here. At a TS, one of the second partial derivates is >0 and all the others are <0 [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:15, 31 May 2017 (BST)}}

=== Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory. ===
[[File:01023138 bestguess.png|1080x720px]]

From the "Internuclear distance vs Time" plot, the best estimate was obtained when rts = 0.9075 Å. With reference to the above Figure, the value was obtained by varying the values of r1 = r2 until the internuclear distance vs time behaviour switches from "oscillating" to just a steady horizontal line. A steady horizontal line means that the internuclear distances are constant with time, and is expected because a transition state will remain in its position indefinitely when the momentum p1 and p2 is zero. {{fontcolor1|gray| Why? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:16, 31 May 2017 (BST)}}

=== Comment on how the ''mep'' and the dynamic trajectory calculated differ. ===
Using r1 = rts + 0.1 and r2 = rts;

In ''mep'', the trajectory follows the valley floor as a non-oscillating straight line because the velocity is reset to zero in each time step, and the trajectory will only move in the direction of the steepest slope (towards the product).

In dynamic, the trajectory follows the valley as an oscilllating line because it takes into accout the initial velocity, which will influence its subsequent motion in the next time step.

{{fontcolor1|gray| Good [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:16, 31 May 2017 (BST)}}

=== Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory. ===
In the following analyses, it is important to point out that the momentum we are adjusting is directly proportional to the kinetic energy required to overcome the activation barrier ( KE = p2/2m).
{| class="wikitable"
!Reaction
!p1
!p2
!Reactive/Unreactive
!Trajectory
!Description
|-
|1
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|Reactive
|[[File:01023138 Case1.png|720x480px]]
|As A approaches B-C, the AB momentum is high enough such that the kinetic energy is sufficient to overcome the activation barrier, thus the trajectory will pass through the transition state to form A-B, with a bond distance of around 0.70. There is greater oscillation once the transition state is passed, implying that the product A-B has more vibrational energy than the reactants.
|-
|2
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|Unreactive
| [[File:01023138 Case2.png|720x480px]]
|As A approaches B-C, the AB momentum is not high enough and there is insufficient kinetic energy to overcome the activation barrier for the trajectory to pass through the transition state to form A-B. The trajectory thus shows an initial decrease in A-B bond length until around 1.1 Å, but increases back to its initial value (to reactants) and is unreactive.
|-
|3
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|Reactive
| [[File:01023138 Case3.png|720x480px]]
|As A approaches B-C, the AB momentum is high enough such that kinetic energy is sufficient to overcome the activation barrier, thus the trajectory will pass through the transition state to form A-B, with a bond distance of around 0.70. Unlike Reaction 1, the product A-B has only slightly more vibrational energy than the reactants as indicated by the greater degree of oscillation once it passes through the transition barrier.
|-
|4
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|Unreactive
| [[File:01023138 Case4.png|720x480px]]
|As A approaches B-C, the AB momentum is high enough such that kinetic energy is sufficient to overcome the activation barrier, however, while the trajectory initially crosses the transition barrier towards the product, it eventually recrosses back towards to reactants, and the reactant is reformed with a much higher vibration energy.
|-
|5
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|Reactive
| [[File:01023138 Case5.png|720x480px]]
|As A approaches B-C, the AB momentum is higher enough such that the kinetic energy is sufficient to overcome the activation barrier. In this case, the trajectory passes through the transition barrier to the products, but recrosses to the reactants and then back to the product A-B, which has a bond length of around 0.70 Å and possess a much greater vibrational energy than the reactants.
|}

=== State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
The transition state theory is used to qualitatively understand how chemical reactions take place. The following are key assumptions made in this theory:
# The theory assumes that the atomic nuclei follow classical mechanics, where sufficient kinetic energy must be present to overcome the activation barrier. This is not true because in quantum mechanics, atoms with insufficient energy can still tunnel through the activation barrier without actually "overcoming" it.
# The theory assumes that barrier recrossing does not occur, as seen in reaction 4 and 5. This means that if a trajectory passes through the activation barrier, it must lead to products. However, from reaction 4, we see that this is not necessarily true - recrossing occured to reform the reactants. This highlights that even if a molecule possesses sufficient kinetic energy to overcome the activation barrier, there are other factors that must be considered for a successful reaction.
# The theory assumes that the intermediates have a sufficiently long lifetime to reach a Boltzmann distribution of energies before forming the products. In reality, intermediates are not long-lived and exist on a timescale of around 10-13 seconds, which is insufficient for the Boltzmann distribution to be reached.
# The theory assumes that the reaction trajectory passes through the transition state (saddle point on the PES) before reaching the product. This is not necessarily the case, and we see in reaction 5 that s possible for trajectories to result in product without having passes through the saddle point.
# The theory only assumes 3 possible states in the reaction - the reactants, transition complex, and the products, anything else is not considered.
Apart from these assumptions, it is important to highlight that the transition state theory may not apply at elevated temperatures because molecules begin to populate higher vibrational modes and result in complex motion which may lead to transition states far away from the lowest energy saddle point.

== Exercise 2: F + H2 system ==

=== Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===
{| class="wikitable"
!Bond
!Bond Strength<ref>Y. R. Luo, ''Comprehensive handbook of chemical bond energies. CRC press.'', 2007.</ref>/kJ mol-1
|-
|H-F
|565
|-
|H-H
|432
|}
For F + H2, the reaction is exothermic as the product energy is lower than the reactant energy as seen in the potential energy surface. This is in agreement with the prediction based on bond strengths, as the H-F bond being formed is stronger than the H-H bond being broken, resulting in a net lowering of energy.

For H + HF, the reaction is endothermic as the product energy is higher than the reactant energy as seen in the potential energy surface. Once again, this is to be expected as the H-H bond being formed is weaker than the H-F bond being broken, resulting in a net raising of energy.

=== Locate the approximate position of the transition state. ===
[[File:01023138 Fhfts.png|none|thumb|1080x720px]]

From the "Internuclear distance vs Time" plot, a constant internuclear distance is obtained (as shown above) when rHH = 0.744 Å and rHF = 1.811 Å.

=== Report the activation energy for both reactions. ===
F + H2 -> HF + H

For this illustration, the reactants are F + H2 and the products are HF + H.
{| class="wikitable"
!Direction of Reaction
!Potential Energy vs Time
!Energy/kJ mol-1
!
|-
|Transition State -> HF + H
|[[File:01023138 Forward.png|720x480px]]
|Transition State = -103.8

Products (HF + H) = -133.9
|
|-
|Transition State -> F + H2
|[[File:01023138 Backward.png|720x480px]]
|Transition State = -103.8

Reactants (F + H2) = -104.0
|
|}

==== Forward Reaction: F + H2 -> HF + H ====
Activation Energy = -103.8 - (-133.9) = 30.1 kcal/mol

==== Backward Reaction: H + HF -> F + H2 ====
Activation Energy = -103.8 - (-104.0) = 0.2 kcal/mol

=== In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally? ===

[[File:01023138_Initialconditions.png|1080x720px]]

In the forward exothermic reaction F + H2 -> H + HF, the reactants posses more potential energy than the products. Because of conservation of energy, during the course of the reaction, the potential energy and is converted to kinetic energy. This reaction energy released in the form of kinetic energy can be converted to vibrational energy and translational energy, and can be detected through infrared chemiluminscence or through temperature measurements.

==== Infrared Chemiluminescence ====
From the above trajectory, it can be seen that the product H-F has a greater vibrational energy than the reactant H-H represented by the greater degree of oscillations. The infrared emission of the vibrationally excited product can be measured using methods such as infrared chemiluminescence <ref>J. C. Polanyi, ''J. Quant. Spectrosc. Radiat. Transf.'', 1963, '''3''', 471–496.</ref>.

==== Temperature Measurements ====
If reaction energy is released as heat (i.e. thermal energy), this can also be measured experimentally by monitoring the increase in temperature as the reaction progresses.

=== Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. ===
For an ''efficient'' reaction, the reactants must not only have sufficient energy, but should have the right distribution of kinetic energy between the vibrational and translational components. Polanyi's empircal rules states for reactions with late transition states, vibrational energy is more effective than translational energy for an efficient reaction, while for reactions with early transition states, translational energy is more effective than vibrational energy for an efficient reaction.

According to Hammond's postulate, endothermic reactions have a late transition state, while exothermic reactions have an early transition state. Thus, his rules can be reframed to mean endothermic reactions prefer vibrational energy, while exothermic reactions prefer translational energy.

{| class="wikitable"
!Direction of Reaction
!Thermodynamics of Reaction
!High Vibrational Energy,
Low Translational Energy
!Low Vibrational Energy,
High Translational Energy
!Analysis using Polanyi's Rule
|-
|Forward (F + H2 -> HF + H)
|Exothermic
|[[File:01023138Exo - low translational, high vibrational.png|320x240px]]

pHF = -0.1 (Low translational momentum)

pHH = 0.8 (High vibrational momentum)
|[[File:01023138Exo - high translational, low vibrational.png|320x240px]]

pHF = -0.8 (High translational momentum)

pHH = 0.1 (Low vibrational momentum)
|In both cases, the total energy is the same. In the first case with low translational energy, the trajectory is unreactive. However, with high translational energy, the trajectory becomes reactive.

This is in line with Polanyi's rule, since translational energy is more important for an effective reaction when the process is exothermic.
|-
|Backaward (H + HF -> F + H2)
|Endothermic
|[[File:01023138Endo - high vibrational, low translational.png|320x240px]]

pHF = 6.5 (High vibrational momentum)

pHH = -1.5 (Low translational momentum)
|[[File:01023138Endo - low vibrational high translational.png|320x240px]]

pHF = 1.5 (Low vibrational momentum)

pHH = -6.5 (High translational momentum)
|In both cases, the total energy is the same. In the first case with high vibrational energy, the trajectory is reactive. As soon as the vibrational energy is lowered in the second case, the trajectory becomes unreactive.

This is in line with Polanyi's rule, since vibrational energy is more important for an effective reaction when the process is endothermic.
|}

== References ==
<references />

MRD:01023138jw7815

2017-05-31T16:16:15Z

== Exercise 1: H + H2 system ==

=== What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface. ===
In a Potential Energy Surface (PES), the first derivative refers to the gradient of a point on the surface; while the second derivative refers to the rate of change of gradient of a point on the surface, also known as the curvature. A positive curvature implies that the gradient changes from negative to zero to positive (minimum point), a negative curvature implies that the gradient changes from positive to zero to negative (maximum point), and zero curvature implies that the gradient changes from negative to zero to negative or positive to zero to positive (saddle point).

The ''gradient'' of the potential energy surface at a minimum and at a transitions structure are both zero.

The minima and transition structures can be differentiated by their ''curvatures'' because the minima has a positive curvature (minimum point), while the transition structure has zero curvature (saddle point).

{{fontcolor1|gray| Not entirely sure what you mean here. At a TS, one of the second partial derivates is >0 and all the others are <0 [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:15, 31 May 2017 (BST)}}

=== Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory. ===
[[File:01023138 bestguess.png|1080x720px]]

From the "Internuclear distance vs Time" plot, the best estimate was obtained when rts = 0.9075 Å. With reference to the above Figure, the value was obtained by varying the values of r1 = r2 until the internuclear distance vs time behaviour switches from "oscillating" to just a steady horizontal line. A steady horizontal line means that the internuclear distances are constant with time, and is expected because a transition state will remain in its position indefinitely when the momentum p1 and p2 is zero. {{fontcolor1|gray| Why? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:16, 31 May 2017 (BST)}}

=== Comment on how the ''mep'' and the dynamic trajectory calculated differ. ===
Using r1 = rts + 0.1 and r2 = rts;

In ''mep'', the trajectory follows the valley floor as a non-oscillating straight line because the velocity is reset to zero in each time step, and the trajectory will only move in the direction of the steepest slope (towards the product).

In dynamic, the trajectory follows the valley as an oscilllating line because it takes into accout the initial velocity, which will influence its subsequent motion in the next time step.

=== Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory. ===
In the following analyses, it is important to point out that the momentum we are adjusting is directly proportional to the kinetic energy required to overcome the activation barrier ( KE = p2/2m).
{| class="wikitable"
!Reaction
!p1
!p2
!Reactive/Unreactive
!Trajectory
!Description
|-
|1
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|Reactive
|[[File:01023138 Case1.png|720x480px]]
|As A approaches B-C, the AB momentum is high enough such that the kinetic energy is sufficient to overcome the activation barrier, thus the trajectory will pass through the transition state to form A-B, with a bond distance of around 0.70. There is greater oscillation once the transition state is passed, implying that the product A-B has more vibrational energy than the reactants.
|-
|2
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|Unreactive
| [[File:01023138 Case2.png|720x480px]]
|As A approaches B-C, the AB momentum is not high enough and there is insufficient kinetic energy to overcome the activation barrier for the trajectory to pass through the transition state to form A-B. The trajectory thus shows an initial decrease in A-B bond length until around 1.1 Å, but increases back to its initial value (to reactants) and is unreactive.
|-
|3
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|Reactive
| [[File:01023138 Case3.png|720x480px]]
|As A approaches B-C, the AB momentum is high enough such that kinetic energy is sufficient to overcome the activation barrier, thus the trajectory will pass through the transition state to form A-B, with a bond distance of around 0.70. Unlike Reaction 1, the product A-B has only slightly more vibrational energy than the reactants as indicated by the greater degree of oscillation once it passes through the transition barrier.
|-
|4
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|Unreactive
| [[File:01023138 Case4.png|720x480px]]
|As A approaches B-C, the AB momentum is high enough such that kinetic energy is sufficient to overcome the activation barrier, however, while the trajectory initially crosses the transition barrier towards the product, it eventually recrosses back towards to reactants, and the reactant is reformed with a much higher vibration energy.
|-
|5
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|Reactive
| [[File:01023138 Case5.png|720x480px]]
|As A approaches B-C, the AB momentum is higher enough such that the kinetic energy is sufficient to overcome the activation barrier. In this case, the trajectory passes through the transition barrier to the products, but recrosses to the reactants and then back to the product A-B, which has a bond length of around 0.70 Å and possess a much greater vibrational energy than the reactants.
|}

=== State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
The transition state theory is used to qualitatively understand how chemical reactions take place. The following are key assumptions made in this theory:
# The theory assumes that the atomic nuclei follow classical mechanics, where sufficient kinetic energy must be present to overcome the activation barrier. This is not true because in quantum mechanics, atoms with insufficient energy can still tunnel through the activation barrier without actually "overcoming" it.
# The theory assumes that barrier recrossing does not occur, as seen in reaction 4 and 5. This means that if a trajectory passes through the activation barrier, it must lead to products. However, from reaction 4, we see that this is not necessarily true - recrossing occured to reform the reactants. This highlights that even if a molecule possesses sufficient kinetic energy to overcome the activation barrier, there are other factors that must be considered for a successful reaction.
# The theory assumes that the intermediates have a sufficiently long lifetime to reach a Boltzmann distribution of energies before forming the products. In reality, intermediates are not long-lived and exist on a timescale of around 10-13 seconds, which is insufficient for the Boltzmann distribution to be reached.
# The theory assumes that the reaction trajectory passes through the transition state (saddle point on the PES) before reaching the product. This is not necessarily the case, and we see in reaction 5 that s possible for trajectories to result in product without having passes through the saddle point.
# The theory only assumes 3 possible states in the reaction - the reactants, transition complex, and the products, anything else is not considered.
Apart from these assumptions, it is important to highlight that the transition state theory may not apply at elevated temperatures because molecules begin to populate higher vibrational modes and result in complex motion which may lead to transition states far away from the lowest energy saddle point.

== Exercise 2: F + H2 system ==

=== Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===
{| class="wikitable"
!Bond
!Bond Strength<ref>Y. R. Luo, ''Comprehensive handbook of chemical bond energies. CRC press.'', 2007.</ref>/kJ mol-1
|-
|H-F
|565
|-
|H-H
|432
|}
For F + H2, the reaction is exothermic as the product energy is lower than the reactant energy as seen in the potential energy surface. This is in agreement with the prediction based on bond strengths, as the H-F bond being formed is stronger than the H-H bond being broken, resulting in a net lowering of energy.

For H + HF, the reaction is endothermic as the product energy is higher than the reactant energy as seen in the potential energy surface. Once again, this is to be expected as the H-H bond being formed is weaker than the H-F bond being broken, resulting in a net raising of energy.

=== Locate the approximate position of the transition state. ===
[[File:01023138 Fhfts.png|none|thumb|1080x720px]]

From the "Internuclear distance vs Time" plot, a constant internuclear distance is obtained (as shown above) when rHH = 0.744 Å and rHF = 1.811 Å.

=== Report the activation energy for both reactions. ===
F + H2 -> HF + H

For this illustration, the reactants are F + H2 and the products are HF + H.
{| class="wikitable"
!Direction of Reaction
!Potential Energy vs Time
!Energy/kJ mol-1
!
|-
|Transition State -> HF + H
|[[File:01023138 Forward.png|720x480px]]
|Transition State = -103.8

Products (HF + H) = -133.9
|
|-
|Transition State -> F + H2
|[[File:01023138 Backward.png|720x480px]]
|Transition State = -103.8

Reactants (F + H2) = -104.0
|
|}

==== Forward Reaction: F + H2 -> HF + H ====
Activation Energy = -103.8 - (-133.9) = 30.1 kcal/mol

==== Backward Reaction: H + HF -> F + H2 ====
Activation Energy = -103.8 - (-104.0) = 0.2 kcal/mol

=== In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally? ===

[[File:01023138_Initialconditions.png|1080x720px]]

In the forward exothermic reaction F + H2 -> H + HF, the reactants posses more potential energy than the products. Because of conservation of energy, during the course of the reaction, the potential energy and is converted to kinetic energy. This reaction energy released in the form of kinetic energy can be converted to vibrational energy and translational energy, and can be detected through infrared chemiluminscence or through temperature measurements.

==== Infrared Chemiluminescence ====
From the above trajectory, it can be seen that the product H-F has a greater vibrational energy than the reactant H-H represented by the greater degree of oscillations. The infrared emission of the vibrationally excited product can be measured using methods such as infrared chemiluminescence <ref>J. C. Polanyi, ''J. Quant. Spectrosc. Radiat. Transf.'', 1963, '''3''', 471–496.</ref>.

==== Temperature Measurements ====
If reaction energy is released as heat (i.e. thermal energy), this can also be measured experimentally by monitoring the increase in temperature as the reaction progresses.

=== Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. ===
For an ''efficient'' reaction, the reactants must not only have sufficient energy, but should have the right distribution of kinetic energy between the vibrational and translational components. Polanyi's empircal rules states for reactions with late transition states, vibrational energy is more effective than translational energy for an efficient reaction, while for reactions with early transition states, translational energy is more effective than vibrational energy for an efficient reaction.

According to Hammond's postulate, endothermic reactions have a late transition state, while exothermic reactions have an early transition state. Thus, his rules can be reframed to mean endothermic reactions prefer vibrational energy, while exothermic reactions prefer translational energy.

{| class="wikitable"
!Direction of Reaction
!Thermodynamics of Reaction
!High Vibrational Energy,
Low Translational Energy
!Low Vibrational Energy,
High Translational Energy
!Analysis using Polanyi's Rule
|-
|Forward (F + H2 -> HF + H)
|Exothermic
|[[File:01023138Exo - low translational, high vibrational.png|320x240px]]

pHF = -0.1 (Low translational momentum)

pHH = 0.8 (High vibrational momentum)
|[[File:01023138Exo - high translational, low vibrational.png|320x240px]]

pHF = -0.8 (High translational momentum)

pHH = 0.1 (Low vibrational momentum)
|In both cases, the total energy is the same. In the first case with low translational energy, the trajectory is unreactive. However, with high translational energy, the trajectory becomes reactive.

This is in line with Polanyi's rule, since translational energy is more important for an effective reaction when the process is exothermic.
|-
|Backaward (H + HF -> F + H2)
|Endothermic
|[[File:01023138Endo - high vibrational, low translational.png|320x240px]]

pHF = 6.5 (High vibrational momentum)

pHH = -1.5 (Low translational momentum)
|[[File:01023138Endo - low vibrational high translational.png|320x240px]]

pHF = 1.5 (Low vibrational momentum)

pHH = -6.5 (High translational momentum)
|In both cases, the total energy is the same. In the first case with high vibrational energy, the trajectory is reactive. As soon as the vibrational energy is lowered in the second case, the trajectory becomes unreactive.

This is in line with Polanyi's rule, since vibrational energy is more important for an effective reaction when the process is endothermic.
|}

== References ==
<references />

MRD:01023138jw7815

2017-05-31T16:15:50Z

== Exercise 1: H + H2 system ==

=== What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface. ===
In a Potential Energy Surface (PES), the first derivative refers to the gradient of a point on the surface; while the second derivative refers to the rate of change of gradient of a point on the surface, also known as the curvature. A positive curvature implies that the gradient changes from negative to zero to positive (minimum point), a negative curvature implies that the gradient changes from positive to zero to negative (maximum point), and zero curvature implies that the gradient changes from negative to zero to negative or positive to zero to positive (saddle point).

The ''gradient'' of the potential energy surface at a minimum and at a transitions structure are both zero.

The minima and transition structures can be differentiated by their ''curvatures'' because the minima has a positive curvature (minimum point), while the transition structure has zero curvature (saddle point).

{{fontcolor1|gray| Not entirely sure what you mean here. At a TS, one of the second partial derivates is >0 and all the others are <0 [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:15, 31 May 2017 (BST)}}

=== Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory. ===
[[File:01023138 bestguess.png|1080x720px]]

From the "Internuclear distance vs Time" plot, the best estimate was obtained when rts = 0.9075 Å. With reference to the above Figure, the value was obtained by varying the values of r1 = r2 until the internuclear distance vs time behaviour switches from "oscillating" to just a steady horizontal line. A steady horizontal line means that the internuclear distances are constant with time, and is expected because a transition state will remain in its position indefinitely when the momentum p1 and p2 is zero.

=== Comment on how the ''mep'' and the dynamic trajectory calculated differ. ===
Using r1 = rts + 0.1 and r2 = rts;

In ''mep'', the trajectory follows the valley floor as a non-oscillating straight line because the velocity is reset to zero in each time step, and the trajectory will only move in the direction of the steepest slope (towards the product).

In dynamic, the trajectory follows the valley as an oscilllating line because it takes into accout the initial velocity, which will influence its subsequent motion in the next time step.

=== Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory. ===
In the following analyses, it is important to point out that the momentum we are adjusting is directly proportional to the kinetic energy required to overcome the activation barrier ( KE = p2/2m).
{| class="wikitable"
!Reaction
!p1
!p2
!Reactive/Unreactive
!Trajectory
!Description
|-
|1
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|Reactive
|[[File:01023138 Case1.png|720x480px]]
|As A approaches B-C, the AB momentum is high enough such that the kinetic energy is sufficient to overcome the activation barrier, thus the trajectory will pass through the transition state to form A-B, with a bond distance of around 0.70. There is greater oscillation once the transition state is passed, implying that the product A-B has more vibrational energy than the reactants.
|-
|2
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|Unreactive
| [[File:01023138 Case2.png|720x480px]]
|As A approaches B-C, the AB momentum is not high enough and there is insufficient kinetic energy to overcome the activation barrier for the trajectory to pass through the transition state to form A-B. The trajectory thus shows an initial decrease in A-B bond length until around 1.1 Å, but increases back to its initial value (to reactants) and is unreactive.
|-
|3
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|Reactive
| [[File:01023138 Case3.png|720x480px]]
|As A approaches B-C, the AB momentum is high enough such that kinetic energy is sufficient to overcome the activation barrier, thus the trajectory will pass through the transition state to form A-B, with a bond distance of around 0.70. Unlike Reaction 1, the product A-B has only slightly more vibrational energy than the reactants as indicated by the greater degree of oscillation once it passes through the transition barrier.
|-
|4
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|Unreactive
| [[File:01023138 Case4.png|720x480px]]
|As A approaches B-C, the AB momentum is high enough such that kinetic energy is sufficient to overcome the activation barrier, however, while the trajectory initially crosses the transition barrier towards the product, it eventually recrosses back towards to reactants, and the reactant is reformed with a much higher vibration energy.
|-
|5
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|Reactive
| [[File:01023138 Case5.png|720x480px]]
|As A approaches B-C, the AB momentum is higher enough such that the kinetic energy is sufficient to overcome the activation barrier. In this case, the trajectory passes through the transition barrier to the products, but recrosses to the reactants and then back to the product A-B, which has a bond length of around 0.70 Å and possess a much greater vibrational energy than the reactants.
|}

=== State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
The transition state theory is used to qualitatively understand how chemical reactions take place. The following are key assumptions made in this theory:
# The theory assumes that the atomic nuclei follow classical mechanics, where sufficient kinetic energy must be present to overcome the activation barrier. This is not true because in quantum mechanics, atoms with insufficient energy can still tunnel through the activation barrier without actually "overcoming" it.
# The theory assumes that barrier recrossing does not occur, as seen in reaction 4 and 5. This means that if a trajectory passes through the activation barrier, it must lead to products. However, from reaction 4, we see that this is not necessarily true - recrossing occured to reform the reactants. This highlights that even if a molecule possesses sufficient kinetic energy to overcome the activation barrier, there are other factors that must be considered for a successful reaction.
# The theory assumes that the intermediates have a sufficiently long lifetime to reach a Boltzmann distribution of energies before forming the products. In reality, intermediates are not long-lived and exist on a timescale of around 10-13 seconds, which is insufficient for the Boltzmann distribution to be reached.
# The theory assumes that the reaction trajectory passes through the transition state (saddle point on the PES) before reaching the product. This is not necessarily the case, and we see in reaction 5 that s possible for trajectories to result in product without having passes through the saddle point.
# The theory only assumes 3 possible states in the reaction - the reactants, transition complex, and the products, anything else is not considered.
Apart from these assumptions, it is important to highlight that the transition state theory may not apply at elevated temperatures because molecules begin to populate higher vibrational modes and result in complex motion which may lead to transition states far away from the lowest energy saddle point.

== Exercise 2: F + H2 system ==

=== Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===
{| class="wikitable"
!Bond
!Bond Strength<ref>Y. R. Luo, ''Comprehensive handbook of chemical bond energies. CRC press.'', 2007.</ref>/kJ mol-1
|-
|H-F
|565
|-
|H-H
|432
|}
For F + H2, the reaction is exothermic as the product energy is lower than the reactant energy as seen in the potential energy surface. This is in agreement with the prediction based on bond strengths, as the H-F bond being formed is stronger than the H-H bond being broken, resulting in a net lowering of energy.

For H + HF, the reaction is endothermic as the product energy is higher than the reactant energy as seen in the potential energy surface. Once again, this is to be expected as the H-H bond being formed is weaker than the H-F bond being broken, resulting in a net raising of energy.

=== Locate the approximate position of the transition state. ===
[[File:01023138 Fhfts.png|none|thumb|1080x720px]]

From the "Internuclear distance vs Time" plot, a constant internuclear distance is obtained (as shown above) when rHH = 0.744 Å and rHF = 1.811 Å.

=== Report the activation energy for both reactions. ===
F + H2 -> HF + H

For this illustration, the reactants are F + H2 and the products are HF + H.
{| class="wikitable"
!Direction of Reaction
!Potential Energy vs Time
!Energy/kJ mol-1
!
|-
|Transition State -> HF + H
|[[File:01023138 Forward.png|720x480px]]
|Transition State = -103.8

Products (HF + H) = -133.9
|
|-
|Transition State -> F + H2
|[[File:01023138 Backward.png|720x480px]]
|Transition State = -103.8

Reactants (F + H2) = -104.0
|
|}

==== Forward Reaction: F + H2 -> HF + H ====
Activation Energy = -103.8 - (-133.9) = 30.1 kcal/mol

==== Backward Reaction: H + HF -> F + H2 ====
Activation Energy = -103.8 - (-104.0) = 0.2 kcal/mol

=== In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally? ===

[[File:01023138_Initialconditions.png|1080x720px]]

In the forward exothermic reaction F + H2 -> H + HF, the reactants posses more potential energy than the products. Because of conservation of energy, during the course of the reaction, the potential energy and is converted to kinetic energy. This reaction energy released in the form of kinetic energy can be converted to vibrational energy and translational energy, and can be detected through infrared chemiluminscence or through temperature measurements.

==== Infrared Chemiluminescence ====
From the above trajectory, it can be seen that the product H-F has a greater vibrational energy than the reactant H-H represented by the greater degree of oscillations. The infrared emission of the vibrationally excited product can be measured using methods such as infrared chemiluminescence <ref>J. C. Polanyi, ''J. Quant. Spectrosc. Radiat. Transf.'', 1963, '''3''', 471–496.</ref>.

==== Temperature Measurements ====
If reaction energy is released as heat (i.e. thermal energy), this can also be measured experimentally by monitoring the increase in temperature as the reaction progresses.

=== Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. ===
For an ''efficient'' reaction, the reactants must not only have sufficient energy, but should have the right distribution of kinetic energy between the vibrational and translational components. Polanyi's empircal rules states for reactions with late transition states, vibrational energy is more effective than translational energy for an efficient reaction, while for reactions with early transition states, translational energy is more effective than vibrational energy for an efficient reaction.

According to Hammond's postulate, endothermic reactions have a late transition state, while exothermic reactions have an early transition state. Thus, his rules can be reframed to mean endothermic reactions prefer vibrational energy, while exothermic reactions prefer translational energy.

{| class="wikitable"
!Direction of Reaction
!Thermodynamics of Reaction
!High Vibrational Energy,
Low Translational Energy
!Low Vibrational Energy,
High Translational Energy
!Analysis using Polanyi's Rule
|-
|Forward (F + H2 -> HF + H)
|Exothermic
|[[File:01023138Exo - low translational, high vibrational.png|320x240px]]

pHF = -0.1 (Low translational momentum)

pHH = 0.8 (High vibrational momentum)
|[[File:01023138Exo - high translational, low vibrational.png|320x240px]]

pHF = -0.8 (High translational momentum)

pHH = 0.1 (Low vibrational momentum)
|In both cases, the total energy is the same. In the first case with low translational energy, the trajectory is unreactive. However, with high translational energy, the trajectory becomes reactive.

This is in line with Polanyi's rule, since translational energy is more important for an effective reaction when the process is exothermic.
|-
|Backaward (H + HF -> F + H2)
|Endothermic
|[[File:01023138Endo - high vibrational, low translational.png|320x240px]]

pHF = 6.5 (High vibrational momentum)

pHH = -1.5 (Low translational momentum)
|[[File:01023138Endo - low vibrational high translational.png|320x240px]]

pHF = 1.5 (Low vibrational momentum)

pHH = -6.5 (High translational momentum)
|In both cases, the total energy is the same. In the first case with high vibrational energy, the trajectory is reactive. As soon as the vibrational energy is lowered in the second case, the trajectory becomes unreactive.

This is in line with Polanyi's rule, since vibrational energy is more important for an effective reaction when the process is endothermic.
|}

== References ==
<references />

Talk:MRD:Yiming Xu

2017-05-31T16:13:41Z

Je714: Created page with "Congratulations, one of the best reports I've seen so far! ~~~~"

Congratulations, one of the best reports I've seen so far!

[[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:13, 31 May 2017 (BST)

MRD:Yiming Xu

2017-05-31T16:13:08Z

Je714: /* Exercise 2: F - H - H system */

= Exercise 1: H + H2 system =
[[File:yx6015_H3_legends.svg|thumb|400px|Definition of the various symbols used.]]
=== Dynamics from the Transition State Region ===
''Question: What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.''

{|style="margin: auto;"
|[[File:Yx6015_overall_minima.png|thumb|400px|'''Figure 1a''': Illustration of the minima "valley"-like area of the potential energy surface (<var>'''rAB'''</var> = 2.30; <var>'''rBC'''</var> = 0.74; <var>'''pA'''</var> = <var>'''pB'''</var> = -2.70)]]
|[[File:Yx6015_overall_transition.png|thumb|400px|'''Figure 1b''': A different perspective illustrating the transition state along the minimum energy path of the potential energy surface (<var>'''rAB'''</var> = 2.30; <var>'''rBC'''</var> = 0.74; <var>'''pA'''</var> = <var>'''pB'''</var> = -2.70)]]
|}

Minima and transition structures can be identified visually from the potential energy surface. As can be seen in '''Fig 1a''', minima appear as a "valley" of low potential energy between two "hills" of higher potential energy. This means that minima show an local upward curvature along parts of the potential energy surface with the lowest energy. The transition state is positioned along the minimum energy path (mep). However, as can be seen from '''Fig 1b''', it appears as a "bump" along the mep "valley". This results in it being a saddle point - with it being the local minimum along one axis and the local maximum along another axis (mep).

Mathematically, both minima and transition structures are at stationary points of the potential energy surface. At stationary points, the following relationship must hold:

<math display="block">
\nabla U = 0
</math>

where <var>'''∇'''</var> is the gradient and <var>'''U'''</var> is the potential energy. In order to determine whether the point found was a local minimum, maximum or a saddle (transition state), the local curvature need to be calculated using the second partial derivative test. For the potential energy surface of two variables <var>'''U(rA, rB)'''</var>, the Hessian matrix can be defined as:

<math display="block">
H(R_A, R_B)=
\begin{pmatrix}
U_{{r_A}{r_A}}(r_A, r_B) & U_{{r_A}{r_B}}(r_A, r_B) \\
U_{{r_B}{r_A}}(r_A, r_B) & U_{{r_B}{r_B}}(r_A, r_B)
\end{pmatrix}
</math>

The determinant of <var>'''H(rA, rB)'''</var>, <var>'''D(rA, rB)'''</var>, determines the nature of the stationary point <var>(a, b)</var>:
* if <var>'''D(a, b)'''</var> > 0 and <var>'''UrArA(a, b)'''</var> > 0, then the point <var>(a, b)</var> is a local minimum;
* if <var>'''D(a, b)'''</var> > 0 and <var>'''UrArA(a, b)'''</var> < 0, then the point <var>(a, b)</var> is a local maximum;
* if <var>'''D(a, b)'''</var> < 0 then the point <var>(a, b)</var> is a saddle point; and
* if <var>'''D(a, b)'''</var> = 0 then another test need to be performed.

From the second partial derivative test, the minima and transition structures can be identified.

{{fontcolor1|gray| Excellent! Good job on using math, and defining your symbols, showing understanding of it! [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:10, 31 May 2017 (BST)}}

=== Trajectories from rA = rB: Locating the Transition State ===
''Question: Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.''

Due to the symmetry of the system, with all three atoms being identical, we expect that <var>'''rts'''</var> will be located at a point such that <var>'''rA'''</var> and <var>'''rB'''</var> are identical. In other words, HB should be located between HA and HC. From the previous section, as the transition state is a stationary point, if the reaction coordinate was located on the transition state ''exactly'' with no momentum, it would be stationary and would not proceed any further. In other words, we can find a suitable <var>'''rts'''</var> at a particular <var>'''rA'''</var> = <var>'''rB'''</var> and <var>'''pA'''</var> = <var>'''pB'''</var> = 0

{|style="margin: auto;"
|[[File:Yx6015_ransition_state_distance_over_time.png|thumb|400px|'''Figure 2a''': Plot of internuclear distance over time, for <var>'''rA'''</var> = <var>'''rB'''</var> = 0.9085406012; <var>'''pA'''</var> = <var>'''pB'''</var> = 0]]
|[[File:Yx6015_ransition_state_distance_over_time_surface_plot.png|thumb|400px|'''Figure 2b''': Plot of the trajectory along the potential energy surface, for <var>'''rA'''</var> = <var>'''rB'''</var> = 0.9085406012; <var>'''pA'''</var> = <var>'''pB'''</var> = 0]]
|}
'''Fig 2a.''' shows a graph of the inter nuclear distances against time. As expected from being located on the stationary point, the distances stay approximately constant over time, with <var>'''rA'''</var> = <var>'''rB'''</var> (<var>'''rA'''</var> is not shown as it was completely overlapped by <var>'''rB'''</var>). The small amount of oscillation was due to symmetric stretching of the molecule, and is along a direction perpendicular to the mep. The nature of the stationary point can be more aptly visualised by '''Fig 2b.''' as the trajectory remain essentially unchanged after 1000 time step.

While <var>'''rts'''</var> used to generate the graphical plot was 0.9085406012, the precision obtained was much higher than what was meaningful due to the limited accuracy of the simulation. Comparing to a literature value of 0.93 , the best obtainable estimate of <var>'''rts'''</var> from this simulation would be 0.9 , an error of 3.3%.<ref>S. Grabowski and R. Hoffmann, ''ChemPhysChem'', 2012, 13, 2286-2288.
</ref>

=== Trajectories from rA = rts+δ, rB = rts ===
Once the transition state has been located, the reaction can be followed from transition state to the products or towards the reagents (due to time symmetry) in order to provide more information on the nature of reaction if only a minimum amount of energy was available.
{|style="margin: auto;"
|[[File:Yx6015_transition_state_mep.png|thumb|400px|'''Figure 3a''': Minimum energy path taken by the particle from the transition state (<var>'''rAB'''</var> = 0.91; <var>'''rBC'''</var> = 0.911; <var>'''pA'''</var> = <var>'''pB'''</var> = 0)]]
|[[File:Yx6015_transition_state_md.png|thumb|400px|'''Figure 3b''': Molecular dynamics trajectory taken by the particle from the transition state (<var>'''rAB'''</var> = 0.91; <var>'''rBC'''</var> = 0.911; <var>'''pA'''</var> = <var>'''pB'''</var> = 0)]]
|}
''Question: Comment on how the mep and the trajectory you just calculated differ.''

Both the mep and the molecular dynamics trajectory start at the exact same location at the same energy. However, they follow a different path "down" the transition state. The molecular dynamics trajectory is the physical trajectory taken by the molecule during the reaction. Compared to the mep, it shows oscillatory motion due to bond vibration. In addition, as the loss of potential energy was converted to kinetic energy and vibrational energy according to the equipartition theorem, with a higher translational kinetic energy, the molecule had moved further along the reaction (away from the transition state) during the same time (10000 steps).

On the other hand, the mep is not really a trajectory taken by the particle. It is an unphysical trajectory that characterises the path of steepest descent for the given potential energy surface. In this simulation, as it was calculated by setting the velocity to be 0 at every time step, it travelled to a lesser extent than that of the molecular dynamics trajectory.

In an actual reaction, the reaction trajectory will follow one provided by molecular dynamics, and may or may not pass through the transition state.

{{fontcolor1|gray| Very good explanation! [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:10, 31 May 2017 (BST)}}

=== Reactive and Unreactive Trajectories ===
Whether a trajectory results in an reaction can be identified by the bond between the atoms before and before interaction. As bonds are not identified explicitly, they can be inferred from the internuclear distances - a bond would have a relatively constant distance over time, with oscillatory motions due to vibration. With initial positions of <var>'''rA'''</var> = 0.74 and <var>'''rB'''</var>= 2.0, which means that HA is bonded with HB, trajectories with the following momenta were investigated.

{|style="margin: auto;" class="wikitable"
!pA
!pB
!colspan="2" | Screenshots
|-
|rowspan="2" | <nowiki>-1.25</nowiki>
|rowspan="2" | <nowiki>-2.5</nowiki>
|[[File:Yx6015_reactive_or_unreactive_4a.png|frameless|400px]]
|[[File:Yx6015_reactive_or_unreactive_4b.png|frameless|400px]]
|-
|style="width:800px;" colspan="2" | '''Figure 4''': This configuration illustrated an reactive trajectory. A straight forward exchange reaction between HC and HB-HA was shown. There was sufficient kinetic energy from HC to overcome the coulombic barrier to reach the transition state. The reaction cross the transition state at about t = 0.5, and resulted in the formation of a HC-HB molecule, and HA was ejected. The ejected HA had less kinetic energy than the incoming HC, and some of the energy was transferred to the vibrational energy of the newly formed HC-HB molecule. (<var>'''rAB'''</var> = 0.74; <var>'''rBC'''</var> = 2.0; <var>'''pA'''</var> = -1.25; <var>'''pB'''</var> = -2.5)
|-
|rowspan="2" | <nowiki>-1.5</nowiki>
|rowspan="2" | <nowiki>-2.0</nowiki>
|[[File:Yx6015_reactive_or_unreactive_5a.png|frameless|400px]]
|[[File:Yx6015_reactive_or_unreactive_5b.png|frameless|400px]]
|-
|style="width:800px;" colspan="2" | '''Figure 5''': This configuration illustrated an unreactive trajectory. There was insufficient energy to reach the transition state as HC did not have sufficient momentum to overcome the coulombic barrier. HC was repelled back to the approaching direction. (<var>'''rAB'''</var> = 0.74; <var>'''rBC'''</var> = 2.0; <var>'''pA'''</var> = -1.5; <var>'''pB'''</var> = -2.0)
|-
|rowspan="2" | <nowiki>-1.5</nowiki>
|rowspan="2" | <nowiki>-2.5</nowiki>
|[[File:Yx6015_reactive_or_unreactive_6a.png|frameless|400px]]
|[[File:Yx6015_reactive_or_unreactive_6b.png|frameless|400px]]
|-
|style="width:800px;" colspan="2" | '''Figure 6''': This configuration illustrated an reactive trajectory. The reaction was similar to that depicted by '''Fig 4.'''. However, with a higher initial <var>'''rA'''</var> momentum, there was more vibration in HB-HA before reaction occurred. The resulting trajectory was similar, however. (<var>'''rAB'''</var> = 0.74; <var>'''rBC'''</var> = 2.0; <var>'''pA'''</var> = -1.5; <var>'''pB'''</var> = -2.5)
|-
|rowspan="2" | <nowiki>-2.5</nowiki>
|rowspan="2" | <nowiki>-5.0</nowiki>
|[[File:Yx6015_reactive_or_unreactive_7a.png|frameless|400px]]
|[[File:Yx6015_reactive_or_unreactive_7b.png|frameless|400px]]
|-
|style="width:800px;" colspan="2" | '''Figure 7''': This configuration illustrated an unreactive trajectory. Unlike the unreactive trajectory depicted in '''Fig 5.''', there is more than sufficient energy to cross over the transition state. However, due to the high initial momentum of HC, it got too close to HB, resulting in high electrostatic forces of repulsion, repelling HC back to its original direction. This is an example of barrier recrossing, as the product did form for a short duration. Part of the kinetic energy of HC was transferred to the HB-HA molecule, resulting in higher vibrational energy as a product of the reaction. Observing the PES, the trajectory deflects off the energy barrier multiple times throughout the reaction, signifying a high energy system in contrast to previously investigated configurations. (<var>'''rAB'''</var> = 0.74; <var>'''rBC'''</var> = 2.0; <var>'''pA'''</var> = -2.5; <var>'''pB'''</var> = -5.0)
|-
|rowspan="2" | <nowiki>-2.5</nowiki>
|rowspan="2" | <nowiki>-5.2</nowiki>
|[[File:Yx6015_reactive_or_unreactive_8a.png|frameless|400px]]
|[[File:Yx6015_reactive_or_unreactive_8b.png|frameless|400px]]
|-
|style="width:800px;" colspan="2" | '''Figure 8''': This configuration illustrated a reactive trajectory. Unlike the trajectory depicted in the PES of '''Fig 7.''', it was deflected in a different direction, resulting in a successful exchange reaction. Barrier recrossing occurred twice in this reaction - the product HC-HB was formed, broken, and then formed again. One possible difference could be due to the slightly higher kinetic energy of HC, which allowed it to come closer to HA than before, and experience a higher attraction force towards the HB-HA molecule. However, an initial HC kinetic energy that is too high would result in a much higher repulsion than can be compensated by the slightly higher attraction and would result in an unreactive trajectory as shown in '''Fig 7.''' (<var>'''rAB'''</var> = 0.74; <var>'''rBC'''</var> = 2.0; <var>'''pA'''</var> = -2.5; <var>'''pB'''</var> = -5.2)
|-
|}

=== Transition State Theory ===
''Question: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''

The Transition State Theory (TST) is an alternative to the Collision Theory, which explains the rate of chemical reactions and provide theoretical basis for the Arrhenius equation, <math> k = A \exp(-\frac{E_a}{RT})</math>. It assumes a special type of equilibrium exists between the reactants and the activated complex before the product is formed, and provides the following rate equation (the Eyring equation):<ref><nowiki>IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). XML on-line corrected version: http://goldbook.iupac.org (2006-) created by M. Nic, J. Jirat, B. Kosata; updates compiled by A. Jenkins. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook</nowiki></ref>

<math display="block">
k = \frac{k_B T}{h} \exp(\frac{\Delta ^\ddagger S ^\circ}{R}) \exp(-\frac{\Delta ^\ddagger H ^\circ}{RT})
</math>

The Eyring equation was developed with two assumptions:<ref>D. Truhlar, B. Garrett and S. Klippenstein, ''The Journal of Physical Chemistry'', 1996, 100, 12771-12800.</ref>
<ol start="1">
<li>The reactants and product space in the PES can be clearly differentiated/divided; and</li>
<li>Trajectories going from reactant space to product space will not go back.</li>
</ol>

In addition, it implicitly assumes that:

<ol start="3">
<li>The reactants are equilibrated in an ensemble; and</li> {{fontcolor1|gray| Does this make sense here? We're just looking at a triatomic collision in isolation -- no ensemble of particles considered... [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:10, 31 May 2017 (BST)}}
<li>The overall reaction is electronically adiabatic.</li>
</ol>From the results of the previous part, it can be seen that assumption 2 is not valid. Even if the transition state had been achieved, the trajectory may still undergoing recrossing to return to the reagent space. This recrossing may even happen if the product was formed momentarily. In addition, multiple recrossing may occur depending on the momenta and angle of the trajectory to give the desired product after an even number of crossings. This effect result in TST overestimating the reaction rate constant as compared to experimental values.

In addition, quantum mechanical effects may also affect the rate of reaction by allowing reactants to tunnel through the activation barrier without having sufficient energy. While this would indicate an underestimation of the rate by TST, the previous effect would be much more significant, leading TST predicting higher reaction rate values when compared with experimental data.

{{fontcolor1|gray| Good! [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:10, 31 May 2017 (BST)}}

= Exercise 2: F - H - H system =
[[File:yx6015_FH2_legends.svg|thumb|400px|Definition of the various symbols used.]]
[[File:yx6015_HFH_legends.svg|thumb|400px|Definition of the various symbols used.]]
=== PES inspection ===
''Question: Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?''

There are two different reaction under investigation. The first reaction (Reaction 1.) is : F + H2 ⟶ HF + H. The second reaction (Reaction 2.) considered is the reverse reaction of the first, namely H + HF ⟶ H2 + F. As they are the reverse reaction of each other, and due to the fact that enthalpy is a state function, if one of them (e.g. Reaction 1.) is exothermic, the reverse reaction (i.e. Reaction 2.) must be endothermic.

The reaction energetics can be determined simply by examining the potential energy surface of the reaction, and identifying the correct viewing perspective.

{|style="margin: auto;"
|[[File:Yx6015_FH2_pes.png|thumb|400px|'''Figure 9a''': Potential energy surface of the reaction. The A-B bond distance refers to '''rHH''', and the BC bond distance refers to '''rFH'''. The top left triangle and the bottom right triangle of the PES is reagent space and product space for Reaction 1 respectively, and vice versa for Reaction 2.]]
|[[File:Yx6015_FH2_energy_profile.png|thumb|400px|'''Figure 9b''': Potential energy of the reaction along '''rFH''', which is representative of the reaction coordinate. The potential energy is the minima for each particular bond distance. HF is shown on the left, and HH is shown on the right.]]
|}

From '''Fig 9a.''' and '''Fig 9b.''', it can be seen that HF is much more stable than that of H2. Taking the difference between the minima at a BC bond distance of 0.92, and the minima of the PES at the extreme right of '''Fig 9b''', it can be calculated that the energy difference between the two states is (-103.9 - -133.9) = 30.0 Kcal/mol = 126 KJ/mol. Thus, ΔE(Reaction 1) = -126 KJ/mol, and ΔE(Reaction 2) = 126 KJ/mol.

The enthalpy of reaction can also be determined by looking at the bond energy (BE) of the bonds broken and bonds formed. In particular:

<math display="block">
\Delta E = \sum \text{BE}_\text{Bonds Broken} - \sum \text{BE}_\text{Bonds Formed}
</math>

For reaction 1 (F + H2 ⟶ HF + H), the bond broken is an H-H bond and the bond formed is an H-F bond. The exothermic energetics of the reaction implies that the H-F bond is stronger than the H-H bond. A stronger bond has higher bond energy. When the bond formed is the stronger bond, it results in a negative ΔE according to the formula above, consistent with our observation. The numerical value can also be verified using bond energies. With an H-H bond energy of 436.0 KJ/mol and an H-F bond energy of 568.5 KJ/mol, <ref>Haynes, W., Bruno, T. and Lide, D. (2015). ''CRC Handbook of Chemistry and Physics''. 97th ed. [Boca Raton, Florida]: CRC Press.</ref> ΔE(Reaction 1) = 436.0 - 568.5 = -132.5 KJ/mol, close to the value obtained from the simulation.

In addition to the two reactions above, another reaction of the F-H-H may occur, which is the collision of an H atom into the F atom of a HF molecule, resulting in an exchange of H atom:

{|style="margin: auto;"
|[[File:Yx6015_HFH_pes.png|thumb|400px|'''Figure 10a''': Potential energy surface of the reaction. A-B and B-C refers to the distance between F and either of the two H atoms.]]
|[[File:Yx6015_HFH_energy_profile.png|thumb|400px|'''Figure 10b''': A different perspective illustrating the transition state along the edge, as well as the most stable structure of the H-F-H system being a HFH molecule.]]
|}

As expected, the PES is symmetric about the y=x diagonal line, implying that H + FH ⟶ HF + H is energetically neutral. In an actual reaction, energy may be transferred in the form of heat or vibration. However, unlike the previous scenario, the most stable molecule is HFH, a linear triatomic molecule. As no bond is broken in this case, the ΔE is negative and the reaction is exothermic.

''Question: Locate the approximate position of the transition state.''
In addition, the transition state can can identified by from '''Fig 9b.''' directly by locating the saddle point. By looking at the correct region of the PES at the correct scale, the approximate location of the transition state can be determined. Further refinement of the approximate location was then carried out in a similar manner as in Exercise 1.

{|style="margin: auto;"
|[[File:Yx6015_FH2_ts_energy.png|thumb|400px|'''Figure 11a''': Approximate transition state from visual inspection of the potential energy surface.]]
|[[File:Yx6015_FH2_ts_distance.png|thumb|400px|'''Figure 11b''': Internuclear distance of the refined transition state over time. (<var>'''rAB'''</var> = 0.74; <var>'''rBC'''</var> = 1.81355; <var>'''pA'''</var> = <var>'''pB'''</var> = 0)]]
|[[File:Yx6015_FH2_minima_search.png|thumb|400px|'''Figure 12''': Illustration of minima searching refinement. (<var>'''rAB'''</var> = 2.5; <var>'''rBC'''</var> = 0.92; <var>'''pA'''</var> = <var>'''pB'''</var> = 0)]]
|}

Due to the highly endothermic nature of the depicted reaction (Reaction 2), the energy of the transition state was very close to the energy of the product. In addition, the coordinate of the transition state is also located close to the product, as informed by Hammond's postulate. This can be seen easily on '''Fig 9b.''', where the transition state was essentially indistinguishable from the product.

From '''Fig 11a.''', the approximate location for the transition was identified as '''<var>rts</var>''' = 1.8. By calculating the internuclear distance of '''<var>rts</var>''' around the approximate location, the refined internuclear distance was found to be '''<var>rts</var>''' = 1.81. Unlike that of r a H3 system, '''<var>rA</var>''' ≠ '''<var>rB</var>''' due to the asymmetry of the system.

''Question: Report the activation energy for both reactions.''

From the PES, the energy for the various states can be found. The energy of the transition state can be found as described previously, while the energy of the minima can be extracted in a similar manner: the approximate minima was found by visual inspection of the PES. The approximate location and energy was then refined by conducting a dynamics simulation at the approximate location, and looking at the change in potential energy over time ('''Fig 12.''').

{| class="wikitable" style="margin: auto;"
!
!'''<var>rHH</var>''' /
!'''<var>rFH</var>''' /
!Energy /Kcal/mol
|-
|HF
|6
|0.92
|<nowiki>-134.03</nowiki>
|-
|[H-H-F]&Dagger;
|0.74
|1.81
|<nowiki>-103.75</nowiki>
|-
|H2
|0.74
|6
|<nowiki>-104.02</nowiki>
|}

The activation energy of a reaction can be found by taking the difference in energy between the transition state and the energy of the reagent:

<math display="block>
E_\text{ea} = E_\text{ts} - E_\text{reagent}
</math>

Thus, the activation energy of reaction 1 (F + H2 ⟶ HF + H) is <var>E</var>ea = (-103.75) - (-104.02) = 0.27 Kcal/mol = 1.1 KJ/mol.

Similarly, for reaction 2 (H + HF ⟶ H2 + F), <var>E</var>ea = (-103.75) - (-134.03) = 30.28 Kcal/mol = 136.7 KJ/mol

{{fontcolor1|gray| Awesome discussion. Clear and to the point, I like it! [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:13, 31 May 2017 (BST)}}

=== Reaction Dynamics ===
{|style="margin: auto;"
|[[File:Yx6015_FH2_reactive_trajectory.png|thumb|400px|'''Figure 13a''': The PES for a reactive trajectory for F + H2 reaction. (<var>'''rAB'''</var> = 0.74; <var>'''rBC'''</var> = 2.0; <var>'''pA'''</var> = -1; <var>'''pB'''</var> = -1.5)]]
|[[File:Yx6015_FH2_reactive_ke.png|thumb|400px|'''Figure 13b''': Kinetic energy of time for the reactive trajectory. (<var>'''rAB'''</var> = 0.74; <var>'''rBC'''</var> = 2.0; <var>'''pA'''</var> = -1; <var>'''pB'''</var> = -1.5)]]
|[[File:Yx6015_final_reactive.png|thumb|400px|'''Figure 14''': A reactive trajectory for reaction 2. (<var>'''rAB'''</var> = 2; <var>'''rBC'''</var> = 0.94; <var>'''pA'''</var> = -2; <var>'''pB'''</var> = 6)]]
|}

''Question: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?''

As energy of conserved, for an exothermic reaction, the reaction energy released was converted from chemical potential energy to kinetic energy. This was seen from '''Fig 13b.''', as the peak kinetic energy was equals to the change of potential energy of the reaction, around 30Kcal/mol. The kinetic energy could be either in the form of translational motion, or vibrational motion, or a combination of both. Translational kinetic energy can be measured through measuring the pressure change of the reaction if carried out in gaseous phase, or by measuring the change in temperature through calorimetry.

Alternatively, the vibrational energy can be probed through spectroscopy. For the HF product, as it is IR-active, IR spectroscopy may be used to characterize its vibrational motion, with ṽo(1→0) = 3962 cm-1.<ref>G. Kuipers, D. Smith and A. Nielsen, ''The Journal of Chemical Physics'', 1956, 25, 275-279.</ref> As the H2 product is non-polar and is IR inactive, Raman spectroscopy may be used, with ṽo(1→0) = 4156 cm-1.<ref>D. Shelton, ''The Journal of Chemical Physics'', 1990, 93, 1491-1495.</ref>

{{fontcolor1|gray| Nice! [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:13, 31 May 2017 (BST)}}

''Question: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.''

Polanyi's Rule relates the efficiency of the reaction to the position of the transition state and the mode of excitation of the exchange reaction of simple molecules.<ref>J. Polanyi and W. Wong, ''The Journal of Chemical Physics'', 1969, 51, 1439-1450.</ref> Specifically, for a reaction with an early transition state, translational kinetic energy is '''more''' efficient than vibrational kinetic energy in promoting the reaction; for a reaction with a late transition state, translational kinetic energy is '''less''' efficient than vibrational kinetic energy in promoting the reaction.<ref>J. Liu, K. Song, W. Hase and S. Anderson, ''Journal of the American Chemical Society'', 2004, 126, 8602-8603.</ref> This can be verified with pairs of simulations with constant total kinetic energy, but with different distributions of kinetic energy.

This can be inferred directly from the simulation. For reaction 1 with early TS ('''Fig 13a.'''), a reactive trajectory generally have vibration momentum to be less than the collision momentum, with recrossing back to reagent occurring with high vibration momentum. However, for reaction 2 with a late TS, the collision momentum needed to be higher than the vibration momentum ('''Fig 14''') in order for the reaction to proceed. This is in line with Polanyi's rule.

{{fontcolor1|gray| Again, really good! [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:13, 31 May 2017 (BST)}}

= References =

MRD:Yiming Xu

2017-05-31T16:10:54Z

Je714: /* Exercise 1: H + H2 system */

= Exercise 1: H + H2 system =
[[File:yx6015_H3_legends.svg|thumb|400px|Definition of the various symbols used.]]
=== Dynamics from the Transition State Region ===
''Question: What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.''

{|style="margin: auto;"
|[[File:Yx6015_overall_minima.png|thumb|400px|'''Figure 1a''': Illustration of the minima "valley"-like area of the potential energy surface (<var>'''rAB'''</var> = 2.30; <var>'''rBC'''</var> = 0.74; <var>'''pA'''</var> = <var>'''pB'''</var> = -2.70)]]
|[[File:Yx6015_overall_transition.png|thumb|400px|'''Figure 1b''': A different perspective illustrating the transition state along the minimum energy path of the potential energy surface (<var>'''rAB'''</var> = 2.30; <var>'''rBC'''</var> = 0.74; <var>'''pA'''</var> = <var>'''pB'''</var> = -2.70)]]
|}

Minima and transition structures can be identified visually from the potential energy surface. As can be seen in '''Fig 1a''', minima appear as a "valley" of low potential energy between two "hills" of higher potential energy. This means that minima show an local upward curvature along parts of the potential energy surface with the lowest energy. The transition state is positioned along the minimum energy path (mep). However, as can be seen from '''Fig 1b''', it appears as a "bump" along the mep "valley". This results in it being a saddle point - with it being the local minimum along one axis and the local maximum along another axis (mep).

Mathematically, both minima and transition structures are at stationary points of the potential energy surface. At stationary points, the following relationship must hold:

<math display="block">
\nabla U = 0
</math>

where <var>'''∇'''</var> is the gradient and <var>'''U'''</var> is the potential energy. In order to determine whether the point found was a local minimum, maximum or a saddle (transition state), the local curvature need to be calculated using the second partial derivative test. For the potential energy surface of two variables <var>'''U(rA, rB)'''</var>, the Hessian matrix can be defined as:

<math display="block">
H(R_A, R_B)=
\begin{pmatrix}
U_{{r_A}{r_A}}(r_A, r_B) & U_{{r_A}{r_B}}(r_A, r_B) \\
U_{{r_B}{r_A}}(r_A, r_B) & U_{{r_B}{r_B}}(r_A, r_B)
\end{pmatrix}
</math>

The determinant of <var>'''H(rA, rB)'''</var>, <var>'''D(rA, rB)'''</var>, determines the nature of the stationary point <var>(a, b)</var>:
* if <var>'''D(a, b)'''</var> > 0 and <var>'''UrArA(a, b)'''</var> > 0, then the point <var>(a, b)</var> is a local minimum;
* if <var>'''D(a, b)'''</var> > 0 and <var>'''UrArA(a, b)'''</var> < 0, then the point <var>(a, b)</var> is a local maximum;
* if <var>'''D(a, b)'''</var> < 0 then the point <var>(a, b)</var> is a saddle point; and
* if <var>'''D(a, b)'''</var> = 0 then another test need to be performed.

From the second partial derivative test, the minima and transition structures can be identified.

{{fontcolor1|gray| Excellent! Good job on using math, and defining your symbols, showing understanding of it! [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:10, 31 May 2017 (BST)}}

=== Trajectories from rA = rB: Locating the Transition State ===
''Question: Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.''

Due to the symmetry of the system, with all three atoms being identical, we expect that <var>'''rts'''</var> will be located at a point such that <var>'''rA'''</var> and <var>'''rB'''</var> are identical. In other words, HB should be located between HA and HC. From the previous section, as the transition state is a stationary point, if the reaction coordinate was located on the transition state ''exactly'' with no momentum, it would be stationary and would not proceed any further. In other words, we can find a suitable <var>'''rts'''</var> at a particular <var>'''rA'''</var> = <var>'''rB'''</var> and <var>'''pA'''</var> = <var>'''pB'''</var> = 0

{|style="margin: auto;"
|[[File:Yx6015_ransition_state_distance_over_time.png|thumb|400px|'''Figure 2a''': Plot of internuclear distance over time, for <var>'''rA'''</var> = <var>'''rB'''</var> = 0.9085406012; <var>'''pA'''</var> = <var>'''pB'''</var> = 0]]
|[[File:Yx6015_ransition_state_distance_over_time_surface_plot.png|thumb|400px|'''Figure 2b''': Plot of the trajectory along the potential energy surface, for <var>'''rA'''</var> = <var>'''rB'''</var> = 0.9085406012; <var>'''pA'''</var> = <var>'''pB'''</var> = 0]]
|}
'''Fig 2a.''' shows a graph of the inter nuclear distances against time. As expected from being located on the stationary point, the distances stay approximately constant over time, with <var>'''rA'''</var> = <var>'''rB'''</var> (<var>'''rA'''</var> is not shown as it was completely overlapped by <var>'''rB'''</var>). The small amount of oscillation was due to symmetric stretching of the molecule, and is along a direction perpendicular to the mep. The nature of the stationary point can be more aptly visualised by '''Fig 2b.''' as the trajectory remain essentially unchanged after 1000 time step.

While <var>'''rts'''</var> used to generate the graphical plot was 0.9085406012, the precision obtained was much higher than what was meaningful due to the limited accuracy of the simulation. Comparing to a literature value of 0.93 , the best obtainable estimate of <var>'''rts'''</var> from this simulation would be 0.9 , an error of 3.3%.<ref>S. Grabowski and R. Hoffmann, ''ChemPhysChem'', 2012, 13, 2286-2288.
</ref>

=== Trajectories from rA = rts+δ, rB = rts ===
Once the transition state has been located, the reaction can be followed from transition state to the products or towards the reagents (due to time symmetry) in order to provide more information on the nature of reaction if only a minimum amount of energy was available.
{|style="margin: auto;"
|[[File:Yx6015_transition_state_mep.png|thumb|400px|'''Figure 3a''': Minimum energy path taken by the particle from the transition state (<var>'''rAB'''</var> = 0.91; <var>'''rBC'''</var> = 0.911; <var>'''pA'''</var> = <var>'''pB'''</var> = 0)]]
|[[File:Yx6015_transition_state_md.png|thumb|400px|'''Figure 3b''': Molecular dynamics trajectory taken by the particle from the transition state (<var>'''rAB'''</var> = 0.91; <var>'''rBC'''</var> = 0.911; <var>'''pA'''</var> = <var>'''pB'''</var> = 0)]]
|}
''Question: Comment on how the mep and the trajectory you just calculated differ.''

Both the mep and the molecular dynamics trajectory start at the exact same location at the same energy. However, they follow a different path "down" the transition state. The molecular dynamics trajectory is the physical trajectory taken by the molecule during the reaction. Compared to the mep, it shows oscillatory motion due to bond vibration. In addition, as the loss of potential energy was converted to kinetic energy and vibrational energy according to the equipartition theorem, with a higher translational kinetic energy, the molecule had moved further along the reaction (away from the transition state) during the same time (10000 steps).

On the other hand, the mep is not really a trajectory taken by the particle. It is an unphysical trajectory that characterises the path of steepest descent for the given potential energy surface. In this simulation, as it was calculated by setting the velocity to be 0 at every time step, it travelled to a lesser extent than that of the molecular dynamics trajectory.

In an actual reaction, the reaction trajectory will follow one provided by molecular dynamics, and may or may not pass through the transition state.

{{fontcolor1|gray| Very good explanation! [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:10, 31 May 2017 (BST)}}

=== Reactive and Unreactive Trajectories ===
Whether a trajectory results in an reaction can be identified by the bond between the atoms before and before interaction. As bonds are not identified explicitly, they can be inferred from the internuclear distances - a bond would have a relatively constant distance over time, with oscillatory motions due to vibration. With initial positions of <var>'''rA'''</var> = 0.74 and <var>'''rB'''</var>= 2.0, which means that HA is bonded with HB, trajectories with the following momenta were investigated.

{|style="margin: auto;" class="wikitable"
!pA
!pB
!colspan="2" | Screenshots
|-
|rowspan="2" | <nowiki>-1.25</nowiki>
|rowspan="2" | <nowiki>-2.5</nowiki>
|[[File:Yx6015_reactive_or_unreactive_4a.png|frameless|400px]]
|[[File:Yx6015_reactive_or_unreactive_4b.png|frameless|400px]]
|-
|style="width:800px;" colspan="2" | '''Figure 4''': This configuration illustrated an reactive trajectory. A straight forward exchange reaction between HC and HB-HA was shown. There was sufficient kinetic energy from HC to overcome the coulombic barrier to reach the transition state. The reaction cross the transition state at about t = 0.5, and resulted in the formation of a HC-HB molecule, and HA was ejected. The ejected HA had less kinetic energy than the incoming HC, and some of the energy was transferred to the vibrational energy of the newly formed HC-HB molecule. (<var>'''rAB'''</var> = 0.74; <var>'''rBC'''</var> = 2.0; <var>'''pA'''</var> = -1.25; <var>'''pB'''</var> = -2.5)
|-
|rowspan="2" | <nowiki>-1.5</nowiki>
|rowspan="2" | <nowiki>-2.0</nowiki>
|[[File:Yx6015_reactive_or_unreactive_5a.png|frameless|400px]]
|[[File:Yx6015_reactive_or_unreactive_5b.png|frameless|400px]]
|-
|style="width:800px;" colspan="2" | '''Figure 5''': This configuration illustrated an unreactive trajectory. There was insufficient energy to reach the transition state as HC did not have sufficient momentum to overcome the coulombic barrier. HC was repelled back to the approaching direction. (<var>'''rAB'''</var> = 0.74; <var>'''rBC'''</var> = 2.0; <var>'''pA'''</var> = -1.5; <var>'''pB'''</var> = -2.0)
|-
|rowspan="2" | <nowiki>-1.5</nowiki>
|rowspan="2" | <nowiki>-2.5</nowiki>
|[[File:Yx6015_reactive_or_unreactive_6a.png|frameless|400px]]
|[[File:Yx6015_reactive_or_unreactive_6b.png|frameless|400px]]
|-
|style="width:800px;" colspan="2" | '''Figure 6''': This configuration illustrated an reactive trajectory. The reaction was similar to that depicted by '''Fig 4.'''. However, with a higher initial <var>'''rA'''</var> momentum, there was more vibration in HB-HA before reaction occurred. The resulting trajectory was similar, however. (<var>'''rAB'''</var> = 0.74; <var>'''rBC'''</var> = 2.0; <var>'''pA'''</var> = -1.5; <var>'''pB'''</var> = -2.5)
|-
|rowspan="2" | <nowiki>-2.5</nowiki>
|rowspan="2" | <nowiki>-5.0</nowiki>
|[[File:Yx6015_reactive_or_unreactive_7a.png|frameless|400px]]
|[[File:Yx6015_reactive_or_unreactive_7b.png|frameless|400px]]
|-
|style="width:800px;" colspan="2" | '''Figure 7''': This configuration illustrated an unreactive trajectory. Unlike the unreactive trajectory depicted in '''Fig 5.''', there is more than sufficient energy to cross over the transition state. However, due to the high initial momentum of HC, it got too close to HB, resulting in high electrostatic forces of repulsion, repelling HC back to its original direction. This is an example of barrier recrossing, as the product did form for a short duration. Part of the kinetic energy of HC was transferred to the HB-HA molecule, resulting in higher vibrational energy as a product of the reaction. Observing the PES, the trajectory deflects off the energy barrier multiple times throughout the reaction, signifying a high energy system in contrast to previously investigated configurations. (<var>'''rAB'''</var> = 0.74; <var>'''rBC'''</var> = 2.0; <var>'''pA'''</var> = -2.5; <var>'''pB'''</var> = -5.0)
|-
|rowspan="2" | <nowiki>-2.5</nowiki>
|rowspan="2" | <nowiki>-5.2</nowiki>
|[[File:Yx6015_reactive_or_unreactive_8a.png|frameless|400px]]
|[[File:Yx6015_reactive_or_unreactive_8b.png|frameless|400px]]
|-
|style="width:800px;" colspan="2" | '''Figure 8''': This configuration illustrated a reactive trajectory. Unlike the trajectory depicted in the PES of '''Fig 7.''', it was deflected in a different direction, resulting in a successful exchange reaction. Barrier recrossing occurred twice in this reaction - the product HC-HB was formed, broken, and then formed again. One possible difference could be due to the slightly higher kinetic energy of HC, which allowed it to come closer to HA than before, and experience a higher attraction force towards the HB-HA molecule. However, an initial HC kinetic energy that is too high would result in a much higher repulsion than can be compensated by the slightly higher attraction and would result in an unreactive trajectory as shown in '''Fig 7.''' (<var>'''rAB'''</var> = 0.74; <var>'''rBC'''</var> = 2.0; <var>'''pA'''</var> = -2.5; <var>'''pB'''</var> = -5.2)
|-
|}

=== Transition State Theory ===
''Question: State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''

The Transition State Theory (TST) is an alternative to the Collision Theory, which explains the rate of chemical reactions and provide theoretical basis for the Arrhenius equation, <math> k = A \exp(-\frac{E_a}{RT})</math>. It assumes a special type of equilibrium exists between the reactants and the activated complex before the product is formed, and provides the following rate equation (the Eyring equation):<ref><nowiki>IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). XML on-line corrected version: http://goldbook.iupac.org (2006-) created by M. Nic, J. Jirat, B. Kosata; updates compiled by A. Jenkins. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook</nowiki></ref>

<math display="block">
k = \frac{k_B T}{h} \exp(\frac{\Delta ^\ddagger S ^\circ}{R}) \exp(-\frac{\Delta ^\ddagger H ^\circ}{RT})
</math>

The Eyring equation was developed with two assumptions:<ref>D. Truhlar, B. Garrett and S. Klippenstein, ''The Journal of Physical Chemistry'', 1996, 100, 12771-12800.</ref>
<ol start="1">
<li>The reactants and product space in the PES can be clearly differentiated/divided; and</li>
<li>Trajectories going from reactant space to product space will not go back.</li>
</ol>

In addition, it implicitly assumes that:

<ol start="3">
<li>The reactants are equilibrated in an ensemble; and</li> {{fontcolor1|gray| Does this make sense here? We're just looking at a triatomic collision in isolation -- no ensemble of particles considered... [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:10, 31 May 2017 (BST)}}
<li>The overall reaction is electronically adiabatic.</li>
</ol>From the results of the previous part, it can be seen that assumption 2 is not valid. Even if the transition state had been achieved, the trajectory may still undergoing recrossing to return to the reagent space. This recrossing may even happen if the product was formed momentarily. In addition, multiple recrossing may occur depending on the momenta and angle of the trajectory to give the desired product after an even number of crossings. This effect result in TST overestimating the reaction rate constant as compared to experimental values.

In addition, quantum mechanical effects may also affect the rate of reaction by allowing reactants to tunnel through the activation barrier without having sufficient energy. While this would indicate an underestimation of the rate by TST, the previous effect would be much more significant, leading TST predicting higher reaction rate values when compared with experimental data.

{{fontcolor1|gray| Good! [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:10, 31 May 2017 (BST)}}

= Exercise 2: F - H - H system =
[[File:yx6015_FH2_legends.svg|thumb|400px|Definition of the various symbols used.]]
[[File:yx6015_HFH_legends.svg|thumb|400px|Definition of the various symbols used.]]
=== PES inspection ===
''Question: Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?''

There are two different reaction under investigation. The first reaction (Reaction 1.) is : F + H2 ⟶ HF + H. The second reaction (Reaction 2.) considered is the reverse reaction of the first, namely H + HF ⟶ H2 + F. As they are the reverse reaction of each other, and due to the fact that enthalpy is a state function, if one of them (e.g. Reaction 1.) is exothermic, the reverse reaction (i.e. Reaction 2.) must be endothermic.

The reaction energetics can be determined simply by examining the potential energy surface of the reaction, and identifying the correct viewing perspective.

{|style="margin: auto;"
|[[File:Yx6015_FH2_pes.png|thumb|400px|'''Figure 9a''': Potential energy surface of the reaction. The A-B bond distance refers to '''rHH''', and the BC bond distance refers to '''rFH'''. The top left triangle and the bottom right triangle of the PES is reagent space and product space for Reaction 1 respectively, and vice versa for Reaction 2.]]
|[[File:Yx6015_FH2_energy_profile.png|thumb|400px|'''Figure 9b''': Potential energy of the reaction along '''rFH''', which is representative of the reaction coordinate. The potential energy is the minima for each particular bond distance. HF is shown on the left, and HH is shown on the right.]]
|}

From '''Fig 9a.''' and '''Fig 9b.''', it can be seen that HF is much more stable than that of H2. Taking the difference between the minima at a BC bond distance of 0.92, and the minima of the PES at the extreme right of '''Fig 9b''', it can be calculated that the energy difference between the two states is (-103.9 - -133.9) = 30.0 Kcal/mol = 126 KJ/mol. Thus, ΔE(Reaction 1) = -126 KJ/mol, and ΔE(Reaction 2) = 126 KJ/mol.

The enthalpy of reaction can also be determined by looking at the bond energy (BE) of the bonds broken and bonds formed. In particular:

<math display="block">
\Delta E = \sum \text{BE}_\text{Bonds Broken} - \sum \text{BE}_\text{Bonds Formed}
</math>

For reaction 1 (F + H2 ⟶ HF + H), the bond broken is an H-H bond and the bond formed is an H-F bond. The exothermic energetics of the reaction implies that the H-F bond is stronger than the H-H bond. A stronger bond has higher bond energy. When the bond formed is the stronger bond, it results in a negative ΔE according to the formula above, consistent with our observation. The numerical value can also be verified using bond energies. With an H-H bond energy of 436.0 KJ/mol and an H-F bond energy of 568.5 KJ/mol, <ref>Haynes, W., Bruno, T. and Lide, D. (2015). ''CRC Handbook of Chemistry and Physics''. 97th ed. [Boca Raton, Florida]: CRC Press.</ref> ΔE(Reaction 1) = 436.0 - 568.5 = -132.5 KJ/mol, close to the value obtained from the simulation.

In addition to the two reactions above, another reaction of the F-H-H may occur, which is the collision of an H atom into the F atom of a HF molecule, resulting in an exchange of H atom:

{|style="margin: auto;"
|[[File:Yx6015_HFH_pes.png|thumb|400px|'''Figure 10a''': Potential energy surface of the reaction. A-B and B-C refers to the distance between F and either of the two H atoms.]]
|[[File:Yx6015_HFH_energy_profile.png|thumb|400px|'''Figure 10b''': A different perspective illustrating the transition state along the edge, as well as the most stable structure of the H-F-H system being a HFH molecule.]]
|}

As expected, the PES is symmetric about the y=x diagonal line, implying that H + FH ⟶ HF + H is energetically neutral. In an actual reaction, energy may be transferred in the form of heat or vibration. However, unlike the previous scenario, the most stable molecule is HFH, a linear triatomic molecule. As no bond is broken in this case, the ΔE is negative and the reaction is exothermic.

''Question: Locate the approximate position of the transition state.''
In addition, the transition state can can identified by from '''Fig 9b.''' directly by locating the saddle point. By looking at the correct region of the PES at the correct scale, the approximate location of the transition state can be determined. Further refinement of the approximate location was then carried out in a similar manner as in Exercise 1.

{|style="margin: auto;"
|[[File:Yx6015_FH2_ts_energy.png|thumb|400px|'''Figure 11a''': Approximate transition state from visual inspection of the potential energy surface.]]
|[[File:Yx6015_FH2_ts_distance.png|thumb|400px|'''Figure 11b''': Internuclear distance of the refined transition state over time. (<var>'''rAB'''</var> = 0.74; <var>'''rBC'''</var> = 1.81355; <var>'''pA'''</var> = <var>'''pB'''</var> = 0)]]
|[[File:Yx6015_FH2_minima_search.png|thumb|400px|'''Figure 12''': Illustration of minima searching refinement. (<var>'''rAB'''</var> = 2.5; <var>'''rBC'''</var> = 0.92; <var>'''pA'''</var> = <var>'''pB'''</var> = 0)]]
|}

Due to the highly endothermic nature of the depicted reaction (Reaction 2), the energy of the transition state was very close to the energy of the product. In addition, the coordinate of the transition state is also located close to the product, as informed by Hammond's postulate. This can be seen easily on '''Fig 9b.''', where the transition state was essentially indistinguishable from the product.

From '''Fig 11a.''', the approximate location for the transition was identified as '''<var>rts</var>''' = 1.8. By calculating the internuclear distance of '''<var>rts</var>''' around the approximate location, the refined internuclear distance was found to be '''<var>rts</var>''' = 1.81. Unlike that of r a H3 system, '''<var>rA</var>''' ≠ '''<var>rB</var>''' due to the asymmetry of the system.

''Question: Report the activation energy for both reactions.''

From the PES, the energy for the various states can be found. The energy of the transition state can be found as described previously, while the energy of the minima can be extracted in a similar manner: the approximate minima was found by visual inspection of the PES. The approximate location and energy was then refined by conducting a dynamics simulation at the approximate location, and looking at the change in potential energy over time ('''Fig 12.''').

{| class="wikitable" style="margin: auto;"
!
!'''<var>rHH</var>''' /
!'''<var>rFH</var>''' /
!Energy /Kcal/mol
|-
|HF
|6
|0.92
|<nowiki>-134.03</nowiki>
|-
|[H-H-F]&Dagger;
|0.74
|1.81
|<nowiki>-103.75</nowiki>
|-
|H2
|0.74
|6
|<nowiki>-104.02</nowiki>
|}

The activation energy of a reaction can be found by taking the difference in energy between the transition state and the energy of the reagent:

<math display="block>
E_\text{ea} = E_\text{ts} - E_\text{reagent}
</math>

Thus, the activation energy of reaction 1 (F + H2 ⟶ HF + H) is <var>E</var>ea = (-103.75) - (-104.02) = 0.27 Kcal/mol = 1.1 KJ/mol.

Similarly, for reaction 2 (H + HF ⟶ H2 + F), <var>E</var>ea = (-103.75) - (-134.03) = 30.28 Kcal/mol = 136.7 KJ/mol

=== Reaction Dynamics ===
{|style="margin: auto;"
|[[File:Yx6015_FH2_reactive_trajectory.png|thumb|400px|'''Figure 13a''': The PES for a reactive trajectory for F + H2 reaction. (<var>'''rAB'''</var> = 0.74; <var>'''rBC'''</var> = 2.0; <var>'''pA'''</var> = -1; <var>'''pB'''</var> = -1.5)]]
|[[File:Yx6015_FH2_reactive_ke.png|thumb|400px|'''Figure 13b''': Kinetic energy of time for the reactive trajectory. (<var>'''rAB'''</var> = 0.74; <var>'''rBC'''</var> = 2.0; <var>'''pA'''</var> = -1; <var>'''pB'''</var> = -1.5)]]
|[[File:Yx6015_final_reactive.png|thumb|400px|'''Figure 14''': A reactive trajectory for reaction 2. (<var>'''rAB'''</var> = 2; <var>'''rBC'''</var> = 0.94; <var>'''pA'''</var> = -2; <var>'''pB'''</var> = 6)]]
|}

''Question: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?''

As energy of conserved, for an exothermic reaction, the reaction energy released was converted from chemical potential energy to kinetic energy. This was seen from '''Fig 13b.''', as the peak kinetic energy was equals to the change of potential energy of the reaction, around 30Kcal/mol. The kinetic energy could be either in the form of translational motion, or vibrational motion, or a combination of both. Translational kinetic energy can be measured through measuring the pressure change of the reaction if carried out in gaseous phase, or by measuring the change in temperature through calorimetry.

Alternatively, the vibrational energy can be probed through spectroscopy. For the HF product, as it is IR-active, IR spectroscopy may be used to characterize its vibrational motion, with ṽo(1→0) = 3962 cm-1.<ref>G. Kuipers, D. Smith and A. Nielsen, ''The Journal of Chemical Physics'', 1956, 25, 275-279.</ref> As the H2 product is non-polar and is IR inactive, Raman spectroscopy may be used, with ṽo(1→0) = 4156 cm-1.<ref>D. Shelton, ''The Journal of Chemical Physics'', 1990, 93, 1491-1495.</ref>

''Question: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.''

Polanyi's Rule relates the efficiency of the reaction to the position of the transition state and the mode of excitation of the exchange reaction of simple molecules.<ref>J. Polanyi and W. Wong, ''The Journal of Chemical Physics'', 1969, 51, 1439-1450.</ref> Specifically, for a reaction with an early transition state, translational kinetic energy is '''more''' efficient than vibrational kinetic energy in promoting the reaction; for a reaction with a late transition state, translational kinetic energy is '''less''' efficient than vibrational kinetic energy in promoting the reaction.<ref>J. Liu, K. Song, W. Hase and S. Anderson, ''Journal of the American Chemical Society'', 2004, 126, 8602-8603.</ref> This can be verified with pairs of simulations with constant total kinetic energy, but with different distributions of kinetic energy.

This can be inferred directly from the simulation. For reaction 1 with early TS ('''Fig 13a.'''), a reactive trajectory generally have vibration momentum to be less than the collision momentum, with recrossing back to reagent occurring with high vibration momentum. However, for reaction 2 with a late TS, the collision momentum needed to be higher than the vibration momentum ('''Fig 14''') in order for the reaction to proceed. This is in line with Polanyi's rule.

= References =

Talk:Yw14115

2017-05-31T16:01:26Z

Je714: Created page with "A Decent report, but missing some answers! It's not totally complete. See my inline comments for more information. ~~~~"

A Decent report, but missing some answers! It's not totally complete. See my inline comments for more information.

[[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:01, 31 May 2017 (BST)

Yw14115

2017-05-31T16:00:42Z

Je714:

== EXERCISE 1: H + H2 system ==
'''{{fontcolor1|blue|What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}'''

The total gradient of the potential energy surface of the minimum are zero i.e ∂V('''r''')/∂'''r'''=0 ( r can be either r1 or r2). {{fontcolor1|gray| Correct [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:00, 31 May 2017 (BST)}}

But on the other hand, at a transition structure, the second derivative shows a positive value for different coordinate ( for both r1 and r2). In the case of using reaction coordinate, the transition structure second derivative shows negative value because it is the minima of the potential well.

{{fontcolor1|gray| Confusing explanation. You first say that at the TS the 2nd derivative is >0 for both r1 and r2 (which is wrong). I don't understand what you mean with the last sentence. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:00, 31 May 2017 (BST)}}

'''{{fontcolor1|blue|Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.}}'''

The best estimation for the position of transition state r1 = r2 = 0.908 Da {{fontcolor1|gray| Da is Dalton units, g/mol? Do you measure the distance in such units? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:00, 31 May 2017 (BST)}}. This is done by using the ' Internuclear Distance vs Time<nowiki>''</nowiki> Graph. At the transition state, there is no oscillation for the trajectory which is shown on the graph. This is because at the transition state the first derivative of ∂u/∂r is zero which is equal to the force. As the net force is zero, there is no movement for the atoms at the transition state. {{fontcolor1|gray| Correct. Small correction, the force is actually = -dV/dr, don't forget the negative sign! [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:00, 31 May 2017 (BST)}}From the graph, it can be seen there is little oscillation of particles i.e the distance between A,B and B,C stays relatively unchanged.[[File:yw14115.png|500px|center|thumb|Figure 1. ' Internuclear Distance vs Time' Graph when r = 0.908 Da.]]'''{{fontcolor1|blue|Comment on how the mep and the trajectory you just calculated differ.}}'''

For the mep, the reaction trajectory is a smooth line whereas oscillation occurs for the dynamics one. This is because the velocity of the atoms are set to zero at each time step which means there is no kinetic energy only potential energy. There is no conversion between potential energy and kinetic energy. Therefore, oscillation is not observed for the molecules. {{fontcolor1|gray| Good! [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:00, 31 May 2017 (BST)}}
[[File:Yw14115ts + 0.01.PNG|500px|center|thumb|Figure 2. Dynamic plot.]]
[[File:Yw14115 mep.PNG|500px|center|thumb|Figure 3. Mep plot.]]'''{{fontcolor1|blue|Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.}}'''
{| class="wikitable"
!p1/Å
!p2/Å
!Reactivity
!Trajectory
!Comment
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5 </nowiki>
|Reactive
|[[File:Yw14115_-1.25_-1.5.png|400px|none|thumb]]
|For the first initial condition,the reaction is reactive which is suggested by the reaction trajectory. The incoming atom has enough kinetic energy to overcome the activation barrier allow it climb up to the saddle point of the potential surface (transition state) and trajectory enters the exist channel.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|Not reactive
|[[File:Yw14115-1.5-2.0.png|400px|none|thumb]]
|For the second initial condition, It is not reactive which can be seen on the graph. The trajectory was bounced off from the barrier and the reactants were regenerated and still oscillating. This is because the incoming atom don't have enough kinetic energy (momentum) to reach the transition state.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|Reactive
|[[File:Yw14115-1.5-2.5.png|400px|none|thumb]]
|The initial condition is similar to the first one, the momentum p1 is not changed significantly which is still is the window which allows reaction to occur.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|No reactive
|[[File:Yw14115-2.5-5.0.png|400px|none|thumb]]
|The forth one is unreactive. The trajectory crosses the transition state region and enter the product channel. This suggests the bond is actually formed between atom A and B but then revert back to the reactant channel. One thing can be noticed is the entering trajectory doesn't follow the lowest potential energy pathway of the transition state region. It climbs up the potential wall and then forms the product. The regenerated reactant oscillate in a greater extant and deviates from the lowest potential pathway. This is because the incoming atom has a lot of kinetic energy. After collision, the initial leaving atom C also has a lot of kinetic energy which can then transfer to potential energy which give rise to the trajectory deviation from the lowest potential energy path way. Due to the high energy of this leaving atom, it can still react with molecule A-B cause the regeneration of the reactants.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|Reactive
|[[File:Yw14115-2.5-5.2.png|400px|none|thumb]]
|This initial condition allows reaction to occur. The trajectory crosses the transition state twice and then move along the product channel.
|}

'''{{fontcolor1|blue|State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}'''

<nowiki> </nowiki>The transition state theory is used to predict the rate of chemical reaction. The transition State theory assume the exist of activated complexes and they are in quasi equilibrium with the reactants even the classical equilibrium is not reached {{fontcolor1|gray| I see this quasi-equilibrium drawback over and over. It's not really applicable here! That is a statistical treatment of TST. In your simulations, a triatomic collision in isolation is considered -- not an ensemble of particles [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:00, 31 May 2017 (BST)}}. The transition state theory predicts when the reactants reach the transition state and forms the activated complex, it can either revert back to the reactants or form the product. In another word, the trajectory of PES is only allowed to cross the transition state once. Transition state theory is different form collision theory in which every successful collision lead to a reaction whereas in this case even the starting reactants have enough energy, the products still might not be formed.

In the experiment, after forming the product, it can still revert back to reactants which is the case of barrier recrossing which is different from the transition state theory. This suggest the rate of reaction in the experiment is lower than the theory predict because many high energy collisions don't lead to successful collision. {{fontcolor1|gray| Good. Recrossing is the biggest point to be made -- it's what you observe in your simulations [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:00, 31 May 2017 (BST)}}

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

=== '''PES Inspection''' ===
{{fontcolor1|blue|Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state.}}

==== For F + H2 ====
It is an exothermic reaction which is suggested by the PES. The product channel has a lower potential energy surface compare to the reactant channel. Fluorine atom has a very high electronegativity.Therefore, H-F can form much stronger bond compare to H-H thus more energy is released for the reaction. The transition position is r1(BC)= 0.745 Da, r2(AB)= 1.811 Da. Compare to H-H-H System in which the transition state occurs between the reactant channel and product channel, the transition state for F+ H-H system occurs rather early in the reactant channel which is called early barrier. This is because the transition state region resemble the reactant structure: The F atom is still far away from the H-H and H-H bond is hardly stretched. High electronegativity of F atom makes it want to abstract the H atom.
[[File:Yw14115 exo.PNG|500px|center|thumb|Figure 8. F+ H-H plot]]
[[File:Yw14115 exo cor.PNG|500px|center|thumb|Figure 9. F+ H-H system plot with transition state position]]

Activation energy for this reaction = 0.2 kcal/mol. it is calculated by using the energy of the transition state and the energy of the initial state.

==== For H + H-F ====
This reaction is the reversible reaction of the previous one which is endothermic. The product channel has a higher potential energy surface compare to the reactant channel. The is because H-F bond is very strong. The energy generated through H-H bond formation can not compensate the energy required to break the H-F bond thus the reaction is endothermic. The transition state position is r1(BC)= 1.811 Da r2(AB)= 0.745 Da. The transition state region resemble the structure of the product, In this case, the H atoms need to be close to each other and F atom is very far away from the H atom. The late barrier is observed for this endothermic reaction. Activation energy = 30.2 kcal/mol.

[[File:Yw14115endo.PNG|500px|center|thumb|Figure 10. H+ H-F plot.]]

[[File:Yw14115 endo cor.PNG|500px|center|thumb|Figure 10. H+ H-F plot with transition state position.]]

=== '''Reaction dynamic''' ===

'''{{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}'''

The following condition has been identified rHH = 0.75 A, rFH = 2 A and pPH = -0.5. When pHH has value between -1 and 0.5. The reactions are successful. When pHH is between -3 and -1 or 0.5 and 3, the reactions are not successful.

The input of momentum determines the total energy of the system, the total energy of the system is conserved throughout the reaction. From the graph below it can be seen the trajectory in the reactant channel has very little oscillation i.e the kinetics energy is low, the potential energy is high. After the reaction has occurred, the oscillation of the trajectory increase in a huge extant. This is because a lot of potential energy has transferred into kinetic energy because the total energy of the system must be conserved. This experimental observation is evident by the following graph. {{fontcolor1|gray| Missing discussion about how you would set up an experiment to confirm this... [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:00, 31 May 2017 (BST)}}

[[File:Yw14115_22.PNG|500px|center]]

'''{{fontcolor1|blue|Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state}}'''

Polanyi's empirical rules state translational energy is most efficient for a exothermic reaction which has a early barrier than vibrational energy. In this case, energy is needed in the reactant channel to cross the barrier. when the relative translational energy is higher than vibrational energy, then there is enough kinetic energy available in the reactant channel which means it can cross the barrier in the reactant channel and the reaction is possible to occur. After crossing the transition state, the trajectory can climb up the potential well and initiate vibrations of product molecule. Therefore, a vibrating trajectory is usually observed in the product channel. If the barrier is in the product channel, the trajectory would just simple bounce off from the potential well, the reaction wouldn't occur. On the other hand, vibrational energy is most efficient for an endothermic reaction which has late barrier. When the relative vibrational energy is higher than the translational energy, the kinetic energy is available along the product channel to go over the late barrier.

In both cases, the total energy should be greater than the energy of the barrier to make the reaction favourable.

{{fontcolor1|gray| You need to show examples of your trajectories for each case! Not complete. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 17:00, 31 May 2017 (BST)}}

Talk:MRD:aoz15page

2017-05-31T15:50:53Z

Je714: Created page with "Good job. See my inline comments for some small corrections. ~~~~"

Good job. See my inline comments for some small corrections.

[[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:50, 31 May 2017 (BST)

MRD:aoz15page

2017-05-31T15:50:33Z

Je714: /* Reaction Dynamics */

==Part 1ː H + H2 system==

=== Dynamics from the Transition State Region ===

{{fontcolor|blue|What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

The total gradient of a potential energy surface has a value of zero both at minimums and at transition structures. Note, for a function of two variables, having a total gradient of zero refers to having a partial derivative of zero with respect to both of its variable separately. i.eː a function f(x,y) will have a stationary point at a location where fx = 0 and fy = 0. This is due to both being stationary points for a function of two variables. However, they are different kinds of stationary points, with the minimum being a minimum and the transition structure a saddle point. Thus if a trajectory is started with no momentum at a position near the minimum point, it will end at the minimum point. However, if a trajectory with no initial momentum is started near the transition structure position, it will end up at either the reactants or products position. Hence this test can be used to determine whether a given point is a minimum or a saddle point for a function of two variables.

{{fontcolor1|gray| I was really enjoying your explanation until the very last part! The point about starting trajectories near the TS is a bit ''ad hoc''. It's a good analogy (and not factually wrong), but not really a rigorous explanation. Since you were correctly discussing total gradients and first partial derivatives, why not extend that to second partial derivatives? Those have information about the curvature of a surface and are actually what's needed to give a mathematically rigorous definition of the TS. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:33, 31 May 2017 (BST)}}

=== Locating the Transition State ===

{{fontcolor|blue|Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.}}

If the collision is taken to be between an atom A and a diatom BC, the transition state occurs when the bond between AB is midway through forming and that between BC is midway through breaking. At this point the system is at maximum potential energy. For a collision H with H2, the potential surface is symmetric, and so the transition state occurs when rAB = rBC. The transition state can thus be located by finding initial distances rAB and rBC such that, when the particles have no initial momentum, their internuclear distances remain constant. Through altering initial separations, an estimate for rrts of 0.908 Angstrom was made. The screenshot for the Internuclear Distances vs Time for this value is shown below (figure 1). It shows that the distances are constant over time. {{fontcolor1|gray| Good. Why are they constant, though? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:33, 31 May 2017 (BST)}}

[[File:Aoz15 transition state position estimate int dist v time.PNG|thumb|centre|600px|Figure 1ː Internuclear distance versus time plot for trajectory used to locate position of transition state.]]

=== Calculating Reaction Path ===

{{fontcolor1|blue|Comment on how the ''mep'' and the trajectory you just calculated differ.}}

The minimum energy path (mep) was found by starting with both initial momenta set to zero, rAB = rrts + 0.01 and rBC = rrts. This results in a trajectory that follows the valley floor of the surface contour to the products. When these initial conditions were run with the MEP calculation type, the following surface plot was obtainedː

[[File:Aoz15 mep surface plot.PNG|thumb|centre|500px|Figure 2ː Surface plot obtained for mep calculation.]]

As expected, this simply shows a smooth trajectory exactly down the valley floor to the products. However, when these same conditions were run with the dynamics calculation type, the trajectory differed and the following plot was producedː

[[File:Aoz15 meptrajectory surface plot.PNG|thumb|centre|500px|Figure 3ː Surface plot for the same initial conditions as for the mep, but performed with the dynamics calculation setting.]]

The mep and trajectory obtained differ in that only the trajectory takes into account vibrational motion; in the mep obtained, the energy stays exactly as the minimum energy of the valley and hence is a smooth line to the products. However, the trajectory takes into account vibrational motion, hence the energy oscillates about the minimum energy whilst reaching the products. This occurs due to the vibrational energy in the breaking/forming bonds, and explains the shape of the line seen in figure 3 above.

If the initial conditions rAB = rrts and rBC = rrts + 0.01 are used, the trajectory still follows the valley floor. However, it ends up at the reactants instead of the products, since it effectively 'rolls' the other way to find an energy minimum.

{{fontcolor1|gray| Fine. What's the Kinetic Energy on a MEP? What does that tell you about the velocities? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:38, 31 May 2017 (BST)}}

=== Reactive and Unreactive Trajectories ===

{{fontcolor1|blue|Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.}}

For initial bond lengths of rAB = 0.74 and rBC = 2.0, the initial momenta were varied and surface plots were found in order to determine which conditions led to reactive trajectories. The results and some discussion are found in the following table.

{| class="wikitable" border="1"
|+
! Surface Plot !! p1!! p2!!Reactive or Unreactive and Brief Discussion
|-
| [[File:Aoz15 reactive or not 1.PNG|centre|400px]] || -1.25 ||-2.5||Reactive, which can be seen by the trajectory passing through the transition state and going to the products.
|-
| [[File:Aoz15 reactive or not 2.PNG|centre|400px]] || -1.5||-2.0||Unreactive, since the path does not successfully pass through the transition state, instead ending at the reactants.
|-
| [[File:Aoz15 reactive or not 3.PNG|centre|400px]] || -1.5||-2.5||Reactive, which can once again be seen by the trajectory passing through the transition state and going to the products. Only difference to the first reaction is a slight increase in overall vibrational energy, which can be seen by greater oscillations in the trajectory.
|-
| [[File:Aoz15 reactive or not 4.PNG|centre|400px]] || -2.5||-5.0||Unreactive, which can be seen by the trajectory ending at the reactants. However, in this case it does pass through the transition state. Therefore, this is a case of barrier recrossing.
|-
| [[File:Aoz15 reactive or not 5.PNG|centre|400px]] || -2.5||-5.2||Reactive, as the reaction proceeds to the products. Here barrier recrossing also occurs, but unlike in the fourth reaction, in this case the reaction still proceeds to products. Hence, multiple barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|blue|State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

There are three main assumptions in Transition State Theory (TST). Before stating them, it is useful to define the terms critical diving surface and activated complex. For a given reaction potential energy surface, the critical diving surface refers to a boundary surface which passes through the surface saddle point. As such, the critical diving surface must be crossed in order for a reaction to go to completion. Any molecule whose nuclear configuration matches a point on the critical diving surface or to any point within a short distance of it is termed an activated complex. So, the saddle point of the potential surface corresponds to the equilibrium structure of the activated complex. With these terms defined, the following are the main assumptions of Transition State Theoryː firstly, that all configurations of molecules which cross the critical diving surface from the reactants side will proceed to become products. Secondly, that all reactant molecules maintain a Boltzmann-distributed energy distribution during the reaction. This leads to the third and final assumption; that the energy distribution of all activated complexes during a reaction will also be a Boltzmann distribution which depends on the system temperature.<ref name=":0">Levine, I. N., ''Physical Chemistry'', McGraw-Hill Publishing, Sixth Edition, 2009, 887-894</ref>
{{fontcolor1|gray| OK. Now, do the last two points make sense here? We're just simulating an isolated triatomic collision -- not an ensemble of particles. So this statistichal treatment is not applicable. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:47, 31 May 2017 (BST)}}

When tested against experimentally obtained values, those found by TST are in most cases within a single order of magnitude, suggesting that the assumptions made are well placed<ref name=":0" />. However, there are certain aspects of reactions which TST doesn't account for. The first is barrier recrossing; since TST assumes that all molecular configurations which pass the critical dividing surface proceed to products, it cannot consider barrier recrossing, the process in which the reaction trajectory crosses the potential barrier but then recrosses it in the opposite direction. As in the fourth simulation above, sometimes barrier recrossing leads to the molecules returning to the reactants and the products not forming at all. In such a case, TST wildly misinterprets the reaction result. Even in cases where barrier recrossing occurs multiple times and the products are formed (as in the fifth simulation above), TST gives a higher reaction rate value than is experimentally obtained, since it assumes the reaction proceeds to completion as soon as the potential barrier is crossed. TST also doesn't take into account quantum effects, such as tunneling. Tunneling occurs when reactant molecules with insufficient energy to bypass the potential barrier pass through the barrier, leading to an exponential decay in their energy. This process occurs thanks to wave-particle duality, and cannot be predicted by TST<ref>P. Atkins, J. de Paula, ''Atkins Physical Chemistry'', Oxford University Press, 10th Edition, Chapter 21</ref>. However, a quantum-mechanical version of the TST was developed and showed good agreement with rate constant values for processes which involved tunneling.<ref name=":0" />.

{{fontcolor1|gray| Good -- just to reiterate, barrier recrossing is the biggest point to be made here. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:47, 31 May 2017 (BST)}}

==Part 2ː F - H - H system==

=== PES Inspection ===

==== Classifying Energetics of Reactions ====

{{fontcolor1|blue|Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?}}

F + H2 → FH + Hː

[[File:Aoz15 F plus H2.PNG|thumb|centre|400px|Figure 4ː Surface plot for the F + H2 → FH + H reaction, showing that the products are at lower energy than the reactants.]]

From this figure it can be seen that the products are lower in energy than the reactants, hence the reaction is exothermic and overall heat is released.

H + HF → H2 + Fː

[[File:Aoz15 FH plus H.PNG|thumb|centre|400px|Figure 5ː Surface plot for the H + HF → H2 + F reaction, showing that the products are at higher energy than the reactants.]]

Conversely, in this case the products are higher in potential energy, and hence the reaction is endothermic.

The following values are the literature bond enthalpies for the H-H and H-F bondsː

H-Hː 435 kJ/mol <ref name=":1">M. H. Stans, ''Bond Dissociation Energies in Simple Molecules'', Washington D.C: National Bureau of Standards, 1970, 28-32</ref>
H-Fː 569 kJ/mol <ref name=":1" />

Using these bond enthalpies, the overall enthalpy change can be found for both reactionsː

F + H2 → FH + Hː 435 - 569 = -134 kJ/mol
H + HF → H2 + Fː 569 - 435 = 134 kJ/mol

This confirms what the potential energy plots showed; that the first reaction is exothermic (overall heat is released) and that the second is endothermic (overall heat is taken in). This is to be expected, since the H-F bond is stronger than the H-H bond, and hence forming the H-F bond is more favourable and results in a larger energy release.

{{fontcolor1|gray| Good use of the PES to prove your points. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:48, 31 May 2017 (BST)}}

==== Finding Transition State Positions ====

{{fontcolor1|blue|Locate the approximate position of the transition state.}}

The transition state was found by initially running the trajectory with no momentum and arbitrary start bond lengths of 1 (see figure 6 below). {{fontcolor1|gray| You could've used Hammond's postulates to make a better initial guess of the TS geometry [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:48, 31 May 2017 (BST)}}At this point, the data cursor was used to find the A-B bond length at the deepest point in the valley, which is also shown in the figure 6 below. This value for the A-B bond length was found to be 0.74 Angstrom. After this, the trajectories were run with A-B bond length as 0.74 and the B-C bond length was varied until the trajectory was a single point, which was taken to be the location of the transition state. This process of trial and error is shown in figure 7-9 below. Not every single run is shown so as not to take up too much space. From this analysis, the following bond lengths were foundː

H-Hː 0.740 Angstrom

H-Fː 1.814 Angstrom

[[File:Aoz15 finding TS 1.PNG|thumb|left|500px|Figure 6ː First plot with no momenta and arbitrary bond lengths. Data cursor shown used to find A-B bond length.]] [[File:Aoz15 finding TS 2.PNG|thumb|centre|500px|Figure 7ː First stage of trial and error in finding B-C bond length.]]

[[File:Aoz15 finding TS 3.PNG|thumb|left|500px|Figure 8ː Second stage of trial and error in finding B-C bond length.]] [[File:Aoz15 finding TS 4.PNG|thumb|centre|500px|Figure 9ː Final surface plot for finding B-C bond length at transition state.]]

==== Finding Activation Energies ====

{{fontcolor1|blue|Report the activation energy for both reactions.}}

The activation energy Ea can be found by finding the energy difference between the reactants and the transition state for each reaction. From the previous part, the energy of the transition state can be found by plotting the potential energy versus time, as shown in figure 10ː

[[File:Aoz15 TS energy.PNG|thumb|centre|600px|Figure 10ː Potential energy versus time plot for the transition state position found in previous section]]

From the figure above, it can be seen that the energy of the transition state is -103.75 kcal/mol

For H-F formation, the minimum energy of the reactants is -104.02 kcal/mol, as seen in the figure below. Hence the activation energy is -103.75 - -104.02 = 0.27 kcal/mol

[[File:Aoz15 reactants energy.PNG|thumb|centre|600px|Figure 11ː Potential energy versus time plot to find activation energy for H-F formation]]

For H-H formation, the minimum energy of the reactants is -133.97, as seen in the figure below. Hence the activation energy is -103.75 - -133.97 = 30.22 kcal/mol

[[File:Aoz15 products1 energy.PNG|thumb|left|600px|Figure 12ː Potential energy versus time plot to find activation energy for H-H formation]] [[File:Aoz15 products2 energy.PNG|thumb|centre|600px|Figure 13ː Potential energy versus time plot to find activation energy for H-F formation, zoomed in to allow finding of minimum energy value.]]

{{fontcolor1|gray| Accurate values with good explanations, GJ! [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:48, 31 May 2017 (BST)}}

=== Reaction Dynamics ===

==== Released Reaction Energy Mechanism Discussion ====

{{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

The figure below shows a reactive trajectory for the F + H2 → FH + Hː reactionː

[[File:Aoz15 successful reaction.PNG|thumb|centre|600px|Figure 14ː Surface plot showing a reactive trajectory for the F + H2 → FH + H reaction.]]

From the surface plot in figure 14 above, it can be seen that the potential energy of the reactants is higher than that of the reactants. This is to be expected since it was earlier determined that the reaction is exothermic. As of such, the potential energy is converted into other forms upon reaction. Due to this simulated reaction occurring with the F atom approaching the H2 molecule from exactly 180°, conversion into rotational energy can be discounted in this scenario. Therefore, the energy is converted into vibrational and translational energy. The conversion into translational energy is hard to discern from the plot above, but the increase in vibrational energy can be seen from the increased size of oscillations about the minimum energy path for the products compared to the reactants.

Experimentally, any heat released (thermal energy) could be measured via calorimetry. Performing IR spectroscopy with a vibrationally unexcited H-F molecule and separately with the vibrationally excited H-F obtained from the reaction would allow the comparison of the peaks, giving the vibrational energy shift between the two.

{{fontcolor1|gray| Fine discussion. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:50, 31 May 2017 (BST)}}

==== Polanyi's Rules Discussion ====

{{fontcolor|blue|Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.}}

Polanyi's rules state that a reaction with a late potential energy barrier (one in which the transition state resembles the products) is more efficiently promoted by vibrational energy than translational energy and that the converse is true for a reaction with an early potential energy barrier (one in which the transition state resembles the reactants)<ref>Z. Zhang, Y. Zhou, D.H. Zhang, ''Theoretical Study of the Validity of the Polanyi Rules for the LateBarrier Cl + CHD3 Reaction'', The Journal of Physical Chemistry, 2012, 3416</ref>.

Exothermic reactions undergo an early transition state, hence via Polanyi's rules would be expected to be more efficiently promoted by translational energy. This would suggest that the F + H2 → FH + H reaction would be promoted more so by translational energy than vibrational. Figures 15 and 16 below show successful and unsuccessful cases confirming Polanyi's rules for this reactionː

[[File:Aoz15 1st polanyi proof.PNG|thumb|left|500px|Figure 15ː Surface plot showing successful F + H2 → FH + H reaction with low initial vibrational energy.]] [[File:Aoz15 2nd polanyi proof.PNG|thumb|centre|500px|Figure 16ː Surface plot showing unsuccessful F + H2 → FH + H reaction with high initial vibrational energy.]]

Conversely, endothermic reactions have a late transition state, so the rules predicts that they will be more efficiently promoted by vibrational energy. This would suggest that the H + HF → H2 + F reaction would be promoted more so by vibrational energy than translational. Figures 17 and 18 below show successful and unsuccessful cases which confirm Polanyi's rules for this reactionː

[[File:Aoz15 3rd polanyi proof.PNG|thumb|left|500px|Figure 17ː Surface plot showing unsuccessful H + HF → H2 + F reaction with low initial vibrational energy.]] [[File:Aoz15 4th polanyi proof.PNG|thumb|centre|500px|Figure 18ː Surface plot showing successful H + HF → H2 + F reaction with high initial vibrational energy.]]

However, Polanyi's rules are not always obeyed, and sometimes act only as a guideline. For instance, figure 19 below shows a successful H + HF → H2 + F reaction with low initial vibrational energy.

[[File:Aoz15 5th polanyi proof.PNG|thumb|centre|600px|Figure 19ː Surface plot showing successful H + HF → H2 + F reaction with low initial vibrational energy.]]

What these figures all together show is that Polanyi's rules can be a good guideline for correct conditions for reactions, since they refer to the efficiency of reactions. I.eː if many reactions with early transition states are run, having high translational energy is likely to result in more successful reactions. The converse is true for late transition states and vibrational energy.

{{fontcolor1|gray| Nice examples [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:50, 31 May 2017 (BST)}}

==References==

MRD:aoz15page

2017-05-31T15:48:59Z

Je714: /* PES Inspection */

==Part 1ː H + H2 system==

=== Dynamics from the Transition State Region ===

{{fontcolor|blue|What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

The total gradient of a potential energy surface has a value of zero both at minimums and at transition structures. Note, for a function of two variables, having a total gradient of zero refers to having a partial derivative of zero with respect to both of its variable separately. i.eː a function f(x,y) will have a stationary point at a location where fx = 0 and fy = 0. This is due to both being stationary points for a function of two variables. However, they are different kinds of stationary points, with the minimum being a minimum and the transition structure a saddle point. Thus if a trajectory is started with no momentum at a position near the minimum point, it will end at the minimum point. However, if a trajectory with no initial momentum is started near the transition structure position, it will end up at either the reactants or products position. Hence this test can be used to determine whether a given point is a minimum or a saddle point for a function of two variables.

{{fontcolor1|gray| I was really enjoying your explanation until the very last part! The point about starting trajectories near the TS is a bit ''ad hoc''. It's a good analogy (and not factually wrong), but not really a rigorous explanation. Since you were correctly discussing total gradients and first partial derivatives, why not extend that to second partial derivatives? Those have information about the curvature of a surface and are actually what's needed to give a mathematically rigorous definition of the TS. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:33, 31 May 2017 (BST)}}

=== Locating the Transition State ===

{{fontcolor|blue|Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.}}

If the collision is taken to be between an atom A and a diatom BC, the transition state occurs when the bond between AB is midway through forming and that between BC is midway through breaking. At this point the system is at maximum potential energy. For a collision H with H2, the potential surface is symmetric, and so the transition state occurs when rAB = rBC. The transition state can thus be located by finding initial distances rAB and rBC such that, when the particles have no initial momentum, their internuclear distances remain constant. Through altering initial separations, an estimate for rrts of 0.908 Angstrom was made. The screenshot for the Internuclear Distances vs Time for this value is shown below (figure 1). It shows that the distances are constant over time. {{fontcolor1|gray| Good. Why are they constant, though? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:33, 31 May 2017 (BST)}}

[[File:Aoz15 transition state position estimate int dist v time.PNG|thumb|centre|600px|Figure 1ː Internuclear distance versus time plot for trajectory used to locate position of transition state.]]

=== Calculating Reaction Path ===

{{fontcolor1|blue|Comment on how the ''mep'' and the trajectory you just calculated differ.}}

The minimum energy path (mep) was found by starting with both initial momenta set to zero, rAB = rrts + 0.01 and rBC = rrts. This results in a trajectory that follows the valley floor of the surface contour to the products. When these initial conditions were run with the MEP calculation type, the following surface plot was obtainedː

[[File:Aoz15 mep surface plot.PNG|thumb|centre|500px|Figure 2ː Surface plot obtained for mep calculation.]]

As expected, this simply shows a smooth trajectory exactly down the valley floor to the products. However, when these same conditions were run with the dynamics calculation type, the trajectory differed and the following plot was producedː

[[File:Aoz15 meptrajectory surface plot.PNG|thumb|centre|500px|Figure 3ː Surface plot for the same initial conditions as for the mep, but performed with the dynamics calculation setting.]]

The mep and trajectory obtained differ in that only the trajectory takes into account vibrational motion; in the mep obtained, the energy stays exactly as the minimum energy of the valley and hence is a smooth line to the products. However, the trajectory takes into account vibrational motion, hence the energy oscillates about the minimum energy whilst reaching the products. This occurs due to the vibrational energy in the breaking/forming bonds, and explains the shape of the line seen in figure 3 above.

If the initial conditions rAB = rrts and rBC = rrts + 0.01 are used, the trajectory still follows the valley floor. However, it ends up at the reactants instead of the products, since it effectively 'rolls' the other way to find an energy minimum.

{{fontcolor1|gray| Fine. What's the Kinetic Energy on a MEP? What does that tell you about the velocities? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:38, 31 May 2017 (BST)}}

=== Reactive and Unreactive Trajectories ===

{{fontcolor1|blue|Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.}}

For initial bond lengths of rAB = 0.74 and rBC = 2.0, the initial momenta were varied and surface plots were found in order to determine which conditions led to reactive trajectories. The results and some discussion are found in the following table.

{| class="wikitable" border="1"
|+
! Surface Plot !! p1!! p2!!Reactive or Unreactive and Brief Discussion
|-
| [[File:Aoz15 reactive or not 1.PNG|centre|400px]] || -1.25 ||-2.5||Reactive, which can be seen by the trajectory passing through the transition state and going to the products.
|-
| [[File:Aoz15 reactive or not 2.PNG|centre|400px]] || -1.5||-2.0||Unreactive, since the path does not successfully pass through the transition state, instead ending at the reactants.
|-
| [[File:Aoz15 reactive or not 3.PNG|centre|400px]] || -1.5||-2.5||Reactive, which can once again be seen by the trajectory passing through the transition state and going to the products. Only difference to the first reaction is a slight increase in overall vibrational energy, which can be seen by greater oscillations in the trajectory.
|-
| [[File:Aoz15 reactive or not 4.PNG|centre|400px]] || -2.5||-5.0||Unreactive, which can be seen by the trajectory ending at the reactants. However, in this case it does pass through the transition state. Therefore, this is a case of barrier recrossing.
|-
| [[File:Aoz15 reactive or not 5.PNG|centre|400px]] || -2.5||-5.2||Reactive, as the reaction proceeds to the products. Here barrier recrossing also occurs, but unlike in the fourth reaction, in this case the reaction still proceeds to products. Hence, multiple barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|blue|State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

There are three main assumptions in Transition State Theory (TST). Before stating them, it is useful to define the terms critical diving surface and activated complex. For a given reaction potential energy surface, the critical diving surface refers to a boundary surface which passes through the surface saddle point. As such, the critical diving surface must be crossed in order for a reaction to go to completion. Any molecule whose nuclear configuration matches a point on the critical diving surface or to any point within a short distance of it is termed an activated complex. So, the saddle point of the potential surface corresponds to the equilibrium structure of the activated complex. With these terms defined, the following are the main assumptions of Transition State Theoryː firstly, that all configurations of molecules which cross the critical diving surface from the reactants side will proceed to become products. Secondly, that all reactant molecules maintain a Boltzmann-distributed energy distribution during the reaction. This leads to the third and final assumption; that the energy distribution of all activated complexes during a reaction will also be a Boltzmann distribution which depends on the system temperature.<ref name=":0">Levine, I. N., ''Physical Chemistry'', McGraw-Hill Publishing, Sixth Edition, 2009, 887-894</ref>
{{fontcolor1|gray| OK. Now, do the last two points make sense here? We're just simulating an isolated triatomic collision -- not an ensemble of particles. So this statistichal treatment is not applicable. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:47, 31 May 2017 (BST)}}

When tested against experimentally obtained values, those found by TST are in most cases within a single order of magnitude, suggesting that the assumptions made are well placed<ref name=":0" />. However, there are certain aspects of reactions which TST doesn't account for. The first is barrier recrossing; since TST assumes that all molecular configurations which pass the critical dividing surface proceed to products, it cannot consider barrier recrossing, the process in which the reaction trajectory crosses the potential barrier but then recrosses it in the opposite direction. As in the fourth simulation above, sometimes barrier recrossing leads to the molecules returning to the reactants and the products not forming at all. In such a case, TST wildly misinterprets the reaction result. Even in cases where barrier recrossing occurs multiple times and the products are formed (as in the fifth simulation above), TST gives a higher reaction rate value than is experimentally obtained, since it assumes the reaction proceeds to completion as soon as the potential barrier is crossed. TST also doesn't take into account quantum effects, such as tunneling. Tunneling occurs when reactant molecules with insufficient energy to bypass the potential barrier pass through the barrier, leading to an exponential decay in their energy. This process occurs thanks to wave-particle duality, and cannot be predicted by TST<ref>P. Atkins, J. de Paula, ''Atkins Physical Chemistry'', Oxford University Press, 10th Edition, Chapter 21</ref>. However, a quantum-mechanical version of the TST was developed and showed good agreement with rate constant values for processes which involved tunneling.<ref name=":0" />.

{{fontcolor1|gray| Good -- just to reiterate, barrier recrossing is the biggest point to be made here. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:47, 31 May 2017 (BST)}}

==Part 2ː F - H - H system==

=== PES Inspection ===

==== Classifying Energetics of Reactions ====

{{fontcolor1|blue|Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?}}

F + H2 → FH + Hː

[[File:Aoz15 F plus H2.PNG|thumb|centre|400px|Figure 4ː Surface plot for the F + H2 → FH + H reaction, showing that the products are at lower energy than the reactants.]]

From this figure it can be seen that the products are lower in energy than the reactants, hence the reaction is exothermic and overall heat is released.

H + HF → H2 + Fː

[[File:Aoz15 FH plus H.PNG|thumb|centre|400px|Figure 5ː Surface plot for the H + HF → H2 + F reaction, showing that the products are at higher energy than the reactants.]]

Conversely, in this case the products are higher in potential energy, and hence the reaction is endothermic.

The following values are the literature bond enthalpies for the H-H and H-F bondsː

H-Hː 435 kJ/mol <ref name=":1">M. H. Stans, ''Bond Dissociation Energies in Simple Molecules'', Washington D.C: National Bureau of Standards, 1970, 28-32</ref>
H-Fː 569 kJ/mol <ref name=":1" />

Using these bond enthalpies, the overall enthalpy change can be found for both reactionsː

F + H2 → FH + Hː 435 - 569 = -134 kJ/mol
H + HF → H2 + Fː 569 - 435 = 134 kJ/mol

This confirms what the potential energy plots showed; that the first reaction is exothermic (overall heat is released) and that the second is endothermic (overall heat is taken in). This is to be expected, since the H-F bond is stronger than the H-H bond, and hence forming the H-F bond is more favourable and results in a larger energy release.

{{fontcolor1|gray| Good use of the PES to prove your points. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:48, 31 May 2017 (BST)}}

==== Finding Transition State Positions ====

{{fontcolor1|blue|Locate the approximate position of the transition state.}}

The transition state was found by initially running the trajectory with no momentum and arbitrary start bond lengths of 1 (see figure 6 below). {{fontcolor1|gray| You could've used Hammond's postulates to make a better initial guess of the TS geometry [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:48, 31 May 2017 (BST)}}At this point, the data cursor was used to find the A-B bond length at the deepest point in the valley, which is also shown in the figure 6 below. This value for the A-B bond length was found to be 0.74 Angstrom. After this, the trajectories were run with A-B bond length as 0.74 and the B-C bond length was varied until the trajectory was a single point, which was taken to be the location of the transition state. This process of trial and error is shown in figure 7-9 below. Not every single run is shown so as not to take up too much space. From this analysis, the following bond lengths were foundː

H-Hː 0.740 Angstrom

H-Fː 1.814 Angstrom

[[File:Aoz15 finding TS 1.PNG|thumb|left|500px|Figure 6ː First plot with no momenta and arbitrary bond lengths. Data cursor shown used to find A-B bond length.]] [[File:Aoz15 finding TS 2.PNG|thumb|centre|500px|Figure 7ː First stage of trial and error in finding B-C bond length.]]

[[File:Aoz15 finding TS 3.PNG|thumb|left|500px|Figure 8ː Second stage of trial and error in finding B-C bond length.]] [[File:Aoz15 finding TS 4.PNG|thumb|centre|500px|Figure 9ː Final surface plot for finding B-C bond length at transition state.]]

==== Finding Activation Energies ====

{{fontcolor1|blue|Report the activation energy for both reactions.}}

The activation energy Ea can be found by finding the energy difference between the reactants and the transition state for each reaction. From the previous part, the energy of the transition state can be found by plotting the potential energy versus time, as shown in figure 10ː

[[File:Aoz15 TS energy.PNG|thumb|centre|600px|Figure 10ː Potential energy versus time plot for the transition state position found in previous section]]

From the figure above, it can be seen that the energy of the transition state is -103.75 kcal/mol

For H-F formation, the minimum energy of the reactants is -104.02 kcal/mol, as seen in the figure below. Hence the activation energy is -103.75 - -104.02 = 0.27 kcal/mol

[[File:Aoz15 reactants energy.PNG|thumb|centre|600px|Figure 11ː Potential energy versus time plot to find activation energy for H-F formation]]

For H-H formation, the minimum energy of the reactants is -133.97, as seen in the figure below. Hence the activation energy is -103.75 - -133.97 = 30.22 kcal/mol

[[File:Aoz15 products1 energy.PNG|thumb|left|600px|Figure 12ː Potential energy versus time plot to find activation energy for H-H formation]] [[File:Aoz15 products2 energy.PNG|thumb|centre|600px|Figure 13ː Potential energy versus time plot to find activation energy for H-F formation, zoomed in to allow finding of minimum energy value.]]

{{fontcolor1|gray| Accurate values with good explanations, GJ! [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:48, 31 May 2017 (BST)}}

=== Reaction Dynamics ===

==== Released Reaction Energy Mechanism Discussion ====

{{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

The figure below shows a reactive trajectory for the F + H2 → FH + Hː reactionː

[[File:Aoz15 successful reaction.PNG|thumb|centre|600px|Figure 14ː Surface plot showing a reactive trajectory for the F + H2 → FH + H reaction.]]

From the surface plot in figure 14 above, it can be seen that the potential energy of the reactants is higher than that of the reactants. This is to be expected since it was earlier determined that the reaction is exothermic. As of such, the potential energy is converted into other forms upon reaction. Due to this simulated reaction occurring with the F atom approaching the H2 molecule from exactly 180°, conversion into rotational energy can be discounted in this scenario. Therefore, the energy is converted into vibrational and translational energy. The conversion into translational energy is hard to discern from the plot above, but the increase in vibrational energy can be seen from the increased size of oscillations about the minimum energy path for the products compared to the reactants.

Experimentally, any heat released (thermal energy) could be measured via calorimetry. Performing IR spectroscopy with a vibrationally unexcited H-F molecule and separately with the vibrationally excited H-F obtained from the reaction would allow the comparison of the peaks, giving the vibrational energy shift between the two.

==== Polanyi's Rules Discussion ====

{{fontcolor|blue|Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.}}

Polanyi's rules state that a reaction with a late potential energy barrier (one in which the transition state resembles the products) is more efficiently promoted by vibrational energy than translational energy and that the converse is true for a reaction with an early potential energy barrier (one in which the transition state resembles the reactants)<ref>Z. Zhang, Y. Zhou, D.H. Zhang, ''Theoretical Study of the Validity of the Polanyi Rules for the LateBarrier Cl + CHD3 Reaction'', The Journal of Physical Chemistry, 2012, 3416</ref>.

Exothermic reactions undergo an early transition state, hence via Polanyi's rules would be expected to be more efficiently promoted by translational energy. This would suggest that the F + H2 → FH + H reaction would be promoted more so by translational energy than vibrational. Figures 15 and 16 below show successful and unsuccessful cases confirming Polanyi's rules for this reactionː

[[File:Aoz15 1st polanyi proof.PNG|thumb|left|500px|Figure 15ː Surface plot showing successful F + H2 → FH + H reaction with low initial vibrational energy.]] [[File:Aoz15 2nd polanyi proof.PNG|thumb|centre|500px|Figure 16ː Surface plot showing unsuccessful F + H2 → FH + H reaction with high initial vibrational energy.]]

Conversely, endothermic reactions have a late transition state, so the rules predicts that they will be more efficiently promoted by vibrational energy. This would suggest that the H + HF → H2 + F reaction would be promoted more so by vibrational energy than translational. Figures 17 and 18 below show successful and unsuccessful cases which confirm Polanyi's rules for this reactionː

[[File:Aoz15 3rd polanyi proof.PNG|thumb|left|500px|Figure 17ː Surface plot showing unsuccessful H + HF → H2 + F reaction with low initial vibrational energy.]] [[File:Aoz15 4th polanyi proof.PNG|thumb|centre|500px|Figure 18ː Surface plot showing successful H + HF → H2 + F reaction with high initial vibrational energy.]]

However, Polanyi's rules are not always obeyed, and sometimes act only as a guideline. For instance, figure 19 below shows a successful H + HF → H2 + F reaction with low initial vibrational energy.

[[File:Aoz15 5th polanyi proof.PNG|thumb|centre|600px|Figure 19ː Surface plot showing successful H + HF → H2 + F reaction with low initial vibrational energy.]]

What these figures all together show is that Polanyi's rules can be a good guideline for correct conditions for reactions, since they refer to the efficiency of reactions. I.eː if many reactions with early transition states are run, having high translational energy is likely to result in more successful reactions. The converse is true for late transition states and vibrational energy.

==References==

MRD:aoz15page

2017-05-31T15:47:22Z

Je714: /* Transition State Theory */

==Part 1ː H + H2 system==

=== Dynamics from the Transition State Region ===

{{fontcolor|blue|What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

The total gradient of a potential energy surface has a value of zero both at minimums and at transition structures. Note, for a function of two variables, having a total gradient of zero refers to having a partial derivative of zero with respect to both of its variable separately. i.eː a function f(x,y) will have a stationary point at a location where fx = 0 and fy = 0. This is due to both being stationary points for a function of two variables. However, they are different kinds of stationary points, with the minimum being a minimum and the transition structure a saddle point. Thus if a trajectory is started with no momentum at a position near the minimum point, it will end at the minimum point. However, if a trajectory with no initial momentum is started near the transition structure position, it will end up at either the reactants or products position. Hence this test can be used to determine whether a given point is a minimum or a saddle point for a function of two variables.

{{fontcolor1|gray| I was really enjoying your explanation until the very last part! The point about starting trajectories near the TS is a bit ''ad hoc''. It's a good analogy (and not factually wrong), but not really a rigorous explanation. Since you were correctly discussing total gradients and first partial derivatives, why not extend that to second partial derivatives? Those have information about the curvature of a surface and are actually what's needed to give a mathematically rigorous definition of the TS. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:33, 31 May 2017 (BST)}}

=== Locating the Transition State ===

{{fontcolor|blue|Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.}}

If the collision is taken to be between an atom A and a diatom BC, the transition state occurs when the bond between AB is midway through forming and that between BC is midway through breaking. At this point the system is at maximum potential energy. For a collision H with H2, the potential surface is symmetric, and so the transition state occurs when rAB = rBC. The transition state can thus be located by finding initial distances rAB and rBC such that, when the particles have no initial momentum, their internuclear distances remain constant. Through altering initial separations, an estimate for rrts of 0.908 Angstrom was made. The screenshot for the Internuclear Distances vs Time for this value is shown below (figure 1). It shows that the distances are constant over time. {{fontcolor1|gray| Good. Why are they constant, though? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:33, 31 May 2017 (BST)}}

[[File:Aoz15 transition state position estimate int dist v time.PNG|thumb|centre|600px|Figure 1ː Internuclear distance versus time plot for trajectory used to locate position of transition state.]]

=== Calculating Reaction Path ===

{{fontcolor1|blue|Comment on how the ''mep'' and the trajectory you just calculated differ.}}

The minimum energy path (mep) was found by starting with both initial momenta set to zero, rAB = rrts + 0.01 and rBC = rrts. This results in a trajectory that follows the valley floor of the surface contour to the products. When these initial conditions were run with the MEP calculation type, the following surface plot was obtainedː

[[File:Aoz15 mep surface plot.PNG|thumb|centre|500px|Figure 2ː Surface plot obtained for mep calculation.]]

As expected, this simply shows a smooth trajectory exactly down the valley floor to the products. However, when these same conditions were run with the dynamics calculation type, the trajectory differed and the following plot was producedː

[[File:Aoz15 meptrajectory surface plot.PNG|thumb|centre|500px|Figure 3ː Surface plot for the same initial conditions as for the mep, but performed with the dynamics calculation setting.]]

The mep and trajectory obtained differ in that only the trajectory takes into account vibrational motion; in the mep obtained, the energy stays exactly as the minimum energy of the valley and hence is a smooth line to the products. However, the trajectory takes into account vibrational motion, hence the energy oscillates about the minimum energy whilst reaching the products. This occurs due to the vibrational energy in the breaking/forming bonds, and explains the shape of the line seen in figure 3 above.

If the initial conditions rAB = rrts and rBC = rrts + 0.01 are used, the trajectory still follows the valley floor. However, it ends up at the reactants instead of the products, since it effectively 'rolls' the other way to find an energy minimum.

{{fontcolor1|gray| Fine. What's the Kinetic Energy on a MEP? What does that tell you about the velocities? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:38, 31 May 2017 (BST)}}

=== Reactive and Unreactive Trajectories ===

{{fontcolor1|blue|Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.}}

For initial bond lengths of rAB = 0.74 and rBC = 2.0, the initial momenta were varied and surface plots were found in order to determine which conditions led to reactive trajectories. The results and some discussion are found in the following table.

{| class="wikitable" border="1"
|+
! Surface Plot !! p1!! p2!!Reactive or Unreactive and Brief Discussion
|-
| [[File:Aoz15 reactive or not 1.PNG|centre|400px]] || -1.25 ||-2.5||Reactive, which can be seen by the trajectory passing through the transition state and going to the products.
|-
| [[File:Aoz15 reactive or not 2.PNG|centre|400px]] || -1.5||-2.0||Unreactive, since the path does not successfully pass through the transition state, instead ending at the reactants.
|-
| [[File:Aoz15 reactive or not 3.PNG|centre|400px]] || -1.5||-2.5||Reactive, which can once again be seen by the trajectory passing through the transition state and going to the products. Only difference to the first reaction is a slight increase in overall vibrational energy, which can be seen by greater oscillations in the trajectory.
|-
| [[File:Aoz15 reactive or not 4.PNG|centre|400px]] || -2.5||-5.0||Unreactive, which can be seen by the trajectory ending at the reactants. However, in this case it does pass through the transition state. Therefore, this is a case of barrier recrossing.
|-
| [[File:Aoz15 reactive or not 5.PNG|centre|400px]] || -2.5||-5.2||Reactive, as the reaction proceeds to the products. Here barrier recrossing also occurs, but unlike in the fourth reaction, in this case the reaction still proceeds to products. Hence, multiple barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|blue|State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

There are three main assumptions in Transition State Theory (TST). Before stating them, it is useful to define the terms critical diving surface and activated complex. For a given reaction potential energy surface, the critical diving surface refers to a boundary surface which passes through the surface saddle point. As such, the critical diving surface must be crossed in order for a reaction to go to completion. Any molecule whose nuclear configuration matches a point on the critical diving surface or to any point within a short distance of it is termed an activated complex. So, the saddle point of the potential surface corresponds to the equilibrium structure of the activated complex. With these terms defined, the following are the main assumptions of Transition State Theoryː firstly, that all configurations of molecules which cross the critical diving surface from the reactants side will proceed to become products. Secondly, that all reactant molecules maintain a Boltzmann-distributed energy distribution during the reaction. This leads to the third and final assumption; that the energy distribution of all activated complexes during a reaction will also be a Boltzmann distribution which depends on the system temperature.<ref name=":0">Levine, I. N., ''Physical Chemistry'', McGraw-Hill Publishing, Sixth Edition, 2009, 887-894</ref>
{{fontcolor1|gray| OK. Now, do the last two points make sense here? We're just simulating an isolated triatomic collision -- not an ensemble of particles. So this statistichal treatment is not applicable. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:47, 31 May 2017 (BST)}}

When tested against experimentally obtained values, those found by TST are in most cases within a single order of magnitude, suggesting that the assumptions made are well placed<ref name=":0" />. However, there are certain aspects of reactions which TST doesn't account for. The first is barrier recrossing; since TST assumes that all molecular configurations which pass the critical dividing surface proceed to products, it cannot consider barrier recrossing, the process in which the reaction trajectory crosses the potential barrier but then recrosses it in the opposite direction. As in the fourth simulation above, sometimes barrier recrossing leads to the molecules returning to the reactants and the products not forming at all. In such a case, TST wildly misinterprets the reaction result. Even in cases where barrier recrossing occurs multiple times and the products are formed (as in the fifth simulation above), TST gives a higher reaction rate value than is experimentally obtained, since it assumes the reaction proceeds to completion as soon as the potential barrier is crossed. TST also doesn't take into account quantum effects, such as tunneling. Tunneling occurs when reactant molecules with insufficient energy to bypass the potential barrier pass through the barrier, leading to an exponential decay in their energy. This process occurs thanks to wave-particle duality, and cannot be predicted by TST<ref>P. Atkins, J. de Paula, ''Atkins Physical Chemistry'', Oxford University Press, 10th Edition, Chapter 21</ref>. However, a quantum-mechanical version of the TST was developed and showed good agreement with rate constant values for processes which involved tunneling.<ref name=":0" />.

{{fontcolor1|gray| Good -- just to reiterate, barrier recrossing is the biggest point to be made here. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:47, 31 May 2017 (BST)}}

==Part 2ː F - H - H system==

=== PES Inspection ===

==== Classifying Energetics of Reactions ====

{{fontcolor1|blue|Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?}}

F + H2 → FH + Hː

[[File:Aoz15 F plus H2.PNG|thumb|centre|400px|Figure 4ː Surface plot for the F + H2 → FH + H reaction, showing that the products are at lower energy than the reactants.]]

From this figure it can be seen that the products are lower in energy than the reactants, hence the reaction is exothermic and overall heat is released.

H + HF → H2 + Fː

[[File:Aoz15 FH plus H.PNG|thumb|centre|400px|Figure 5ː Surface plot for the H + HF → H2 + F reaction, showing that the products are at higher energy than the reactants.]]

Conversely, in this case the products are higher in potential energy, and hence the reaction is endothermic.

The following values are the literature bond enthalpies for the H-H and H-F bondsː

H-Hː 435 kJ/mol <ref name=":1">M. H. Stans, ''Bond Dissociation Energies in Simple Molecules'', Washington D.C: National Bureau of Standards, 1970, 28-32</ref>
H-Fː 569 kJ/mol <ref name=":1" />

Using these bond enthalpies, the overall enthalpy change can be found for both reactionsː

F + H2 → FH + Hː 435 - 569 = -134 kJ/mol
H + HF → H2 + Fː 569 - 435 = 134 kJ/mol

This confirms what the potential energy plots showed; that the first reaction is exothermic (overall heat is released) and that the second is endothermic (overall heat is taken in). This is to be expected, since the H-F bond is stronger than the H-H bond, and hence forming the H-F bond is more favourable and results in a larger energy release.

==== Finding Transition State Positions ====

{{fontcolor1|blue|Locate the approximate position of the transition state.}}

The transition state was found by initially running the trajectory with no momentum and arbitrary start bond lengths of 1 (see figure 6 below). At this point, the data cursor was used to find the A-B bond length at the deepest point in the valley, which is also shown in the figure 6 below. This value for the A-B bond length was found to be 0.74 Angstrom. After this, the trajectories were run with A-B bond length as 0.74 and the B-C bond length was varied until the trajectory was a single point, which was taken to be the location of the transition state. This process of trial and error is shown in figure 7-9 below. Not every single run is shown so as not to take up too much space. From this analysis, the following bond lengths were foundː

H-Hː 0.740 Angstrom

H-Fː 1.814 Angstrom

[[File:Aoz15 finding TS 1.PNG|thumb|left|500px|Figure 6ː First plot with no momenta and arbitrary bond lengths. Data cursor shown used to find A-B bond length.]] [[File:Aoz15 finding TS 2.PNG|thumb|centre|500px|Figure 7ː First stage of trial and error in finding B-C bond length.]]

[[File:Aoz15 finding TS 3.PNG|thumb|left|500px|Figure 8ː Second stage of trial and error in finding B-C bond length.]] [[File:Aoz15 finding TS 4.PNG|thumb|centre|500px|Figure 9ː Final surface plot for finding B-C bond length at transition state.]]

==== Finding Activation Energies ====

{{fontcolor1|blue|Report the activation energy for both reactions.}}

The activation energy Ea can be found by finding the energy difference between the reactants and the transition state for each reaction. From the previous part, the energy of the transition state can be found by plotting the potential energy versus time, as shown in figure 10ː

[[File:Aoz15 TS energy.PNG|thumb|centre|600px|Figure 10ː Potential energy versus time plot for the transition state position found in previous section]]

From the figure above, it can be seen that the energy of the transition state is -103.75 kcal/mol

For H-F formation, the minimum energy of the reactants is -104.02 kcal/mol, as seen in the figure below. Hence the activation energy is -103.75 - -104.02 = 0.27 kcal/mol

[[File:Aoz15 reactants energy.PNG|thumb|centre|600px|Figure 11ː Potential energy versus time plot to find activation energy for H-F formation]]

For H-H formation, the minimum energy of the reactants is -133.97, as seen in the figure below. Hence the activation energy is -103.75 - -133.97 = 30.22 kcal/mol

[[File:Aoz15 products1 energy.PNG|thumb|left|600px|Figure 12ː Potential energy versus time plot to find activation energy for H-H formation]] [[File:Aoz15 products2 energy.PNG|thumb|centre|600px|Figure 13ː Potential energy versus time plot to find activation energy for H-F formation, zoomed in to allow finding of minimum energy value.]]

=== Reaction Dynamics ===

==== Released Reaction Energy Mechanism Discussion ====

{{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

The figure below shows a reactive trajectory for the F + H2 → FH + Hː reactionː

[[File:Aoz15 successful reaction.PNG|thumb|centre|600px|Figure 14ː Surface plot showing a reactive trajectory for the F + H2 → FH + H reaction.]]

From the surface plot in figure 14 above, it can be seen that the potential energy of the reactants is higher than that of the reactants. This is to be expected since it was earlier determined that the reaction is exothermic. As of such, the potential energy is converted into other forms upon reaction. Due to this simulated reaction occurring with the F atom approaching the H2 molecule from exactly 180°, conversion into rotational energy can be discounted in this scenario. Therefore, the energy is converted into vibrational and translational energy. The conversion into translational energy is hard to discern from the plot above, but the increase in vibrational energy can be seen from the increased size of oscillations about the minimum energy path for the products compared to the reactants.

Experimentally, any heat released (thermal energy) could be measured via calorimetry. Performing IR spectroscopy with a vibrationally unexcited H-F molecule and separately with the vibrationally excited H-F obtained from the reaction would allow the comparison of the peaks, giving the vibrational energy shift between the two.

==== Polanyi's Rules Discussion ====

{{fontcolor|blue|Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.}}

Polanyi's rules state that a reaction with a late potential energy barrier (one in which the transition state resembles the products) is more efficiently promoted by vibrational energy than translational energy and that the converse is true for a reaction with an early potential energy barrier (one in which the transition state resembles the reactants)<ref>Z. Zhang, Y. Zhou, D.H. Zhang, ''Theoretical Study of the Validity of the Polanyi Rules for the LateBarrier Cl + CHD3 Reaction'', The Journal of Physical Chemistry, 2012, 3416</ref>.

Exothermic reactions undergo an early transition state, hence via Polanyi's rules would be expected to be more efficiently promoted by translational energy. This would suggest that the F + H2 → FH + H reaction would be promoted more so by translational energy than vibrational. Figures 15 and 16 below show successful and unsuccessful cases confirming Polanyi's rules for this reactionː

[[File:Aoz15 1st polanyi proof.PNG|thumb|left|500px|Figure 15ː Surface plot showing successful F + H2 → FH + H reaction with low initial vibrational energy.]] [[File:Aoz15 2nd polanyi proof.PNG|thumb|centre|500px|Figure 16ː Surface plot showing unsuccessful F + H2 → FH + H reaction with high initial vibrational energy.]]

Conversely, endothermic reactions have a late transition state, so the rules predicts that they will be more efficiently promoted by vibrational energy. This would suggest that the H + HF → H2 + F reaction would be promoted more so by vibrational energy than translational. Figures 17 and 18 below show successful and unsuccessful cases which confirm Polanyi's rules for this reactionː

[[File:Aoz15 3rd polanyi proof.PNG|thumb|left|500px|Figure 17ː Surface plot showing unsuccessful H + HF → H2 + F reaction with low initial vibrational energy.]] [[File:Aoz15 4th polanyi proof.PNG|thumb|centre|500px|Figure 18ː Surface plot showing successful H + HF → H2 + F reaction with high initial vibrational energy.]]

However, Polanyi's rules are not always obeyed, and sometimes act only as a guideline. For instance, figure 19 below shows a successful H + HF → H2 + F reaction with low initial vibrational energy.

[[File:Aoz15 5th polanyi proof.PNG|thumb|centre|600px|Figure 19ː Surface plot showing successful H + HF → H2 + F reaction with low initial vibrational energy.]]

What these figures all together show is that Polanyi's rules can be a good guideline for correct conditions for reactions, since they refer to the efficiency of reactions. I.eː if many reactions with early transition states are run, having high translational energy is likely to result in more successful reactions. The converse is true for late transition states and vibrational energy.

==References==

MRD:aoz15page

2017-05-31T15:38:33Z

Je714: /* Calculating Reaction Path */

==Part 1ː H + H2 system==

=== Dynamics from the Transition State Region ===

{{fontcolor|blue|What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

The total gradient of a potential energy surface has a value of zero both at minimums and at transition structures. Note, for a function of two variables, having a total gradient of zero refers to having a partial derivative of zero with respect to both of its variable separately. i.eː a function f(x,y) will have a stationary point at a location where fx = 0 and fy = 0. This is due to both being stationary points for a function of two variables. However, they are different kinds of stationary points, with the minimum being a minimum and the transition structure a saddle point. Thus if a trajectory is started with no momentum at a position near the minimum point, it will end at the minimum point. However, if a trajectory with no initial momentum is started near the transition structure position, it will end up at either the reactants or products position. Hence this test can be used to determine whether a given point is a minimum or a saddle point for a function of two variables.

{{fontcolor1|gray| I was really enjoying your explanation until the very last part! The point about starting trajectories near the TS is a bit ''ad hoc''. It's a good analogy (and not factually wrong), but not really a rigorous explanation. Since you were correctly discussing total gradients and first partial derivatives, why not extend that to second partial derivatives? Those have information about the curvature of a surface and are actually what's needed to give a mathematically rigorous definition of the TS. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:33, 31 May 2017 (BST)}}

=== Locating the Transition State ===

{{fontcolor|blue|Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.}}

If the collision is taken to be between an atom A and a diatom BC, the transition state occurs when the bond between AB is midway through forming and that between BC is midway through breaking. At this point the system is at maximum potential energy. For a collision H with H2, the potential surface is symmetric, and so the transition state occurs when rAB = rBC. The transition state can thus be located by finding initial distances rAB and rBC such that, when the particles have no initial momentum, their internuclear distances remain constant. Through altering initial separations, an estimate for rrts of 0.908 Angstrom was made. The screenshot for the Internuclear Distances vs Time for this value is shown below (figure 1). It shows that the distances are constant over time. {{fontcolor1|gray| Good. Why are they constant, though? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:33, 31 May 2017 (BST)}}

[[File:Aoz15 transition state position estimate int dist v time.PNG|thumb|centre|600px|Figure 1ː Internuclear distance versus time plot for trajectory used to locate position of transition state.]]

=== Calculating Reaction Path ===

{{fontcolor1|blue|Comment on how the ''mep'' and the trajectory you just calculated differ.}}

The minimum energy path (mep) was found by starting with both initial momenta set to zero, rAB = rrts + 0.01 and rBC = rrts. This results in a trajectory that follows the valley floor of the surface contour to the products. When these initial conditions were run with the MEP calculation type, the following surface plot was obtainedː

[[File:Aoz15 mep surface plot.PNG|thumb|centre|500px|Figure 2ː Surface plot obtained for mep calculation.]]

As expected, this simply shows a smooth trajectory exactly down the valley floor to the products. However, when these same conditions were run with the dynamics calculation type, the trajectory differed and the following plot was producedː

[[File:Aoz15 meptrajectory surface plot.PNG|thumb|centre|500px|Figure 3ː Surface plot for the same initial conditions as for the mep, but performed with the dynamics calculation setting.]]

The mep and trajectory obtained differ in that only the trajectory takes into account vibrational motion; in the mep obtained, the energy stays exactly as the minimum energy of the valley and hence is a smooth line to the products. However, the trajectory takes into account vibrational motion, hence the energy oscillates about the minimum energy whilst reaching the products. This occurs due to the vibrational energy in the breaking/forming bonds, and explains the shape of the line seen in figure 3 above.

If the initial conditions rAB = rrts and rBC = rrts + 0.01 are used, the trajectory still follows the valley floor. However, it ends up at the reactants instead of the products, since it effectively 'rolls' the other way to find an energy minimum.

{{fontcolor1|gray| Fine. What's the Kinetic Energy on a MEP? What does that tell you about the velocities? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:38, 31 May 2017 (BST)}}

=== Reactive and Unreactive Trajectories ===

{{fontcolor1|blue|Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.}}

For initial bond lengths of rAB = 0.74 and rBC = 2.0, the initial momenta were varied and surface plots were found in order to determine which conditions led to reactive trajectories. The results and some discussion are found in the following table.

{| class="wikitable" border="1"
|+
! Surface Plot !! p1!! p2!!Reactive or Unreactive and Brief Discussion
|-
| [[File:Aoz15 reactive or not 1.PNG|centre|400px]] || -1.25 ||-2.5||Reactive, which can be seen by the trajectory passing through the transition state and going to the products.
|-
| [[File:Aoz15 reactive or not 2.PNG|centre|400px]] || -1.5||-2.0||Unreactive, since the path does not successfully pass through the transition state, instead ending at the reactants.
|-
| [[File:Aoz15 reactive or not 3.PNG|centre|400px]] || -1.5||-2.5||Reactive, which can once again be seen by the trajectory passing through the transition state and going to the products. Only difference to the first reaction is a slight increase in overall vibrational energy, which can be seen by greater oscillations in the trajectory.
|-
| [[File:Aoz15 reactive or not 4.PNG|centre|400px]] || -2.5||-5.0||Unreactive, which can be seen by the trajectory ending at the reactants. However, in this case it does pass through the transition state. Therefore, this is a case of barrier recrossing.
|-
| [[File:Aoz15 reactive or not 5.PNG|centre|400px]] || -2.5||-5.2||Reactive, as the reaction proceeds to the products. Here barrier recrossing also occurs, but unlike in the fourth reaction, in this case the reaction still proceeds to products. Hence, multiple barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|blue|State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

There are three main assumptions in Transition State Theory (TST). Before stating them, it is useful to define the terms critical diving surface and activated complex. For a given reaction potential energy surface, the critical diving surface refers to a boundary surface which passes through the surface saddle point. As such, the critical diving surface must be crossed in order for a reaction to go to completion. Any molecule whose nuclear configuration matches a point on the critical diving surface or to any point within a short distance of it is termed an activated complex. So, the saddle point of the potential surface corresponds to the equilibrium structure of the activated complex. With these terms defined, the following are the main assumptions of Transition State Theoryː firstly, that all configurations of molecules which cross the critical diving surface from the reactants side will proceed to become products. Secondly, that all reactant molecules maintain a Boltzmann-distributed energy distribution during the reaction. This leads to the third and final assumption; that the energy distribution of all activated complexes during a reaction will also be a Boltzmann distribution which depends on the system temperature.<ref name=":0">Levine, I. N., ''Physical Chemistry'', McGraw-Hill Publishing, Sixth Edition, 2009, 887-894</ref>

When tested against experimentally obtained values, those found by TST are in most cases within a single order of magnitude, suggesting that the assumptions made are well placed<ref name=":0" />. However, there are certain aspects of reactions which TST doesn't account for. The first is barrier recrossing; since TST assumes that all molecular configurations which pass the critical dividing surface proceed to products, it cannot consider barrier recrossing, the process in which the reaction trajectory crosses the potential barrier but then recrosses it in the opposite direction. As in the fourth simulation above, sometimes barrier recrossing leads to the molecules returning to the reactants and the products not forming at all. In such a case, TST wildly misinterprets the reaction result. Even in cases where barrier recrossing occurs multiple times and the products are formed (as in the fifth simulation above), TST gives a higher reaction rate value than is experimentally obtained, since it assumes the reaction proceeds to completion as soon as the potential barrier is crossed. TST also doesn't take into account quantum effects, such as tunneling. Tunneling occurs when reactant molecules with insufficient energy to bypass the potential barrier pass through the barrier, leading to an exponential decay in their energy. This process occurs thanks to wave-particle duality, and cannot be predicted by TST<ref>P. Atkins, J. de Paula, ''Atkins Physical Chemistry'', Oxford University Press, 10th Edition, Chapter 21</ref>. However, a quantum-mechanical version of the TST was developed and showed good agreement with rate constant values for processes which involved tunneling.<ref name=":0" />.

==Part 2ː F - H - H system==

=== PES Inspection ===

==== Classifying Energetics of Reactions ====

{{fontcolor1|blue|Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?}}

F + H2 → FH + Hː

[[File:Aoz15 F plus H2.PNG|thumb|centre|400px|Figure 4ː Surface plot for the F + H2 → FH + H reaction, showing that the products are at lower energy than the reactants.]]

From this figure it can be seen that the products are lower in energy than the reactants, hence the reaction is exothermic and overall heat is released.

H + HF → H2 + Fː

[[File:Aoz15 FH plus H.PNG|thumb|centre|400px|Figure 5ː Surface plot for the H + HF → H2 + F reaction, showing that the products are at higher energy than the reactants.]]

Conversely, in this case the products are higher in potential energy, and hence the reaction is endothermic.

The following values are the literature bond enthalpies for the H-H and H-F bondsː

H-Hː 435 kJ/mol <ref name=":1">M. H. Stans, ''Bond Dissociation Energies in Simple Molecules'', Washington D.C: National Bureau of Standards, 1970, 28-32</ref>
H-Fː 569 kJ/mol <ref name=":1" />

Using these bond enthalpies, the overall enthalpy change can be found for both reactionsː

F + H2 → FH + Hː 435 - 569 = -134 kJ/mol
H + HF → H2 + Fː 569 - 435 = 134 kJ/mol

This confirms what the potential energy plots showed; that the first reaction is exothermic (overall heat is released) and that the second is endothermic (overall heat is taken in). This is to be expected, since the H-F bond is stronger than the H-H bond, and hence forming the H-F bond is more favourable and results in a larger energy release.

==== Finding Transition State Positions ====

{{fontcolor1|blue|Locate the approximate position of the transition state.}}

The transition state was found by initially running the trajectory with no momentum and arbitrary start bond lengths of 1 (see figure 6 below). At this point, the data cursor was used to find the A-B bond length at the deepest point in the valley, which is also shown in the figure 6 below. This value for the A-B bond length was found to be 0.74 Angstrom. After this, the trajectories were run with A-B bond length as 0.74 and the B-C bond length was varied until the trajectory was a single point, which was taken to be the location of the transition state. This process of trial and error is shown in figure 7-9 below. Not every single run is shown so as not to take up too much space. From this analysis, the following bond lengths were foundː

H-Hː 0.740 Angstrom

H-Fː 1.814 Angstrom

[[File:Aoz15 finding TS 1.PNG|thumb|left|500px|Figure 6ː First plot with no momenta and arbitrary bond lengths. Data cursor shown used to find A-B bond length.]] [[File:Aoz15 finding TS 2.PNG|thumb|centre|500px|Figure 7ː First stage of trial and error in finding B-C bond length.]]

[[File:Aoz15 finding TS 3.PNG|thumb|left|500px|Figure 8ː Second stage of trial and error in finding B-C bond length.]] [[File:Aoz15 finding TS 4.PNG|thumb|centre|500px|Figure 9ː Final surface plot for finding B-C bond length at transition state.]]

==== Finding Activation Energies ====

{{fontcolor1|blue|Report the activation energy for both reactions.}}

The activation energy Ea can be found by finding the energy difference between the reactants and the transition state for each reaction. From the previous part, the energy of the transition state can be found by plotting the potential energy versus time, as shown in figure 10ː

[[File:Aoz15 TS energy.PNG|thumb|centre|600px|Figure 10ː Potential energy versus time plot for the transition state position found in previous section]]

From the figure above, it can be seen that the energy of the transition state is -103.75 kcal/mol

For H-F formation, the minimum energy of the reactants is -104.02 kcal/mol, as seen in the figure below. Hence the activation energy is -103.75 - -104.02 = 0.27 kcal/mol

[[File:Aoz15 reactants energy.PNG|thumb|centre|600px|Figure 11ː Potential energy versus time plot to find activation energy for H-F formation]]

For H-H formation, the minimum energy of the reactants is -133.97, as seen in the figure below. Hence the activation energy is -103.75 - -133.97 = 30.22 kcal/mol

[[File:Aoz15 products1 energy.PNG|thumb|left|600px|Figure 12ː Potential energy versus time plot to find activation energy for H-H formation]] [[File:Aoz15 products2 energy.PNG|thumb|centre|600px|Figure 13ː Potential energy versus time plot to find activation energy for H-F formation, zoomed in to allow finding of minimum energy value.]]

=== Reaction Dynamics ===

==== Released Reaction Energy Mechanism Discussion ====

{{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

The figure below shows a reactive trajectory for the F + H2 → FH + Hː reactionː

[[File:Aoz15 successful reaction.PNG|thumb|centre|600px|Figure 14ː Surface plot showing a reactive trajectory for the F + H2 → FH + H reaction.]]

From the surface plot in figure 14 above, it can be seen that the potential energy of the reactants is higher than that of the reactants. This is to be expected since it was earlier determined that the reaction is exothermic. As of such, the potential energy is converted into other forms upon reaction. Due to this simulated reaction occurring with the F atom approaching the H2 molecule from exactly 180°, conversion into rotational energy can be discounted in this scenario. Therefore, the energy is converted into vibrational and translational energy. The conversion into translational energy is hard to discern from the plot above, but the increase in vibrational energy can be seen from the increased size of oscillations about the minimum energy path for the products compared to the reactants.

Experimentally, any heat released (thermal energy) could be measured via calorimetry. Performing IR spectroscopy with a vibrationally unexcited H-F molecule and separately with the vibrationally excited H-F obtained from the reaction would allow the comparison of the peaks, giving the vibrational energy shift between the two.

==== Polanyi's Rules Discussion ====

{{fontcolor|blue|Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.}}

Polanyi's rules state that a reaction with a late potential energy barrier (one in which the transition state resembles the products) is more efficiently promoted by vibrational energy than translational energy and that the converse is true for a reaction with an early potential energy barrier (one in which the transition state resembles the reactants)<ref>Z. Zhang, Y. Zhou, D.H. Zhang, ''Theoretical Study of the Validity of the Polanyi Rules for the LateBarrier Cl + CHD3 Reaction'', The Journal of Physical Chemistry, 2012, 3416</ref>.

Exothermic reactions undergo an early transition state, hence via Polanyi's rules would be expected to be more efficiently promoted by translational energy. This would suggest that the F + H2 → FH + H reaction would be promoted more so by translational energy than vibrational. Figures 15 and 16 below show successful and unsuccessful cases confirming Polanyi's rules for this reactionː

[[File:Aoz15 1st polanyi proof.PNG|thumb|left|500px|Figure 15ː Surface plot showing successful F + H2 → FH + H reaction with low initial vibrational energy.]] [[File:Aoz15 2nd polanyi proof.PNG|thumb|centre|500px|Figure 16ː Surface plot showing unsuccessful F + H2 → FH + H reaction with high initial vibrational energy.]]

Conversely, endothermic reactions have a late transition state, so the rules predicts that they will be more efficiently promoted by vibrational energy. This would suggest that the H + HF → H2 + F reaction would be promoted more so by vibrational energy than translational. Figures 17 and 18 below show successful and unsuccessful cases which confirm Polanyi's rules for this reactionː

[[File:Aoz15 3rd polanyi proof.PNG|thumb|left|500px|Figure 17ː Surface plot showing unsuccessful H + HF → H2 + F reaction with low initial vibrational energy.]] [[File:Aoz15 4th polanyi proof.PNG|thumb|centre|500px|Figure 18ː Surface plot showing successful H + HF → H2 + F reaction with high initial vibrational energy.]]

However, Polanyi's rules are not always obeyed, and sometimes act only as a guideline. For instance, figure 19 below shows a successful H + HF → H2 + F reaction with low initial vibrational energy.

[[File:Aoz15 5th polanyi proof.PNG|thumb|centre|600px|Figure 19ː Surface plot showing successful H + HF → H2 + F reaction with low initial vibrational energy.]]

What these figures all together show is that Polanyi's rules can be a good guideline for correct conditions for reactions, since they refer to the efficiency of reactions. I.eː if many reactions with early transition states are run, having high translational energy is likely to result in more successful reactions. The converse is true for late transition states and vibrational energy.

==References==

MRD:aoz15page

2017-05-31T15:33:44Z

Je714: /* Locating the Transition State */

==Part 1ː H + H2 system==

=== Dynamics from the Transition State Region ===

{{fontcolor|blue|What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

The total gradient of a potential energy surface has a value of zero both at minimums and at transition structures. Note, for a function of two variables, having a total gradient of zero refers to having a partial derivative of zero with respect to both of its variable separately. i.eː a function f(x,y) will have a stationary point at a location where fx = 0 and fy = 0. This is due to both being stationary points for a function of two variables. However, they are different kinds of stationary points, with the minimum being a minimum and the transition structure a saddle point. Thus if a trajectory is started with no momentum at a position near the minimum point, it will end at the minimum point. However, if a trajectory with no initial momentum is started near the transition structure position, it will end up at either the reactants or products position. Hence this test can be used to determine whether a given point is a minimum or a saddle point for a function of two variables.

{{fontcolor1|gray| I was really enjoying your explanation until the very last part! The point about starting trajectories near the TS is a bit ''ad hoc''. It's a good analogy (and not factually wrong), but not really a rigorous explanation. Since you were correctly discussing total gradients and first partial derivatives, why not extend that to second partial derivatives? Those have information about the curvature of a surface and are actually what's needed to give a mathematically rigorous definition of the TS. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:33, 31 May 2017 (BST)}}

=== Locating the Transition State ===

{{fontcolor|blue|Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.}}

If the collision is taken to be between an atom A and a diatom BC, the transition state occurs when the bond between AB is midway through forming and that between BC is midway through breaking. At this point the system is at maximum potential energy. For a collision H with H2, the potential surface is symmetric, and so the transition state occurs when rAB = rBC. The transition state can thus be located by finding initial distances rAB and rBC such that, when the particles have no initial momentum, their internuclear distances remain constant. Through altering initial separations, an estimate for rrts of 0.908 Angstrom was made. The screenshot for the Internuclear Distances vs Time for this value is shown below (figure 1). It shows that the distances are constant over time. {{fontcolor1|gray| Good. Why are they constant, though? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:33, 31 May 2017 (BST)}}

[[File:Aoz15 transition state position estimate int dist v time.PNG|thumb|centre|600px|Figure 1ː Internuclear distance versus time plot for trajectory used to locate position of transition state.]]

=== Calculating Reaction Path ===

{{fontcolor1|blue|Comment on how the ''mep'' and the trajectory you just calculated differ.}}

The minimum energy path (mep) was found by starting with both initial momenta set to zero, rAB = rrts + 0.01 and rBC = rrts. This results in a trajectory that follows the valley floor of the surface contour to the products. When these initial conditions were run with the MEP calculation type, the following surface plot was obtainedː

[[File:Aoz15 mep surface plot.PNG|thumb|centre|500px|Figure 2ː Surface plot obtained for mep calculation.]]

As expected, this simply shows a smooth trajectory exactly down the valley floor to the products. However, when these same conditions were run with the dynamics calculation type, the trajectory differed and the following plot was producedː

[[File:Aoz15 meptrajectory surface plot.PNG|thumb|centre|500px|Figure 3ː Surface plot for the same initial conditions as for the mep, but performed with the dynamics calculation setting.]]

The mep and trajectory obtained differ in that only the trajectory takes into account vibrational motion; in the mep obtained, the energy stays exactly as the minimum energy of the valley and hence is a smooth line to the products. However, the trajectory takes into account vibrational motion, hence the energy oscillates about the minimum energy whilst reaching the products. This occurs due to the vibrational energy in the breaking/forming bonds, and explains the shape of the line seen in figure 3 above.

If the initial conditions rAB = rrts and rBC = rrts + 0.01 are used, the trajectory still follows the valley floor. However, it ends up at the reactants instead of the products, since it effectively 'rolls' the other way to find an energy minimum.

=== Reactive and Unreactive Trajectories ===

{{fontcolor1|blue|Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.}}

For initial bond lengths of rAB = 0.74 and rBC = 2.0, the initial momenta were varied and surface plots were found in order to determine which conditions led to reactive trajectories. The results and some discussion are found in the following table.

{| class="wikitable" border="1"
|+
! Surface Plot !! p1!! p2!!Reactive or Unreactive and Brief Discussion
|-
| [[File:Aoz15 reactive or not 1.PNG|centre|400px]] || -1.25 ||-2.5||Reactive, which can be seen by the trajectory passing through the transition state and going to the products.
|-
| [[File:Aoz15 reactive or not 2.PNG|centre|400px]] || -1.5||-2.0||Unreactive, since the path does not successfully pass through the transition state, instead ending at the reactants.
|-
| [[File:Aoz15 reactive or not 3.PNG|centre|400px]] || -1.5||-2.5||Reactive, which can once again be seen by the trajectory passing through the transition state and going to the products. Only difference to the first reaction is a slight increase in overall vibrational energy, which can be seen by greater oscillations in the trajectory.
|-
| [[File:Aoz15 reactive or not 4.PNG|centre|400px]] || -2.5||-5.0||Unreactive, which can be seen by the trajectory ending at the reactants. However, in this case it does pass through the transition state. Therefore, this is a case of barrier recrossing.
|-
| [[File:Aoz15 reactive or not 5.PNG|centre|400px]] || -2.5||-5.2||Reactive, as the reaction proceeds to the products. Here barrier recrossing also occurs, but unlike in the fourth reaction, in this case the reaction still proceeds to products. Hence, multiple barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|blue|State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

There are three main assumptions in Transition State Theory (TST). Before stating them, it is useful to define the terms critical diving surface and activated complex. For a given reaction potential energy surface, the critical diving surface refers to a boundary surface which passes through the surface saddle point. As such, the critical diving surface must be crossed in order for a reaction to go to completion. Any molecule whose nuclear configuration matches a point on the critical diving surface or to any point within a short distance of it is termed an activated complex. So, the saddle point of the potential surface corresponds to the equilibrium structure of the activated complex. With these terms defined, the following are the main assumptions of Transition State Theoryː firstly, that all configurations of molecules which cross the critical diving surface from the reactants side will proceed to become products. Secondly, that all reactant molecules maintain a Boltzmann-distributed energy distribution during the reaction. This leads to the third and final assumption; that the energy distribution of all activated complexes during a reaction will also be a Boltzmann distribution which depends on the system temperature.<ref name=":0">Levine, I. N., ''Physical Chemistry'', McGraw-Hill Publishing, Sixth Edition, 2009, 887-894</ref>

When tested against experimentally obtained values, those found by TST are in most cases within a single order of magnitude, suggesting that the assumptions made are well placed<ref name=":0" />. However, there are certain aspects of reactions which TST doesn't account for. The first is barrier recrossing; since TST assumes that all molecular configurations which pass the critical dividing surface proceed to products, it cannot consider barrier recrossing, the process in which the reaction trajectory crosses the potential barrier but then recrosses it in the opposite direction. As in the fourth simulation above, sometimes barrier recrossing leads to the molecules returning to the reactants and the products not forming at all. In such a case, TST wildly misinterprets the reaction result. Even in cases where barrier recrossing occurs multiple times and the products are formed (as in the fifth simulation above), TST gives a higher reaction rate value than is experimentally obtained, since it assumes the reaction proceeds to completion as soon as the potential barrier is crossed. TST also doesn't take into account quantum effects, such as tunneling. Tunneling occurs when reactant molecules with insufficient energy to bypass the potential barrier pass through the barrier, leading to an exponential decay in their energy. This process occurs thanks to wave-particle duality, and cannot be predicted by TST<ref>P. Atkins, J. de Paula, ''Atkins Physical Chemistry'', Oxford University Press, 10th Edition, Chapter 21</ref>. However, a quantum-mechanical version of the TST was developed and showed good agreement with rate constant values for processes which involved tunneling.<ref name=":0" />.

==Part 2ː F - H - H system==

=== PES Inspection ===

==== Classifying Energetics of Reactions ====

{{fontcolor1|blue|Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?}}

F + H2 → FH + Hː

[[File:Aoz15 F plus H2.PNG|thumb|centre|400px|Figure 4ː Surface plot for the F + H2 → FH + H reaction, showing that the products are at lower energy than the reactants.]]

From this figure it can be seen that the products are lower in energy than the reactants, hence the reaction is exothermic and overall heat is released.

H + HF → H2 + Fː

[[File:Aoz15 FH plus H.PNG|thumb|centre|400px|Figure 5ː Surface plot for the H + HF → H2 + F reaction, showing that the products are at higher energy than the reactants.]]

Conversely, in this case the products are higher in potential energy, and hence the reaction is endothermic.

The following values are the literature bond enthalpies for the H-H and H-F bondsː

H-Hː 435 kJ/mol <ref name=":1">M. H. Stans, ''Bond Dissociation Energies in Simple Molecules'', Washington D.C: National Bureau of Standards, 1970, 28-32</ref>
H-Fː 569 kJ/mol <ref name=":1" />

Using these bond enthalpies, the overall enthalpy change can be found for both reactionsː

F + H2 → FH + Hː 435 - 569 = -134 kJ/mol
H + HF → H2 + Fː 569 - 435 = 134 kJ/mol

This confirms what the potential energy plots showed; that the first reaction is exothermic (overall heat is released) and that the second is endothermic (overall heat is taken in). This is to be expected, since the H-F bond is stronger than the H-H bond, and hence forming the H-F bond is more favourable and results in a larger energy release.

==== Finding Transition State Positions ====

{{fontcolor1|blue|Locate the approximate position of the transition state.}}

The transition state was found by initially running the trajectory with no momentum and arbitrary start bond lengths of 1 (see figure 6 below). At this point, the data cursor was used to find the A-B bond length at the deepest point in the valley, which is also shown in the figure 6 below. This value for the A-B bond length was found to be 0.74 Angstrom. After this, the trajectories were run with A-B bond length as 0.74 and the B-C bond length was varied until the trajectory was a single point, which was taken to be the location of the transition state. This process of trial and error is shown in figure 7-9 below. Not every single run is shown so as not to take up too much space. From this analysis, the following bond lengths were foundː

H-Hː 0.740 Angstrom

H-Fː 1.814 Angstrom

[[File:Aoz15 finding TS 1.PNG|thumb|left|500px|Figure 6ː First plot with no momenta and arbitrary bond lengths. Data cursor shown used to find A-B bond length.]] [[File:Aoz15 finding TS 2.PNG|thumb|centre|500px|Figure 7ː First stage of trial and error in finding B-C bond length.]]

[[File:Aoz15 finding TS 3.PNG|thumb|left|500px|Figure 8ː Second stage of trial and error in finding B-C bond length.]] [[File:Aoz15 finding TS 4.PNG|thumb|centre|500px|Figure 9ː Final surface plot for finding B-C bond length at transition state.]]

==== Finding Activation Energies ====

{{fontcolor1|blue|Report the activation energy for both reactions.}}

The activation energy Ea can be found by finding the energy difference between the reactants and the transition state for each reaction. From the previous part, the energy of the transition state can be found by plotting the potential energy versus time, as shown in figure 10ː

[[File:Aoz15 TS energy.PNG|thumb|centre|600px|Figure 10ː Potential energy versus time plot for the transition state position found in previous section]]

From the figure above, it can be seen that the energy of the transition state is -103.75 kcal/mol

For H-F formation, the minimum energy of the reactants is -104.02 kcal/mol, as seen in the figure below. Hence the activation energy is -103.75 - -104.02 = 0.27 kcal/mol

[[File:Aoz15 reactants energy.PNG|thumb|centre|600px|Figure 11ː Potential energy versus time plot to find activation energy for H-F formation]]

For H-H formation, the minimum energy of the reactants is -133.97, as seen in the figure below. Hence the activation energy is -103.75 - -133.97 = 30.22 kcal/mol

[[File:Aoz15 products1 energy.PNG|thumb|left|600px|Figure 12ː Potential energy versus time plot to find activation energy for H-H formation]] [[File:Aoz15 products2 energy.PNG|thumb|centre|600px|Figure 13ː Potential energy versus time plot to find activation energy for H-F formation, zoomed in to allow finding of minimum energy value.]]

=== Reaction Dynamics ===

==== Released Reaction Energy Mechanism Discussion ====

{{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

The figure below shows a reactive trajectory for the F + H2 → FH + Hː reactionː

[[File:Aoz15 successful reaction.PNG|thumb|centre|600px|Figure 14ː Surface plot showing a reactive trajectory for the F + H2 → FH + H reaction.]]

From the surface plot in figure 14 above, it can be seen that the potential energy of the reactants is higher than that of the reactants. This is to be expected since it was earlier determined that the reaction is exothermic. As of such, the potential energy is converted into other forms upon reaction. Due to this simulated reaction occurring with the F atom approaching the H2 molecule from exactly 180°, conversion into rotational energy can be discounted in this scenario. Therefore, the energy is converted into vibrational and translational energy. The conversion into translational energy is hard to discern from the plot above, but the increase in vibrational energy can be seen from the increased size of oscillations about the minimum energy path for the products compared to the reactants.

Experimentally, any heat released (thermal energy) could be measured via calorimetry. Performing IR spectroscopy with a vibrationally unexcited H-F molecule and separately with the vibrationally excited H-F obtained from the reaction would allow the comparison of the peaks, giving the vibrational energy shift between the two.

==== Polanyi's Rules Discussion ====

{{fontcolor|blue|Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.}}

Polanyi's rules state that a reaction with a late potential energy barrier (one in which the transition state resembles the products) is more efficiently promoted by vibrational energy than translational energy and that the converse is true for a reaction with an early potential energy barrier (one in which the transition state resembles the reactants)<ref>Z. Zhang, Y. Zhou, D.H. Zhang, ''Theoretical Study of the Validity of the Polanyi Rules for the LateBarrier Cl + CHD3 Reaction'', The Journal of Physical Chemistry, 2012, 3416</ref>.

Exothermic reactions undergo an early transition state, hence via Polanyi's rules would be expected to be more efficiently promoted by translational energy. This would suggest that the F + H2 → FH + H reaction would be promoted more so by translational energy than vibrational. Figures 15 and 16 below show successful and unsuccessful cases confirming Polanyi's rules for this reactionː

[[File:Aoz15 1st polanyi proof.PNG|thumb|left|500px|Figure 15ː Surface plot showing successful F + H2 → FH + H reaction with low initial vibrational energy.]] [[File:Aoz15 2nd polanyi proof.PNG|thumb|centre|500px|Figure 16ː Surface plot showing unsuccessful F + H2 → FH + H reaction with high initial vibrational energy.]]

Conversely, endothermic reactions have a late transition state, so the rules predicts that they will be more efficiently promoted by vibrational energy. This would suggest that the H + HF → H2 + F reaction would be promoted more so by vibrational energy than translational. Figures 17 and 18 below show successful and unsuccessful cases which confirm Polanyi's rules for this reactionː

[[File:Aoz15 3rd polanyi proof.PNG|thumb|left|500px|Figure 17ː Surface plot showing unsuccessful H + HF → H2 + F reaction with low initial vibrational energy.]] [[File:Aoz15 4th polanyi proof.PNG|thumb|centre|500px|Figure 18ː Surface plot showing successful H + HF → H2 + F reaction with high initial vibrational energy.]]

However, Polanyi's rules are not always obeyed, and sometimes act only as a guideline. For instance, figure 19 below shows a successful H + HF → H2 + F reaction with low initial vibrational energy.

[[File:Aoz15 5th polanyi proof.PNG|thumb|centre|600px|Figure 19ː Surface plot showing successful H + HF → H2 + F reaction with low initial vibrational energy.]]

What these figures all together show is that Polanyi's rules can be a good guideline for correct conditions for reactions, since they refer to the efficiency of reactions. I.eː if many reactions with early transition states are run, having high translational energy is likely to result in more successful reactions. The converse is true for late transition states and vibrational energy.

==References==

MRD:aoz15page

2017-05-31T15:33:01Z

Je714: /* Dynamics from the Transition State Region */

==Part 1ː H + H2 system==

=== Dynamics from the Transition State Region ===

{{fontcolor|blue|What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

The total gradient of a potential energy surface has a value of zero both at minimums and at transition structures. Note, for a function of two variables, having a total gradient of zero refers to having a partial derivative of zero with respect to both of its variable separately. i.eː a function f(x,y) will have a stationary point at a location where fx = 0 and fy = 0. This is due to both being stationary points for a function of two variables. However, they are different kinds of stationary points, with the minimum being a minimum and the transition structure a saddle point. Thus if a trajectory is started with no momentum at a position near the minimum point, it will end at the minimum point. However, if a trajectory with no initial momentum is started near the transition structure position, it will end up at either the reactants or products position. Hence this test can be used to determine whether a given point is a minimum or a saddle point for a function of two variables.

{{fontcolor1|gray| I was really enjoying your explanation until the very last part! The point about starting trajectories near the TS is a bit ''ad hoc''. It's a good analogy (and not factually wrong), but not really a rigorous explanation. Since you were correctly discussing total gradients and first partial derivatives, why not extend that to second partial derivatives? Those have information about the curvature of a surface and are actually what's needed to give a mathematically rigorous definition of the TS. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:33, 31 May 2017 (BST)}}

=== Locating the Transition State ===

{{fontcolor|blue|Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.}}

If the collision is taken to be between an atom A and a diatom BC, the transition state occurs when the bond between AB is midway through forming and that between BC is midway through breaking. At this point the system is at maximum potential energy. For a collision H with H2, the potential surface is symmetric, and so the transition state occurs when rAB = rBC. The transition state can thus be located by finding initial distances rAB and rBC such that, when the particles have no initial momentum, their internuclear distances remain constant. Through altering initial separations, an estimate for rrts of 0.908 Angstrom was made. The screenshot for the Internuclear Distances vs Time for this value is shown below (figure 1). It shows that the distances are constant over time.

[[File:Aoz15 transition state position estimate int dist v time.PNG|thumb|centre|600px|Figure 1ː Internuclear distance versus time plot for trajectory used to locate position of transition state.]]

=== Calculating Reaction Path ===

{{fontcolor1|blue|Comment on how the ''mep'' and the trajectory you just calculated differ.}}

The minimum energy path (mep) was found by starting with both initial momenta set to zero, rAB = rrts + 0.01 and rBC = rrts. This results in a trajectory that follows the valley floor of the surface contour to the products. When these initial conditions were run with the MEP calculation type, the following surface plot was obtainedː

[[File:Aoz15 mep surface plot.PNG|thumb|centre|500px|Figure 2ː Surface plot obtained for mep calculation.]]

As expected, this simply shows a smooth trajectory exactly down the valley floor to the products. However, when these same conditions were run with the dynamics calculation type, the trajectory differed and the following plot was producedː

[[File:Aoz15 meptrajectory surface plot.PNG|thumb|centre|500px|Figure 3ː Surface plot for the same initial conditions as for the mep, but performed with the dynamics calculation setting.]]

The mep and trajectory obtained differ in that only the trajectory takes into account vibrational motion; in the mep obtained, the energy stays exactly as the minimum energy of the valley and hence is a smooth line to the products. However, the trajectory takes into account vibrational motion, hence the energy oscillates about the minimum energy whilst reaching the products. This occurs due to the vibrational energy in the breaking/forming bonds, and explains the shape of the line seen in figure 3 above.

If the initial conditions rAB = rrts and rBC = rrts + 0.01 are used, the trajectory still follows the valley floor. However, it ends up at the reactants instead of the products, since it effectively 'rolls' the other way to find an energy minimum.

=== Reactive and Unreactive Trajectories ===

{{fontcolor1|blue|Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.}}

For initial bond lengths of rAB = 0.74 and rBC = 2.0, the initial momenta were varied and surface plots were found in order to determine which conditions led to reactive trajectories. The results and some discussion are found in the following table.

{| class="wikitable" border="1"
|+
! Surface Plot !! p1!! p2!!Reactive or Unreactive and Brief Discussion
|-
| [[File:Aoz15 reactive or not 1.PNG|centre|400px]] || -1.25 ||-2.5||Reactive, which can be seen by the trajectory passing through the transition state and going to the products.
|-
| [[File:Aoz15 reactive or not 2.PNG|centre|400px]] || -1.5||-2.0||Unreactive, since the path does not successfully pass through the transition state, instead ending at the reactants.
|-
| [[File:Aoz15 reactive or not 3.PNG|centre|400px]] || -1.5||-2.5||Reactive, which can once again be seen by the trajectory passing through the transition state and going to the products. Only difference to the first reaction is a slight increase in overall vibrational energy, which can be seen by greater oscillations in the trajectory.
|-
| [[File:Aoz15 reactive or not 4.PNG|centre|400px]] || -2.5||-5.0||Unreactive, which can be seen by the trajectory ending at the reactants. However, in this case it does pass through the transition state. Therefore, this is a case of barrier recrossing.
|-
| [[File:Aoz15 reactive or not 5.PNG|centre|400px]] || -2.5||-5.2||Reactive, as the reaction proceeds to the products. Here barrier recrossing also occurs, but unlike in the fourth reaction, in this case the reaction still proceeds to products. Hence, multiple barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|blue|State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

There are three main assumptions in Transition State Theory (TST). Before stating them, it is useful to define the terms critical diving surface and activated complex. For a given reaction potential energy surface, the critical diving surface refers to a boundary surface which passes through the surface saddle point. As such, the critical diving surface must be crossed in order for a reaction to go to completion. Any molecule whose nuclear configuration matches a point on the critical diving surface or to any point within a short distance of it is termed an activated complex. So, the saddle point of the potential surface corresponds to the equilibrium structure of the activated complex. With these terms defined, the following are the main assumptions of Transition State Theoryː firstly, that all configurations of molecules which cross the critical diving surface from the reactants side will proceed to become products. Secondly, that all reactant molecules maintain a Boltzmann-distributed energy distribution during the reaction. This leads to the third and final assumption; that the energy distribution of all activated complexes during a reaction will also be a Boltzmann distribution which depends on the system temperature.<ref name=":0">Levine, I. N., ''Physical Chemistry'', McGraw-Hill Publishing, Sixth Edition, 2009, 887-894</ref>

When tested against experimentally obtained values, those found by TST are in most cases within a single order of magnitude, suggesting that the assumptions made are well placed<ref name=":0" />. However, there are certain aspects of reactions which TST doesn't account for. The first is barrier recrossing; since TST assumes that all molecular configurations which pass the critical dividing surface proceed to products, it cannot consider barrier recrossing, the process in which the reaction trajectory crosses the potential barrier but then recrosses it in the opposite direction. As in the fourth simulation above, sometimes barrier recrossing leads to the molecules returning to the reactants and the products not forming at all. In such a case, TST wildly misinterprets the reaction result. Even in cases where barrier recrossing occurs multiple times and the products are formed (as in the fifth simulation above), TST gives a higher reaction rate value than is experimentally obtained, since it assumes the reaction proceeds to completion as soon as the potential barrier is crossed. TST also doesn't take into account quantum effects, such as tunneling. Tunneling occurs when reactant molecules with insufficient energy to bypass the potential barrier pass through the barrier, leading to an exponential decay in their energy. This process occurs thanks to wave-particle duality, and cannot be predicted by TST<ref>P. Atkins, J. de Paula, ''Atkins Physical Chemistry'', Oxford University Press, 10th Edition, Chapter 21</ref>. However, a quantum-mechanical version of the TST was developed and showed good agreement with rate constant values for processes which involved tunneling.<ref name=":0" />.

==Part 2ː F - H - H system==

=== PES Inspection ===

==== Classifying Energetics of Reactions ====

{{fontcolor1|blue|Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?}}

F + H2 → FH + Hː

[[File:Aoz15 F plus H2.PNG|thumb|centre|400px|Figure 4ː Surface plot for the F + H2 → FH + H reaction, showing that the products are at lower energy than the reactants.]]

From this figure it can be seen that the products are lower in energy than the reactants, hence the reaction is exothermic and overall heat is released.

H + HF → H2 + Fː

[[File:Aoz15 FH plus H.PNG|thumb|centre|400px|Figure 5ː Surface plot for the H + HF → H2 + F reaction, showing that the products are at higher energy than the reactants.]]

Conversely, in this case the products are higher in potential energy, and hence the reaction is endothermic.

The following values are the literature bond enthalpies for the H-H and H-F bondsː

H-Hː 435 kJ/mol <ref name=":1">M. H. Stans, ''Bond Dissociation Energies in Simple Molecules'', Washington D.C: National Bureau of Standards, 1970, 28-32</ref>
H-Fː 569 kJ/mol <ref name=":1" />

Using these bond enthalpies, the overall enthalpy change can be found for both reactionsː

F + H2 → FH + Hː 435 - 569 = -134 kJ/mol
H + HF → H2 + Fː 569 - 435 = 134 kJ/mol

This confirms what the potential energy plots showed; that the first reaction is exothermic (overall heat is released) and that the second is endothermic (overall heat is taken in). This is to be expected, since the H-F bond is stronger than the H-H bond, and hence forming the H-F bond is more favourable and results in a larger energy release.

==== Finding Transition State Positions ====

{{fontcolor1|blue|Locate the approximate position of the transition state.}}

The transition state was found by initially running the trajectory with no momentum and arbitrary start bond lengths of 1 (see figure 6 below). At this point, the data cursor was used to find the A-B bond length at the deepest point in the valley, which is also shown in the figure 6 below. This value for the A-B bond length was found to be 0.74 Angstrom. After this, the trajectories were run with A-B bond length as 0.74 and the B-C bond length was varied until the trajectory was a single point, which was taken to be the location of the transition state. This process of trial and error is shown in figure 7-9 below. Not every single run is shown so as not to take up too much space. From this analysis, the following bond lengths were foundː

H-Hː 0.740 Angstrom

H-Fː 1.814 Angstrom

[[File:Aoz15 finding TS 1.PNG|thumb|left|500px|Figure 6ː First plot with no momenta and arbitrary bond lengths. Data cursor shown used to find A-B bond length.]] [[File:Aoz15 finding TS 2.PNG|thumb|centre|500px|Figure 7ː First stage of trial and error in finding B-C bond length.]]

[[File:Aoz15 finding TS 3.PNG|thumb|left|500px|Figure 8ː Second stage of trial and error in finding B-C bond length.]] [[File:Aoz15 finding TS 4.PNG|thumb|centre|500px|Figure 9ː Final surface plot for finding B-C bond length at transition state.]]

==== Finding Activation Energies ====

{{fontcolor1|blue|Report the activation energy for both reactions.}}

The activation energy Ea can be found by finding the energy difference between the reactants and the transition state for each reaction. From the previous part, the energy of the transition state can be found by plotting the potential energy versus time, as shown in figure 10ː

[[File:Aoz15 TS energy.PNG|thumb|centre|600px|Figure 10ː Potential energy versus time plot for the transition state position found in previous section]]

From the figure above, it can be seen that the energy of the transition state is -103.75 kcal/mol

For H-F formation, the minimum energy of the reactants is -104.02 kcal/mol, as seen in the figure below. Hence the activation energy is -103.75 - -104.02 = 0.27 kcal/mol

[[File:Aoz15 reactants energy.PNG|thumb|centre|600px|Figure 11ː Potential energy versus time plot to find activation energy for H-F formation]]

For H-H formation, the minimum energy of the reactants is -133.97, as seen in the figure below. Hence the activation energy is -103.75 - -133.97 = 30.22 kcal/mol

[[File:Aoz15 products1 energy.PNG|thumb|left|600px|Figure 12ː Potential energy versus time plot to find activation energy for H-H formation]] [[File:Aoz15 products2 energy.PNG|thumb|centre|600px|Figure 13ː Potential energy versus time plot to find activation energy for H-F formation, zoomed in to allow finding of minimum energy value.]]

=== Reaction Dynamics ===

==== Released Reaction Energy Mechanism Discussion ====

{{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

The figure below shows a reactive trajectory for the F + H2 → FH + Hː reactionː

[[File:Aoz15 successful reaction.PNG|thumb|centre|600px|Figure 14ː Surface plot showing a reactive trajectory for the F + H2 → FH + H reaction.]]

From the surface plot in figure 14 above, it can be seen that the potential energy of the reactants is higher than that of the reactants. This is to be expected since it was earlier determined that the reaction is exothermic. As of such, the potential energy is converted into other forms upon reaction. Due to this simulated reaction occurring with the F atom approaching the H2 molecule from exactly 180°, conversion into rotational energy can be discounted in this scenario. Therefore, the energy is converted into vibrational and translational energy. The conversion into translational energy is hard to discern from the plot above, but the increase in vibrational energy can be seen from the increased size of oscillations about the minimum energy path for the products compared to the reactants.

Experimentally, any heat released (thermal energy) could be measured via calorimetry. Performing IR spectroscopy with a vibrationally unexcited H-F molecule and separately with the vibrationally excited H-F obtained from the reaction would allow the comparison of the peaks, giving the vibrational energy shift between the two.

==== Polanyi's Rules Discussion ====

{{fontcolor|blue|Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.}}

Polanyi's rules state that a reaction with a late potential energy barrier (one in which the transition state resembles the products) is more efficiently promoted by vibrational energy than translational energy and that the converse is true for a reaction with an early potential energy barrier (one in which the transition state resembles the reactants)<ref>Z. Zhang, Y. Zhou, D.H. Zhang, ''Theoretical Study of the Validity of the Polanyi Rules for the LateBarrier Cl + CHD3 Reaction'', The Journal of Physical Chemistry, 2012, 3416</ref>.

Exothermic reactions undergo an early transition state, hence via Polanyi's rules would be expected to be more efficiently promoted by translational energy. This would suggest that the F + H2 → FH + H reaction would be promoted more so by translational energy than vibrational. Figures 15 and 16 below show successful and unsuccessful cases confirming Polanyi's rules for this reactionː

[[File:Aoz15 1st polanyi proof.PNG|thumb|left|500px|Figure 15ː Surface plot showing successful F + H2 → FH + H reaction with low initial vibrational energy.]] [[File:Aoz15 2nd polanyi proof.PNG|thumb|centre|500px|Figure 16ː Surface plot showing unsuccessful F + H2 → FH + H reaction with high initial vibrational energy.]]

Conversely, endothermic reactions have a late transition state, so the rules predicts that they will be more efficiently promoted by vibrational energy. This would suggest that the H + HF → H2 + F reaction would be promoted more so by vibrational energy than translational. Figures 17 and 18 below show successful and unsuccessful cases which confirm Polanyi's rules for this reactionː

[[File:Aoz15 3rd polanyi proof.PNG|thumb|left|500px|Figure 17ː Surface plot showing unsuccessful H + HF → H2 + F reaction with low initial vibrational energy.]] [[File:Aoz15 4th polanyi proof.PNG|thumb|centre|500px|Figure 18ː Surface plot showing successful H + HF → H2 + F reaction with high initial vibrational energy.]]

However, Polanyi's rules are not always obeyed, and sometimes act only as a guideline. For instance, figure 19 below shows a successful H + HF → H2 + F reaction with low initial vibrational energy.

[[File:Aoz15 5th polanyi proof.PNG|thumb|centre|600px|Figure 19ː Surface plot showing successful H + HF → H2 + F reaction with low initial vibrational energy.]]

What these figures all together show is that Polanyi's rules can be a good guideline for correct conditions for reactions, since they refer to the efficiency of reactions. I.eː if many reactions with early transition states are run, having high translational energy is likely to result in more successful reactions. The converse is true for late transition states and vibrational energy.

==References==

Talk:MRD:CPYX

2017-05-31T15:28:27Z

Je714: Created page with "A very thorough report. Congratulations. ~~~~"

A very thorough report. Congratulations.

[[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:28, 31 May 2017 (BST)

MRD:CPYX

2017-05-31T15:25:24Z

Je714: /* Comparison of Transition State Theory predictions with experimental values */

== H + H2 system ==

=== Transition state region and position ===

The total gradient of the potential energy surface is 0 both at a minimum and at a transition structure. At a minimum, the gradient is negative before that point and positive after that point. The transition state is the maximum point on the minimum energy path, which means that it is at a relative minimum along 1 direction and at a relative maximum along the crossing direction. This point is known as a saddle point, as it is neither a minimum nor a maximum.

{{fontcolor1|gray| Good. Some discussion using second partial derivatives would have been nicer, though. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:17, 31 May 2017 (BST)}}

The transition state position rts is estimated to be 0.90775 Å (5.s.f.). When the initial conditions of r1 = r2 = 0.90775 Å and p1 = p2 = 0 are tested, the system does not seem to vibrate and the bond distances remain at 0.90775 Å until t=8.63, when atom C starts to move away from the newly-formed vibrating molecule AB. This is shown in the plot of internuclear distances against time below. The above initial conditions at time values of 0 < t < 8.63 also produced a trajectory where all the atoms remained at their initial position. This means that the trajectory only stays on the ridge when the bonds have zero momenta and this is the optimal distance between each of the 3 H atoms when they are all bonded to one another in the transition state.

{{fontcolor1|gray| The TS is a mathematical construct. In practice, you can always run a sufficiently long simulation in which the system will fall off towards the products or reactants. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:17, 31 May 2017 (BST)}}

[[File:CPYX_H2+H_TS_1.PNG|400px|thumb|center|"Internuclear Distances vs Time" plot for H2 + H]]

=== Comparison between the reaction path/mep and the trajectory using the same initial conditions ===

{{fontcolor1|gray| Thorough explanation, GJ. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:18, 31 May 2017 (BST)}}

A mep and a trajectory with its calculation type as Dynamics were run separately using the initial conditions of r1 = rBC = 0.91775 Å, r2 = rAB = 0.90775 Å and p1 = p2 = 0 and compared. All atoms A, B and C refer to H atoms and all internuclear distances, time and internuclear momenta values which do not have to be very accurate are reported to 2.d.p.

If the initial conditions r1 = rBC = 0.90775 Å and r2 = rAB = 0.91775 Å were used instead, the trajectory will be a reflection of the original trajectory in the diagonal connecting the origin and the corner corresponding to the maximum internuclear distance values. In the internuclear distance and internuclear momentum vs time plots, the lines for the new A-B bond will be the same as the lines for the original B-C bond and the lines for the new B-C bond will be the same as the lines for the original A-B bond. The shapes of the plots will otherwise be the same.

{| class="wikitable" border="1"
|+ Comparison between mep and trajectory
! Differences !! Mep (20 000 steps) !! Trajectory (300 steps)
|-
| rowspan="2" | Contour plot || [[File:CPYX_H2+H_contour_mep.PNG|500px]] || [[File:CPYX_H2+H_contour_1.PNG|500px]]
|-
| colspan="2" | In both plots above, the products of molecule AB and atom C are formed, and atom C moves away from molecule AB. The mep is much smoother than the trajectory due to the much greater vibrational energy of the new molecule AB in the trajectory.
|-
| Internuclear distance vs time plot || [[File:CPYX_H2+H_mep_internuclear_distance_vs_time.PNG|500px]] || [[File:CPYX_H2+H_trajectory_internuclear_distance_vs_time.PNG|500px]]
|-
| Explanation || The final internuclear distances recorded are r1(100)=2.53 Å and r2(100)=0.74 Å. The value of r1(100) is much smaller than the value of r1(1.5) in the trajectory, as the velocity in the mep is set to 0 in each time step, unlike the trajectory. This means that atom C has very little translational kinetic energy in the mep and moves away from molecule AB much more slowly. After t=3.00, the value of r1 increases with time and the increase in the value of r1 decreases with time, as the distance between atom C and molecule AB increases and there is less repulsion between their electrons. The value of r2 stays constant at 0.74 Å, which is the H - H bond length, and vibrations are not observed due to the zero velocity and kinetic energy in this H - H bond. || The final internuclear distances recorded are r1(1.5)=5.28 Å and r2 ranges from 0.73 Å to 0.76 Å at large values of t. The products have reasonable non-zero amounts of translational and vibrational energy due to the release of energy from the transition state structure. After t=0.40, the value of r1 increases linearly with time while the value of r2 stays approximately constant, averaging around 0.74 Å, which is the H - H bond length.
|-
| Internuclear momentum vs time plot || [[File:CPYX_H2+H_mep_internuclear_momenta_vs_time.PNG|500px]] || [[File:CPYX_H2+H_trajectory_internuclear_momenta_vs_time.PNG|500px]]
|-
| Explanation || When obtaining the mep, the velocity is set to 0 in each time step. Since momentum pi =μvi where μ is mass and v is velocity, the internuclear momenta at all time values is 0. || Initially, both internuclear momenta are 0 and become slightly negative as the 3 hydrogen atoms move closer together, at a velocity that decreases the atomic distance, to form a transition state structure. After the transition state structure forms, the AB internuclear momentum remains negative as the new molecule AB is formed and the internuclear distance rAB decreases to a value of around the H - H bond length. However, the BC internuclear momentum becomes positive after t=0.11 and increases steadily as atom C moves away from the new molecule AB, converging to a value of 2.48 at large values of t, so p1(1.5)=2.48. After the molecule AB is formed, the AB internuclear momentum increases to a positive value and oscillates between 0.91 and 1.57 at large values of t due to vibrations in the H - H bond. This means that the final average internuclear momenta p2(1.5)=1.24.
|}

A calculation with initial positions at the final positions of the trajectory obtained above and the initial momenta the same as the final momenta in the trajectory obtained above but with their signs reversed, was performed. This means that rAB, rBC, AB momentum and BC momentum were set to 0.74 Å, 5.28 Å, -1.24 and -2.48 respectively. The results of the calculation are shown below.

{| class="wikitable" border="1"
! Contour plot !! Internuclear distance vs time plot !! Internuclear momentum vs time plot (extended to 500 steps/ t=2.50)
|-
| [[File:CPYX_H2+H_contour_2.PNG|400px]] || [[File:CPYX_H2+H_trajectory_internuclear_distance_vs_time_reversed.PNG|400px]] || [[File:CPYX_H2+H_internuclear_distance_vs_time_reversed.PNG|400px]]
|-
| The shape of this plot is very similar to the shape of the earlier trajectory, except that this trajectory moves in the opposite direction and circles around the region of 0.70 Å < rAB < 0.90 Å and 0.90 Å < rBC < 1.10 Å. This trajectory is very smooth, as most of the energy in this system is translational energy. The value of rAB, which corresponds to the vibrational energy of the H - H bond in molecule AB, is 0.74 Å, which is the H - H bond length, so molecule AB has very little vibrational energy. This shape shows that the trajectory is unreactive and will revert back to reactants AB and C, as the initial conditions do not provide enough kinetic energy to overcome the activation barrier. The maximum amount of potential energy reached by this system is lower than the potential energy of the transition state structure.
|| The shape of this plot is roughly the reverse of the plot obtained with the earlier trajectory. The value of rBC decreases linearly with time while the value of rAB stays approximately constant with vibrations. From t=1.20 to t=1.50, a transition state structure between atom C and molecule AB is attempted but not formed as the maximum value of rAB of 0.87 Å is reached at t=1.37 and the minimum value of rBC of 0.96 Å is reached at t=1.33. These values are not close to one another as well as to at the value of rts which is 0.90775 Å. Thus, atom C moves away from the molecule AB which remains intact, so rBC increases and rAB decreases to its original value of 0.74 Å. || The internuclear momenta remain relatively constant initially and become less negative after t=0.5 as the 3 H atoms move closer together more slowly, with a less negative velocity, due to repulsions between their electrons. pAB becomes positive at t=1.24 as rAB increases in an attempt to form a transition state structure. pAB becomes negative again at t=1.39 as rAB decreases to its original value of 0.74 Å. pAB becomes positive again at t=1.51, after molecule AB is re-formed, and oscillates between 0.70 and 1.71 at large values of t due to H - H bond vibrations. pBC becomes positive at t=1.31 and continues to increase as rBC increases and atom C moves away from molecule AB.
|}

=== Reactive and unreactive trajectories ===

In the following table, the initial positions are r1 = rBC = 0.74 Å and r2 = rAB = 2.00 Å. The momentum values are recorded in the table and the reactivity of their trajectories is shown.

{| class="wikitable" border=1
! p1 !! p2 !! Screenshot of trajectory !! Internuclear distance vs time plot !! Reactive/unreactive
|-
| -1.25 || -2.5 || [[File:CPYX_H2+H_contour_trajectory_1.PNG|300px]] || [[File:CPYX_H2+H_trajectory_1.PNG|300px]] || Reactive
The trajectory and internuclear distance vs time plot show that atom A moves towards molecule BC and collides with it. A transition state structure is formed at t=0.43, where rAB = rBC ≈ 0.90 Å, which is close to the rts of 0.90755 Å found earlier. Even though the magnitudes of the internuclear momenta appear to be small, this system has sufficient kinetic energy to overcome the activation barrier. This is because the system is driven much more by translational energy than by vibrational energy, as shown in the smooth trajectory. A bond forms between atoms A and B and the B-C bond breaks. rAB decreases until the optimum bond distance of 0.74 Å is reached at t=0.58, and the molecule continues to vibrate slightly, within the contour lines. rBC increases almost linearly with time as atom C moves away from molecule AB at a constant velocity.
|-
| -1.5 || -2.0 || [[File:CPYX_H2+H_contour_trajectory_2.PNG|300px]] || [[File:CPYX_H2+H_trajectory_2.PNG|300px]]
|| Unreactive
These plots show atom A initially moving towards a slightly vibrating molecule BC. After a minimum internuclear distance rAB of 1.12 Å is reached at t=0.44, a transition state structure is not formed and rAB increases linearly with time as atom A moves away from the intact molecule BC at a constant velocity. The magnitude of the momentum p2 is not high enough for atom A to move towards molecule BC with sufficient translational kinetic energy to overcome the activation barrier and form a transition state structure. In the trajectory shown in the contour plot, the system does not have enough energy for the trajectory to cross the contour line to reach a higher energy transition state structure. The transition state internuclear distance rts of 0.90775 Å is lower than the minimum value of rAB observed. The molecule BC remains intact and rBC is nearly constant throughout the whole trajectory, oscillating between 0.71 Å and 0.77 Å, apart from bond stretching vibrations and a very slight increase in value from t=0.26 to t=0.67 to a peak of 0.81 Å at t=0.47. This is due to the weakening of the H - H bond in molecule BC as atom A approaches near atom B in molecule BC.
|-
| -1.5 || -2.5 || [[File:CPYX_H2+H_contour_trajectory_3.PNG|300px]] || [[File:CPYX_H2+H_trajectory_3.PNG|300px]] || Reactive
An increase in the initial magnitude of the AB momentum by 0.5 causes the trajectory to change from unreactive to reactive. This magnitude pAB, which corresponds to the relative translational energy of atom A, is the same as the magnitude pAB in the first trajectory in this section. As atom A moves towards a vibrating molecule BC, they collide to form a transition state structure at t=0.43, where rAB = rBC ≈ 0.89 Å. This value is close to the rts reported above of 0.90775 Å and this system has sufficient kinetic energy to overcome the activation barrier. A bond forms between atoms A and B to form a new molecule AB and the H - H bond in molecule BC breaks. rAB decreases until an optimum bond distance of 0.74 Å, which is the H -H bond length, is reached at t=0.59, and the molecule AB continues to vibrate such that rAB oscillates between 0.71 Å and 0.78 Å, within the contour lines, at large values of t. rBC increases almost linearly with time as atom C moves away from molecule AB at a constant velocity. Comparing this result with the previous rows shows that the magnitude of p2 (pAB) and hence the translational energy of the atom A is more important in determining the reactivity of a trajectory than the magnitude of p1 which reflects the vibrational energy of the molecule BC.
|-
| -2.5 || -5.0 || [[File:CPYX_H2+H_contour_trajectory_4.PNG|300px]] || [[File:CPYX_H2+H_trajectory_4.PNG|300px]] || Unreactive
Although atom A and molecule BC have high magnitudes of their momentum and kinetic energy, the transition state structure formed does not result in the formation of products AB and C, but the system reverts back to the reactants. Initially, atom A moves towards molecule BC at a high velocity and a transition state structure containing all 3 hydrogen atoms is formed very early on. After t=0.16, where rAB = rBC ≈ 0.82 Å, rAB < rBC. At t=0.21 and t=0.46, rAB have minima at 0.65 Å and 0.57 Å respectively. At t=0.34, rAB reaches a maximum in the transition state structure of 1.11 Å, but it is still smaller than the rBC value of 1.18 Å. Meanwhile, from t=0, rBC increases from its initial value of 0.74 Å to its maximum value of 1.37 Å at t=0.46. As rBC decreases from its maximum value and rAB increases from its minimum value, the 2 lines representing the bond distances intersect again at t=0.54, where rAB = rBC = 1.09 Å. With a high kinetic energy in this system, atom A and molecule BC orbit around each other after collision, so they approach each other in random directions and all memory of the initial approach direction is lost. Atom A and molecule BC also rotate as they spend much time around the transition state, so they eventually approach each other in directions, rotational states and vibrational states that do not lead to the formation of products. This is why this trajectory is unreactive and molecule AB and atom C do not form. The excess energy is converted into vibrational energy in molecule BC as well as a relative translational energy in atom A instead. At large values of t, rBC oscillates between 0.56 Å and 1.01 Å, crossing a few contours in the trajectory, as molecule BC is in a vibrationally excited state. rAB increases almost linearly with time as atom A moves away from molecule BC at a constant velocity, offset by the strong H - H bond stretching vibrations in molecule BC.
|-
| -2.5 || -5.2 || [[File:CPYX_H2+H_contour_trajectory_5.PNG|300px]] || [[File:CPYX_H2+H_trajectory_5.PNG|300px]]
|| Reactive
Although this internuclear distance vs time plot appears to be very similar to the above plot and the initial conditions are almost similar, the products of molecule AB and atom C are formed and the system does not revert back to the reactants. Initially, atom A moves towards molecule BC at a high velocity, so rAB decreases and rBC increases with time. At t=0.16, rAB = rBC = 0.82 Å. Atom A continues to move towards molecule BC until t=0.20, when rAB reaches a minimum at 0.64 Å. Atom A then "bounces off" molecule BC slightly. At t=0.32, rAB = rBC = 1.13 Å, which is very close to the maximum value of rBC reached in this transition state structure of 1.14 Å at t=0.33. rBC then decreases to a minimum value of 0.57 Å at t=0.46 while rAB increases to a maximum value of 1.37 Å at t=0.45. After that, rAB decreases and rBC increases with time and at t=0.54, rAB = rBC ≈ 1.07 Å. As rAB decreases further, the transition state structure is resolved, molecule AB is formed and atom C moves away from the new molecule AB and rBC increases. At large values of t, rAB oscillates between 0.56 Å and 1.02 Å, crossing a few contours in the trajectory, as molecule AB is in a vibrationally excited state. rBC increases almost linearly with time as atom C moves away from molecule AB at a constant velocity, offset by the strong H - H bond stretching vibrations in molecule AB. This trajectory is reactive as even though atom A and molecule BC spend much time around the transition state structure and orbit around each other and rotate, they eventually approach each other at the correct angle, vibrational and rotational states with sufficient translational energy, leading to product formation.
|}

=== Comparison of Transition State Theory predictions with experimental values ===

The main assumptions of Transition State Theory are that an activated complex is in equilibrium with the reactants, the rate of formation of products from this activated complex depends on the rate at which it passes through a transition state and every trajectory that passes through the transition state structure is reactive.<ref name="Atkins" />

From the results obtained in the previous part, the main factor influencing the outcome of the first 3 trajectories is the amount of kinetic energy present in the system, as the magnitudes of the internuclear momenta are relatively low. If the reactants have enough kinetic energy to reach the transition state and cross the activation barrier, products will be formed and the trajectory will be reactive. If the reactants do not have enough kinetic energy, the transition state will not be reached and products will not be formed. The reactive trajectories in this category all pass through the transition state structure and the 2nd trajectory which is unreactive does not pass through the transition state structure as the reactants do not have enough kinetic energy. These results show that the assumptions of the Transition State Theory are largely valid. Thus, Transition State Theory predictions for reaction rate values will be similar to experimental reaction rate values when the reactants approach each other at reasonably low levels of translational and vibrational energy.

The last 2 trajectories from the previous part have a very high kinetic energy in the system, as the magnitudes of internuclear momenta are high and around 2 times the values in the previous trajectories. According to the assumptions of the Transition State Theory, they both have enough kinetic energy so they should both pass through the transition state and form products. However, the 4th trajectory is unreactive while the last trajectory is reactive, even though their trajectories both show that they pass through a transition state. This means that not every trajectory that passes through the transition state structure is reactive and some trajectories that pass through the transition state structure will cross the barrier again and revert back to the reactants. This is because the orientation and direction in which the reactants approach each other, as well as their rotational and vibrational states, can influence the success of their collision, and this is the main factor influencing the outcome of these trajectories. Thus, Transition State Theory predictions for reaction rate values will be an overestimate and will be higher than experimental rate values as Transition State Theory assumes that collisions at the transition state will always be successful and this is not always the case.

{{fontcolor1|gray| Good. Minor point: the limitations regarding the equilibrium between TS and reactants are not really applicable here, since we're simulating a triatomic collision in isolation -- and not an ensemble of particles. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:25, 31 May 2017 (BST)}}

== F - H - H system ==

=== Energetics of the F + H2 and H + HF reactions ===

The F + H2 reaction is exothermic and the H + HF reaction is endothermic.

In the F + H2 reaction, a H - H bond is broken and a H - F bond is formed. Bond breaking is endothermic while bond forming is exothermic. The H - F bond is very strong (565 kJ/mol) due to the polarized nature of the H - F covalent bond and the high electronegativity difference between H, which has a moderate electronegativity, and F, which has a very high electronegativity. In contrast, the H - H bond is non-polar, so it is much weaker (432 kJ/mol) than the H - F bond. As the formation of the H - F bond releases more energy than the amount of energy used in the dissociation of the H - H bond, the F + H2 reaction releases energy to the surroundings in the form of heat and is exothermic.

In the H + HF reaction, a strong H - F bond is broken and a weaker H - H bond is formed. As the dissociation of the H - F bond requires more energy than the energy released through the formation of the H - H bond, this reaction takes in energy from the surroundings in the form of heat and is endothermic.

=== Transition state position and activation energies ===

When A is a F atom and B and C are both H atoms, the transition state is located at approximately rAB = 1.8100 Å and rBC = 0.74649 Å (5.s.f.). The plot of internuclear distance vs time where AB and BC both have zero momenta is shown below.

[[File:CPYX_FHH_TS_3.PNG|500px|thumb|center|"Internuclear Distances vs Time" plot for F-H-H]]

From this plot, the bond distances rAB and rBC remain constant until t=6.5. This means that these bond distances are at their optimum values in the transition state when all 3 atoms are bonded, and this structure stays on the ridge on the trajectory and does not fall off into reactants or products.

The activation energies for the F + H2 and H + HF reactions can be calculated using the energies of the reactants for each reaction and the energy of the transition state. These energy values can be found from the contour plot and they are recorded below.

Energy of H + H-F: -133.9 Kcal/mol

Energy of F + H-H: -103.9 Kcal/mol

Energy at the transition state: -103.7 Kcal/mol

These values can be used to find the activation energies for the H + HF and the F + H2 reactions, which are reported below.

Activation energy for H + HF reaction = -103.7 - (-103.9) = 0.2 Kcal/mol

Activation energy for F + H2 reaction = -103.7 - (-133.9) = 30.2 Kcal/mol

Other methods can be used to find the activation energies for these reactions. A mep using the initial conditions of rFH = 1.82 Å, rHH = 0.74649 Å, pFH = pHH = 0 was performed. The potential energy vs time plot shown below shows an initial energy of -103.75 Kcal/mol (2.d.p.), which corresponds to the energy of the transition state, and a final energy after 900 000 steps of -104.01 Kcal/mol (2.d.p.) which is the energy of F + H2. Thus, a better estimate of the activation energy for the F + H2 reaction is -103.75 - (-104.01) = 0.26 Kcal/mol.

[[File:CPYX_FHH_PE_vs_time_3.PNG|500px|thumb|center|Potential energy vs time plot for to find the activation energy for F + H2]]

A mep using the initial conditions of rFH = 1.80 Å, rHH = 0.74649 Å, pFH = pHH = 0 was performed. The potential energy vs time plot shown below shows an initial energy of -103.76 Kcal/mol (2.d.p.), the energy of the transition state, and a final energy of -133.92 Kcal/mol (2.d.p.), which is the energy of H + HF. Thus, a better estimate of the activation energy for the H + HF reaction is -103.76 - (-133.92) = 30.16 Kcal/mol.

[[File:CPYX_FHH_PE_vs_time_1.PNG|500px|thumb|center|Potential energy vs time plot for to find the activation energy for H + HF]]

=== Reaction dynamics ===

==== Reactive trajectories for the F + H2 reaction ====

A set of initial conditions that results in a reactive trajectory for the F + H2 reaction are rHF = 2.00 Å, rHH = 0.74 Å, pHF = -2.00 and pHH = -1.25. The reaction energy is released as kinetic energy, in the form of the relative translational energy of the H atom away from the HF molecule as well as vibrational energy of the HF molecule itself. This is confirmed by the contour plot and the plot of internuclear momentum vs time shown below.

{| class="wikitable"
| [[File:CPYX_FHH_contour_2.PNG|500px|thumb|center|Contour plot showing that trajectory is reactive]] || [[File:CPYX_FHH_internuclear_momentum_vs_time_1.PNG|500px|thumb|center|"Internuclear Momentum vs Time" plot]]
|}

From the plots above, the H2 molecule has some vibrational energy initially, as shown in both the plots above where the trajectory and initial internuclear momentum pHH oscillate. The F atom moves towards the H2 molecule to form a transition state structure and then a HF molecule and a H atom. From the contour plot, the newly-formed HF molecule vibrates between energy levels quite high above the minimum energy path, and this vibration is much more extensive than the initial H2 molecule. In the initial part of the trajectory, even with vibrations present in the H2 molecule, the contour lines are not crossed until the transition state structure dissociates to form products. The internuclear momentum pHF also oscillates between -5.98 and 9.23, which is a large range, at large values of t. This is because the newly-formed HF molecule is in a vibrationally excited state. The internuclear momentum pHH at large values of t is a constant but rather low positive value of 1.83, as the H atom leaves the HF molecule with some but not much translational energy. From the contour plot, the products contain more vibrational energy in the HF molecule than translational energy in the leaving H atom.

This can be confirmed experimentally by using infrared chemiluminescence, where vibrationally excited molecules emit infrared radiation as they return to the ground state. Reactants with particular vibrational and translational energy levels can be shot at each other to react and the populations of the vibrational states of the HF molecules formed can be determined by measuring and comparing the intensities of the peaks in the infrared emission spectrum.<ref name="Atkins" />

Setting rHH = 0.74 Å, rFH = 2.00 Å and pFH = -0.5, the values of pHH in the range -3 to 3 that result in reactive trajectories do not fall into a few particular ranges, but range across the whole spectrum. However, low magnitudes of pHH from -0.90 to 0.32 result in reactive trajectories. Outside this range of values, for trajectories with even a small range of pHH values, their general reactivity cannot be meaningfully predicted. Changing the internuclear distance rFH only slightly can also make an unreactive trajectory reactive or make a reactive trajectory unreactive.

This is because the reaction of F + H2 is an exothermic reaction which requires a very small amount of energy of 0.26 Kcal/mol to achieve a transition state structure. Thus, the main factor determining the reactivity of a trajectory for this reaction is not the amount of kinetic energy in this system, but the direction and orientation of approach, the rotational states and the vibrational states of the reactants. In this system, setting a low initial magnitude of pHH which corresponds to very little kinetic energy in the H - H bond allows the F atom to approach and collide with the H2 molecule in a linear configuration easily. The products of a H atom and a vibrationally excited HF molecule will be formed easily as energy is released in the form of vibrational energy in the HF molecule as the products are formed. However, setting a large value of pHH means that the amount of kinetic energy in the system is high, allowing the F atom and the HF molecule to orbit around each other after collision, so they approach each other in random directions and all memory of the initial approach direction is lost. The reactants also rotate if they spend much time around the transition state, so they may or may not eventually approach each other in directions, rotational states and vibrational states that lead to the formation of products.

After setting up the initial conditions of rHH = 0.74 Å, rFH = 2.00 Å, pFH = -0.8 and pHH = 0.1, the trajectory obtained is reactive. The overall energy of the system is considerably reduced at these low momenta, so the reactants approach each other in a linear configuration that results in product formation. The contour plot and plot of internuclear distance vs time for this trajectory are shown below.

{| class="wikitable"
| [[File:CPYX_FHH_contour_3.PNG|500px|thumb|center|Contour plot showing that trajectory is reactive]] || [[File:CPYX_FHH_internuclear_distance_vs_time_1.PNG|500px|thumb|center|"Internuclear Distance vs Time" plot showing that the trajectory is reactive]]
|}

From the plot above, rFH decreases and rHH oscillates initially due to vibrations in the H2 molecule. At t=0.47, rFH=1.81 Å and the transition state structure starts to form. rHH starts to increase slowly and it increases more quickly from t=1.11 and rHH = rFH = 1.03 Å at t=1.25. At t=1.30, rFH reaches a minimum of 0.76 Å and rHH reaches a maximum of 1.29 Å. rFH increases and rHH decreases and at t=1.35, rHH = rFH = 1.06 Å. At t=1.52, rFH reaches a maximum of 1.34 Å and at t=1.57, rHH reaches a minimum in the transition state of 0.74 Å. rHH then increases almost linearly with time as a H atom originally from the H - H bond moves away from the newly-formed HF molecule, offset by the strong H - F bond stretching vibrations. At large values of t, rFH oscillates between 0.75 Å and 1.19 Å due to the H - F bond stretching vibrations. The newly-formed HF molecule is in a vibrationally excited state, as the trajectory crosses s few contours as it oscillates and rFH oscillates over a large range of values.

==== Reactive trajectories for the H + HF reaction ====

For this reaction, setting rFH = 0.92 Å, rHH = 2.5 Å, pFH = 0.1 and pHH = -10 results in a reactive trajectory.

{| class="wikitable"
| [[File:CPYX_FHH_contour_4.PNG|500px|thumb|center|Contour plot showing that trajectory is reactive]] || [[File:CPYX_FHH_internuclear_distance_vs_time_2.PNG|500px|thumb|center|"Internuclear Distance vs Time" plot showing that the trajectory is reactive]]
|}

In this reaction, unlike the previous reaction, both the reactants and products do not have much vibrational energy, but have more translational energy. This is shown in the smooth curves in the plots above. A high negative initial value of pHH and a much lower initial value of pFH were set. Initially, the H atom moves towards the HF molecule at a high negative velocity through a path that does not have the minimum potential energy, the H - F bond in the molecule weakens and rFH increases to a maximum of 1.33 Å at t=0.08. The H atom moves closer and collides with the HF molecule, such that the minimum value of rHH of 0.56 Å is reached very early on, at t=0.16. Kinetic energy is converted to potential energy in the HF molecule, and rFH decreases to a minimum value of 0.67 Å at t=0.21 as rHH increases to 1.01 Å. From the contour plot, this is the point on the trajectory where the system has the highest energy of ≈ -40 Kcal/mol. Potential energy is converted into kinetic energy when the trajectory moves away from this point and the kinetic energy is later converted into potential energy to overcome the activation barrier and form products. rHH decreases after reaching a maximum value of 1.03 Å at t=0.22, as a new H2 molecule is formed. At large values of t, the H2 molecule vibrates slightly, within the contour lines, and rHH averages at around 0.74 Å, the H – H bond length. The F atom has a high translational energy and moves away from the new H2 molecule at a high positive constant velocity such that rFH increases very quickly and reaches a value of 12.00 Å at t=1.50.

Using the same initial conditions as above but with pFH = 3.5 and pHH = -7.5 also results in a reactive trajectory, as shown by the contour plot and the internuclear distance vs time plot below. A slightly less negative, but still high, initial value of pHH and a higher initial value of pFH were set as compared to the previous reaction.

{| class="wikitable"
| [[File:CPYX_FHH_contour_1.PNG|500px|thumb|center|Contour plot showing that the trajectory is reactive]] || [[File:CPYX_FHH_internuclear_distance_vs_time_4.PNG|500px|thumb|center|"Internuclear Distance vs Time" plot showing that the trajectory is reactive]]
|}

In this reaction, the products have more vibrational energy and slightly less translational energy than the previous reaction, while the reactants still have much more translational energy than vibrational energy. Initially, the H atom moves towards the HF molecule at a high negative velocity through a path that does not have the minimum potential energy, the H - F bond in the molecule weakens and rFH increases to a maximum of 1.42 Å at t=0.10. The H atom moves closer and collides with the HF molecule, such that the minimum value of rHH of 0.54 Å is reached very early on, at t=0.18. Kinetic energy is converted to potential energy in the HF molecule, and rFH decreases to a minimum value of 0.67 Å at t=0.25 as rHH increases to 1.22 Å. From the contour plot, this is the point on the trajectory where the system has the highest energy of ≈ -50 Kcal/mol. This energy value obtained is more negative and not as high as the value obtained for the previous reaction, as the magnitude of the internuclear momentum pHH and the relative translational energy of the H atom is lower than in the previous reaction, so the H atom moves towards the HF molecule with less kinetic energy that can be converted into potential energy. The potential energy is converted back into kinetic energy when the trajectory moves away from this point, and the kinetic energy is later converted back into potential energy to overcome the activation barrier and form products. rHH decreases as a new H2 molecule is formed. At large values of t, the H2 molecule vibrates and rHH averages at around 0.74 Å, the H – H bond length. The vibrational energy of the H2 molecule is greater than in the previous reaction, as the oscillations in the plots are stronger and the trajectory crosses a contour line at each peak, meaning that the H2 molecule is in a vibrationally excited state. The F atom has a high translational energy and moves away from the new H2 molecule at a high positive constant velocity such that rFH increases very quickly and reaches a value of 10.29 Å at t=1.50, which is still not as high as the value obtained in the previous reaction with a greater translational energy contribution.

By changing the initial H - F bond length to increase the energy of the H - F vibration, a reactive trajectory cannot be obtained without decreasing the magnitude of pHH, and this means that this reaction is driven mainly by the translational energy of the incoming H atom, much more than the vibrational energy of the HF molecule.

==== Differences between the F + H2 and H + HF reactions ====

A reactive trajectory for the F + H2 reaction usually involves a low amount of translational energy in the reactants and generates a HF molecule in a vibrationally excited state and a H atom that has little translational energy. On the other hand, a reactive trajectory for the H + HF reaction involves a high amount of translational energy in the incoming H atom and generates a H2 molecule that may or may not be vibrationally excited and a F atom that has a very high translational energy.

For the exothermic F + H2 reaction, the activation barrier is very small and the transition state is reached early on in the trajectory, so the incoming F atom only needs very little translational energy to collide and react successfully with the H2 molecule, and the main factors determining the reactivity of the trajectory are the direction and orientation of approach and the states of the reactants. As this reaction is exothermic, energy is released, mainly as vibrational energy in the new HF molecule. On the other hand, for the endothermic H + HF reaction, the energy barrier of -30.162 Kcal/mol is not small and has to be overcome for a reaction to be successful. The transition state is also reached late in the trajectory. This explains the need for the H atom to approach the HF molecule with a high translational kinetic energy in reactive trajectories.

<references>
<ref name="Atkins">Atkins P and de Paula J. Physical Chemistry, 9th edition. Oxford: Oxford University Press;2009.</ref>
</references>

MRD:CPYX

2017-05-31T15:18:49Z

Je714: /* Comparison between the reaction path/mep and the trajectory using the same initial conditions */

== H + H2 system ==

=== Transition state region and position ===

The total gradient of the potential energy surface is 0 both at a minimum and at a transition structure. At a minimum, the gradient is negative before that point and positive after that point. The transition state is the maximum point on the minimum energy path, which means that it is at a relative minimum along 1 direction and at a relative maximum along the crossing direction. This point is known as a saddle point, as it is neither a minimum nor a maximum.

{{fontcolor1|gray| Good. Some discussion using second partial derivatives would have been nicer, though. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:17, 31 May 2017 (BST)}}

The transition state position rts is estimated to be 0.90775 Å (5.s.f.). When the initial conditions of r1 = r2 = 0.90775 Å and p1 = p2 = 0 are tested, the system does not seem to vibrate and the bond distances remain at 0.90775 Å until t=8.63, when atom C starts to move away from the newly-formed vibrating molecule AB. This is shown in the plot of internuclear distances against time below. The above initial conditions at time values of 0 < t < 8.63 also produced a trajectory where all the atoms remained at their initial position. This means that the trajectory only stays on the ridge when the bonds have zero momenta and this is the optimal distance between each of the 3 H atoms when they are all bonded to one another in the transition state.

{{fontcolor1|gray| The TS is a mathematical construct. In practice, you can always run a sufficiently long simulation in which the system will fall off towards the products or reactants. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:17, 31 May 2017 (BST)}}

[[File:CPYX_H2+H_TS_1.PNG|400px|thumb|center|"Internuclear Distances vs Time" plot for H2 + H]]

=== Comparison between the reaction path/mep and the trajectory using the same initial conditions ===

{{fontcolor1|gray| Thorough explanation, GJ. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:18, 31 May 2017 (BST)}}

A mep and a trajectory with its calculation type as Dynamics were run separately using the initial conditions of r1 = rBC = 0.91775 Å, r2 = rAB = 0.90775 Å and p1 = p2 = 0 and compared. All atoms A, B and C refer to H atoms and all internuclear distances, time and internuclear momenta values which do not have to be very accurate are reported to 2.d.p.

If the initial conditions r1 = rBC = 0.90775 Å and r2 = rAB = 0.91775 Å were used instead, the trajectory will be a reflection of the original trajectory in the diagonal connecting the origin and the corner corresponding to the maximum internuclear distance values. In the internuclear distance and internuclear momentum vs time plots, the lines for the new A-B bond will be the same as the lines for the original B-C bond and the lines for the new B-C bond will be the same as the lines for the original A-B bond. The shapes of the plots will otherwise be the same.

{| class="wikitable" border="1"
|+ Comparison between mep and trajectory
! Differences !! Mep (20 000 steps) !! Trajectory (300 steps)
|-
| rowspan="2" | Contour plot || [[File:CPYX_H2+H_contour_mep.PNG|500px]] || [[File:CPYX_H2+H_contour_1.PNG|500px]]
|-
| colspan="2" | In both plots above, the products of molecule AB and atom C are formed, and atom C moves away from molecule AB. The mep is much smoother than the trajectory due to the much greater vibrational energy of the new molecule AB in the trajectory.
|-
| Internuclear distance vs time plot || [[File:CPYX_H2+H_mep_internuclear_distance_vs_time.PNG|500px]] || [[File:CPYX_H2+H_trajectory_internuclear_distance_vs_time.PNG|500px]]
|-
| Explanation || The final internuclear distances recorded are r1(100)=2.53 Å and r2(100)=0.74 Å. The value of r1(100) is much smaller than the value of r1(1.5) in the trajectory, as the velocity in the mep is set to 0 in each time step, unlike the trajectory. This means that atom C has very little translational kinetic energy in the mep and moves away from molecule AB much more slowly. After t=3.00, the value of r1 increases with time and the increase in the value of r1 decreases with time, as the distance between atom C and molecule AB increases and there is less repulsion between their electrons. The value of r2 stays constant at 0.74 Å, which is the H - H bond length, and vibrations are not observed due to the zero velocity and kinetic energy in this H - H bond. || The final internuclear distances recorded are r1(1.5)=5.28 Å and r2 ranges from 0.73 Å to 0.76 Å at large values of t. The products have reasonable non-zero amounts of translational and vibrational energy due to the release of energy from the transition state structure. After t=0.40, the value of r1 increases linearly with time while the value of r2 stays approximately constant, averaging around 0.74 Å, which is the H - H bond length.
|-
| Internuclear momentum vs time plot || [[File:CPYX_H2+H_mep_internuclear_momenta_vs_time.PNG|500px]] || [[File:CPYX_H2+H_trajectory_internuclear_momenta_vs_time.PNG|500px]]
|-
| Explanation || When obtaining the mep, the velocity is set to 0 in each time step. Since momentum pi =μvi where μ is mass and v is velocity, the internuclear momenta at all time values is 0. || Initially, both internuclear momenta are 0 and become slightly negative as the 3 hydrogen atoms move closer together, at a velocity that decreases the atomic distance, to form a transition state structure. After the transition state structure forms, the AB internuclear momentum remains negative as the new molecule AB is formed and the internuclear distance rAB decreases to a value of around the H - H bond length. However, the BC internuclear momentum becomes positive after t=0.11 and increases steadily as atom C moves away from the new molecule AB, converging to a value of 2.48 at large values of t, so p1(1.5)=2.48. After the molecule AB is formed, the AB internuclear momentum increases to a positive value and oscillates between 0.91 and 1.57 at large values of t due to vibrations in the H - H bond. This means that the final average internuclear momenta p2(1.5)=1.24.
|}

A calculation with initial positions at the final positions of the trajectory obtained above and the initial momenta the same as the final momenta in the trajectory obtained above but with their signs reversed, was performed. This means that rAB, rBC, AB momentum and BC momentum were set to 0.74 Å, 5.28 Å, -1.24 and -2.48 respectively. The results of the calculation are shown below.

{| class="wikitable" border="1"
! Contour plot !! Internuclear distance vs time plot !! Internuclear momentum vs time plot (extended to 500 steps/ t=2.50)
|-
| [[File:CPYX_H2+H_contour_2.PNG|400px]] || [[File:CPYX_H2+H_trajectory_internuclear_distance_vs_time_reversed.PNG|400px]] || [[File:CPYX_H2+H_internuclear_distance_vs_time_reversed.PNG|400px]]
|-
| The shape of this plot is very similar to the shape of the earlier trajectory, except that this trajectory moves in the opposite direction and circles around the region of 0.70 Å < rAB < 0.90 Å and 0.90 Å < rBC < 1.10 Å. This trajectory is very smooth, as most of the energy in this system is translational energy. The value of rAB, which corresponds to the vibrational energy of the H - H bond in molecule AB, is 0.74 Å, which is the H - H bond length, so molecule AB has very little vibrational energy. This shape shows that the trajectory is unreactive and will revert back to reactants AB and C, as the initial conditions do not provide enough kinetic energy to overcome the activation barrier. The maximum amount of potential energy reached by this system is lower than the potential energy of the transition state structure.
|| The shape of this plot is roughly the reverse of the plot obtained with the earlier trajectory. The value of rBC decreases linearly with time while the value of rAB stays approximately constant with vibrations. From t=1.20 to t=1.50, a transition state structure between atom C and molecule AB is attempted but not formed as the maximum value of rAB of 0.87 Å is reached at t=1.37 and the minimum value of rBC of 0.96 Å is reached at t=1.33. These values are not close to one another as well as to at the value of rts which is 0.90775 Å. Thus, atom C moves away from the molecule AB which remains intact, so rBC increases and rAB decreases to its original value of 0.74 Å. || The internuclear momenta remain relatively constant initially and become less negative after t=0.5 as the 3 H atoms move closer together more slowly, with a less negative velocity, due to repulsions between their electrons. pAB becomes positive at t=1.24 as rAB increases in an attempt to form a transition state structure. pAB becomes negative again at t=1.39 as rAB decreases to its original value of 0.74 Å. pAB becomes positive again at t=1.51, after molecule AB is re-formed, and oscillates between 0.70 and 1.71 at large values of t due to H - H bond vibrations. pBC becomes positive at t=1.31 and continues to increase as rBC increases and atom C moves away from molecule AB.
|}

=== Reactive and unreactive trajectories ===

In the following table, the initial positions are r1 = rBC = 0.74 Å and r2 = rAB = 2.00 Å. The momentum values are recorded in the table and the reactivity of their trajectories is shown.

{| class="wikitable" border=1
! p1 !! p2 !! Screenshot of trajectory !! Internuclear distance vs time plot !! Reactive/unreactive
|-
| -1.25 || -2.5 || [[File:CPYX_H2+H_contour_trajectory_1.PNG|300px]] || [[File:CPYX_H2+H_trajectory_1.PNG|300px]] || Reactive
The trajectory and internuclear distance vs time plot show that atom A moves towards molecule BC and collides with it. A transition state structure is formed at t=0.43, where rAB = rBC ≈ 0.90 Å, which is close to the rts of 0.90755 Å found earlier. Even though the magnitudes of the internuclear momenta appear to be small, this system has sufficient kinetic energy to overcome the activation barrier. This is because the system is driven much more by translational energy than by vibrational energy, as shown in the smooth trajectory. A bond forms between atoms A and B and the B-C bond breaks. rAB decreases until the optimum bond distance of 0.74 Å is reached at t=0.58, and the molecule continues to vibrate slightly, within the contour lines. rBC increases almost linearly with time as atom C moves away from molecule AB at a constant velocity.
|-
| -1.5 || -2.0 || [[File:CPYX_H2+H_contour_trajectory_2.PNG|300px]] || [[File:CPYX_H2+H_trajectory_2.PNG|300px]]
|| Unreactive
These plots show atom A initially moving towards a slightly vibrating molecule BC. After a minimum internuclear distance rAB of 1.12 Å is reached at t=0.44, a transition state structure is not formed and rAB increases linearly with time as atom A moves away from the intact molecule BC at a constant velocity. The magnitude of the momentum p2 is not high enough for atom A to move towards molecule BC with sufficient translational kinetic energy to overcome the activation barrier and form a transition state structure. In the trajectory shown in the contour plot, the system does not have enough energy for the trajectory to cross the contour line to reach a higher energy transition state structure. The transition state internuclear distance rts of 0.90775 Å is lower than the minimum value of rAB observed. The molecule BC remains intact and rBC is nearly constant throughout the whole trajectory, oscillating between 0.71 Å and 0.77 Å, apart from bond stretching vibrations and a very slight increase in value from t=0.26 to t=0.67 to a peak of 0.81 Å at t=0.47. This is due to the weakening of the H - H bond in molecule BC as atom A approaches near atom B in molecule BC.
|-
| -1.5 || -2.5 || [[File:CPYX_H2+H_contour_trajectory_3.PNG|300px]] || [[File:CPYX_H2+H_trajectory_3.PNG|300px]] || Reactive
An increase in the initial magnitude of the AB momentum by 0.5 causes the trajectory to change from unreactive to reactive. This magnitude pAB, which corresponds to the relative translational energy of atom A, is the same as the magnitude pAB in the first trajectory in this section. As atom A moves towards a vibrating molecule BC, they collide to form a transition state structure at t=0.43, where rAB = rBC ≈ 0.89 Å. This value is close to the rts reported above of 0.90775 Å and this system has sufficient kinetic energy to overcome the activation barrier. A bond forms between atoms A and B to form a new molecule AB and the H - H bond in molecule BC breaks. rAB decreases until an optimum bond distance of 0.74 Å, which is the H -H bond length, is reached at t=0.59, and the molecule AB continues to vibrate such that rAB oscillates between 0.71 Å and 0.78 Å, within the contour lines, at large values of t. rBC increases almost linearly with time as atom C moves away from molecule AB at a constant velocity. Comparing this result with the previous rows shows that the magnitude of p2 (pAB) and hence the translational energy of the atom A is more important in determining the reactivity of a trajectory than the magnitude of p1 which reflects the vibrational energy of the molecule BC.
|-
| -2.5 || -5.0 || [[File:CPYX_H2+H_contour_trajectory_4.PNG|300px]] || [[File:CPYX_H2+H_trajectory_4.PNG|300px]] || Unreactive
Although atom A and molecule BC have high magnitudes of their momentum and kinetic energy, the transition state structure formed does not result in the formation of products AB and C, but the system reverts back to the reactants. Initially, atom A moves towards molecule BC at a high velocity and a transition state structure containing all 3 hydrogen atoms is formed very early on. After t=0.16, where rAB = rBC ≈ 0.82 Å, rAB < rBC. At t=0.21 and t=0.46, rAB have minima at 0.65 Å and 0.57 Å respectively. At t=0.34, rAB reaches a maximum in the transition state structure of 1.11 Å, but it is still smaller than the rBC value of 1.18 Å. Meanwhile, from t=0, rBC increases from its initial value of 0.74 Å to its maximum value of 1.37 Å at t=0.46. As rBC decreases from its maximum value and rAB increases from its minimum value, the 2 lines representing the bond distances intersect again at t=0.54, where rAB = rBC = 1.09 Å. With a high kinetic energy in this system, atom A and molecule BC orbit around each other after collision, so they approach each other in random directions and all memory of the initial approach direction is lost. Atom A and molecule BC also rotate as they spend much time around the transition state, so they eventually approach each other in directions, rotational states and vibrational states that do not lead to the formation of products. This is why this trajectory is unreactive and molecule AB and atom C do not form. The excess energy is converted into vibrational energy in molecule BC as well as a relative translational energy in atom A instead. At large values of t, rBC oscillates between 0.56 Å and 1.01 Å, crossing a few contours in the trajectory, as molecule BC is in a vibrationally excited state. rAB increases almost linearly with time as atom A moves away from molecule BC at a constant velocity, offset by the strong H - H bond stretching vibrations in molecule BC.
|-
| -2.5 || -5.2 || [[File:CPYX_H2+H_contour_trajectory_5.PNG|300px]] || [[File:CPYX_H2+H_trajectory_5.PNG|300px]]
|| Reactive
Although this internuclear distance vs time plot appears to be very similar to the above plot and the initial conditions are almost similar, the products of molecule AB and atom C are formed and the system does not revert back to the reactants. Initially, atom A moves towards molecule BC at a high velocity, so rAB decreases and rBC increases with time. At t=0.16, rAB = rBC = 0.82 Å. Atom A continues to move towards molecule BC until t=0.20, when rAB reaches a minimum at 0.64 Å. Atom A then "bounces off" molecule BC slightly. At t=0.32, rAB = rBC = 1.13 Å, which is very close to the maximum value of rBC reached in this transition state structure of 1.14 Å at t=0.33. rBC then decreases to a minimum value of 0.57 Å at t=0.46 while rAB increases to a maximum value of 1.37 Å at t=0.45. After that, rAB decreases and rBC increases with time and at t=0.54, rAB = rBC ≈ 1.07 Å. As rAB decreases further, the transition state structure is resolved, molecule AB is formed and atom C moves away from the new molecule AB and rBC increases. At large values of t, rAB oscillates between 0.56 Å and 1.02 Å, crossing a few contours in the trajectory, as molecule AB is in a vibrationally excited state. rBC increases almost linearly with time as atom C moves away from molecule AB at a constant velocity, offset by the strong H - H bond stretching vibrations in molecule AB. This trajectory is reactive as even though atom A and molecule BC spend much time around the transition state structure and orbit around each other and rotate, they eventually approach each other at the correct angle, vibrational and rotational states with sufficient translational energy, leading to product formation.
|}

=== Comparison of Transition State Theory predictions with experimental values ===

The main assumptions of Transition State Theory are that an activated complex is in equilibrium with the reactants, the rate of formation of products from this activated complex depends on the rate at which it passes through a transition state and every trajectory that passes through the transition state structure is reactive.<ref name="Atkins" />

From the results obtained in the previous part, the main factor influencing the outcome of the first 3 trajectories is the amount of kinetic energy present in the system, as the magnitudes of the internuclear momenta are relatively low. If the reactants have enough kinetic energy to reach the transition state and cross the activation barrier, products will be formed and the trajectory will be reactive. If the reactants do not have enough kinetic energy, the transition state will not be reached and products will not be formed. The reactive trajectories in this category all pass through the transition state structure and the 2nd trajectory which is unreactive does not pass through the transition state structure as the reactants do not have enough kinetic energy. These results show that the assumptions of the Transition State Theory are largely valid. Thus, Transition State Theory predictions for reaction rate values will be similar to experimental reaction rate values when the reactants approach each other at reasonably low levels of translational and vibrational energy.

The last 2 trajectories from the previous part have a very high kinetic energy in the system, as the magnitudes of internuclear momenta are high and around 2 times the values in the previous trajectories. According to the assumptions of the Transition State Theory, they both have enough kinetic energy so they should both pass through the transition state and form products. However, the 4th trajectory is unreactive while the last trajectory is reactive, even though their trajectories both show that they pass through a transition state. This means that not every trajectory that passes through the transition state structure is reactive and some trajectories that pass through the transition state structure will cross the barrier again and revert back to the reactants. This is because the orientation and direction in which the reactants approach each other, as well as their rotational and vibrational states, can influence the success of their collision, and this is the main factor influencing the outcome of these trajectories. Thus, Transition State Theory predictions for reaction rate values will be an overestimate and will be higher than experimental rate values as Transition State Theory assumes that collisions at the transition state will always be successful and this is not always the case.

== F - H - H system ==

=== Energetics of the F + H2 and H + HF reactions ===

The F + H2 reaction is exothermic and the H + HF reaction is endothermic.

In the F + H2 reaction, a H - H bond is broken and a H - F bond is formed. Bond breaking is endothermic while bond forming is exothermic. The H - F bond is very strong (565 kJ/mol) due to the polarized nature of the H - F covalent bond and the high electronegativity difference between H, which has a moderate electronegativity, and F, which has a very high electronegativity. In contrast, the H - H bond is non-polar, so it is much weaker (432 kJ/mol) than the H - F bond. As the formation of the H - F bond releases more energy than the amount of energy used in the dissociation of the H - H bond, the F + H2 reaction releases energy to the surroundings in the form of heat and is exothermic.

In the H + HF reaction, a strong H - F bond is broken and a weaker H - H bond is formed. As the dissociation of the H - F bond requires more energy than the energy released through the formation of the H - H bond, this reaction takes in energy from the surroundings in the form of heat and is endothermic.

=== Transition state position and activation energies ===

When A is a F atom and B and C are both H atoms, the transition state is located at approximately rAB = 1.8100 Å and rBC = 0.74649 Å (5.s.f.). The plot of internuclear distance vs time where AB and BC both have zero momenta is shown below.

[[File:CPYX_FHH_TS_3.PNG|500px|thumb|center|"Internuclear Distances vs Time" plot for F-H-H]]

From this plot, the bond distances rAB and rBC remain constant until t=6.5. This means that these bond distances are at their optimum values in the transition state when all 3 atoms are bonded, and this structure stays on the ridge on the trajectory and does not fall off into reactants or products.

The activation energies for the F + H2 and H + HF reactions can be calculated using the energies of the reactants for each reaction and the energy of the transition state. These energy values can be found from the contour plot and they are recorded below.

Energy of H + H-F: -133.9 Kcal/mol

Energy of F + H-H: -103.9 Kcal/mol

Energy at the transition state: -103.7 Kcal/mol

These values can be used to find the activation energies for the H + HF and the F + H2 reactions, which are reported below.

Activation energy for H + HF reaction = -103.7 - (-103.9) = 0.2 Kcal/mol

Activation energy for F + H2 reaction = -103.7 - (-133.9) = 30.2 Kcal/mol

Other methods can be used to find the activation energies for these reactions. A mep using the initial conditions of rFH = 1.82 Å, rHH = 0.74649 Å, pFH = pHH = 0 was performed. The potential energy vs time plot shown below shows an initial energy of -103.75 Kcal/mol (2.d.p.), which corresponds to the energy of the transition state, and a final energy after 900 000 steps of -104.01 Kcal/mol (2.d.p.) which is the energy of F + H2. Thus, a better estimate of the activation energy for the F + H2 reaction is -103.75 - (-104.01) = 0.26 Kcal/mol.

[[File:CPYX_FHH_PE_vs_time_3.PNG|500px|thumb|center|Potential energy vs time plot for to find the activation energy for F + H2]]

A mep using the initial conditions of rFH = 1.80 Å, rHH = 0.74649 Å, pFH = pHH = 0 was performed. The potential energy vs time plot shown below shows an initial energy of -103.76 Kcal/mol (2.d.p.), the energy of the transition state, and a final energy of -133.92 Kcal/mol (2.d.p.), which is the energy of H + HF. Thus, a better estimate of the activation energy for the H + HF reaction is -103.76 - (-133.92) = 30.16 Kcal/mol.

[[File:CPYX_FHH_PE_vs_time_1.PNG|500px|thumb|center|Potential energy vs time plot for to find the activation energy for H + HF]]

=== Reaction dynamics ===

==== Reactive trajectories for the F + H2 reaction ====

A set of initial conditions that results in a reactive trajectory for the F + H2 reaction are rHF = 2.00 Å, rHH = 0.74 Å, pHF = -2.00 and pHH = -1.25. The reaction energy is released as kinetic energy, in the form of the relative translational energy of the H atom away from the HF molecule as well as vibrational energy of the HF molecule itself. This is confirmed by the contour plot and the plot of internuclear momentum vs time shown below.

{| class="wikitable"
| [[File:CPYX_FHH_contour_2.PNG|500px|thumb|center|Contour plot showing that trajectory is reactive]] || [[File:CPYX_FHH_internuclear_momentum_vs_time_1.PNG|500px|thumb|center|"Internuclear Momentum vs Time" plot]]
|}

From the plots above, the H2 molecule has some vibrational energy initially, as shown in both the plots above where the trajectory and initial internuclear momentum pHH oscillate. The F atom moves towards the H2 molecule to form a transition state structure and then a HF molecule and a H atom. From the contour plot, the newly-formed HF molecule vibrates between energy levels quite high above the minimum energy path, and this vibration is much more extensive than the initial H2 molecule. In the initial part of the trajectory, even with vibrations present in the H2 molecule, the contour lines are not crossed until the transition state structure dissociates to form products. The internuclear momentum pHF also oscillates between -5.98 and 9.23, which is a large range, at large values of t. This is because the newly-formed HF molecule is in a vibrationally excited state. The internuclear momentum pHH at large values of t is a constant but rather low positive value of 1.83, as the H atom leaves the HF molecule with some but not much translational energy. From the contour plot, the products contain more vibrational energy in the HF molecule than translational energy in the leaving H atom.

This can be confirmed experimentally by using infrared chemiluminescence, where vibrationally excited molecules emit infrared radiation as they return to the ground state. Reactants with particular vibrational and translational energy levels can be shot at each other to react and the populations of the vibrational states of the HF molecules formed can be determined by measuring and comparing the intensities of the peaks in the infrared emission spectrum.<ref name="Atkins" />

Setting rHH = 0.74 Å, rFH = 2.00 Å and pFH = -0.5, the values of pHH in the range -3 to 3 that result in reactive trajectories do not fall into a few particular ranges, but range across the whole spectrum. However, low magnitudes of pHH from -0.90 to 0.32 result in reactive trajectories. Outside this range of values, for trajectories with even a small range of pHH values, their general reactivity cannot be meaningfully predicted. Changing the internuclear distance rFH only slightly can also make an unreactive trajectory reactive or make a reactive trajectory unreactive.

This is because the reaction of F + H2 is an exothermic reaction which requires a very small amount of energy of 0.26 Kcal/mol to achieve a transition state structure. Thus, the main factor determining the reactivity of a trajectory for this reaction is not the amount of kinetic energy in this system, but the direction and orientation of approach, the rotational states and the vibrational states of the reactants. In this system, setting a low initial magnitude of pHH which corresponds to very little kinetic energy in the H - H bond allows the F atom to approach and collide with the H2 molecule in a linear configuration easily. The products of a H atom and a vibrationally excited HF molecule will be formed easily as energy is released in the form of vibrational energy in the HF molecule as the products are formed. However, setting a large value of pHH means that the amount of kinetic energy in the system is high, allowing the F atom and the HF molecule to orbit around each other after collision, so they approach each other in random directions and all memory of the initial approach direction is lost. The reactants also rotate if they spend much time around the transition state, so they may or may not eventually approach each other in directions, rotational states and vibrational states that lead to the formation of products.

After setting up the initial conditions of rHH = 0.74 Å, rFH = 2.00 Å, pFH = -0.8 and pHH = 0.1, the trajectory obtained is reactive. The overall energy of the system is considerably reduced at these low momenta, so the reactants approach each other in a linear configuration that results in product formation. The contour plot and plot of internuclear distance vs time for this trajectory are shown below.

{| class="wikitable"
| [[File:CPYX_FHH_contour_3.PNG|500px|thumb|center|Contour plot showing that trajectory is reactive]] || [[File:CPYX_FHH_internuclear_distance_vs_time_1.PNG|500px|thumb|center|"Internuclear Distance vs Time" plot showing that the trajectory is reactive]]
|}

From the plot above, rFH decreases and rHH oscillates initially due to vibrations in the H2 molecule. At t=0.47, rFH=1.81 Å and the transition state structure starts to form. rHH starts to increase slowly and it increases more quickly from t=1.11 and rHH = rFH = 1.03 Å at t=1.25. At t=1.30, rFH reaches a minimum of 0.76 Å and rHH reaches a maximum of 1.29 Å. rFH increases and rHH decreases and at t=1.35, rHH = rFH = 1.06 Å. At t=1.52, rFH reaches a maximum of 1.34 Å and at t=1.57, rHH reaches a minimum in the transition state of 0.74 Å. rHH then increases almost linearly with time as a H atom originally from the H - H bond moves away from the newly-formed HF molecule, offset by the strong H - F bond stretching vibrations. At large values of t, rFH oscillates between 0.75 Å and 1.19 Å due to the H - F bond stretching vibrations. The newly-formed HF molecule is in a vibrationally excited state, as the trajectory crosses s few contours as it oscillates and rFH oscillates over a large range of values.

==== Reactive trajectories for the H + HF reaction ====

For this reaction, setting rFH = 0.92 Å, rHH = 2.5 Å, pFH = 0.1 and pHH = -10 results in a reactive trajectory.

{| class="wikitable"
| [[File:CPYX_FHH_contour_4.PNG|500px|thumb|center|Contour plot showing that trajectory is reactive]] || [[File:CPYX_FHH_internuclear_distance_vs_time_2.PNG|500px|thumb|center|"Internuclear Distance vs Time" plot showing that the trajectory is reactive]]
|}

In this reaction, unlike the previous reaction, both the reactants and products do not have much vibrational energy, but have more translational energy. This is shown in the smooth curves in the plots above. A high negative initial value of pHH and a much lower initial value of pFH were set. Initially, the H atom moves towards the HF molecule at a high negative velocity through a path that does not have the minimum potential energy, the H - F bond in the molecule weakens and rFH increases to a maximum of 1.33 Å at t=0.08. The H atom moves closer and collides with the HF molecule, such that the minimum value of rHH of 0.56 Å is reached very early on, at t=0.16. Kinetic energy is converted to potential energy in the HF molecule, and rFH decreases to a minimum value of 0.67 Å at t=0.21 as rHH increases to 1.01 Å. From the contour plot, this is the point on the trajectory where the system has the highest energy of ≈ -40 Kcal/mol. Potential energy is converted into kinetic energy when the trajectory moves away from this point and the kinetic energy is later converted into potential energy to overcome the activation barrier and form products. rHH decreases after reaching a maximum value of 1.03 Å at t=0.22, as a new H2 molecule is formed. At large values of t, the H2 molecule vibrates slightly, within the contour lines, and rHH averages at around 0.74 Å, the H – H bond length. The F atom has a high translational energy and moves away from the new H2 molecule at a high positive constant velocity such that rFH increases very quickly and reaches a value of 12.00 Å at t=1.50.

Using the same initial conditions as above but with pFH = 3.5 and pHH = -7.5 also results in a reactive trajectory, as shown by the contour plot and the internuclear distance vs time plot below. A slightly less negative, but still high, initial value of pHH and a higher initial value of pFH were set as compared to the previous reaction.

{| class="wikitable"
| [[File:CPYX_FHH_contour_1.PNG|500px|thumb|center|Contour plot showing that the trajectory is reactive]] || [[File:CPYX_FHH_internuclear_distance_vs_time_4.PNG|500px|thumb|center|"Internuclear Distance vs Time" plot showing that the trajectory is reactive]]
|}

In this reaction, the products have more vibrational energy and slightly less translational energy than the previous reaction, while the reactants still have much more translational energy than vibrational energy. Initially, the H atom moves towards the HF molecule at a high negative velocity through a path that does not have the minimum potential energy, the H - F bond in the molecule weakens and rFH increases to a maximum of 1.42 Å at t=0.10. The H atom moves closer and collides with the HF molecule, such that the minimum value of rHH of 0.54 Å is reached very early on, at t=0.18. Kinetic energy is converted to potential energy in the HF molecule, and rFH decreases to a minimum value of 0.67 Å at t=0.25 as rHH increases to 1.22 Å. From the contour plot, this is the point on the trajectory where the system has the highest energy of ≈ -50 Kcal/mol. This energy value obtained is more negative and not as high as the value obtained for the previous reaction, as the magnitude of the internuclear momentum pHH and the relative translational energy of the H atom is lower than in the previous reaction, so the H atom moves towards the HF molecule with less kinetic energy that can be converted into potential energy. The potential energy is converted back into kinetic energy when the trajectory moves away from this point, and the kinetic energy is later converted back into potential energy to overcome the activation barrier and form products. rHH decreases as a new H2 molecule is formed. At large values of t, the H2 molecule vibrates and rHH averages at around 0.74 Å, the H – H bond length. The vibrational energy of the H2 molecule is greater than in the previous reaction, as the oscillations in the plots are stronger and the trajectory crosses a contour line at each peak, meaning that the H2 molecule is in a vibrationally excited state. The F atom has a high translational energy and moves away from the new H2 molecule at a high positive constant velocity such that rFH increases very quickly and reaches a value of 10.29 Å at t=1.50, which is still not as high as the value obtained in the previous reaction with a greater translational energy contribution.

By changing the initial H - F bond length to increase the energy of the H - F vibration, a reactive trajectory cannot be obtained without decreasing the magnitude of pHH, and this means that this reaction is driven mainly by the translational energy of the incoming H atom, much more than the vibrational energy of the HF molecule.

==== Differences between the F + H2 and H + HF reactions ====

A reactive trajectory for the F + H2 reaction usually involves a low amount of translational energy in the reactants and generates a HF molecule in a vibrationally excited state and a H atom that has little translational energy. On the other hand, a reactive trajectory for the H + HF reaction involves a high amount of translational energy in the incoming H atom and generates a H2 molecule that may or may not be vibrationally excited and a F atom that has a very high translational energy.

For the exothermic F + H2 reaction, the activation barrier is very small and the transition state is reached early on in the trajectory, so the incoming F atom only needs very little translational energy to collide and react successfully with the H2 molecule, and the main factors determining the reactivity of the trajectory are the direction and orientation of approach and the states of the reactants. As this reaction is exothermic, energy is released, mainly as vibrational energy in the new HF molecule. On the other hand, for the endothermic H + HF reaction, the energy barrier of -30.162 Kcal/mol is not small and has to be overcome for a reaction to be successful. The transition state is also reached late in the trajectory. This explains the need for the H atom to approach the HF molecule with a high translational kinetic energy in reactive trajectories.

<references>
<ref name="Atkins">Atkins P and de Paula J. Physical Chemistry, 9th edition. Oxford: Oxford University Press;2009.</ref>
</references>

MRD:CPYX

2017-05-31T15:17:53Z

Je714: /* Transition state region and position */

== H + H2 system ==

=== Transition state region and position ===

The total gradient of the potential energy surface is 0 both at a minimum and at a transition structure. At a minimum, the gradient is negative before that point and positive after that point. The transition state is the maximum point on the minimum energy path, which means that it is at a relative minimum along 1 direction and at a relative maximum along the crossing direction. This point is known as a saddle point, as it is neither a minimum nor a maximum.

{{fontcolor1|gray| Good. Some discussion using second partial derivatives would have been nicer, though. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:17, 31 May 2017 (BST)}}

The transition state position rts is estimated to be 0.90775 Å (5.s.f.). When the initial conditions of r1 = r2 = 0.90775 Å and p1 = p2 = 0 are tested, the system does not seem to vibrate and the bond distances remain at 0.90775 Å until t=8.63, when atom C starts to move away from the newly-formed vibrating molecule AB. This is shown in the plot of internuclear distances against time below. The above initial conditions at time values of 0 < t < 8.63 also produced a trajectory where all the atoms remained at their initial position. This means that the trajectory only stays on the ridge when the bonds have zero momenta and this is the optimal distance between each of the 3 H atoms when they are all bonded to one another in the transition state.

{{fontcolor1|gray| The TS is a mathematical construct. In practice, you can always run a sufficiently long simulation in which the system will fall off towards the products or reactants. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:17, 31 May 2017 (BST)}}

[[File:CPYX_H2+H_TS_1.PNG|400px|thumb|center|"Internuclear Distances vs Time" plot for H2 + H]]

=== Comparison between the reaction path/mep and the trajectory using the same initial conditions ===

A mep and a trajectory with its calculation type as Dynamics were run separately using the initial conditions of r1 = rBC = 0.91775 Å, r2 = rAB = 0.90775 Å and p1 = p2 = 0 and compared. All atoms A, B and C refer to H atoms and all internuclear distances, time and internuclear momenta values which do not have to be very accurate are reported to 2.d.p.

If the initial conditions r1 = rBC = 0.90775 Å and r2 = rAB = 0.91775 Å were used instead, the trajectory will be a reflection of the original trajectory in the diagonal connecting the origin and the corner corresponding to the maximum internuclear distance values. In the internuclear distance and internuclear momentum vs time plots, the lines for the new A-B bond will be the same as the lines for the original B-C bond and the lines for the new B-C bond will be the same as the lines for the original A-B bond. The shapes of the plots will otherwise be the same.

{| class="wikitable" border="1"
|+ Comparison between mep and trajectory
! Differences !! Mep (20 000 steps) !! Trajectory (300 steps)
|-
| rowspan="2" | Contour plot || [[File:CPYX_H2+H_contour_mep.PNG|500px]] || [[File:CPYX_H2+H_contour_1.PNG|500px]]
|-
| colspan="2" | In both plots above, the products of molecule AB and atom C are formed, and atom C moves away from molecule AB. The mep is much smoother than the trajectory due to the much greater vibrational energy of the new molecule AB in the trajectory.
|-
| Internuclear distance vs time plot || [[File:CPYX_H2+H_mep_internuclear_distance_vs_time.PNG|500px]] || [[File:CPYX_H2+H_trajectory_internuclear_distance_vs_time.PNG|500px]]
|-
| Explanation || The final internuclear distances recorded are r1(100)=2.53 Å and r2(100)=0.74 Å. The value of r1(100) is much smaller than the value of r1(1.5) in the trajectory, as the velocity in the mep is set to 0 in each time step, unlike the trajectory. This means that atom C has very little translational kinetic energy in the mep and moves away from molecule AB much more slowly. After t=3.00, the value of r1 increases with time and the increase in the value of r1 decreases with time, as the distance between atom C and molecule AB increases and there is less repulsion between their electrons. The value of r2 stays constant at 0.74 Å, which is the H - H bond length, and vibrations are not observed due to the zero velocity and kinetic energy in this H - H bond. || The final internuclear distances recorded are r1(1.5)=5.28 Å and r2 ranges from 0.73 Å to 0.76 Å at large values of t. The products have reasonable non-zero amounts of translational and vibrational energy due to the release of energy from the transition state structure. After t=0.40, the value of r1 increases linearly with time while the value of r2 stays approximately constant, averaging around 0.74 Å, which is the H - H bond length.
|-
| Internuclear momentum vs time plot || [[File:CPYX_H2+H_mep_internuclear_momenta_vs_time.PNG|500px]] || [[File:CPYX_H2+H_trajectory_internuclear_momenta_vs_time.PNG|500px]]
|-
| Explanation || When obtaining the mep, the velocity is set to 0 in each time step. Since momentum pi =μvi where μ is mass and v is velocity, the internuclear momenta at all time values is 0. || Initially, both internuclear momenta are 0 and become slightly negative as the 3 hydrogen atoms move closer together, at a velocity that decreases the atomic distance, to form a transition state structure. After the transition state structure forms, the AB internuclear momentum remains negative as the new molecule AB is formed and the internuclear distance rAB decreases to a value of around the H - H bond length. However, the BC internuclear momentum becomes positive after t=0.11 and increases steadily as atom C moves away from the new molecule AB, converging to a value of 2.48 at large values of t, so p1(1.5)=2.48. After the molecule AB is formed, the AB internuclear momentum increases to a positive value and oscillates between 0.91 and 1.57 at large values of t due to vibrations in the H - H bond. This means that the final average internuclear momenta p2(1.5)=1.24.
|}

A calculation with initial positions at the final positions of the trajectory obtained above and the initial momenta the same as the final momenta in the trajectory obtained above but with their signs reversed, was performed. This means that rAB, rBC, AB momentum and BC momentum were set to 0.74 Å, 5.28 Å, -1.24 and -2.48 respectively. The results of the calculation are shown below.

{| class="wikitable" border="1"
! Contour plot !! Internuclear distance vs time plot !! Internuclear momentum vs time plot (extended to 500 steps/ t=2.50)
|-
| [[File:CPYX_H2+H_contour_2.PNG|400px]] || [[File:CPYX_H2+H_trajectory_internuclear_distance_vs_time_reversed.PNG|400px]] || [[File:CPYX_H2+H_internuclear_distance_vs_time_reversed.PNG|400px]]
|-
| The shape of this plot is very similar to the shape of the earlier trajectory, except that this trajectory moves in the opposite direction and circles around the region of 0.70 Å < rAB < 0.90 Å and 0.90 Å < rBC < 1.10 Å. This trajectory is very smooth, as most of the energy in this system is translational energy. The value of rAB, which corresponds to the vibrational energy of the H - H bond in molecule AB, is 0.74 Å, which is the H - H bond length, so molecule AB has very little vibrational energy. This shape shows that the trajectory is unreactive and will revert back to reactants AB and C, as the initial conditions do not provide enough kinetic energy to overcome the activation barrier. The maximum amount of potential energy reached by this system is lower than the potential energy of the transition state structure.
|| The shape of this plot is roughly the reverse of the plot obtained with the earlier trajectory. The value of rBC decreases linearly with time while the value of rAB stays approximately constant with vibrations. From t=1.20 to t=1.50, a transition state structure between atom C and molecule AB is attempted but not formed as the maximum value of rAB of 0.87 Å is reached at t=1.37 and the minimum value of rBC of 0.96 Å is reached at t=1.33. These values are not close to one another as well as to at the value of rts which is 0.90775 Å. Thus, atom C moves away from the molecule AB which remains intact, so rBC increases and rAB decreases to its original value of 0.74 Å. || The internuclear momenta remain relatively constant initially and become less negative after t=0.5 as the 3 H atoms move closer together more slowly, with a less negative velocity, due to repulsions between their electrons. pAB becomes positive at t=1.24 as rAB increases in an attempt to form a transition state structure. pAB becomes negative again at t=1.39 as rAB decreases to its original value of 0.74 Å. pAB becomes positive again at t=1.51, after molecule AB is re-formed, and oscillates between 0.70 and 1.71 at large values of t due to H - H bond vibrations. pBC becomes positive at t=1.31 and continues to increase as rBC increases and atom C moves away from molecule AB.
|}

=== Reactive and unreactive trajectories ===

In the following table, the initial positions are r1 = rBC = 0.74 Å and r2 = rAB = 2.00 Å. The momentum values are recorded in the table and the reactivity of their trajectories is shown.

{| class="wikitable" border=1
! p1 !! p2 !! Screenshot of trajectory !! Internuclear distance vs time plot !! Reactive/unreactive
|-
| -1.25 || -2.5 || [[File:CPYX_H2+H_contour_trajectory_1.PNG|300px]] || [[File:CPYX_H2+H_trajectory_1.PNG|300px]] || Reactive
The trajectory and internuclear distance vs time plot show that atom A moves towards molecule BC and collides with it. A transition state structure is formed at t=0.43, where rAB = rBC ≈ 0.90 Å, which is close to the rts of 0.90755 Å found earlier. Even though the magnitudes of the internuclear momenta appear to be small, this system has sufficient kinetic energy to overcome the activation barrier. This is because the system is driven much more by translational energy than by vibrational energy, as shown in the smooth trajectory. A bond forms between atoms A and B and the B-C bond breaks. rAB decreases until the optimum bond distance of 0.74 Å is reached at t=0.58, and the molecule continues to vibrate slightly, within the contour lines. rBC increases almost linearly with time as atom C moves away from molecule AB at a constant velocity.
|-
| -1.5 || -2.0 || [[File:CPYX_H2+H_contour_trajectory_2.PNG|300px]] || [[File:CPYX_H2+H_trajectory_2.PNG|300px]]
|| Unreactive
These plots show atom A initially moving towards a slightly vibrating molecule BC. After a minimum internuclear distance rAB of 1.12 Å is reached at t=0.44, a transition state structure is not formed and rAB increases linearly with time as atom A moves away from the intact molecule BC at a constant velocity. The magnitude of the momentum p2 is not high enough for atom A to move towards molecule BC with sufficient translational kinetic energy to overcome the activation barrier and form a transition state structure. In the trajectory shown in the contour plot, the system does not have enough energy for the trajectory to cross the contour line to reach a higher energy transition state structure. The transition state internuclear distance rts of 0.90775 Å is lower than the minimum value of rAB observed. The molecule BC remains intact and rBC is nearly constant throughout the whole trajectory, oscillating between 0.71 Å and 0.77 Å, apart from bond stretching vibrations and a very slight increase in value from t=0.26 to t=0.67 to a peak of 0.81 Å at t=0.47. This is due to the weakening of the H - H bond in molecule BC as atom A approaches near atom B in molecule BC.
|-
| -1.5 || -2.5 || [[File:CPYX_H2+H_contour_trajectory_3.PNG|300px]] || [[File:CPYX_H2+H_trajectory_3.PNG|300px]] || Reactive
An increase in the initial magnitude of the AB momentum by 0.5 causes the trajectory to change from unreactive to reactive. This magnitude pAB, which corresponds to the relative translational energy of atom A, is the same as the magnitude pAB in the first trajectory in this section. As atom A moves towards a vibrating molecule BC, they collide to form a transition state structure at t=0.43, where rAB = rBC ≈ 0.89 Å. This value is close to the rts reported above of 0.90775 Å and this system has sufficient kinetic energy to overcome the activation barrier. A bond forms between atoms A and B to form a new molecule AB and the H - H bond in molecule BC breaks. rAB decreases until an optimum bond distance of 0.74 Å, which is the H -H bond length, is reached at t=0.59, and the molecule AB continues to vibrate such that rAB oscillates between 0.71 Å and 0.78 Å, within the contour lines, at large values of t. rBC increases almost linearly with time as atom C moves away from molecule AB at a constant velocity. Comparing this result with the previous rows shows that the magnitude of p2 (pAB) and hence the translational energy of the atom A is more important in determining the reactivity of a trajectory than the magnitude of p1 which reflects the vibrational energy of the molecule BC.
|-
| -2.5 || -5.0 || [[File:CPYX_H2+H_contour_trajectory_4.PNG|300px]] || [[File:CPYX_H2+H_trajectory_4.PNG|300px]] || Unreactive
Although atom A and molecule BC have high magnitudes of their momentum and kinetic energy, the transition state structure formed does not result in the formation of products AB and C, but the system reverts back to the reactants. Initially, atom A moves towards molecule BC at a high velocity and a transition state structure containing all 3 hydrogen atoms is formed very early on. After t=0.16, where rAB = rBC ≈ 0.82 Å, rAB < rBC. At t=0.21 and t=0.46, rAB have minima at 0.65 Å and 0.57 Å respectively. At t=0.34, rAB reaches a maximum in the transition state structure of 1.11 Å, but it is still smaller than the rBC value of 1.18 Å. Meanwhile, from t=0, rBC increases from its initial value of 0.74 Å to its maximum value of 1.37 Å at t=0.46. As rBC decreases from its maximum value and rAB increases from its minimum value, the 2 lines representing the bond distances intersect again at t=0.54, where rAB = rBC = 1.09 Å. With a high kinetic energy in this system, atom A and molecule BC orbit around each other after collision, so they approach each other in random directions and all memory of the initial approach direction is lost. Atom A and molecule BC also rotate as they spend much time around the transition state, so they eventually approach each other in directions, rotational states and vibrational states that do not lead to the formation of products. This is why this trajectory is unreactive and molecule AB and atom C do not form. The excess energy is converted into vibrational energy in molecule BC as well as a relative translational energy in atom A instead. At large values of t, rBC oscillates between 0.56 Å and 1.01 Å, crossing a few contours in the trajectory, as molecule BC is in a vibrationally excited state. rAB increases almost linearly with time as atom A moves away from molecule BC at a constant velocity, offset by the strong H - H bond stretching vibrations in molecule BC.
|-
| -2.5 || -5.2 || [[File:CPYX_H2+H_contour_trajectory_5.PNG|300px]] || [[File:CPYX_H2+H_trajectory_5.PNG|300px]]
|| Reactive
Although this internuclear distance vs time plot appears to be very similar to the above plot and the initial conditions are almost similar, the products of molecule AB and atom C are formed and the system does not revert back to the reactants. Initially, atom A moves towards molecule BC at a high velocity, so rAB decreases and rBC increases with time. At t=0.16, rAB = rBC = 0.82 Å. Atom A continues to move towards molecule BC until t=0.20, when rAB reaches a minimum at 0.64 Å. Atom A then "bounces off" molecule BC slightly. At t=0.32, rAB = rBC = 1.13 Å, which is very close to the maximum value of rBC reached in this transition state structure of 1.14 Å at t=0.33. rBC then decreases to a minimum value of 0.57 Å at t=0.46 while rAB increases to a maximum value of 1.37 Å at t=0.45. After that, rAB decreases and rBC increases with time and at t=0.54, rAB = rBC ≈ 1.07 Å. As rAB decreases further, the transition state structure is resolved, molecule AB is formed and atom C moves away from the new molecule AB and rBC increases. At large values of t, rAB oscillates between 0.56 Å and 1.02 Å, crossing a few contours in the trajectory, as molecule AB is in a vibrationally excited state. rBC increases almost linearly with time as atom C moves away from molecule AB at a constant velocity, offset by the strong H - H bond stretching vibrations in molecule AB. This trajectory is reactive as even though atom A and molecule BC spend much time around the transition state structure and orbit around each other and rotate, they eventually approach each other at the correct angle, vibrational and rotational states with sufficient translational energy, leading to product formation.
|}

=== Comparison of Transition State Theory predictions with experimental values ===

The main assumptions of Transition State Theory are that an activated complex is in equilibrium with the reactants, the rate of formation of products from this activated complex depends on the rate at which it passes through a transition state and every trajectory that passes through the transition state structure is reactive.<ref name="Atkins" />

From the results obtained in the previous part, the main factor influencing the outcome of the first 3 trajectories is the amount of kinetic energy present in the system, as the magnitudes of the internuclear momenta are relatively low. If the reactants have enough kinetic energy to reach the transition state and cross the activation barrier, products will be formed and the trajectory will be reactive. If the reactants do not have enough kinetic energy, the transition state will not be reached and products will not be formed. The reactive trajectories in this category all pass through the transition state structure and the 2nd trajectory which is unreactive does not pass through the transition state structure as the reactants do not have enough kinetic energy. These results show that the assumptions of the Transition State Theory are largely valid. Thus, Transition State Theory predictions for reaction rate values will be similar to experimental reaction rate values when the reactants approach each other at reasonably low levels of translational and vibrational energy.

The last 2 trajectories from the previous part have a very high kinetic energy in the system, as the magnitudes of internuclear momenta are high and around 2 times the values in the previous trajectories. According to the assumptions of the Transition State Theory, they both have enough kinetic energy so they should both pass through the transition state and form products. However, the 4th trajectory is unreactive while the last trajectory is reactive, even though their trajectories both show that they pass through a transition state. This means that not every trajectory that passes through the transition state structure is reactive and some trajectories that pass through the transition state structure will cross the barrier again and revert back to the reactants. This is because the orientation and direction in which the reactants approach each other, as well as their rotational and vibrational states, can influence the success of their collision, and this is the main factor influencing the outcome of these trajectories. Thus, Transition State Theory predictions for reaction rate values will be an overestimate and will be higher than experimental rate values as Transition State Theory assumes that collisions at the transition state will always be successful and this is not always the case.

== F - H - H system ==

=== Energetics of the F + H2 and H + HF reactions ===

The F + H2 reaction is exothermic and the H + HF reaction is endothermic.

In the F + H2 reaction, a H - H bond is broken and a H - F bond is formed. Bond breaking is endothermic while bond forming is exothermic. The H - F bond is very strong (565 kJ/mol) due to the polarized nature of the H - F covalent bond and the high electronegativity difference between H, which has a moderate electronegativity, and F, which has a very high electronegativity. In contrast, the H - H bond is non-polar, so it is much weaker (432 kJ/mol) than the H - F bond. As the formation of the H - F bond releases more energy than the amount of energy used in the dissociation of the H - H bond, the F + H2 reaction releases energy to the surroundings in the form of heat and is exothermic.

In the H + HF reaction, a strong H - F bond is broken and a weaker H - H bond is formed. As the dissociation of the H - F bond requires more energy than the energy released through the formation of the H - H bond, this reaction takes in energy from the surroundings in the form of heat and is endothermic.

=== Transition state position and activation energies ===

When A is a F atom and B and C are both H atoms, the transition state is located at approximately rAB = 1.8100 Å and rBC = 0.74649 Å (5.s.f.). The plot of internuclear distance vs time where AB and BC both have zero momenta is shown below.

[[File:CPYX_FHH_TS_3.PNG|500px|thumb|center|"Internuclear Distances vs Time" plot for F-H-H]]

From this plot, the bond distances rAB and rBC remain constant until t=6.5. This means that these bond distances are at their optimum values in the transition state when all 3 atoms are bonded, and this structure stays on the ridge on the trajectory and does not fall off into reactants or products.

The activation energies for the F + H2 and H + HF reactions can be calculated using the energies of the reactants for each reaction and the energy of the transition state. These energy values can be found from the contour plot and they are recorded below.

Energy of H + H-F: -133.9 Kcal/mol

Energy of F + H-H: -103.9 Kcal/mol

Energy at the transition state: -103.7 Kcal/mol

These values can be used to find the activation energies for the H + HF and the F + H2 reactions, which are reported below.

Activation energy for H + HF reaction = -103.7 - (-103.9) = 0.2 Kcal/mol

Activation energy for F + H2 reaction = -103.7 - (-133.9) = 30.2 Kcal/mol

Other methods can be used to find the activation energies for these reactions. A mep using the initial conditions of rFH = 1.82 Å, rHH = 0.74649 Å, pFH = pHH = 0 was performed. The potential energy vs time plot shown below shows an initial energy of -103.75 Kcal/mol (2.d.p.), which corresponds to the energy of the transition state, and a final energy after 900 000 steps of -104.01 Kcal/mol (2.d.p.) which is the energy of F + H2. Thus, a better estimate of the activation energy for the F + H2 reaction is -103.75 - (-104.01) = 0.26 Kcal/mol.

[[File:CPYX_FHH_PE_vs_time_3.PNG|500px|thumb|center|Potential energy vs time plot for to find the activation energy for F + H2]]

A mep using the initial conditions of rFH = 1.80 Å, rHH = 0.74649 Å, pFH = pHH = 0 was performed. The potential energy vs time plot shown below shows an initial energy of -103.76 Kcal/mol (2.d.p.), the energy of the transition state, and a final energy of -133.92 Kcal/mol (2.d.p.), which is the energy of H + HF. Thus, a better estimate of the activation energy for the H + HF reaction is -103.76 - (-133.92) = 30.16 Kcal/mol.

[[File:CPYX_FHH_PE_vs_time_1.PNG|500px|thumb|center|Potential energy vs time plot for to find the activation energy for H + HF]]

=== Reaction dynamics ===

==== Reactive trajectories for the F + H2 reaction ====

A set of initial conditions that results in a reactive trajectory for the F + H2 reaction are rHF = 2.00 Å, rHH = 0.74 Å, pHF = -2.00 and pHH = -1.25. The reaction energy is released as kinetic energy, in the form of the relative translational energy of the H atom away from the HF molecule as well as vibrational energy of the HF molecule itself. This is confirmed by the contour plot and the plot of internuclear momentum vs time shown below.

{| class="wikitable"
| [[File:CPYX_FHH_contour_2.PNG|500px|thumb|center|Contour plot showing that trajectory is reactive]] || [[File:CPYX_FHH_internuclear_momentum_vs_time_1.PNG|500px|thumb|center|"Internuclear Momentum vs Time" plot]]
|}

From the plots above, the H2 molecule has some vibrational energy initially, as shown in both the plots above where the trajectory and initial internuclear momentum pHH oscillate. The F atom moves towards the H2 molecule to form a transition state structure and then a HF molecule and a H atom. From the contour plot, the newly-formed HF molecule vibrates between energy levels quite high above the minimum energy path, and this vibration is much more extensive than the initial H2 molecule. In the initial part of the trajectory, even with vibrations present in the H2 molecule, the contour lines are not crossed until the transition state structure dissociates to form products. The internuclear momentum pHF also oscillates between -5.98 and 9.23, which is a large range, at large values of t. This is because the newly-formed HF molecule is in a vibrationally excited state. The internuclear momentum pHH at large values of t is a constant but rather low positive value of 1.83, as the H atom leaves the HF molecule with some but not much translational energy. From the contour plot, the products contain more vibrational energy in the HF molecule than translational energy in the leaving H atom.

This can be confirmed experimentally by using infrared chemiluminescence, where vibrationally excited molecules emit infrared radiation as they return to the ground state. Reactants with particular vibrational and translational energy levels can be shot at each other to react and the populations of the vibrational states of the HF molecules formed can be determined by measuring and comparing the intensities of the peaks in the infrared emission spectrum.<ref name="Atkins" />

Setting rHH = 0.74 Å, rFH = 2.00 Å and pFH = -0.5, the values of pHH in the range -3 to 3 that result in reactive trajectories do not fall into a few particular ranges, but range across the whole spectrum. However, low magnitudes of pHH from -0.90 to 0.32 result in reactive trajectories. Outside this range of values, for trajectories with even a small range of pHH values, their general reactivity cannot be meaningfully predicted. Changing the internuclear distance rFH only slightly can also make an unreactive trajectory reactive or make a reactive trajectory unreactive.

This is because the reaction of F + H2 is an exothermic reaction which requires a very small amount of energy of 0.26 Kcal/mol to achieve a transition state structure. Thus, the main factor determining the reactivity of a trajectory for this reaction is not the amount of kinetic energy in this system, but the direction and orientation of approach, the rotational states and the vibrational states of the reactants. In this system, setting a low initial magnitude of pHH which corresponds to very little kinetic energy in the H - H bond allows the F atom to approach and collide with the H2 molecule in a linear configuration easily. The products of a H atom and a vibrationally excited HF molecule will be formed easily as energy is released in the form of vibrational energy in the HF molecule as the products are formed. However, setting a large value of pHH means that the amount of kinetic energy in the system is high, allowing the F atom and the HF molecule to orbit around each other after collision, so they approach each other in random directions and all memory of the initial approach direction is lost. The reactants also rotate if they spend much time around the transition state, so they may or may not eventually approach each other in directions, rotational states and vibrational states that lead to the formation of products.

After setting up the initial conditions of rHH = 0.74 Å, rFH = 2.00 Å, pFH = -0.8 and pHH = 0.1, the trajectory obtained is reactive. The overall energy of the system is considerably reduced at these low momenta, so the reactants approach each other in a linear configuration that results in product formation. The contour plot and plot of internuclear distance vs time for this trajectory are shown below.

{| class="wikitable"
| [[File:CPYX_FHH_contour_3.PNG|500px|thumb|center|Contour plot showing that trajectory is reactive]] || [[File:CPYX_FHH_internuclear_distance_vs_time_1.PNG|500px|thumb|center|"Internuclear Distance vs Time" plot showing that the trajectory is reactive]]
|}

From the plot above, rFH decreases and rHH oscillates initially due to vibrations in the H2 molecule. At t=0.47, rFH=1.81 Å and the transition state structure starts to form. rHH starts to increase slowly and it increases more quickly from t=1.11 and rHH = rFH = 1.03 Å at t=1.25. At t=1.30, rFH reaches a minimum of 0.76 Å and rHH reaches a maximum of 1.29 Å. rFH increases and rHH decreases and at t=1.35, rHH = rFH = 1.06 Å. At t=1.52, rFH reaches a maximum of 1.34 Å and at t=1.57, rHH reaches a minimum in the transition state of 0.74 Å. rHH then increases almost linearly with time as a H atom originally from the H - H bond moves away from the newly-formed HF molecule, offset by the strong H - F bond stretching vibrations. At large values of t, rFH oscillates between 0.75 Å and 1.19 Å due to the H - F bond stretching vibrations. The newly-formed HF molecule is in a vibrationally excited state, as the trajectory crosses s few contours as it oscillates and rFH oscillates over a large range of values.

==== Reactive trajectories for the H + HF reaction ====

For this reaction, setting rFH = 0.92 Å, rHH = 2.5 Å, pFH = 0.1 and pHH = -10 results in a reactive trajectory.

{| class="wikitable"
| [[File:CPYX_FHH_contour_4.PNG|500px|thumb|center|Contour plot showing that trajectory is reactive]] || [[File:CPYX_FHH_internuclear_distance_vs_time_2.PNG|500px|thumb|center|"Internuclear Distance vs Time" plot showing that the trajectory is reactive]]
|}

In this reaction, unlike the previous reaction, both the reactants and products do not have much vibrational energy, but have more translational energy. This is shown in the smooth curves in the plots above. A high negative initial value of pHH and a much lower initial value of pFH were set. Initially, the H atom moves towards the HF molecule at a high negative velocity through a path that does not have the minimum potential energy, the H - F bond in the molecule weakens and rFH increases to a maximum of 1.33 Å at t=0.08. The H atom moves closer and collides with the HF molecule, such that the minimum value of rHH of 0.56 Å is reached very early on, at t=0.16. Kinetic energy is converted to potential energy in the HF molecule, and rFH decreases to a minimum value of 0.67 Å at t=0.21 as rHH increases to 1.01 Å. From the contour plot, this is the point on the trajectory where the system has the highest energy of ≈ -40 Kcal/mol. Potential energy is converted into kinetic energy when the trajectory moves away from this point and the kinetic energy is later converted into potential energy to overcome the activation barrier and form products. rHH decreases after reaching a maximum value of 1.03 Å at t=0.22, as a new H2 molecule is formed. At large values of t, the H2 molecule vibrates slightly, within the contour lines, and rHH averages at around 0.74 Å, the H – H bond length. The F atom has a high translational energy and moves away from the new H2 molecule at a high positive constant velocity such that rFH increases very quickly and reaches a value of 12.00 Å at t=1.50.

Using the same initial conditions as above but with pFH = 3.5 and pHH = -7.5 also results in a reactive trajectory, as shown by the contour plot and the internuclear distance vs time plot below. A slightly less negative, but still high, initial value of pHH and a higher initial value of pFH were set as compared to the previous reaction.

{| class="wikitable"
| [[File:CPYX_FHH_contour_1.PNG|500px|thumb|center|Contour plot showing that the trajectory is reactive]] || [[File:CPYX_FHH_internuclear_distance_vs_time_4.PNG|500px|thumb|center|"Internuclear Distance vs Time" plot showing that the trajectory is reactive]]
|}

In this reaction, the products have more vibrational energy and slightly less translational energy than the previous reaction, while the reactants still have much more translational energy than vibrational energy. Initially, the H atom moves towards the HF molecule at a high negative velocity through a path that does not have the minimum potential energy, the H - F bond in the molecule weakens and rFH increases to a maximum of 1.42 Å at t=0.10. The H atom moves closer and collides with the HF molecule, such that the minimum value of rHH of 0.54 Å is reached very early on, at t=0.18. Kinetic energy is converted to potential energy in the HF molecule, and rFH decreases to a minimum value of 0.67 Å at t=0.25 as rHH increases to 1.22 Å. From the contour plot, this is the point on the trajectory where the system has the highest energy of ≈ -50 Kcal/mol. This energy value obtained is more negative and not as high as the value obtained for the previous reaction, as the magnitude of the internuclear momentum pHH and the relative translational energy of the H atom is lower than in the previous reaction, so the H atom moves towards the HF molecule with less kinetic energy that can be converted into potential energy. The potential energy is converted back into kinetic energy when the trajectory moves away from this point, and the kinetic energy is later converted back into potential energy to overcome the activation barrier and form products. rHH decreases as a new H2 molecule is formed. At large values of t, the H2 molecule vibrates and rHH averages at around 0.74 Å, the H – H bond length. The vibrational energy of the H2 molecule is greater than in the previous reaction, as the oscillations in the plots are stronger and the trajectory crosses a contour line at each peak, meaning that the H2 molecule is in a vibrationally excited state. The F atom has a high translational energy and moves away from the new H2 molecule at a high positive constant velocity such that rFH increases very quickly and reaches a value of 10.29 Å at t=1.50, which is still not as high as the value obtained in the previous reaction with a greater translational energy contribution.

By changing the initial H - F bond length to increase the energy of the H - F vibration, a reactive trajectory cannot be obtained without decreasing the magnitude of pHH, and this means that this reaction is driven mainly by the translational energy of the incoming H atom, much more than the vibrational energy of the HF molecule.

==== Differences between the F + H2 and H + HF reactions ====

A reactive trajectory for the F + H2 reaction usually involves a low amount of translational energy in the reactants and generates a HF molecule in a vibrationally excited state and a H atom that has little translational energy. On the other hand, a reactive trajectory for the H + HF reaction involves a high amount of translational energy in the incoming H atom and generates a H2 molecule that may or may not be vibrationally excited and a F atom that has a very high translational energy.

For the exothermic F + H2 reaction, the activation barrier is very small and the transition state is reached early on in the trajectory, so the incoming F atom only needs very little translational energy to collide and react successfully with the H2 molecule, and the main factors determining the reactivity of the trajectory are the direction and orientation of approach and the states of the reactants. As this reaction is exothermic, energy is released, mainly as vibrational energy in the new HF molecule. On the other hand, for the endothermic H + HF reaction, the energy barrier of -30.162 Kcal/mol is not small and has to be overcome for a reaction to be successful. The transition state is also reached late in the trajectory. This explains the need for the H atom to approach the HF molecule with a high translational kinetic energy in reactive trajectories.

<references>
<ref name="Atkins">Atkins P and de Paula J. Physical Chemistry, 9th edition. Oxford: Oxford University Press;2009.</ref>
</references>

Talk:MRD:wfbfmicou

2017-05-31T15:15:18Z

Je714: Created page with "Good job. A Clear report, concise and showing you understand the concepts. Careful with your first answer, you should perhaps elaborate a bit more to show understanding. Pleas..."

Good job. A Clear report, concise and showing you understand the concepts. Careful with your first answer, you should perhaps elaborate a bit more to show understanding. Please see my inline comments for minor corrections.

[[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:15, 31 May 2017 (BST)

MRD:wfbfmicou

2017-05-31T15:14:18Z

===What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.===

At both a minimum and at a transition structure, the total gradient of the potential energy surface is zero. The points can be distinguished by their second derivative; the transition state is a saddle point. For a function <math>f\left(x,y\right)</math>, at a stationary point <math>\left(a,b\right)</math> consider the discriminant <math>D\equiv f_{xx}f_{yy}-f_{xy}^{2}</math>:

If <math>D>0</math>, and <math>f_{xx}\left(a,b\right)>0</math>, the point (a,b) is a local minimum.

If <math>D>0</math>, and <math>f_{xx}\left(a,b\right)<0</math>, the point (a,b) is a local maximum.

If <math>D<0</math>, the point (a,b) is a saddle point.

If <math>D=0</math>, higher order tests are required.

{{fontcolor1|gray| I assume you've copied this over from wikipedia (https://www.wikiwand.com/en/Second_partial_derivative_test). It's not wrong, but:
1) Cite your sources
2) D is the determinant of the Hessian matrix
[[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:05, 31 May 2017 (BST)}}

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.===

rts = 0.90775 Å

[[File:wm1415TS_position_internuclear_distance-time.png|500px]]

The internuclear distances remain effectively constant {{fontcolor1|gray| Why? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:06, 31 May 2017 (BST)}}: this represents the metastable transition state.

===Comment on how the ''mep'' and the dynamic trajectories differ.===
[[File:wm1415Mep_vs_dynamic.png|500px|thumb|left|MEP Calculation]]
[[File:wm1415Dynamic_vs_mep.png|500px|thumb|right|Dynamic Calculation]]

{{fontcolor1|gray| Good description [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:07, 31 May 2017 (BST)}}

In the MEP calculation, the kinetic energy of the particles is reset at each time step. The calculation begins at a structure similar to the transition state, with the H1-H2 distance slightly larger than the TS structure. As a result, the trajectory of the reaction in the MEP calculation follows the valley of the potential energy surface, without oscillating, to the reactants: H1 and H2-H3.
In the dynamic calculation, the inertia of the particles is taken into account: the velocities are not reset to zero at each step. Hence, vibrations are taken into account: the trajectory of the dynamic calculation also ends at the reactants, but oscillations are observed in this process.

===How do internuclear distances and momenta vary with time, when a reaction coordinate on either side of the transition state is chosen?===
[[File:wm1415Internuc_dist_time_r1.png|500px|thumb|left|Internuclear dist. vs. time - r1 = rts+δ, r2 = rts]]
[[File:wm1415Internuc_dist_time_r2.png|500px|thumb|right|Internuclear dist. vs. time - r1 = rts, r2 = rts+δ]]

[[File:wm1415Internuc_momenta_time_r1.png|500px|thumb|left|Internuclear momenta vs. time - r1 = rts+δ, r2 = rts]]
[[File:wm1415Internuc_momenta_time_r2.png|500px|thumb|right|Internuclear momenta vs. time - r1 = rts, r2 = rts+δ]]

The transition state is symmetrical: changing the initial position from (r1 = rts+δ, r2 = rts) to (r1 = rts, r2 = rts+δ) simply results in the formation of H1-H2 + H3 instead of H1 + H2-H3. The internuclear momenta and distance-time relationships are identical, with labels swapped.

===Reactive and unreactive trajectories===
{| class="wikitable" border=1
! p1 !! p2
!Outcome
!Description
|-
| -1.25 || -2.5
|[[File:wm1415Case1.png|200px|thumb|Reactive]]
|The H1-H2 molecule initially has no vibrational energy: the initial kinetic energy is purely translational energy of the approaching H3 atom. The H3 atom has sufficient kinetic energy to lead to the formation of the transition state and the products are formed: this is a reactive trajectory. The resulting H2-H3 molecule is vibrationally excited.
|-
| -1.5 || -2.0
|[[File:wm1415Case2.png|200px|thumb|Unreactive]]
|The H1-H2 molecule is initially vibrating: this causes the oscillations in the A-B (H3-H2) distance. H3 slows down as it approaches H2, converting kinetic energy to potential energy. For this set of initial conditions, the kinetic energy of H3 is insufficient to form the transition state and hence the products are not formed. H3 is eventually expelled by the H2-H1 molecule: this trajectory is unreactive.
|-
| -1.5 || -2.5
|[[File:wm1415Case3.png|200px|thumb|Reactive]]
|This reactive trajectory is similar to the first case. The H1-H2 molecule is initially vibrating with small amplitudes. H3 has sufficient kinetic energy to form the transition state and to expel H1 from the molecule. Following the reaction, the H3-H2 molecule is vibrationally excited.
|-
| -2.5 || -5.0
|[[File:wm1415Case4.png|200px|thumb|Unreactive]]
|The H2-H1 molecule initially has no vibrational energy. The approaching H3 atom has sufficient kinetic energy to reach the transition state: however, this particular set of initial conditions leads to barrier recrossing - the trajectory is unreactive. The large energies involved - far larger than required to reach the transition state - lead to a more unpredictable outcome. In this case, the transition state is so vibrationally excited that H3 is ejected from the activated complex and the system returns to the reactants channel, with the H2-H1 molecule vibrationally excited as a result.
|-
| -2.5 || -5.2
|[[File:wm1415Case5.png|200px|thumb|Reactive]]
|This set of initial conditions is similar to the previous example, except in this case the barrier recrossing leads to the ejection of the H1 atom from activated complex, giving the vibrationally excited H2-H3 molecule in the products channel.
|}

===State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for rate values compare with experimental values?===

Transition State Theory (TST) is a model that simplifies the computational complexity of calculating trajectories along a potential energy surface. TST requires knowledge of the PES in the reactant and product channels, as well as at the transition state1. A boundary is chosen - known as the critical dividing surface1 - which intersects the saddle point of the potential energy surface. The main assumptions of TST of ideal gas reactions are:

- "All supermolecules that cross the critical dividing surface from the reactant side become products."1

- "During the reaction, the Boltzmann distribution of energy is maintained for the reactant molecules."1

- "The supermolecules crossing the critical surface from the reactant side have a Boltzmann distribution of energy corresponding to the temperature of the reacting system."1

{{fontcolor1|gray| The 2nd and 3rd point are from a statistical point of view of TST. Not really applicable here, since you're only looking at a triatomic collision in isolation --- not at an ensemble of molecules [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:10, 31 May 2017 (BST)}}

As seen in the fourth example above, the assumption that all supermolecules with energy greater than the transition state lead to a reaction breaks down. In the fourth and fifth examples, barrier recrossing is observed in the trajectory calculations. Barrier recrossing is a phenomenon that is not accounted for in the transition state model. Therefore, transition state theory would predict larger rates of reaction than observed experimentally.

{{fontcolor1|gray| Good. This is the main point: it's what you're observing with your previous calculations. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:10, 31 May 2017 (BST)}}

Neither TST nor the model used in this investigation account for a further complication - the effect of quantum tunnelling. For small particles that do not classically have the energy to overcome a potential barrier, there is a certain probability of quantum tunnelling through the barrier (which depends on the energy difference, tunnelling length, and mass). Therefore, in the example of H3 + H2-H1, if the initial conditions lead to a collision which has insufficient energy to reach the transition state, tunnelling through the potential barrier to the other side of the PES and hence to the products is feasible. This phenomenon would lead to a small increase in the expected rate of reaction - however, the barrier recrossing effect dominates overall and hence we would still expect slower rates than calculated in TST.

{{fontcolor1|gray| What is a supermolecule? Good discussion of QM and acknowledging that it's not taking into account in these simulations [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:10, 31 May 2017 (BST)}}

===Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions.===

F + H2
[[File:wm1415F_H2_exothermic.png|500 px|thumb|left|Exothermic PES]]
[[File:wm1415F_H2_transition_state.png|500 px|thumb|right|Approximate location of transition state]]
[[File:Wm1415_exo_reactants.png|500 px|thumb|left|MEP calculation from a distorted transition state to the reactants.]]
[[File:Wm1415_exo_products.png|500 px|thumb|right|MEP calculation from a distorted transition state to the products.]]

The reactants channel is higher in energy than the products channel: the reaction is clearly exothermic. Formation of a H-F bond releases more energy than the formation of a H-H bond: the H-F bond is stronger. Since the activation energy is small, the energies of the reactants and the transition states are similar: by Hammond's postulate, the transition state will resemble the structure of the reactants. Thus we expect the transition state to be located near the reactants channel: this can be seen above. {{fontcolor1|gray| Good reasoning [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:12, 31 May 2017 (BST)}}
To find the energy of the reactants, an MEP calculation was run from a structure nearing the transition state, with the F-H bond slightly stretched, to produce a potential energy vs. time graph. Similarly. the energy of the products was found from a structure nearing the transition state, with the F-H bond slightly compressed.

Energy of the reactants ≈ -103.95 kcal/mol

Energy of the transition state ≈ -103.75 kcal/mol

Energy of the products ≈ -133.85 kcal/mol

Activation energy ≈ 0.2 kcal/mol

Energy change of reaction ≈ -29.90 kcal/mol (negative because exothermic - release of energy)

{{fontcolor1|gray| Good discussion [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:12, 31 May 2017 (BST)}}

H + HF
[[File:wm1415H_HF_endothermic.png|500 px|thumb|left|Endothermic PES]]
[[File:Wm1415H_HF_transition_state.png|500 px|thumb|right|Approximate location of transition state]]
[[File:Wm1415_endo_reactants.png|500 px|thumb|left|MEP calculation from a distorted transition state to the reactants.]]
[[File:Wm1415_endo_products.png|500 px|thumb|right|MEP calculation from a distorted transition state to the products.]]

The reactants channel in the reverse reaction is lower in energy than the products channel: the reaction is endothermic. By Hammond's postulate, the transition state should resemble the structure of the products - the location of the transition state is very close to the products channel. Potential energy calculations against time were carried out for completeness - but since it is the reverse reaction of the above situation, it is clear that the energy calculations should be symmetrical.

Energy of the reactants ≈ -133.85 kcal/mol

Energy of the transition state ≈ -103.75 kcal/mol

Energy of the products ≈ -103.85 kcal/mol

Activation energy ≈ 30.0 kcal/mol

Energy change of reaction ≈ +29.90 kcal/mol (positive because endothermic)

===F + H2: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?===
[[File:Wm1415_energy_conservation.png|500 px|thumb|left|Reaction trajectory: potential energy surface.]]
[[File:Wm1415_energy_conservation_momenta.png|500 px|thumb|right|F + H2: internuclear momenta vs. time]]

Following the reaction, the H-F molecule formed is vibrationally excited. This can be seen from the very large oscillations in the potential energy diagram and the internuclear momenta vs. time graph. The energy released in the reaction has been converted into vibrational kinetic energy - heat. The amount of heat energy dissipated by the reaction could be measured by calorimetry.
The vibrational excitation of the H-F bond could be further confirmed by IR spectroscopy by comparing the IR spectrum obtained from the vibrationally excited products to the IR spectrum of a HF molecule at room temperature. At ambient temperatures, only the ground vibrational state is substantially occupied, as the vibrational temperature of the H-F bond is very large. Therefore, the transitions observed in the infrared spectrum of HF would be from the v = 0 to v = 1 state (and perhaps some v = 1 to v = 2 transitions). The lower vibrational levels closely follow the simple harmonic oscillator model, therefore the transitions occurring at room temperature would all involve the same energy and only a single line would be observed in the spectrum.
The simple harmonic oscillator model breaks down at high vibrational levels for diatomic molecules. The energy spacings become progressively smaller until they converge at the dissociation limit, where the bond breaks due to the large vibrational energy. The HF molecule produced in this reaction is vibrationally excited, and its infrared spectrum would involve energy transitions that would match energy spacings in the anharmonic oscillator model. Therefore, in comparing its spectrum to room temperature, there would be a shift to many lower frequency absorptions, that would eventually converge to the dissociation limit (although the convergence point may not be experimentally observed, as the vibrational energy would most likely be dissipated to the environment such that the bond formed in the reaction wouldn't break).

{{fontcolor1|gray| Good discussion! [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:13, 31 May 2017 (BST)}}

===Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.===

{{fontcolor1|gray| Good job giving examples [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:14, 31 May 2017 (BST)}}

F + H2

The Polanyi rules refer to how the distribution of kinetic energy in the transition state affects whether a trajectory is reactive or unreactive. A higher proportion of vibrational energy promotes reactions with a late transition state, whereas a higher proportion of translational energy promotes reactions with an early transition state2.
{| class="wikitable" border=1
! pFH !! pHH
!Outcome
|-
| -0.5 || -2.8
|[[File:Wm1415mom-2.8.png|200 px|thumb|Reactive]]
|-
| -0.5 || -2.7
|[[File:Wm1415mom-2.7.png|200 px|thumb|Unreactive]]
|-
| -0.5 || 2.7
|[[File:Wm1415mom2.7.png|200 px|thumb|Unreactive]]
|-
| -0.5 || 3.0
|[[File:Wm1415mom3.0.png|200 px|thumb|Reactive]]
|-
| -0.8 || 0.1
|[[File:Wm1415lowmom.png|200 px|thumb|Reactive]]
|}
The F + H2 reaction is exothermic, with a very early transition state, as shown above. Thus, according to the empirical Polanyi's rules, a higher proportion of translational energy would result in a reactive trajectory. This is clearly seen in the last example, where a significantly lower energy system results in a reactive trajectory: the kinetic energy is largely translational (there are only small oscillations in the reactants channel).
In the first four cases, most of the kinetic energy in the system is in the form of bond vibrations. With pHH = 3.0 and -3.0, the trajectory is reactive; small deviations in pHH, however, result in unreactive trajectories. The system has a very large amount of energy compared to the activation energy, thus the outcomes of the trajectories become increasingly unpredictable.

H + HF

The H + HF reaction is endothermic, with a very late transition state, as shown above. Thus, according to the empirical Polanyi's rules, a higher proportion of vibrational energy would result in a reactive trajectory. Consider the following initial conditions:
{| class="wikitable" border=1
! pFH !! pHH
!Description
!Outcome
|-
| -9.5 || -10
|High translational, low vibrational
|[[File:Wm1415_high_kinetic_energy.png|400 px|thumb|left]][[File:Wm1415_high_kinetic_energy2.png|400 px|thumb|right]]
|-
| -10 || -0.01
|High vibrational, low translational
|[[File:Wm1415_high_vibrational_energy.png|400 px|thumb|left]][[File:Wm1415_high_vibrational_energy2.png|400 px|thumb|right]]
|}

In the first case, where the H atom approaches the transition state with a high ratio of translational:vibrational kinetic energy, the trajectory is unreactive. It is clearly seen from the PES that the transition state will not be reached with purely translational motion: the transition state will just decay back to the reactants channel.

In the second example, there is a high ratio of vibrational:translational kinetic energy resulting in a reactive trajectory. The high vibrational energy result in the bond being closer to breaking, and hence the transition state is both easier to form and more likely to result in the formation of products. This is a demonstration of Polanyi's rules in action - but not a proof. The rules are only empirical and often fail to predict how systems behave2.

=== References ===

1 I. N. Levine, ''Physical Chemistry'', McGraw-Hill, Boston, Mass., 6th edn, 2009, pp. 893

2 Z. Zhang, Y. Zhou, D. H. Zhang, G. Czakó and J. M. Bowman, ''J. Phys. Chem. Lett.'', 2012, 3, 3416–3419.

MRD:wfbfmicou

2017-05-31T15:13:25Z

Je714: /* F + H2: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally? */

===What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.===

At both a minimum and at a transition structure, the total gradient of the potential energy surface is zero. The points can be distinguished by their second derivative; the transition state is a saddle point. For a function <math>f\left(x,y\right)</math>, at a stationary point <math>\left(a,b\right)</math> consider the discriminant <math>D\equiv f_{xx}f_{yy}-f_{xy}^{2}</math>:

If <math>D>0</math>, and <math>f_{xx}\left(a,b\right)>0</math>, the point (a,b) is a local minimum.

If <math>D>0</math>, and <math>f_{xx}\left(a,b\right)<0</math>, the point (a,b) is a local maximum.

If <math>D<0</math>, the point (a,b) is a saddle point.

If <math>D=0</math>, higher order tests are required.

{{fontcolor1|gray| I assume you've copied this over from wikipedia (https://www.wikiwand.com/en/Second_partial_derivative_test). It's not wrong, but:
1) Cite your sources
2) D is the determinant of the Hessian matrix
[[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:05, 31 May 2017 (BST)}}

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.===

rts = 0.90775 Å

[[File:wm1415TS_position_internuclear_distance-time.png|500px]]

The internuclear distances remain effectively constant {{fontcolor1|gray| Why? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:06, 31 May 2017 (BST)}}: this represents the metastable transition state.

===Comment on how the ''mep'' and the dynamic trajectories differ.===
[[File:wm1415Mep_vs_dynamic.png|500px|thumb|left|MEP Calculation]]
[[File:wm1415Dynamic_vs_mep.png|500px|thumb|right|Dynamic Calculation]]

{{fontcolor1|gray| Good description [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:07, 31 May 2017 (BST)}}

In the MEP calculation, the kinetic energy of the particles is reset at each time step. The calculation begins at a structure similar to the transition state, with the H1-H2 distance slightly larger than the TS structure. As a result, the trajectory of the reaction in the MEP calculation follows the valley of the potential energy surface, without oscillating, to the reactants: H1 and H2-H3.
In the dynamic calculation, the inertia of the particles is taken into account: the velocities are not reset to zero at each step. Hence, vibrations are taken into account: the trajectory of the dynamic calculation also ends at the reactants, but oscillations are observed in this process.

===How do internuclear distances and momenta vary with time, when a reaction coordinate on either side of the transition state is chosen?===
[[File:wm1415Internuc_dist_time_r1.png|500px|thumb|left|Internuclear dist. vs. time - r1 = rts+δ, r2 = rts]]
[[File:wm1415Internuc_dist_time_r2.png|500px|thumb|right|Internuclear dist. vs. time - r1 = rts, r2 = rts+δ]]

[[File:wm1415Internuc_momenta_time_r1.png|500px|thumb|left|Internuclear momenta vs. time - r1 = rts+δ, r2 = rts]]
[[File:wm1415Internuc_momenta_time_r2.png|500px|thumb|right|Internuclear momenta vs. time - r1 = rts, r2 = rts+δ]]

The transition state is symmetrical: changing the initial position from (r1 = rts+δ, r2 = rts) to (r1 = rts, r2 = rts+δ) simply results in the formation of H1-H2 + H3 instead of H1 + H2-H3. The internuclear momenta and distance-time relationships are identical, with labels swapped.

===Reactive and unreactive trajectories===
{| class="wikitable" border=1
! p1 !! p2
!Outcome
!Description
|-
| -1.25 || -2.5
|[[File:wm1415Case1.png|200px|thumb|Reactive]]
|The H1-H2 molecule initially has no vibrational energy: the initial kinetic energy is purely translational energy of the approaching H3 atom. The H3 atom has sufficient kinetic energy to lead to the formation of the transition state and the products are formed: this is a reactive trajectory. The resulting H2-H3 molecule is vibrationally excited.
|-
| -1.5 || -2.0
|[[File:wm1415Case2.png|200px|thumb|Unreactive]]
|The H1-H2 molecule is initially vibrating: this causes the oscillations in the A-B (H3-H2) distance. H3 slows down as it approaches H2, converting kinetic energy to potential energy. For this set of initial conditions, the kinetic energy of H3 is insufficient to form the transition state and hence the products are not formed. H3 is eventually expelled by the H2-H1 molecule: this trajectory is unreactive.
|-
| -1.5 || -2.5
|[[File:wm1415Case3.png|200px|thumb|Reactive]]
|This reactive trajectory is similar to the first case. The H1-H2 molecule is initially vibrating with small amplitudes. H3 has sufficient kinetic energy to form the transition state and to expel H1 from the molecule. Following the reaction, the H3-H2 molecule is vibrationally excited.
|-
| -2.5 || -5.0
|[[File:wm1415Case4.png|200px|thumb|Unreactive]]
|The H2-H1 molecule initially has no vibrational energy. The approaching H3 atom has sufficient kinetic energy to reach the transition state: however, this particular set of initial conditions leads to barrier recrossing - the trajectory is unreactive. The large energies involved - far larger than required to reach the transition state - lead to a more unpredictable outcome. In this case, the transition state is so vibrationally excited that H3 is ejected from the activated complex and the system returns to the reactants channel, with the H2-H1 molecule vibrationally excited as a result.
|-
| -2.5 || -5.2
|[[File:wm1415Case5.png|200px|thumb|Reactive]]
|This set of initial conditions is similar to the previous example, except in this case the barrier recrossing leads to the ejection of the H1 atom from activated complex, giving the vibrationally excited H2-H3 molecule in the products channel.
|}

===State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for rate values compare with experimental values?===

Transition State Theory (TST) is a model that simplifies the computational complexity of calculating trajectories along a potential energy surface. TST requires knowledge of the PES in the reactant and product channels, as well as at the transition state1. A boundary is chosen - known as the critical dividing surface1 - which intersects the saddle point of the potential energy surface. The main assumptions of TST of ideal gas reactions are:

- "All supermolecules that cross the critical dividing surface from the reactant side become products."1

- "During the reaction, the Boltzmann distribution of energy is maintained for the reactant molecules."1

- "The supermolecules crossing the critical surface from the reactant side have a Boltzmann distribution of energy corresponding to the temperature of the reacting system."1

{{fontcolor1|gray| The 2nd and 3rd point are from a statistical point of view of TST. Not really applicable here, since you're only looking at a triatomic collision in isolation --- not at an ensemble of molecules [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:10, 31 May 2017 (BST)}}

As seen in the fourth example above, the assumption that all supermolecules with energy greater than the transition state lead to a reaction breaks down. In the fourth and fifth examples, barrier recrossing is observed in the trajectory calculations. Barrier recrossing is a phenomenon that is not accounted for in the transition state model. Therefore, transition state theory would predict larger rates of reaction than observed experimentally.

{{fontcolor1|gray| Good. This is the main point: it's what you're observing with your previous calculations. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:10, 31 May 2017 (BST)}}

Neither TST nor the model used in this investigation account for a further complication - the effect of quantum tunnelling. For small particles that do not classically have the energy to overcome a potential barrier, there is a certain probability of quantum tunnelling through the barrier (which depends on the energy difference, tunnelling length, and mass). Therefore, in the example of H3 + H2-H1, if the initial conditions lead to a collision which has insufficient energy to reach the transition state, tunnelling through the potential barrier to the other side of the PES and hence to the products is feasible. This phenomenon would lead to a small increase in the expected rate of reaction - however, the barrier recrossing effect dominates overall and hence we would still expect slower rates than calculated in TST.

{{fontcolor1|gray| What is a supermolecule? Good discussion of QM and acknowledging that it's not taking into account in these simulations [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:10, 31 May 2017 (BST)}}

===Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions.===

F + H2
[[File:wm1415F_H2_exothermic.png|500 px|thumb|left|Exothermic PES]]
[[File:wm1415F_H2_transition_state.png|500 px|thumb|right|Approximate location of transition state]]
[[File:Wm1415_exo_reactants.png|500 px|thumb|left|MEP calculation from a distorted transition state to the reactants.]]
[[File:Wm1415_exo_products.png|500 px|thumb|right|MEP calculation from a distorted transition state to the products.]]

The reactants channel is higher in energy than the products channel: the reaction is clearly exothermic. Formation of a H-F bond releases more energy than the formation of a H-H bond: the H-F bond is stronger. Since the activation energy is small, the energies of the reactants and the transition states are similar: by Hammond's postulate, the transition state will resemble the structure of the reactants. Thus we expect the transition state to be located near the reactants channel: this can be seen above. {{fontcolor1|gray| Good reasoning [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:12, 31 May 2017 (BST)}}
To find the energy of the reactants, an MEP calculation was run from a structure nearing the transition state, with the F-H bond slightly stretched, to produce a potential energy vs. time graph. Similarly. the energy of the products was found from a structure nearing the transition state, with the F-H bond slightly compressed.

Energy of the reactants ≈ -103.95 kcal/mol

Energy of the transition state ≈ -103.75 kcal/mol

Energy of the products ≈ -133.85 kcal/mol

Activation energy ≈ 0.2 kcal/mol

Energy change of reaction ≈ -29.90 kcal/mol (negative because exothermic - release of energy)

{{fontcolor1|gray| Good discussion [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:12, 31 May 2017 (BST)}}

H + HF
[[File:wm1415H_HF_endothermic.png|500 px|thumb|left|Endothermic PES]]
[[File:Wm1415H_HF_transition_state.png|500 px|thumb|right|Approximate location of transition state]]
[[File:Wm1415_endo_reactants.png|500 px|thumb|left|MEP calculation from a distorted transition state to the reactants.]]
[[File:Wm1415_endo_products.png|500 px|thumb|right|MEP calculation from a distorted transition state to the products.]]

The reactants channel in the reverse reaction is lower in energy than the products channel: the reaction is endothermic. By Hammond's postulate, the transition state should resemble the structure of the products - the location of the transition state is very close to the products channel. Potential energy calculations against time were carried out for completeness - but since it is the reverse reaction of the above situation, it is clear that the energy calculations should be symmetrical.

Energy of the reactants ≈ -133.85 kcal/mol

Energy of the transition state ≈ -103.75 kcal/mol

Energy of the products ≈ -103.85 kcal/mol

Activation energy ≈ 30.0 kcal/mol

Energy change of reaction ≈ +29.90 kcal/mol (positive because endothermic)

===F + H2: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?===
[[File:Wm1415_energy_conservation.png|500 px|thumb|left|Reaction trajectory: potential energy surface.]]
[[File:Wm1415_energy_conservation_momenta.png|500 px|thumb|right|F + H2: internuclear momenta vs. time]]

Following the reaction, the H-F molecule formed is vibrationally excited. This can be seen from the very large oscillations in the potential energy diagram and the internuclear momenta vs. time graph. The energy released in the reaction has been converted into vibrational kinetic energy - heat. The amount of heat energy dissipated by the reaction could be measured by calorimetry.
The vibrational excitation of the H-F bond could be further confirmed by IR spectroscopy by comparing the IR spectrum obtained from the vibrationally excited products to the IR spectrum of a HF molecule at room temperature. At ambient temperatures, only the ground vibrational state is substantially occupied, as the vibrational temperature of the H-F bond is very large. Therefore, the transitions observed in the infrared spectrum of HF would be from the v = 0 to v = 1 state (and perhaps some v = 1 to v = 2 transitions). The lower vibrational levels closely follow the simple harmonic oscillator model, therefore the transitions occurring at room temperature would all involve the same energy and only a single line would be observed in the spectrum.
The simple harmonic oscillator model breaks down at high vibrational levels for diatomic molecules. The energy spacings become progressively smaller until they converge at the dissociation limit, where the bond breaks due to the large vibrational energy. The HF molecule produced in this reaction is vibrationally excited, and its infrared spectrum would involve energy transitions that would match energy spacings in the anharmonic oscillator model. Therefore, in comparing its spectrum to room temperature, there would be a shift to many lower frequency absorptions, that would eventually converge to the dissociation limit (although the convergence point may not be experimentally observed, as the vibrational energy would most likely be dissipated to the environment such that the bond formed in the reaction wouldn't break).

{{fontcolor1|gray| Good discussion! [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:13, 31 May 2017 (BST)}}

===Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.===

F + H2

The Polanyi rules refer to how the distribution of kinetic energy in the transition state affects whether a trajectory is reactive or unreactive. A higher proportion of vibrational energy promotes reactions with a late transition state, whereas a higher proportion of translational energy promotes reactions with an early transition state2.
{| class="wikitable" border=1
! pFH !! pHH
!Outcome
|-
| -0.5 || -2.8
|[[File:Wm1415mom-2.8.png|200 px|thumb|Reactive]]
|-
| -0.5 || -2.7
|[[File:Wm1415mom-2.7.png|200 px|thumb|Unreactive]]
|-
| -0.5 || 2.7
|[[File:Wm1415mom2.7.png|200 px|thumb|Unreactive]]
|-
| -0.5 || 3.0
|[[File:Wm1415mom3.0.png|200 px|thumb|Reactive]]
|-
| -0.8 || 0.1
|[[File:Wm1415lowmom.png|200 px|thumb|Reactive]]
|}
The F + H2 reaction is exothermic, with a very early transition state, as shown above. Thus, according to the empirical Polanyi's rules, a higher proportion of translational energy would result in a reactive trajectory. This is clearly seen in the last example, where a significantly lower energy system results in a reactive trajectory: the kinetic energy is largely translational (there are only small oscillations in the reactants channel).
In the first four cases, most of the kinetic energy in the system is in the form of bond vibrations. With pHH = 3.0 and -3.0, the trajectory is reactive; small deviations in pHH, however, result in unreactive trajectories. The system has a very large amount of energy compared to the activation energy, thus the outcomes of the trajectories become increasingly unpredictable.

H + HF

The H + HF reaction is endothermic, with a very late transition state, as shown above. Thus, according to the empirical Polanyi's rules, a higher proportion of vibrational energy would result in a reactive trajectory. Consider the following initial conditions:
{| class="wikitable" border=1
! pFH !! pHH
!Description
!Outcome
|-
| -9.5 || -10
|High translational, low vibrational
|[[File:Wm1415_high_kinetic_energy.png|400 px|thumb|left]][[File:Wm1415_high_kinetic_energy2.png|400 px|thumb|right]]
|-
| -10 || -0.01
|High vibrational, low translational
|[[File:Wm1415_high_vibrational_energy.png|400 px|thumb|left]][[File:Wm1415_high_vibrational_energy2.png|400 px|thumb|right]]
|}

In the first case, where the H atom approaches the transition state with a high ratio of translational:vibrational kinetic energy, the trajectory is unreactive. It is clearly seen from the PES that the transition state will not be reached with purely translational motion: the transition state will just decay back to the reactants channel.

In the second example, there is a high ratio of vibrational:translational kinetic energy resulting in a reactive trajectory. The high vibrational energy result in the bond being closer to breaking, and hence the transition state is both easier to form and more likely to result in the formation of products. This is a demonstration of Polanyi's rules in action - but not a proof. The rules are only empirical and often fail to predict how systems behave2.

=== References ===

1 I. N. Levine, ''Physical Chemistry'', McGraw-Hill, Boston, Mass., 6th edn, 2009, pp. 893

2 Z. Zhang, Y. Zhou, D. H. Zhang, G. Czakó and J. M. Bowman, ''J. Phys. Chem. Lett.'', 2012, 3, 3416–3419.

MRD:wfbfmicou

2017-05-31T15:12:39Z

Je714: /* Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the act...

===What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.===

At both a minimum and at a transition structure, the total gradient of the potential energy surface is zero. The points can be distinguished by their second derivative; the transition state is a saddle point. For a function <math>f\left(x,y\right)</math>, at a stationary point <math>\left(a,b\right)</math> consider the discriminant <math>D\equiv f_{xx}f_{yy}-f_{xy}^{2}</math>:

If <math>D>0</math>, and <math>f_{xx}\left(a,b\right)>0</math>, the point (a,b) is a local minimum.

If <math>D>0</math>, and <math>f_{xx}\left(a,b\right)<0</math>, the point (a,b) is a local maximum.

If <math>D<0</math>, the point (a,b) is a saddle point.

If <math>D=0</math>, higher order tests are required.

{{fontcolor1|gray| I assume you've copied this over from wikipedia (https://www.wikiwand.com/en/Second_partial_derivative_test). It's not wrong, but:
1) Cite your sources
2) D is the determinant of the Hessian matrix
[[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:05, 31 May 2017 (BST)}}

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.===

rts = 0.90775 Å

[[File:wm1415TS_position_internuclear_distance-time.png|500px]]

The internuclear distances remain effectively constant {{fontcolor1|gray| Why? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:06, 31 May 2017 (BST)}}: this represents the metastable transition state.

===Comment on how the ''mep'' and the dynamic trajectories differ.===
[[File:wm1415Mep_vs_dynamic.png|500px|thumb|left|MEP Calculation]]
[[File:wm1415Dynamic_vs_mep.png|500px|thumb|right|Dynamic Calculation]]

{{fontcolor1|gray| Good description [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:07, 31 May 2017 (BST)}}

In the MEP calculation, the kinetic energy of the particles is reset at each time step. The calculation begins at a structure similar to the transition state, with the H1-H2 distance slightly larger than the TS structure. As a result, the trajectory of the reaction in the MEP calculation follows the valley of the potential energy surface, without oscillating, to the reactants: H1 and H2-H3.
In the dynamic calculation, the inertia of the particles is taken into account: the velocities are not reset to zero at each step. Hence, vibrations are taken into account: the trajectory of the dynamic calculation also ends at the reactants, but oscillations are observed in this process.

===How do internuclear distances and momenta vary with time, when a reaction coordinate on either side of the transition state is chosen?===
[[File:wm1415Internuc_dist_time_r1.png|500px|thumb|left|Internuclear dist. vs. time - r1 = rts+δ, r2 = rts]]
[[File:wm1415Internuc_dist_time_r2.png|500px|thumb|right|Internuclear dist. vs. time - r1 = rts, r2 = rts+δ]]

[[File:wm1415Internuc_momenta_time_r1.png|500px|thumb|left|Internuclear momenta vs. time - r1 = rts+δ, r2 = rts]]
[[File:wm1415Internuc_momenta_time_r2.png|500px|thumb|right|Internuclear momenta vs. time - r1 = rts, r2 = rts+δ]]

The transition state is symmetrical: changing the initial position from (r1 = rts+δ, r2 = rts) to (r1 = rts, r2 = rts+δ) simply results in the formation of H1-H2 + H3 instead of H1 + H2-H3. The internuclear momenta and distance-time relationships are identical, with labels swapped.

===Reactive and unreactive trajectories===
{| class="wikitable" border=1
! p1 !! p2
!Outcome
!Description
|-
| -1.25 || -2.5
|[[File:wm1415Case1.png|200px|thumb|Reactive]]
|The H1-H2 molecule initially has no vibrational energy: the initial kinetic energy is purely translational energy of the approaching H3 atom. The H3 atom has sufficient kinetic energy to lead to the formation of the transition state and the products are formed: this is a reactive trajectory. The resulting H2-H3 molecule is vibrationally excited.
|-
| -1.5 || -2.0
|[[File:wm1415Case2.png|200px|thumb|Unreactive]]
|The H1-H2 molecule is initially vibrating: this causes the oscillations in the A-B (H3-H2) distance. H3 slows down as it approaches H2, converting kinetic energy to potential energy. For this set of initial conditions, the kinetic energy of H3 is insufficient to form the transition state and hence the products are not formed. H3 is eventually expelled by the H2-H1 molecule: this trajectory is unreactive.
|-
| -1.5 || -2.5
|[[File:wm1415Case3.png|200px|thumb|Reactive]]
|This reactive trajectory is similar to the first case. The H1-H2 molecule is initially vibrating with small amplitudes. H3 has sufficient kinetic energy to form the transition state and to expel H1 from the molecule. Following the reaction, the H3-H2 molecule is vibrationally excited.
|-
| -2.5 || -5.0
|[[File:wm1415Case4.png|200px|thumb|Unreactive]]
|The H2-H1 molecule initially has no vibrational energy. The approaching H3 atom has sufficient kinetic energy to reach the transition state: however, this particular set of initial conditions leads to barrier recrossing - the trajectory is unreactive. The large energies involved - far larger than required to reach the transition state - lead to a more unpredictable outcome. In this case, the transition state is so vibrationally excited that H3 is ejected from the activated complex and the system returns to the reactants channel, with the H2-H1 molecule vibrationally excited as a result.
|-
| -2.5 || -5.2
|[[File:wm1415Case5.png|200px|thumb|Reactive]]
|This set of initial conditions is similar to the previous example, except in this case the barrier recrossing leads to the ejection of the H1 atom from activated complex, giving the vibrationally excited H2-H3 molecule in the products channel.
|}

===State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for rate values compare with experimental values?===

Transition State Theory (TST) is a model that simplifies the computational complexity of calculating trajectories along a potential energy surface. TST requires knowledge of the PES in the reactant and product channels, as well as at the transition state1. A boundary is chosen - known as the critical dividing surface1 - which intersects the saddle point of the potential energy surface. The main assumptions of TST of ideal gas reactions are:

- "All supermolecules that cross the critical dividing surface from the reactant side become products."1

- "During the reaction, the Boltzmann distribution of energy is maintained for the reactant molecules."1

- "The supermolecules crossing the critical surface from the reactant side have a Boltzmann distribution of energy corresponding to the temperature of the reacting system."1

{{fontcolor1|gray| The 2nd and 3rd point are from a statistical point of view of TST. Not really applicable here, since you're only looking at a triatomic collision in isolation --- not at an ensemble of molecules [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:10, 31 May 2017 (BST)}}

As seen in the fourth example above, the assumption that all supermolecules with energy greater than the transition state lead to a reaction breaks down. In the fourth and fifth examples, barrier recrossing is observed in the trajectory calculations. Barrier recrossing is a phenomenon that is not accounted for in the transition state model. Therefore, transition state theory would predict larger rates of reaction than observed experimentally.

{{fontcolor1|gray| Good. This is the main point: it's what you're observing with your previous calculations. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:10, 31 May 2017 (BST)}}

Neither TST nor the model used in this investigation account for a further complication - the effect of quantum tunnelling. For small particles that do not classically have the energy to overcome a potential barrier, there is a certain probability of quantum tunnelling through the barrier (which depends on the energy difference, tunnelling length, and mass). Therefore, in the example of H3 + H2-H1, if the initial conditions lead to a collision which has insufficient energy to reach the transition state, tunnelling through the potential barrier to the other side of the PES and hence to the products is feasible. This phenomenon would lead to a small increase in the expected rate of reaction - however, the barrier recrossing effect dominates overall and hence we would still expect slower rates than calculated in TST.

{{fontcolor1|gray| What is a supermolecule? Good discussion of QM and acknowledging that it's not taking into account in these simulations [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:10, 31 May 2017 (BST)}}

===Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions.===

F + H2
[[File:wm1415F_H2_exothermic.png|500 px|thumb|left|Exothermic PES]]
[[File:wm1415F_H2_transition_state.png|500 px|thumb|right|Approximate location of transition state]]
[[File:Wm1415_exo_reactants.png|500 px|thumb|left|MEP calculation from a distorted transition state to the reactants.]]
[[File:Wm1415_exo_products.png|500 px|thumb|right|MEP calculation from a distorted transition state to the products.]]

The reactants channel is higher in energy than the products channel: the reaction is clearly exothermic. Formation of a H-F bond releases more energy than the formation of a H-H bond: the H-F bond is stronger. Since the activation energy is small, the energies of the reactants and the transition states are similar: by Hammond's postulate, the transition state will resemble the structure of the reactants. Thus we expect the transition state to be located near the reactants channel: this can be seen above. {{fontcolor1|gray| Good reasoning [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:12, 31 May 2017 (BST)}}
To find the energy of the reactants, an MEP calculation was run from a structure nearing the transition state, with the F-H bond slightly stretched, to produce a potential energy vs. time graph. Similarly. the energy of the products was found from a structure nearing the transition state, with the F-H bond slightly compressed.

Energy of the reactants ≈ -103.95 kcal/mol

Energy of the transition state ≈ -103.75 kcal/mol

Energy of the products ≈ -133.85 kcal/mol

Activation energy ≈ 0.2 kcal/mol

Energy change of reaction ≈ -29.90 kcal/mol (negative because exothermic - release of energy)

{{fontcolor1|gray| Good discussion [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:12, 31 May 2017 (BST)}}

H + HF
[[File:wm1415H_HF_endothermic.png|500 px|thumb|left|Endothermic PES]]
[[File:Wm1415H_HF_transition_state.png|500 px|thumb|right|Approximate location of transition state]]
[[File:Wm1415_endo_reactants.png|500 px|thumb|left|MEP calculation from a distorted transition state to the reactants.]]
[[File:Wm1415_endo_products.png|500 px|thumb|right|MEP calculation from a distorted transition state to the products.]]

The reactants channel in the reverse reaction is lower in energy than the products channel: the reaction is endothermic. By Hammond's postulate, the transition state should resemble the structure of the products - the location of the transition state is very close to the products channel. Potential energy calculations against time were carried out for completeness - but since it is the reverse reaction of the above situation, it is clear that the energy calculations should be symmetrical.

Energy of the reactants ≈ -133.85 kcal/mol

Energy of the transition state ≈ -103.75 kcal/mol

Energy of the products ≈ -103.85 kcal/mol

Activation energy ≈ 30.0 kcal/mol

Energy change of reaction ≈ +29.90 kcal/mol (positive because endothermic)

===F + H2: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?===
[[File:Wm1415_energy_conservation.png|500 px|thumb|left|Reaction trajectory: potential energy surface.]]
[[File:Wm1415_energy_conservation_momenta.png|500 px|thumb|right|F + H2: internuclear momenta vs. time]]

Following the reaction, the H-F molecule formed is vibrationally excited. This can be seen from the very large oscillations in the potential energy diagram and the internuclear momenta vs. time graph. The energy released in the reaction has been converted into vibrational kinetic energy - heat. The amount of heat energy dissipated by the reaction could be measured by calorimetry.
The vibrational excitation of the H-F bond could be further confirmed by IR spectroscopy by comparing the IR spectrum obtained from the vibrationally excited products to the IR spectrum of a HF molecule at room temperature. At ambient temperatures, only the ground vibrational state is substantially occupied, as the vibrational temperature of the H-F bond is very large. Therefore, the transitions observed in the infrared spectrum of HF would be from the v = 0 to v = 1 state (and perhaps some v = 1 to v = 2 transitions). The lower vibrational levels closely follow the simple harmonic oscillator model, therefore the transitions occurring at room temperature would all involve the same energy and only a single line would be observed in the spectrum.
The simple harmonic oscillator model breaks down at high vibrational levels for diatomic molecules. The energy spacings become progressively smaller until they converge at the dissociation limit, where the bond breaks due to the large vibrational energy. The HF molecule produced in this reaction is vibrationally excited, and its infrared spectrum would involve energy transitions that would match energy spacings in the anharmonic oscillator model. Therefore, in comparing its spectrum to room temperature, there would be a shift to many lower frequency absorptions, that would eventually converge to the dissociation limit (although the convergence point may not be experimentally observed, as the vibrational energy would most likely be dissipated to the environment such that the bond formed in the reaction wouldn't break).

===Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.===

F + H2

The Polanyi rules refer to how the distribution of kinetic energy in the transition state affects whether a trajectory is reactive or unreactive. A higher proportion of vibrational energy promotes reactions with a late transition state, whereas a higher proportion of translational energy promotes reactions with an early transition state2.
{| class="wikitable" border=1
! pFH !! pHH
!Outcome
|-
| -0.5 || -2.8
|[[File:Wm1415mom-2.8.png|200 px|thumb|Reactive]]
|-
| -0.5 || -2.7
|[[File:Wm1415mom-2.7.png|200 px|thumb|Unreactive]]
|-
| -0.5 || 2.7
|[[File:Wm1415mom2.7.png|200 px|thumb|Unreactive]]
|-
| -0.5 || 3.0
|[[File:Wm1415mom3.0.png|200 px|thumb|Reactive]]
|-
| -0.8 || 0.1
|[[File:Wm1415lowmom.png|200 px|thumb|Reactive]]
|}
The F + H2 reaction is exothermic, with a very early transition state, as shown above. Thus, according to the empirical Polanyi's rules, a higher proportion of translational energy would result in a reactive trajectory. This is clearly seen in the last example, where a significantly lower energy system results in a reactive trajectory: the kinetic energy is largely translational (there are only small oscillations in the reactants channel).
In the first four cases, most of the kinetic energy in the system is in the form of bond vibrations. With pHH = 3.0 and -3.0, the trajectory is reactive; small deviations in pHH, however, result in unreactive trajectories. The system has a very large amount of energy compared to the activation energy, thus the outcomes of the trajectories become increasingly unpredictable.

H + HF

The H + HF reaction is endothermic, with a very late transition state, as shown above. Thus, according to the empirical Polanyi's rules, a higher proportion of vibrational energy would result in a reactive trajectory. Consider the following initial conditions:
{| class="wikitable" border=1
! pFH !! pHH
!Description
!Outcome
|-
| -9.5 || -10
|High translational, low vibrational
|[[File:Wm1415_high_kinetic_energy.png|400 px|thumb|left]][[File:Wm1415_high_kinetic_energy2.png|400 px|thumb|right]]
|-
| -10 || -0.01
|High vibrational, low translational
|[[File:Wm1415_high_vibrational_energy.png|400 px|thumb|left]][[File:Wm1415_high_vibrational_energy2.png|400 px|thumb|right]]
|}

In the first case, where the H atom approaches the transition state with a high ratio of translational:vibrational kinetic energy, the trajectory is unreactive. It is clearly seen from the PES that the transition state will not be reached with purely translational motion: the transition state will just decay back to the reactants channel.

In the second example, there is a high ratio of vibrational:translational kinetic energy resulting in a reactive trajectory. The high vibrational energy result in the bond being closer to breaking, and hence the transition state is both easier to form and more likely to result in the formation of products. This is a demonstration of Polanyi's rules in action - but not a proof. The rules are only empirical and often fail to predict how systems behave2.

=== References ===

1 I. N. Levine, ''Physical Chemistry'', McGraw-Hill, Boston, Mass., 6th edn, 2009, pp. 893

2 Z. Zhang, Y. Zhou, D. H. Zhang, G. Czakó and J. M. Bowman, ''J. Phys. Chem. Lett.'', 2012, 3, 3416–3419.

MRD:wfbfmicou

2017-05-31T15:10:36Z

Je714: /* State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for rate values compare with experimental values? */

===What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.===

At both a minimum and at a transition structure, the total gradient of the potential energy surface is zero. The points can be distinguished by their second derivative; the transition state is a saddle point. For a function <math>f\left(x,y\right)</math>, at a stationary point <math>\left(a,b\right)</math> consider the discriminant <math>D\equiv f_{xx}f_{yy}-f_{xy}^{2}</math>:

If <math>D>0</math>, and <math>f_{xx}\left(a,b\right)>0</math>, the point (a,b) is a local minimum.

If <math>D>0</math>, and <math>f_{xx}\left(a,b\right)<0</math>, the point (a,b) is a local maximum.

If <math>D<0</math>, the point (a,b) is a saddle point.

If <math>D=0</math>, higher order tests are required.

{{fontcolor1|gray| I assume you've copied this over from wikipedia (https://www.wikiwand.com/en/Second_partial_derivative_test). It's not wrong, but:
1) Cite your sources
2) D is the determinant of the Hessian matrix
[[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:05, 31 May 2017 (BST)}}

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.===

rts = 0.90775 Å

[[File:wm1415TS_position_internuclear_distance-time.png|500px]]

The internuclear distances remain effectively constant {{fontcolor1|gray| Why? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:06, 31 May 2017 (BST)}}: this represents the metastable transition state.

===Comment on how the ''mep'' and the dynamic trajectories differ.===
[[File:wm1415Mep_vs_dynamic.png|500px|thumb|left|MEP Calculation]]
[[File:wm1415Dynamic_vs_mep.png|500px|thumb|right|Dynamic Calculation]]

{{fontcolor1|gray| Good description [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:07, 31 May 2017 (BST)}}

In the MEP calculation, the kinetic energy of the particles is reset at each time step. The calculation begins at a structure similar to the transition state, with the H1-H2 distance slightly larger than the TS structure. As a result, the trajectory of the reaction in the MEP calculation follows the valley of the potential energy surface, without oscillating, to the reactants: H1 and H2-H3.
In the dynamic calculation, the inertia of the particles is taken into account: the velocities are not reset to zero at each step. Hence, vibrations are taken into account: the trajectory of the dynamic calculation also ends at the reactants, but oscillations are observed in this process.

===How do internuclear distances and momenta vary with time, when a reaction coordinate on either side of the transition state is chosen?===
[[File:wm1415Internuc_dist_time_r1.png|500px|thumb|left|Internuclear dist. vs. time - r1 = rts+δ, r2 = rts]]
[[File:wm1415Internuc_dist_time_r2.png|500px|thumb|right|Internuclear dist. vs. time - r1 = rts, r2 = rts+δ]]

[[File:wm1415Internuc_momenta_time_r1.png|500px|thumb|left|Internuclear momenta vs. time - r1 = rts+δ, r2 = rts]]
[[File:wm1415Internuc_momenta_time_r2.png|500px|thumb|right|Internuclear momenta vs. time - r1 = rts, r2 = rts+δ]]

The transition state is symmetrical: changing the initial position from (r1 = rts+δ, r2 = rts) to (r1 = rts, r2 = rts+δ) simply results in the formation of H1-H2 + H3 instead of H1 + H2-H3. The internuclear momenta and distance-time relationships are identical, with labels swapped.

===Reactive and unreactive trajectories===
{| class="wikitable" border=1
! p1 !! p2
!Outcome
!Description
|-
| -1.25 || -2.5
|[[File:wm1415Case1.png|200px|thumb|Reactive]]
|The H1-H2 molecule initially has no vibrational energy: the initial kinetic energy is purely translational energy of the approaching H3 atom. The H3 atom has sufficient kinetic energy to lead to the formation of the transition state and the products are formed: this is a reactive trajectory. The resulting H2-H3 molecule is vibrationally excited.
|-
| -1.5 || -2.0
|[[File:wm1415Case2.png|200px|thumb|Unreactive]]
|The H1-H2 molecule is initially vibrating: this causes the oscillations in the A-B (H3-H2) distance. H3 slows down as it approaches H2, converting kinetic energy to potential energy. For this set of initial conditions, the kinetic energy of H3 is insufficient to form the transition state and hence the products are not formed. H3 is eventually expelled by the H2-H1 molecule: this trajectory is unreactive.
|-
| -1.5 || -2.5
|[[File:wm1415Case3.png|200px|thumb|Reactive]]
|This reactive trajectory is similar to the first case. The H1-H2 molecule is initially vibrating with small amplitudes. H3 has sufficient kinetic energy to form the transition state and to expel H1 from the molecule. Following the reaction, the H3-H2 molecule is vibrationally excited.
|-
| -2.5 || -5.0
|[[File:wm1415Case4.png|200px|thumb|Unreactive]]
|The H2-H1 molecule initially has no vibrational energy. The approaching H3 atom has sufficient kinetic energy to reach the transition state: however, this particular set of initial conditions leads to barrier recrossing - the trajectory is unreactive. The large energies involved - far larger than required to reach the transition state - lead to a more unpredictable outcome. In this case, the transition state is so vibrationally excited that H3 is ejected from the activated complex and the system returns to the reactants channel, with the H2-H1 molecule vibrationally excited as a result.
|-
| -2.5 || -5.2
|[[File:wm1415Case5.png|200px|thumb|Reactive]]
|This set of initial conditions is similar to the previous example, except in this case the barrier recrossing leads to the ejection of the H1 atom from activated complex, giving the vibrationally excited H2-H3 molecule in the products channel.
|}

===State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for rate values compare with experimental values?===

Transition State Theory (TST) is a model that simplifies the computational complexity of calculating trajectories along a potential energy surface. TST requires knowledge of the PES in the reactant and product channels, as well as at the transition state1. A boundary is chosen - known as the critical dividing surface1 - which intersects the saddle point of the potential energy surface. The main assumptions of TST of ideal gas reactions are:

- "All supermolecules that cross the critical dividing surface from the reactant side become products."1

- "During the reaction, the Boltzmann distribution of energy is maintained for the reactant molecules."1

- "The supermolecules crossing the critical surface from the reactant side have a Boltzmann distribution of energy corresponding to the temperature of the reacting system."1

{{fontcolor1|gray| The 2nd and 3rd point are from a statistical point of view of TST. Not really applicable here, since you're only looking at a triatomic collision in isolation --- not at an ensemble of molecules [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:10, 31 May 2017 (BST)}}

As seen in the fourth example above, the assumption that all supermolecules with energy greater than the transition state lead to a reaction breaks down. In the fourth and fifth examples, barrier recrossing is observed in the trajectory calculations. Barrier recrossing is a phenomenon that is not accounted for in the transition state model. Therefore, transition state theory would predict larger rates of reaction than observed experimentally.

{{fontcolor1|gray| Good. This is the main point: it's what you're observing with your previous calculations. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:10, 31 May 2017 (BST)}}

Neither TST nor the model used in this investigation account for a further complication - the effect of quantum tunnelling. For small particles that do not classically have the energy to overcome a potential barrier, there is a certain probability of quantum tunnelling through the barrier (which depends on the energy difference, tunnelling length, and mass). Therefore, in the example of H3 + H2-H1, if the initial conditions lead to a collision which has insufficient energy to reach the transition state, tunnelling through the potential barrier to the other side of the PES and hence to the products is feasible. This phenomenon would lead to a small increase in the expected rate of reaction - however, the barrier recrossing effect dominates overall and hence we would still expect slower rates than calculated in TST.

{{fontcolor1|gray| What is a supermolecule? Good discussion of QM and acknowledging that it's not taking into account in these simulations [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:10, 31 May 2017 (BST)}}

===Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions.===

F + H2
[[File:wm1415F_H2_exothermic.png|500 px|thumb|left|Exothermic PES]]
[[File:wm1415F_H2_transition_state.png|500 px|thumb|right|Approximate location of transition state]]
[[File:Wm1415_exo_reactants.png|500 px|thumb|left|MEP calculation from a distorted transition state to the reactants.]]
[[File:Wm1415_exo_products.png|500 px|thumb|right|MEP calculation from a distorted transition state to the products.]]

The reactants channel is higher in energy than the products channel: the reaction is clearly exothermic. Formation of a H-F bond releases more energy than the formation of a H-H bond: the H-F bond is stronger. Since the activation energy is small, the energies of the reactants and the transition states are similar: by Hammond's postulate, the transition state will resemble the structure of the reactants. Thus we expect the transition state to be located near the reactants channel: this can be seen above.
To find the energy of the reactants, an MEP calculation was run from a structure nearing the transition state, with the F-H bond slightly stretched, to produce a potential energy vs. time graph. Similarly. the energy of the products was found from a structure nearing the transition state, with the F-H bond slightly compressed.

Energy of the reactants ≈ -103.95 kcal/mol

Energy of the transition state ≈ -103.75 kcal/mol

Energy of the products ≈ -133.85 kcal/mol

Activation energy ≈ 0.2 kcal/mol

Energy change of reaction ≈ -29.90 kcal/mol (negative because exothermic - release of energy)

H + HF
[[File:wm1415H_HF_endothermic.png|500 px|thumb|left|Endothermic PES]]
[[File:Wm1415H_HF_transition_state.png|500 px|thumb|right|Approximate location of transition state]]
[[File:Wm1415_endo_reactants.png|500 px|thumb|left|MEP calculation from a distorted transition state to the reactants.]]
[[File:Wm1415_endo_products.png|500 px|thumb|right|MEP calculation from a distorted transition state to the products.]]

The reactants channel in the reverse reaction is lower in energy than the products channel: the reaction is endothermic. By Hammond's postulate, the transition state should resemble the structure of the products - the location of the transition state is very close to the products channel. Potential energy calculations against time were carried out for completeness - but since it is the reverse reaction of the above situation, it is clear that the energy calculations should be symmetrical.

Energy of the reactants ≈ -133.85 kcal/mol

Energy of the transition state ≈ -103.75 kcal/mol

Energy of the products ≈ -103.85 kcal/mol

Activation energy ≈ 30.0 kcal/mol

Energy change of reaction ≈ +29.90 kcal/mol (positive because endothermic)

===F + H2: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?===
[[File:Wm1415_energy_conservation.png|500 px|thumb|left|Reaction trajectory: potential energy surface.]]
[[File:Wm1415_energy_conservation_momenta.png|500 px|thumb|right|F + H2: internuclear momenta vs. time]]

Following the reaction, the H-F molecule formed is vibrationally excited. This can be seen from the very large oscillations in the potential energy diagram and the internuclear momenta vs. time graph. The energy released in the reaction has been converted into vibrational kinetic energy - heat. The amount of heat energy dissipated by the reaction could be measured by calorimetry.
The vibrational excitation of the H-F bond could be further confirmed by IR spectroscopy by comparing the IR spectrum obtained from the vibrationally excited products to the IR spectrum of a HF molecule at room temperature. At ambient temperatures, only the ground vibrational state is substantially occupied, as the vibrational temperature of the H-F bond is very large. Therefore, the transitions observed in the infrared spectrum of HF would be from the v = 0 to v = 1 state (and perhaps some v = 1 to v = 2 transitions). The lower vibrational levels closely follow the simple harmonic oscillator model, therefore the transitions occurring at room temperature would all involve the same energy and only a single line would be observed in the spectrum.
The simple harmonic oscillator model breaks down at high vibrational levels for diatomic molecules. The energy spacings become progressively smaller until they converge at the dissociation limit, where the bond breaks due to the large vibrational energy. The HF molecule produced in this reaction is vibrationally excited, and its infrared spectrum would involve energy transitions that would match energy spacings in the anharmonic oscillator model. Therefore, in comparing its spectrum to room temperature, there would be a shift to many lower frequency absorptions, that would eventually converge to the dissociation limit (although the convergence point may not be experimentally observed, as the vibrational energy would most likely be dissipated to the environment such that the bond formed in the reaction wouldn't break).

===Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.===

F + H2

The Polanyi rules refer to how the distribution of kinetic energy in the transition state affects whether a trajectory is reactive or unreactive. A higher proportion of vibrational energy promotes reactions with a late transition state, whereas a higher proportion of translational energy promotes reactions with an early transition state2.
{| class="wikitable" border=1
! pFH !! pHH
!Outcome
|-
| -0.5 || -2.8
|[[File:Wm1415mom-2.8.png|200 px|thumb|Reactive]]
|-
| -0.5 || -2.7
|[[File:Wm1415mom-2.7.png|200 px|thumb|Unreactive]]
|-
| -0.5 || 2.7
|[[File:Wm1415mom2.7.png|200 px|thumb|Unreactive]]
|-
| -0.5 || 3.0
|[[File:Wm1415mom3.0.png|200 px|thumb|Reactive]]
|-
| -0.8 || 0.1
|[[File:Wm1415lowmom.png|200 px|thumb|Reactive]]
|}
The F + H2 reaction is exothermic, with a very early transition state, as shown above. Thus, according to the empirical Polanyi's rules, a higher proportion of translational energy would result in a reactive trajectory. This is clearly seen in the last example, where a significantly lower energy system results in a reactive trajectory: the kinetic energy is largely translational (there are only small oscillations in the reactants channel).
In the first four cases, most of the kinetic energy in the system is in the form of bond vibrations. With pHH = 3.0 and -3.0, the trajectory is reactive; small deviations in pHH, however, result in unreactive trajectories. The system has a very large amount of energy compared to the activation energy, thus the outcomes of the trajectories become increasingly unpredictable.

H + HF

The H + HF reaction is endothermic, with a very late transition state, as shown above. Thus, according to the empirical Polanyi's rules, a higher proportion of vibrational energy would result in a reactive trajectory. Consider the following initial conditions:
{| class="wikitable" border=1
! pFH !! pHH
!Description
!Outcome
|-
| -9.5 || -10
|High translational, low vibrational
|[[File:Wm1415_high_kinetic_energy.png|400 px|thumb|left]][[File:Wm1415_high_kinetic_energy2.png|400 px|thumb|right]]
|-
| -10 || -0.01
|High vibrational, low translational
|[[File:Wm1415_high_vibrational_energy.png|400 px|thumb|left]][[File:Wm1415_high_vibrational_energy2.png|400 px|thumb|right]]
|}

In the first case, where the H atom approaches the transition state with a high ratio of translational:vibrational kinetic energy, the trajectory is unreactive. It is clearly seen from the PES that the transition state will not be reached with purely translational motion: the transition state will just decay back to the reactants channel.

In the second example, there is a high ratio of vibrational:translational kinetic energy resulting in a reactive trajectory. The high vibrational energy result in the bond being closer to breaking, and hence the transition state is both easier to form and more likely to result in the formation of products. This is a demonstration of Polanyi's rules in action - but not a proof. The rules are only empirical and often fail to predict how systems behave2.

=== References ===

1 I. N. Levine, ''Physical Chemistry'', McGraw-Hill, Boston, Mass., 6th edn, 2009, pp. 893

2 Z. Zhang, Y. Zhou, D. H. Zhang, G. Czakó and J. M. Bowman, ''J. Phys. Chem. Lett.'', 2012, 3, 3416–3419.

MRD:wfbfmicou

2017-05-31T15:07:05Z

Je714: /* Comment on how the mep and the dynamic trajectories differ. */

===What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.===

At both a minimum and at a transition structure, the total gradient of the potential energy surface is zero. The points can be distinguished by their second derivative; the transition state is a saddle point. For a function <math>f\left(x,y\right)</math>, at a stationary point <math>\left(a,b\right)</math> consider the discriminant <math>D\equiv f_{xx}f_{yy}-f_{xy}^{2}</math>:

If <math>D>0</math>, and <math>f_{xx}\left(a,b\right)>0</math>, the point (a,b) is a local minimum.

If <math>D>0</math>, and <math>f_{xx}\left(a,b\right)<0</math>, the point (a,b) is a local maximum.

If <math>D<0</math>, the point (a,b) is a saddle point.

If <math>D=0</math>, higher order tests are required.

{{fontcolor1|gray| I assume you've copied this over from wikipedia (https://www.wikiwand.com/en/Second_partial_derivative_test). It's not wrong, but:
1) Cite your sources
2) D is the determinant of the Hessian matrix
[[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:05, 31 May 2017 (BST)}}

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.===

rts = 0.90775 Å

[[File:wm1415TS_position_internuclear_distance-time.png|500px]]

The internuclear distances remain effectively constant {{fontcolor1|gray| Why? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:06, 31 May 2017 (BST)}}: this represents the metastable transition state.

===Comment on how the ''mep'' and the dynamic trajectories differ.===
[[File:wm1415Mep_vs_dynamic.png|500px|thumb|left|MEP Calculation]]
[[File:wm1415Dynamic_vs_mep.png|500px|thumb|right|Dynamic Calculation]]

{{fontcolor1|gray| Good description [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:07, 31 May 2017 (BST)}}

In the MEP calculation, the kinetic energy of the particles is reset at each time step. The calculation begins at a structure similar to the transition state, with the H1-H2 distance slightly larger than the TS structure. As a result, the trajectory of the reaction in the MEP calculation follows the valley of the potential energy surface, without oscillating, to the reactants: H1 and H2-H3.
In the dynamic calculation, the inertia of the particles is taken into account: the velocities are not reset to zero at each step. Hence, vibrations are taken into account: the trajectory of the dynamic calculation also ends at the reactants, but oscillations are observed in this process.

===How do internuclear distances and momenta vary with time, when a reaction coordinate on either side of the transition state is chosen?===
[[File:wm1415Internuc_dist_time_r1.png|500px|thumb|left|Internuclear dist. vs. time - r1 = rts+δ, r2 = rts]]
[[File:wm1415Internuc_dist_time_r2.png|500px|thumb|right|Internuclear dist. vs. time - r1 = rts, r2 = rts+δ]]

[[File:wm1415Internuc_momenta_time_r1.png|500px|thumb|left|Internuclear momenta vs. time - r1 = rts+δ, r2 = rts]]
[[File:wm1415Internuc_momenta_time_r2.png|500px|thumb|right|Internuclear momenta vs. time - r1 = rts, r2 = rts+δ]]

The transition state is symmetrical: changing the initial position from (r1 = rts+δ, r2 = rts) to (r1 = rts, r2 = rts+δ) simply results in the formation of H1-H2 + H3 instead of H1 + H2-H3. The internuclear momenta and distance-time relationships are identical, with labels swapped.

===Reactive and unreactive trajectories===
{| class="wikitable" border=1
! p1 !! p2
!Outcome
!Description
|-
| -1.25 || -2.5
|[[File:wm1415Case1.png|200px|thumb|Reactive]]
|The H1-H2 molecule initially has no vibrational energy: the initial kinetic energy is purely translational energy of the approaching H3 atom. The H3 atom has sufficient kinetic energy to lead to the formation of the transition state and the products are formed: this is a reactive trajectory. The resulting H2-H3 molecule is vibrationally excited.
|-
| -1.5 || -2.0
|[[File:wm1415Case2.png|200px|thumb|Unreactive]]
|The H1-H2 molecule is initially vibrating: this causes the oscillations in the A-B (H3-H2) distance. H3 slows down as it approaches H2, converting kinetic energy to potential energy. For this set of initial conditions, the kinetic energy of H3 is insufficient to form the transition state and hence the products are not formed. H3 is eventually expelled by the H2-H1 molecule: this trajectory is unreactive.
|-
| -1.5 || -2.5
|[[File:wm1415Case3.png|200px|thumb|Reactive]]
|This reactive trajectory is similar to the first case. The H1-H2 molecule is initially vibrating with small amplitudes. H3 has sufficient kinetic energy to form the transition state and to expel H1 from the molecule. Following the reaction, the H3-H2 molecule is vibrationally excited.
|-
| -2.5 || -5.0
|[[File:wm1415Case4.png|200px|thumb|Unreactive]]
|The H2-H1 molecule initially has no vibrational energy. The approaching H3 atom has sufficient kinetic energy to reach the transition state: however, this particular set of initial conditions leads to barrier recrossing - the trajectory is unreactive. The large energies involved - far larger than required to reach the transition state - lead to a more unpredictable outcome. In this case, the transition state is so vibrationally excited that H3 is ejected from the activated complex and the system returns to the reactants channel, with the H2-H1 molecule vibrationally excited as a result.
|-
| -2.5 || -5.2
|[[File:wm1415Case5.png|200px|thumb|Reactive]]
|This set of initial conditions is similar to the previous example, except in this case the barrier recrossing leads to the ejection of the H1 atom from activated complex, giving the vibrationally excited H2-H3 molecule in the products channel.
|}

===State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for rate values compare with experimental values?===

Transition State Theory (TST) is a model that simplifies the computational complexity of calculating trajectories along a potential energy surface. TST requires knowledge of the PES in the reactant and product channels, as well as at the transition state1. A boundary is chosen - known as the critical dividing surface1 - which intersects the saddle point of the potential energy surface. The main assumptions of TST of ideal gas reactions are:

- "All supermolecules that cross the critical dividing surface from the reactant side become products."1

- "During the reaction, the Boltzmann distribution of energy is maintained for the reactant molecules."1

- "The supermolecules crossing the critical surface from the reactant side have a Boltzmann distribution of energy corresponding to the temperature of the reacting system."1

As seen in the fourth example above, the assumption that all supermolecules with energy greater than the transition state lead to a reaction breaks down. In the fourth and fifth examples, barrier recrossing is observed in the trajectory calculations. Barrier recrossing is a phenomenon that is not accounted for in the transition state model. Therefore, transition state theory would predict larger rates of reaction than observed experimentally.

Neither TST nor the model used in this investigation account for a further complication - the effect of quantum tunnelling. For small particles that do not classically have the energy to overcome a potential barrier, there is a certain probability of quantum tunnelling through the barrier (which depends on the energy difference, tunnelling length, and mass). Therefore, in the example of H3 + H2-H1, if the initial conditions lead to a collision which has insufficient energy to reach the transition state, tunnelling through the potential barrier to the other side of the PES and hence to the products is feasible. This phenomenon would lead to a small increase in the expected rate of reaction - however, the barrier recrossing effect dominates overall and hence we would still expect slower rates than calculated in TST.

===Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions.===

F + H2
[[File:wm1415F_H2_exothermic.png|500 px|thumb|left|Exothermic PES]]
[[File:wm1415F_H2_transition_state.png|500 px|thumb|right|Approximate location of transition state]]
[[File:Wm1415_exo_reactants.png|500 px|thumb|left|MEP calculation from a distorted transition state to the reactants.]]
[[File:Wm1415_exo_products.png|500 px|thumb|right|MEP calculation from a distorted transition state to the products.]]

The reactants channel is higher in energy than the products channel: the reaction is clearly exothermic. Formation of a H-F bond releases more energy than the formation of a H-H bond: the H-F bond is stronger. Since the activation energy is small, the energies of the reactants and the transition states are similar: by Hammond's postulate, the transition state will resemble the structure of the reactants. Thus we expect the transition state to be located near the reactants channel: this can be seen above.
To find the energy of the reactants, an MEP calculation was run from a structure nearing the transition state, with the F-H bond slightly stretched, to produce a potential energy vs. time graph. Similarly. the energy of the products was found from a structure nearing the transition state, with the F-H bond slightly compressed.

Energy of the reactants ≈ -103.95 kcal/mol

Energy of the transition state ≈ -103.75 kcal/mol

Energy of the products ≈ -133.85 kcal/mol

Activation energy ≈ 0.2 kcal/mol

Energy change of reaction ≈ -29.90 kcal/mol (negative because exothermic - release of energy)

H + HF
[[File:wm1415H_HF_endothermic.png|500 px|thumb|left|Endothermic PES]]
[[File:Wm1415H_HF_transition_state.png|500 px|thumb|right|Approximate location of transition state]]
[[File:Wm1415_endo_reactants.png|500 px|thumb|left|MEP calculation from a distorted transition state to the reactants.]]
[[File:Wm1415_endo_products.png|500 px|thumb|right|MEP calculation from a distorted transition state to the products.]]

The reactants channel in the reverse reaction is lower in energy than the products channel: the reaction is endothermic. By Hammond's postulate, the transition state should resemble the structure of the products - the location of the transition state is very close to the products channel. Potential energy calculations against time were carried out for completeness - but since it is the reverse reaction of the above situation, it is clear that the energy calculations should be symmetrical.

Energy of the reactants ≈ -133.85 kcal/mol

Energy of the transition state ≈ -103.75 kcal/mol

Energy of the products ≈ -103.85 kcal/mol

Activation energy ≈ 30.0 kcal/mol

Energy change of reaction ≈ +29.90 kcal/mol (positive because endothermic)

===F + H2: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?===
[[File:Wm1415_energy_conservation.png|500 px|thumb|left|Reaction trajectory: potential energy surface.]]
[[File:Wm1415_energy_conservation_momenta.png|500 px|thumb|right|F + H2: internuclear momenta vs. time]]

Following the reaction, the H-F molecule formed is vibrationally excited. This can be seen from the very large oscillations in the potential energy diagram and the internuclear momenta vs. time graph. The energy released in the reaction has been converted into vibrational kinetic energy - heat. The amount of heat energy dissipated by the reaction could be measured by calorimetry.
The vibrational excitation of the H-F bond could be further confirmed by IR spectroscopy by comparing the IR spectrum obtained from the vibrationally excited products to the IR spectrum of a HF molecule at room temperature. At ambient temperatures, only the ground vibrational state is substantially occupied, as the vibrational temperature of the H-F bond is very large. Therefore, the transitions observed in the infrared spectrum of HF would be from the v = 0 to v = 1 state (and perhaps some v = 1 to v = 2 transitions). The lower vibrational levels closely follow the simple harmonic oscillator model, therefore the transitions occurring at room temperature would all involve the same energy and only a single line would be observed in the spectrum.
The simple harmonic oscillator model breaks down at high vibrational levels for diatomic molecules. The energy spacings become progressively smaller until they converge at the dissociation limit, where the bond breaks due to the large vibrational energy. The HF molecule produced in this reaction is vibrationally excited, and its infrared spectrum would involve energy transitions that would match energy spacings in the anharmonic oscillator model. Therefore, in comparing its spectrum to room temperature, there would be a shift to many lower frequency absorptions, that would eventually converge to the dissociation limit (although the convergence point may not be experimentally observed, as the vibrational energy would most likely be dissipated to the environment such that the bond formed in the reaction wouldn't break).

===Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.===

F + H2

The Polanyi rules refer to how the distribution of kinetic energy in the transition state affects whether a trajectory is reactive or unreactive. A higher proportion of vibrational energy promotes reactions with a late transition state, whereas a higher proportion of translational energy promotes reactions with an early transition state2.
{| class="wikitable" border=1
! pFH !! pHH
!Outcome
|-
| -0.5 || -2.8
|[[File:Wm1415mom-2.8.png|200 px|thumb|Reactive]]
|-
| -0.5 || -2.7
|[[File:Wm1415mom-2.7.png|200 px|thumb|Unreactive]]
|-
| -0.5 || 2.7
|[[File:Wm1415mom2.7.png|200 px|thumb|Unreactive]]
|-
| -0.5 || 3.0
|[[File:Wm1415mom3.0.png|200 px|thumb|Reactive]]
|-
| -0.8 || 0.1
|[[File:Wm1415lowmom.png|200 px|thumb|Reactive]]
|}
The F + H2 reaction is exothermic, with a very early transition state, as shown above. Thus, according to the empirical Polanyi's rules, a higher proportion of translational energy would result in a reactive trajectory. This is clearly seen in the last example, where a significantly lower energy system results in a reactive trajectory: the kinetic energy is largely translational (there are only small oscillations in the reactants channel).
In the first four cases, most of the kinetic energy in the system is in the form of bond vibrations. With pHH = 3.0 and -3.0, the trajectory is reactive; small deviations in pHH, however, result in unreactive trajectories. The system has a very large amount of energy compared to the activation energy, thus the outcomes of the trajectories become increasingly unpredictable.

H + HF

The H + HF reaction is endothermic, with a very late transition state, as shown above. Thus, according to the empirical Polanyi's rules, a higher proportion of vibrational energy would result in a reactive trajectory. Consider the following initial conditions:
{| class="wikitable" border=1
! pFH !! pHH
!Description
!Outcome
|-
| -9.5 || -10
|High translational, low vibrational
|[[File:Wm1415_high_kinetic_energy.png|400 px|thumb|left]][[File:Wm1415_high_kinetic_energy2.png|400 px|thumb|right]]
|-
| -10 || -0.01
|High vibrational, low translational
|[[File:Wm1415_high_vibrational_energy.png|400 px|thumb|left]][[File:Wm1415_high_vibrational_energy2.png|400 px|thumb|right]]
|}

In the first case, where the H atom approaches the transition state with a high ratio of translational:vibrational kinetic energy, the trajectory is unreactive. It is clearly seen from the PES that the transition state will not be reached with purely translational motion: the transition state will just decay back to the reactants channel.

In the second example, there is a high ratio of vibrational:translational kinetic energy resulting in a reactive trajectory. The high vibrational energy result in the bond being closer to breaking, and hence the transition state is both easier to form and more likely to result in the formation of products. This is a demonstration of Polanyi's rules in action - but not a proof. The rules are only empirical and often fail to predict how systems behave2.

=== References ===

1 I. N. Levine, ''Physical Chemistry'', McGraw-Hill, Boston, Mass., 6th edn, 2009, pp. 893

2 Z. Zhang, Y. Zhou, D. H. Zhang, G. Czakó and J. M. Bowman, ''J. Phys. Chem. Lett.'', 2012, 3, 3416–3419.

MRD:wfbfmicou

2017-05-31T15:06:30Z

===What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.===

At both a minimum and at a transition structure, the total gradient of the potential energy surface is zero. The points can be distinguished by their second derivative; the transition state is a saddle point. For a function <math>f\left(x,y\right)</math>, at a stationary point <math>\left(a,b\right)</math> consider the discriminant <math>D\equiv f_{xx}f_{yy}-f_{xy}^{2}</math>:

If <math>D>0</math>, and <math>f_{xx}\left(a,b\right)>0</math>, the point (a,b) is a local minimum.

If <math>D>0</math>, and <math>f_{xx}\left(a,b\right)<0</math>, the point (a,b) is a local maximum.

If <math>D<0</math>, the point (a,b) is a saddle point.

If <math>D=0</math>, higher order tests are required.

{{fontcolor1|gray| I assume you've copied this over from wikipedia (https://www.wikiwand.com/en/Second_partial_derivative_test). It's not wrong, but:
1) Cite your sources
2) D is the determinant of the Hessian matrix
[[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:05, 31 May 2017 (BST)}}

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.===

rts = 0.90775 Å

[[File:wm1415TS_position_internuclear_distance-time.png|500px]]

The internuclear distances remain effectively constant {{fontcolor1|gray| Why? [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:06, 31 May 2017 (BST)}}: this represents the metastable transition state.

===Comment on how the ''mep'' and the dynamic trajectories differ.===
[[File:wm1415Mep_vs_dynamic.png|500px|thumb|left|MEP Calculation]]
[[File:wm1415Dynamic_vs_mep.png|500px|thumb|right|Dynamic Calculation]]

In the MEP calculation, the kinetic energy of the particles is reset at each time step. The calculation begins at a structure similar to the transition state, with the H1-H2 distance slightly larger than the TS structure. As a result, the trajectory of the reaction in the MEP calculation follows the valley of the potential energy surface, without oscillating, to the reactants: H1 and H2-H3.
In the dynamic calculation, the inertia of the particles is taken into account: the velocities are not reset to zero at each step. Hence, vibrations are taken into account: the trajectory of the dynamic calculation also ends at the reactants, but oscillations are observed in this process.

===How do internuclear distances and momenta vary with time, when a reaction coordinate on either side of the transition state is chosen?===
[[File:wm1415Internuc_dist_time_r1.png|500px|thumb|left|Internuclear dist. vs. time - r1 = rts+δ, r2 = rts]]
[[File:wm1415Internuc_dist_time_r2.png|500px|thumb|right|Internuclear dist. vs. time - r1 = rts, r2 = rts+δ]]

[[File:wm1415Internuc_momenta_time_r1.png|500px|thumb|left|Internuclear momenta vs. time - r1 = rts+δ, r2 = rts]]
[[File:wm1415Internuc_momenta_time_r2.png|500px|thumb|right|Internuclear momenta vs. time - r1 = rts, r2 = rts+δ]]

The transition state is symmetrical: changing the initial position from (r1 = rts+δ, r2 = rts) to (r1 = rts, r2 = rts+δ) simply results in the formation of H1-H2 + H3 instead of H1 + H2-H3. The internuclear momenta and distance-time relationships are identical, with labels swapped.

===Reactive and unreactive trajectories===
{| class="wikitable" border=1
! p1 !! p2
!Outcome
!Description
|-
| -1.25 || -2.5
|[[File:wm1415Case1.png|200px|thumb|Reactive]]
|The H1-H2 molecule initially has no vibrational energy: the initial kinetic energy is purely translational energy of the approaching H3 atom. The H3 atom has sufficient kinetic energy to lead to the formation of the transition state and the products are formed: this is a reactive trajectory. The resulting H2-H3 molecule is vibrationally excited.
|-
| -1.5 || -2.0
|[[File:wm1415Case2.png|200px|thumb|Unreactive]]
|The H1-H2 molecule is initially vibrating: this causes the oscillations in the A-B (H3-H2) distance. H3 slows down as it approaches H2, converting kinetic energy to potential energy. For this set of initial conditions, the kinetic energy of H3 is insufficient to form the transition state and hence the products are not formed. H3 is eventually expelled by the H2-H1 molecule: this trajectory is unreactive.
|-
| -1.5 || -2.5
|[[File:wm1415Case3.png|200px|thumb|Reactive]]
|This reactive trajectory is similar to the first case. The H1-H2 molecule is initially vibrating with small amplitudes. H3 has sufficient kinetic energy to form the transition state and to expel H1 from the molecule. Following the reaction, the H3-H2 molecule is vibrationally excited.
|-
| -2.5 || -5.0
|[[File:wm1415Case4.png|200px|thumb|Unreactive]]
|The H2-H1 molecule initially has no vibrational energy. The approaching H3 atom has sufficient kinetic energy to reach the transition state: however, this particular set of initial conditions leads to barrier recrossing - the trajectory is unreactive. The large energies involved - far larger than required to reach the transition state - lead to a more unpredictable outcome. In this case, the transition state is so vibrationally excited that H3 is ejected from the activated complex and the system returns to the reactants channel, with the H2-H1 molecule vibrationally excited as a result.
|-
| -2.5 || -5.2
|[[File:wm1415Case5.png|200px|thumb|Reactive]]
|This set of initial conditions is similar to the previous example, except in this case the barrier recrossing leads to the ejection of the H1 atom from activated complex, giving the vibrationally excited H2-H3 molecule in the products channel.
|}

===State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for rate values compare with experimental values?===

Transition State Theory (TST) is a model that simplifies the computational complexity of calculating trajectories along a potential energy surface. TST requires knowledge of the PES in the reactant and product channels, as well as at the transition state1. A boundary is chosen - known as the critical dividing surface1 - which intersects the saddle point of the potential energy surface. The main assumptions of TST of ideal gas reactions are:

- "All supermolecules that cross the critical dividing surface from the reactant side become products."1

- "During the reaction, the Boltzmann distribution of energy is maintained for the reactant molecules."1

- "The supermolecules crossing the critical surface from the reactant side have a Boltzmann distribution of energy corresponding to the temperature of the reacting system."1

As seen in the fourth example above, the assumption that all supermolecules with energy greater than the transition state lead to a reaction breaks down. In the fourth and fifth examples, barrier recrossing is observed in the trajectory calculations. Barrier recrossing is a phenomenon that is not accounted for in the transition state model. Therefore, transition state theory would predict larger rates of reaction than observed experimentally.

Neither TST nor the model used in this investigation account for a further complication - the effect of quantum tunnelling. For small particles that do not classically have the energy to overcome a potential barrier, there is a certain probability of quantum tunnelling through the barrier (which depends on the energy difference, tunnelling length, and mass). Therefore, in the example of H3 + H2-H1, if the initial conditions lead to a collision which has insufficient energy to reach the transition state, tunnelling through the potential barrier to the other side of the PES and hence to the products is feasible. This phenomenon would lead to a small increase in the expected rate of reaction - however, the barrier recrossing effect dominates overall and hence we would still expect slower rates than calculated in TST.

===Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions.===

F + H2
[[File:wm1415F_H2_exothermic.png|500 px|thumb|left|Exothermic PES]]
[[File:wm1415F_H2_transition_state.png|500 px|thumb|right|Approximate location of transition state]]
[[File:Wm1415_exo_reactants.png|500 px|thumb|left|MEP calculation from a distorted transition state to the reactants.]]
[[File:Wm1415_exo_products.png|500 px|thumb|right|MEP calculation from a distorted transition state to the products.]]

The reactants channel is higher in energy than the products channel: the reaction is clearly exothermic. Formation of a H-F bond releases more energy than the formation of a H-H bond: the H-F bond is stronger. Since the activation energy is small, the energies of the reactants and the transition states are similar: by Hammond's postulate, the transition state will resemble the structure of the reactants. Thus we expect the transition state to be located near the reactants channel: this can be seen above.
To find the energy of the reactants, an MEP calculation was run from a structure nearing the transition state, with the F-H bond slightly stretched, to produce a potential energy vs. time graph. Similarly. the energy of the products was found from a structure nearing the transition state, with the F-H bond slightly compressed.

Energy of the reactants ≈ -103.95 kcal/mol

Energy of the transition state ≈ -103.75 kcal/mol

Energy of the products ≈ -133.85 kcal/mol

Activation energy ≈ 0.2 kcal/mol

Energy change of reaction ≈ -29.90 kcal/mol (negative because exothermic - release of energy)

H + HF
[[File:wm1415H_HF_endothermic.png|500 px|thumb|left|Endothermic PES]]
[[File:Wm1415H_HF_transition_state.png|500 px|thumb|right|Approximate location of transition state]]
[[File:Wm1415_endo_reactants.png|500 px|thumb|left|MEP calculation from a distorted transition state to the reactants.]]
[[File:Wm1415_endo_products.png|500 px|thumb|right|MEP calculation from a distorted transition state to the products.]]

The reactants channel in the reverse reaction is lower in energy than the products channel: the reaction is endothermic. By Hammond's postulate, the transition state should resemble the structure of the products - the location of the transition state is very close to the products channel. Potential energy calculations against time were carried out for completeness - but since it is the reverse reaction of the above situation, it is clear that the energy calculations should be symmetrical.

Energy of the reactants ≈ -133.85 kcal/mol

Energy of the transition state ≈ -103.75 kcal/mol

Energy of the products ≈ -103.85 kcal/mol

Activation energy ≈ 30.0 kcal/mol

Energy change of reaction ≈ +29.90 kcal/mol (positive because endothermic)

===F + H2: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?===
[[File:Wm1415_energy_conservation.png|500 px|thumb|left|Reaction trajectory: potential energy surface.]]
[[File:Wm1415_energy_conservation_momenta.png|500 px|thumb|right|F + H2: internuclear momenta vs. time]]

Following the reaction, the H-F molecule formed is vibrationally excited. This can be seen from the very large oscillations in the potential energy diagram and the internuclear momenta vs. time graph. The energy released in the reaction has been converted into vibrational kinetic energy - heat. The amount of heat energy dissipated by the reaction could be measured by calorimetry.
The vibrational excitation of the H-F bond could be further confirmed by IR spectroscopy by comparing the IR spectrum obtained from the vibrationally excited products to the IR spectrum of a HF molecule at room temperature. At ambient temperatures, only the ground vibrational state is substantially occupied, as the vibrational temperature of the H-F bond is very large. Therefore, the transitions observed in the infrared spectrum of HF would be from the v = 0 to v = 1 state (and perhaps some v = 1 to v = 2 transitions). The lower vibrational levels closely follow the simple harmonic oscillator model, therefore the transitions occurring at room temperature would all involve the same energy and only a single line would be observed in the spectrum.
The simple harmonic oscillator model breaks down at high vibrational levels for diatomic molecules. The energy spacings become progressively smaller until they converge at the dissociation limit, where the bond breaks due to the large vibrational energy. The HF molecule produced in this reaction is vibrationally excited, and its infrared spectrum would involve energy transitions that would match energy spacings in the anharmonic oscillator model. Therefore, in comparing its spectrum to room temperature, there would be a shift to many lower frequency absorptions, that would eventually converge to the dissociation limit (although the convergence point may not be experimentally observed, as the vibrational energy would most likely be dissipated to the environment such that the bond formed in the reaction wouldn't break).

===Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.===

F + H2

The Polanyi rules refer to how the distribution of kinetic energy in the transition state affects whether a trajectory is reactive or unreactive. A higher proportion of vibrational energy promotes reactions with a late transition state, whereas a higher proportion of translational energy promotes reactions with an early transition state2.
{| class="wikitable" border=1
! pFH !! pHH
!Outcome
|-
| -0.5 || -2.8
|[[File:Wm1415mom-2.8.png|200 px|thumb|Reactive]]
|-
| -0.5 || -2.7
|[[File:Wm1415mom-2.7.png|200 px|thumb|Unreactive]]
|-
| -0.5 || 2.7
|[[File:Wm1415mom2.7.png|200 px|thumb|Unreactive]]
|-
| -0.5 || 3.0
|[[File:Wm1415mom3.0.png|200 px|thumb|Reactive]]
|-
| -0.8 || 0.1
|[[File:Wm1415lowmom.png|200 px|thumb|Reactive]]
|}
The F + H2 reaction is exothermic, with a very early transition state, as shown above. Thus, according to the empirical Polanyi's rules, a higher proportion of translational energy would result in a reactive trajectory. This is clearly seen in the last example, where a significantly lower energy system results in a reactive trajectory: the kinetic energy is largely translational (there are only small oscillations in the reactants channel).
In the first four cases, most of the kinetic energy in the system is in the form of bond vibrations. With pHH = 3.0 and -3.0, the trajectory is reactive; small deviations in pHH, however, result in unreactive trajectories. The system has a very large amount of energy compared to the activation energy, thus the outcomes of the trajectories become increasingly unpredictable.

H + HF

The H + HF reaction is endothermic, with a very late transition state, as shown above. Thus, according to the empirical Polanyi's rules, a higher proportion of vibrational energy would result in a reactive trajectory. Consider the following initial conditions:
{| class="wikitable" border=1
! pFH !! pHH
!Description
!Outcome
|-
| -9.5 || -10
|High translational, low vibrational
|[[File:Wm1415_high_kinetic_energy.png|400 px|thumb|left]][[File:Wm1415_high_kinetic_energy2.png|400 px|thumb|right]]
|-
| -10 || -0.01
|High vibrational, low translational
|[[File:Wm1415_high_vibrational_energy.png|400 px|thumb|left]][[File:Wm1415_high_vibrational_energy2.png|400 px|thumb|right]]
|}

In the first case, where the H atom approaches the transition state with a high ratio of translational:vibrational kinetic energy, the trajectory is unreactive. It is clearly seen from the PES that the transition state will not be reached with purely translational motion: the transition state will just decay back to the reactants channel.

In the second example, there is a high ratio of vibrational:translational kinetic energy resulting in a reactive trajectory. The high vibrational energy result in the bond being closer to breaking, and hence the transition state is both easier to form and more likely to result in the formation of products. This is a demonstration of Polanyi's rules in action - but not a proof. The rules are only empirical and often fail to predict how systems behave2.

=== References ===

1 I. N. Levine, ''Physical Chemistry'', McGraw-Hill, Boston, Mass., 6th edn, 2009, pp. 893

2 Z. Zhang, Y. Zhou, D. H. Zhang, G. Czakó and J. M. Bowman, ''J. Phys. Chem. Lett.'', 2012, 3, 3416–3419.

MRD:wfbfmicou

2017-05-31T15:05:58Z

===What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.===

At both a minimum and at a transition structure, the total gradient of the potential energy surface is zero. The points can be distinguished by their second derivative; the transition state is a saddle point. For a function <math>f\left(x,y\right)</math>, at a stationary point <math>\left(a,b\right)</math> consider the discriminant <math>D\equiv f_{xx}f_{yy}-f_{xy}^{2}</math>:

If <math>D>0</math>, and <math>f_{xx}\left(a,b\right)>0</math>, the point (a,b) is a local minimum.

If <math>D>0</math>, and <math>f_{xx}\left(a,b\right)<0</math>, the point (a,b) is a local maximum.

If <math>D<0</math>, the point (a,b) is a saddle point.

If <math>D=0</math>, higher order tests are required.

{{fontcolor1|gray| I assume you've copied this over from wikipedia (https://www.wikiwand.com/en/Second_partial_derivative_test). It's not wrong, but:
1) Cite your sources
2) D is the determinant of the Hessian matrix
[[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:05, 31 May 2017 (BST)}}

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.===

rts = 0.90775 Å

[[File:wm1415TS_position_internuclear_distance-time.png|500px]]

The internuclear distances remain effectively constant: this represents the metastable transition state.

===Comment on how the ''mep'' and the dynamic trajectories differ.===
[[File:wm1415Mep_vs_dynamic.png|500px|thumb|left|MEP Calculation]]
[[File:wm1415Dynamic_vs_mep.png|500px|thumb|right|Dynamic Calculation]]

In the MEP calculation, the kinetic energy of the particles is reset at each time step. The calculation begins at a structure similar to the transition state, with the H1-H2 distance slightly larger than the TS structure. As a result, the trajectory of the reaction in the MEP calculation follows the valley of the potential energy surface, without oscillating, to the reactants: H1 and H2-H3.
In the dynamic calculation, the inertia of the particles is taken into account: the velocities are not reset to zero at each step. Hence, vibrations are taken into account: the trajectory of the dynamic calculation also ends at the reactants, but oscillations are observed in this process.

===How do internuclear distances and momenta vary with time, when a reaction coordinate on either side of the transition state is chosen?===
[[File:wm1415Internuc_dist_time_r1.png|500px|thumb|left|Internuclear dist. vs. time - r1 = rts+δ, r2 = rts]]
[[File:wm1415Internuc_dist_time_r2.png|500px|thumb|right|Internuclear dist. vs. time - r1 = rts, r2 = rts+δ]]

[[File:wm1415Internuc_momenta_time_r1.png|500px|thumb|left|Internuclear momenta vs. time - r1 = rts+δ, r2 = rts]]
[[File:wm1415Internuc_momenta_time_r2.png|500px|thumb|right|Internuclear momenta vs. time - r1 = rts, r2 = rts+δ]]

The transition state is symmetrical: changing the initial position from (r1 = rts+δ, r2 = rts) to (r1 = rts, r2 = rts+δ) simply results in the formation of H1-H2 + H3 instead of H1 + H2-H3. The internuclear momenta and distance-time relationships are identical, with labels swapped.

===Reactive and unreactive trajectories===
{| class="wikitable" border=1
! p1 !! p2
!Outcome
!Description
|-
| -1.25 || -2.5
|[[File:wm1415Case1.png|200px|thumb|Reactive]]
|The H1-H2 molecule initially has no vibrational energy: the initial kinetic energy is purely translational energy of the approaching H3 atom. The H3 atom has sufficient kinetic energy to lead to the formation of the transition state and the products are formed: this is a reactive trajectory. The resulting H2-H3 molecule is vibrationally excited.
|-
| -1.5 || -2.0
|[[File:wm1415Case2.png|200px|thumb|Unreactive]]
|The H1-H2 molecule is initially vibrating: this causes the oscillations in the A-B (H3-H2) distance. H3 slows down as it approaches H2, converting kinetic energy to potential energy. For this set of initial conditions, the kinetic energy of H3 is insufficient to form the transition state and hence the products are not formed. H3 is eventually expelled by the H2-H1 molecule: this trajectory is unreactive.
|-
| -1.5 || -2.5
|[[File:wm1415Case3.png|200px|thumb|Reactive]]
|This reactive trajectory is similar to the first case. The H1-H2 molecule is initially vibrating with small amplitudes. H3 has sufficient kinetic energy to form the transition state and to expel H1 from the molecule. Following the reaction, the H3-H2 molecule is vibrationally excited.
|-
| -2.5 || -5.0
|[[File:wm1415Case4.png|200px|thumb|Unreactive]]
|The H2-H1 molecule initially has no vibrational energy. The approaching H3 atom has sufficient kinetic energy to reach the transition state: however, this particular set of initial conditions leads to barrier recrossing - the trajectory is unreactive. The large energies involved - far larger than required to reach the transition state - lead to a more unpredictable outcome. In this case, the transition state is so vibrationally excited that H3 is ejected from the activated complex and the system returns to the reactants channel, with the H2-H1 molecule vibrationally excited as a result.
|-
| -2.5 || -5.2
|[[File:wm1415Case5.png|200px|thumb|Reactive]]
|This set of initial conditions is similar to the previous example, except in this case the barrier recrossing leads to the ejection of the H1 atom from activated complex, giving the vibrationally excited H2-H3 molecule in the products channel.
|}

===State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for rate values compare with experimental values?===

Transition State Theory (TST) is a model that simplifies the computational complexity of calculating trajectories along a potential energy surface. TST requires knowledge of the PES in the reactant and product channels, as well as at the transition state1. A boundary is chosen - known as the critical dividing surface1 - which intersects the saddle point of the potential energy surface. The main assumptions of TST of ideal gas reactions are:

- "All supermolecules that cross the critical dividing surface from the reactant side become products."1

- "During the reaction, the Boltzmann distribution of energy is maintained for the reactant molecules."1

- "The supermolecules crossing the critical surface from the reactant side have a Boltzmann distribution of energy corresponding to the temperature of the reacting system."1

As seen in the fourth example above, the assumption that all supermolecules with energy greater than the transition state lead to a reaction breaks down. In the fourth and fifth examples, barrier recrossing is observed in the trajectory calculations. Barrier recrossing is a phenomenon that is not accounted for in the transition state model. Therefore, transition state theory would predict larger rates of reaction than observed experimentally.

Neither TST nor the model used in this investigation account for a further complication - the effect of quantum tunnelling. For small particles that do not classically have the energy to overcome a potential barrier, there is a certain probability of quantum tunnelling through the barrier (which depends on the energy difference, tunnelling length, and mass). Therefore, in the example of H3 + H2-H1, if the initial conditions lead to a collision which has insufficient energy to reach the transition state, tunnelling through the potential barrier to the other side of the PES and hence to the products is feasible. This phenomenon would lead to a small increase in the expected rate of reaction - however, the barrier recrossing effect dominates overall and hence we would still expect slower rates than calculated in TST.

===Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions.===

F + H2
[[File:wm1415F_H2_exothermic.png|500 px|thumb|left|Exothermic PES]]
[[File:wm1415F_H2_transition_state.png|500 px|thumb|right|Approximate location of transition state]]
[[File:Wm1415_exo_reactants.png|500 px|thumb|left|MEP calculation from a distorted transition state to the reactants.]]
[[File:Wm1415_exo_products.png|500 px|thumb|right|MEP calculation from a distorted transition state to the products.]]

The reactants channel is higher in energy than the products channel: the reaction is clearly exothermic. Formation of a H-F bond releases more energy than the formation of a H-H bond: the H-F bond is stronger. Since the activation energy is small, the energies of the reactants and the transition states are similar: by Hammond's postulate, the transition state will resemble the structure of the reactants. Thus we expect the transition state to be located near the reactants channel: this can be seen above.
To find the energy of the reactants, an MEP calculation was run from a structure nearing the transition state, with the F-H bond slightly stretched, to produce a potential energy vs. time graph. Similarly. the energy of the products was found from a structure nearing the transition state, with the F-H bond slightly compressed.

Energy of the reactants ≈ -103.95 kcal/mol

Energy of the transition state ≈ -103.75 kcal/mol

Energy of the products ≈ -133.85 kcal/mol

Activation energy ≈ 0.2 kcal/mol

Energy change of reaction ≈ -29.90 kcal/mol (negative because exothermic - release of energy)

H + HF
[[File:wm1415H_HF_endothermic.png|500 px|thumb|left|Endothermic PES]]
[[File:Wm1415H_HF_transition_state.png|500 px|thumb|right|Approximate location of transition state]]
[[File:Wm1415_endo_reactants.png|500 px|thumb|left|MEP calculation from a distorted transition state to the reactants.]]
[[File:Wm1415_endo_products.png|500 px|thumb|right|MEP calculation from a distorted transition state to the products.]]

The reactants channel in the reverse reaction is lower in energy than the products channel: the reaction is endothermic. By Hammond's postulate, the transition state should resemble the structure of the products - the location of the transition state is very close to the products channel. Potential energy calculations against time were carried out for completeness - but since it is the reverse reaction of the above situation, it is clear that the energy calculations should be symmetrical.

Energy of the reactants ≈ -133.85 kcal/mol

Energy of the transition state ≈ -103.75 kcal/mol

Energy of the products ≈ -103.85 kcal/mol

Activation energy ≈ 30.0 kcal/mol

Energy change of reaction ≈ +29.90 kcal/mol (positive because endothermic)

===F + H2: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?===
[[File:Wm1415_energy_conservation.png|500 px|thumb|left|Reaction trajectory: potential energy surface.]]
[[File:Wm1415_energy_conservation_momenta.png|500 px|thumb|right|F + H2: internuclear momenta vs. time]]

Following the reaction, the H-F molecule formed is vibrationally excited. This can be seen from the very large oscillations in the potential energy diagram and the internuclear momenta vs. time graph. The energy released in the reaction has been converted into vibrational kinetic energy - heat. The amount of heat energy dissipated by the reaction could be measured by calorimetry.
The vibrational excitation of the H-F bond could be further confirmed by IR spectroscopy by comparing the IR spectrum obtained from the vibrationally excited products to the IR spectrum of a HF molecule at room temperature. At ambient temperatures, only the ground vibrational state is substantially occupied, as the vibrational temperature of the H-F bond is very large. Therefore, the transitions observed in the infrared spectrum of HF would be from the v = 0 to v = 1 state (and perhaps some v = 1 to v = 2 transitions). The lower vibrational levels closely follow the simple harmonic oscillator model, therefore the transitions occurring at room temperature would all involve the same energy and only a single line would be observed in the spectrum.
The simple harmonic oscillator model breaks down at high vibrational levels for diatomic molecules. The energy spacings become progressively smaller until they converge at the dissociation limit, where the bond breaks due to the large vibrational energy. The HF molecule produced in this reaction is vibrationally excited, and its infrared spectrum would involve energy transitions that would match energy spacings in the anharmonic oscillator model. Therefore, in comparing its spectrum to room temperature, there would be a shift to many lower frequency absorptions, that would eventually converge to the dissociation limit (although the convergence point may not be experimentally observed, as the vibrational energy would most likely be dissipated to the environment such that the bond formed in the reaction wouldn't break).

===Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.===

F + H2

The Polanyi rules refer to how the distribution of kinetic energy in the transition state affects whether a trajectory is reactive or unreactive. A higher proportion of vibrational energy promotes reactions with a late transition state, whereas a higher proportion of translational energy promotes reactions with an early transition state2.
{| class="wikitable" border=1
! pFH !! pHH
!Outcome
|-
| -0.5 || -2.8
|[[File:Wm1415mom-2.8.png|200 px|thumb|Reactive]]
|-
| -0.5 || -2.7
|[[File:Wm1415mom-2.7.png|200 px|thumb|Unreactive]]
|-
| -0.5 || 2.7
|[[File:Wm1415mom2.7.png|200 px|thumb|Unreactive]]
|-
| -0.5 || 3.0
|[[File:Wm1415mom3.0.png|200 px|thumb|Reactive]]
|-
| -0.8 || 0.1
|[[File:Wm1415lowmom.png|200 px|thumb|Reactive]]
|}
The F + H2 reaction is exothermic, with a very early transition state, as shown above. Thus, according to the empirical Polanyi's rules, a higher proportion of translational energy would result in a reactive trajectory. This is clearly seen in the last example, where a significantly lower energy system results in a reactive trajectory: the kinetic energy is largely translational (there are only small oscillations in the reactants channel).
In the first four cases, most of the kinetic energy in the system is in the form of bond vibrations. With pHH = 3.0 and -3.0, the trajectory is reactive; small deviations in pHH, however, result in unreactive trajectories. The system has a very large amount of energy compared to the activation energy, thus the outcomes of the trajectories become increasingly unpredictable.

H + HF

The H + HF reaction is endothermic, with a very late transition state, as shown above. Thus, according to the empirical Polanyi's rules, a higher proportion of vibrational energy would result in a reactive trajectory. Consider the following initial conditions:
{| class="wikitable" border=1
! pFH !! pHH
!Description
!Outcome
|-
| -9.5 || -10
|High translational, low vibrational
|[[File:Wm1415_high_kinetic_energy.png|400 px|thumb|left]][[File:Wm1415_high_kinetic_energy2.png|400 px|thumb|right]]
|-
| -10 || -0.01
|High vibrational, low translational
|[[File:Wm1415_high_vibrational_energy.png|400 px|thumb|left]][[File:Wm1415_high_vibrational_energy2.png|400 px|thumb|right]]
|}

In the first case, where the H atom approaches the transition state with a high ratio of translational:vibrational kinetic energy, the trajectory is unreactive. It is clearly seen from the PES that the transition state will not be reached with purely translational motion: the transition state will just decay back to the reactants channel.

In the second example, there is a high ratio of vibrational:translational kinetic energy resulting in a reactive trajectory. The high vibrational energy result in the bond being closer to breaking, and hence the transition state is both easier to form and more likely to result in the formation of products. This is a demonstration of Polanyi's rules in action - but not a proof. The rules are only empirical and often fail to predict how systems behave2.

=== References ===

1 I. N. Levine, ''Physical Chemistry'', McGraw-Hill, Boston, Mass., 6th edn, 2009, pp. 893

2 Z. Zhang, Y. Zhou, D. H. Zhang, G. Czakó and J. M. Bowman, ''J. Phys. Chem. Lett.'', 2012, 3, 3416–3419.

MRD:wfbfmicou

2017-05-31T15:05:38Z

===What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.===

At both a minimum and at a transition structure, the total gradient of the potential energy surface is zero. The points can be distinguished by their second derivative; the transition state is a saddle point. For a function <math>f\left(x,y\right)</math>, at a stationary point <math>\left(a,b\right)</math> consider the discriminant <math>D\equiv f_{xx}f_{yy}-f_{xy}^{2}</math>:

If <math>D>0</math>, and <math>f_{xx}\left(a,b\right)>0</math>, the point (a,b) is a local minimum.

If <math>D>0</math>, and <math>f_{xx}\left(a,b\right)<0</math>, the point (a,b) is a local maximum.

If <math>D<0</math>, the point (a,b) is a saddle point.

If <math>D=0</math>, higher order tests are required.

{{fontcolor1|gray| I assume you've copied this over from wikipedia (https://www.wikiwand.com/en/Second_partial_derivative_test). It's not wrong, but:

1) Cite your sources
2) D is the determinant of the Hessian matrix

[[User:Je714|Je714]] ([[User talk:Je714|talk]]) 16:05, 31 May 2017 (BST)}}

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.===

rts = 0.90775 Å

[[File:wm1415TS_position_internuclear_distance-time.png|500px]]

The internuclear distances remain effectively constant: this represents the metastable transition state.

===Comment on how the ''mep'' and the dynamic trajectories differ.===
[[File:wm1415Mep_vs_dynamic.png|500px|thumb|left|MEP Calculation]]
[[File:wm1415Dynamic_vs_mep.png|500px|thumb|right|Dynamic Calculation]]

In the MEP calculation, the kinetic energy of the particles is reset at each time step. The calculation begins at a structure similar to the transition state, with the H1-H2 distance slightly larger than the TS structure. As a result, the trajectory of the reaction in the MEP calculation follows the valley of the potential energy surface, without oscillating, to the reactants: H1 and H2-H3.
In the dynamic calculation, the inertia of the particles is taken into account: the velocities are not reset to zero at each step. Hence, vibrations are taken into account: the trajectory of the dynamic calculation also ends at the reactants, but oscillations are observed in this process.

===How do internuclear distances and momenta vary with time, when a reaction coordinate on either side of the transition state is chosen?===
[[File:wm1415Internuc_dist_time_r1.png|500px|thumb|left|Internuclear dist. vs. time - r1 = rts+δ, r2 = rts]]
[[File:wm1415Internuc_dist_time_r2.png|500px|thumb|right|Internuclear dist. vs. time - r1 = rts, r2 = rts+δ]]

[[File:wm1415Internuc_momenta_time_r1.png|500px|thumb|left|Internuclear momenta vs. time - r1 = rts+δ, r2 = rts]]
[[File:wm1415Internuc_momenta_time_r2.png|500px|thumb|right|Internuclear momenta vs. time - r1 = rts, r2 = rts+δ]]

The transition state is symmetrical: changing the initial position from (r1 = rts+δ, r2 = rts) to (r1 = rts, r2 = rts+δ) simply results in the formation of H1-H2 + H3 instead of H1 + H2-H3. The internuclear momenta and distance-time relationships are identical, with labels swapped.

===Reactive and unreactive trajectories===
{| class="wikitable" border=1
! p1 !! p2
!Outcome
!Description
|-
| -1.25 || -2.5
|[[File:wm1415Case1.png|200px|thumb|Reactive]]
|The H1-H2 molecule initially has no vibrational energy: the initial kinetic energy is purely translational energy of the approaching H3 atom. The H3 atom has sufficient kinetic energy to lead to the formation of the transition state and the products are formed: this is a reactive trajectory. The resulting H2-H3 molecule is vibrationally excited.
|-
| -1.5 || -2.0
|[[File:wm1415Case2.png|200px|thumb|Unreactive]]
|The H1-H2 molecule is initially vibrating: this causes the oscillations in the A-B (H3-H2) distance. H3 slows down as it approaches H2, converting kinetic energy to potential energy. For this set of initial conditions, the kinetic energy of H3 is insufficient to form the transition state and hence the products are not formed. H3 is eventually expelled by the H2-H1 molecule: this trajectory is unreactive.
|-
| -1.5 || -2.5
|[[File:wm1415Case3.png|200px|thumb|Reactive]]
|This reactive trajectory is similar to the first case. The H1-H2 molecule is initially vibrating with small amplitudes. H3 has sufficient kinetic energy to form the transition state and to expel H1 from the molecule. Following the reaction, the H3-H2 molecule is vibrationally excited.
|-
| -2.5 || -5.0
|[[File:wm1415Case4.png|200px|thumb|Unreactive]]
|The H2-H1 molecule initially has no vibrational energy. The approaching H3 atom has sufficient kinetic energy to reach the transition state: however, this particular set of initial conditions leads to barrier recrossing - the trajectory is unreactive. The large energies involved - far larger than required to reach the transition state - lead to a more unpredictable outcome. In this case, the transition state is so vibrationally excited that H3 is ejected from the activated complex and the system returns to the reactants channel, with the H2-H1 molecule vibrationally excited as a result.
|-
| -2.5 || -5.2
|[[File:wm1415Case5.png|200px|thumb|Reactive]]
|This set of initial conditions is similar to the previous example, except in this case the barrier recrossing leads to the ejection of the H1 atom from activated complex, giving the vibrationally excited H2-H3 molecule in the products channel.
|}

===State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for rate values compare with experimental values?===

Transition State Theory (TST) is a model that simplifies the computational complexity of calculating trajectories along a potential energy surface. TST requires knowledge of the PES in the reactant and product channels, as well as at the transition state1. A boundary is chosen - known as the critical dividing surface1 - which intersects the saddle point of the potential energy surface. The main assumptions of TST of ideal gas reactions are:

- "All supermolecules that cross the critical dividing surface from the reactant side become products."1

- "During the reaction, the Boltzmann distribution of energy is maintained for the reactant molecules."1

- "The supermolecules crossing the critical surface from the reactant side have a Boltzmann distribution of energy corresponding to the temperature of the reacting system."1

As seen in the fourth example above, the assumption that all supermolecules with energy greater than the transition state lead to a reaction breaks down. In the fourth and fifth examples, barrier recrossing is observed in the trajectory calculations. Barrier recrossing is a phenomenon that is not accounted for in the transition state model. Therefore, transition state theory would predict larger rates of reaction than observed experimentally.

Neither TST nor the model used in this investigation account for a further complication - the effect of quantum tunnelling. For small particles that do not classically have the energy to overcome a potential barrier, there is a certain probability of quantum tunnelling through the barrier (which depends on the energy difference, tunnelling length, and mass). Therefore, in the example of H3 + H2-H1, if the initial conditions lead to a collision which has insufficient energy to reach the transition state, tunnelling through the potential barrier to the other side of the PES and hence to the products is feasible. This phenomenon would lead to a small increase in the expected rate of reaction - however, the barrier recrossing effect dominates overall and hence we would still expect slower rates than calculated in TST.

===Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions.===

F + H2
[[File:wm1415F_H2_exothermic.png|500 px|thumb|left|Exothermic PES]]
[[File:wm1415F_H2_transition_state.png|500 px|thumb|right|Approximate location of transition state]]
[[File:Wm1415_exo_reactants.png|500 px|thumb|left|MEP calculation from a distorted transition state to the reactants.]]
[[File:Wm1415_exo_products.png|500 px|thumb|right|MEP calculation from a distorted transition state to the products.]]

The reactants channel is higher in energy than the products channel: the reaction is clearly exothermic. Formation of a H-F bond releases more energy than the formation of a H-H bond: the H-F bond is stronger. Since the activation energy is small, the energies of the reactants and the transition states are similar: by Hammond's postulate, the transition state will resemble the structure of the reactants. Thus we expect the transition state to be located near the reactants channel: this can be seen above.
To find the energy of the reactants, an MEP calculation was run from a structure nearing the transition state, with the F-H bond slightly stretched, to produce a potential energy vs. time graph. Similarly. the energy of the products was found from a structure nearing the transition state, with the F-H bond slightly compressed.

Energy of the reactants ≈ -103.95 kcal/mol

Energy of the transition state ≈ -103.75 kcal/mol

Energy of the products ≈ -133.85 kcal/mol

Activation energy ≈ 0.2 kcal/mol

Energy change of reaction ≈ -29.90 kcal/mol (negative because exothermic - release of energy)

H + HF
[[File:wm1415H_HF_endothermic.png|500 px|thumb|left|Endothermic PES]]
[[File:Wm1415H_HF_transition_state.png|500 px|thumb|right|Approximate location of transition state]]
[[File:Wm1415_endo_reactants.png|500 px|thumb|left|MEP calculation from a distorted transition state to the reactants.]]
[[File:Wm1415_endo_products.png|500 px|thumb|right|MEP calculation from a distorted transition state to the products.]]

The reactants channel in the reverse reaction is lower in energy than the products channel: the reaction is endothermic. By Hammond's postulate, the transition state should resemble the structure of the products - the location of the transition state is very close to the products channel. Potential energy calculations against time were carried out for completeness - but since it is the reverse reaction of the above situation, it is clear that the energy calculations should be symmetrical.

Energy of the reactants ≈ -133.85 kcal/mol

Energy of the transition state ≈ -103.75 kcal/mol

Energy of the products ≈ -103.85 kcal/mol

Activation energy ≈ 30.0 kcal/mol

Energy change of reaction ≈ +29.90 kcal/mol (positive because endothermic)

===F + H2: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?===
[[File:Wm1415_energy_conservation.png|500 px|thumb|left|Reaction trajectory: potential energy surface.]]
[[File:Wm1415_energy_conservation_momenta.png|500 px|thumb|right|F + H2: internuclear momenta vs. time]]

Following the reaction, the H-F molecule formed is vibrationally excited. This can be seen from the very large oscillations in the potential energy diagram and the internuclear momenta vs. time graph. The energy released in the reaction has been converted into vibrational kinetic energy - heat. The amount of heat energy dissipated by the reaction could be measured by calorimetry.
The vibrational excitation of the H-F bond could be further confirmed by IR spectroscopy by comparing the IR spectrum obtained from the vibrationally excited products to the IR spectrum of a HF molecule at room temperature. At ambient temperatures, only the ground vibrational state is substantially occupied, as the vibrational temperature of the H-F bond is very large. Therefore, the transitions observed in the infrared spectrum of HF would be from the v = 0 to v = 1 state (and perhaps some v = 1 to v = 2 transitions). The lower vibrational levels closely follow the simple harmonic oscillator model, therefore the transitions occurring at room temperature would all involve the same energy and only a single line would be observed in the spectrum.
The simple harmonic oscillator model breaks down at high vibrational levels for diatomic molecules. The energy spacings become progressively smaller until they converge at the dissociation limit, where the bond breaks due to the large vibrational energy. The HF molecule produced in this reaction is vibrationally excited, and its infrared spectrum would involve energy transitions that would match energy spacings in the anharmonic oscillator model. Therefore, in comparing its spectrum to room temperature, there would be a shift to many lower frequency absorptions, that would eventually converge to the dissociation limit (although the convergence point may not be experimentally observed, as the vibrational energy would most likely be dissipated to the environment such that the bond formed in the reaction wouldn't break).

===Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.===

F + H2

The Polanyi rules refer to how the distribution of kinetic energy in the transition state affects whether a trajectory is reactive or unreactive. A higher proportion of vibrational energy promotes reactions with a late transition state, whereas a higher proportion of translational energy promotes reactions with an early transition state2.
{| class="wikitable" border=1
! pFH !! pHH
!Outcome
|-
| -0.5 || -2.8
|[[File:Wm1415mom-2.8.png|200 px|thumb|Reactive]]
|-
| -0.5 || -2.7
|[[File:Wm1415mom-2.7.png|200 px|thumb|Unreactive]]
|-
| -0.5 || 2.7
|[[File:Wm1415mom2.7.png|200 px|thumb|Unreactive]]
|-
| -0.5 || 3.0
|[[File:Wm1415mom3.0.png|200 px|thumb|Reactive]]
|-
| -0.8 || 0.1
|[[File:Wm1415lowmom.png|200 px|thumb|Reactive]]
|}
The F + H2 reaction is exothermic, with a very early transition state, as shown above. Thus, according to the empirical Polanyi's rules, a higher proportion of translational energy would result in a reactive trajectory. This is clearly seen in the last example, where a significantly lower energy system results in a reactive trajectory: the kinetic energy is largely translational (there are only small oscillations in the reactants channel).
In the first four cases, most of the kinetic energy in the system is in the form of bond vibrations. With pHH = 3.0 and -3.0, the trajectory is reactive; small deviations in pHH, however, result in unreactive trajectories. The system has a very large amount of energy compared to the activation energy, thus the outcomes of the trajectories become increasingly unpredictable.

H + HF

The H + HF reaction is endothermic, with a very late transition state, as shown above. Thus, according to the empirical Polanyi's rules, a higher proportion of vibrational energy would result in a reactive trajectory. Consider the following initial conditions:
{| class="wikitable" border=1
! pFH !! pHH
!Description
!Outcome
|-
| -9.5 || -10
|High translational, low vibrational
|[[File:Wm1415_high_kinetic_energy.png|400 px|thumb|left]][[File:Wm1415_high_kinetic_energy2.png|400 px|thumb|right]]
|-
| -10 || -0.01
|High vibrational, low translational
|[[File:Wm1415_high_vibrational_energy.png|400 px|thumb|left]][[File:Wm1415_high_vibrational_energy2.png|400 px|thumb|right]]
|}

In the first case, where the H atom approaches the transition state with a high ratio of translational:vibrational kinetic energy, the trajectory is unreactive. It is clearly seen from the PES that the transition state will not be reached with purely translational motion: the transition state will just decay back to the reactants channel.

In the second example, there is a high ratio of vibrational:translational kinetic energy resulting in a reactive trajectory. The high vibrational energy result in the bond being closer to breaking, and hence the transition state is both easier to form and more likely to result in the formation of products. This is a demonstration of Polanyi's rules in action - but not a proof. The rules are only empirical and often fail to predict how systems behave2.

=== References ===

1 I. N. Levine, ''Physical Chemistry'', McGraw-Hill, Boston, Mass., 6th edn, 2009, pp. 893

2 Z. Zhang, Y. Zhou, D. H. Zhang, G. Czakó and J. M. Bowman, ''J. Phys. Chem. Lett.'', 2012, 3, 3416–3419.

MRD:Stella

2017-05-31T14:42:54Z

Je714: /* EXERCISE 1: H + H2 system */

== EXERCISE 1: H + H2 system ==

=== Dynamics from the transition state region ===
*What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Molecular reaction dynamics is a critical technique in the investigation of the collisions between molecules. It enhances the understanding at the molecular level mechanism of elementary chemical and physical processes by considering and controlling the parameters specifying the initial state of the reactants, followed by observation on how it affects the trajectory of the reaction.

While modern computational methods typically involves quantum calculations and analyses, this study aims to achieve a fundamental understanding of molecular dynamics through a classical approach.

A calculation of a linear triatomic H + H2 system was performed, where r1 = 0.74, r2 = 2.30, p1 = 0.0 and p2 = -2.7.

An illustration of the reaction trajectory is illustrated in Figure 1a, and the potential energy of the system against time is displayed in Figure 1b.
{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none; width:60%"
|-
| | [[File:Syl815 Surface Plot of H-H-H trajectory.PNG|center|400px]]
| [[File:Syl 815 Potential Energy Graph of H-H-H trajectory.PNG|center|400px]]
|-
| style = "text-align:center;"| Figure 1a. The surface plot of H1–H2.
| style = "text-align:center;"| Figure 1b. Graph of the potential energy of the system against t.
|}

Each point along the potential energy surface (Figure 1a) is a representation of the energy for a specific geometrical configuration of the atoms/molecules involved, which are expressed by the distances r1 and r2. The motion of the atom could be scrutinised by determining the forces acting on the atoms.

This is determined by taking the first derivative, as expressed in Eq. 1:
{| align = "center" style="width:90%;background-color: white; border-spacing: 0px;"
| width = 30%|
| width = 30%|<math display ="block"> \bold{F} = -\nabla \text{V} = -({\partial\over\partial r_1} + {\partial\over\partial r_2})</math>
| style="text-align:right" width = 30%|(1)
|}

If '''''F''''' ≠ 0, the atom would move in the direction of the force, until the force acting on the atom is zero, and the potential energy of the system minimised.

A preliminary observation of the reaction trajectory of H3 and H1–H2 establishes that the reaction proceeds along a path of minimum potential energy (near the base of the potential well), which is clearly featured in Figure 2a.

{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none; width:60%"
|-
| [[File: Syl815 Potential Energy Graph of H-H-H minima Transition State.png|400px|center]]
| [[File: Syl815 Potential Energy Graph of H-H-H trajectory Reaction trajectory.PNG|400px]]
|-
| Figure 2a. Surface plot featuring the energy potential of the reaction coordinate H1–H2 and H3. It is along the path of minima in this well.
| Figure 2b. Variation of potential energy along the reaction coordinate of H1–H2 and H3.
|}

A salient feature is the slight sinusoidal-like pattern of the reaction trajectory. This could be attributed to the molecular vibrations at the equilibrium state of molecules, as displayed in Figure 1a and Figure 2a. A graph of the internuclear momentum against time is presented in Figure 3 for reference.

[[File:Syl815_Graph_of_Internuclear_Momenta_against_Time.PNG|center|upright=2.5|thumb|Figure 3. A graph of the internuclear momentum against time t ]]

The sinusoidal pattern of the internuclear momentum for H1–H2 from t = 0 and t = 0.325, for instance, is indicative of molecular vibrations. This is corroborated by the sinusoidal pattern of the H1–H3 internuclear momentum curve after t = 0.695, after the formation of the H1–H3 bond. The highly symmetrical oscillatory nature of the graph is characteristic of symmetric molecular vibrations, and this is further corroborated by the plateau of the H1–H2 internuclear momentum at the same time, corresponding to an absence of vibrations between the now free H1 and H2 atoms.

An inspection of the reaction coordinate reveals a distinct maxima as exhibited in Figure 2b, corresponding to the transition state. The transition state is characterised as the region which separates the reactants from the products, and is associated with the highest potential energy along the reaction coordinate. In this case, this is the region where the system most resembles a triatomic molecule, where r1 = r2.

In the three-dimensional visualisation, stable species appear as minima on the potential energy surface, where the deepest well corresponds to the most stable species. On the other hand, transition states correlate with the saddle-point in the potential energy surface. It is a maximum in one dimension, which is along the reaction coordinate, and minimum in all other dimensions, as illustrated in Figure 2b and Figure 2a respectively.

{{fontcolor1|gray| Up to this point, all the above was not really related to the question. Perhaps you could've moved this to an introductory section. The question was merely asking between the mathematical distinction between a minima and a transition state in a '''generic''' PES. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 15:20, 31 May 2017 (BST)}}

Therefore, the total gradient, '''''F''''' at the minima, and the transition state, is equal to 0, and therefore, it cannot be used as a distinguishing factor between the two.

However, a second order partial derivative can be performed to determine the nature of the local extremum, as illustrated in '''Eq. (2)''':

{| align = "center" style="width:90%;background-color: white; border-spacing: 0px;"
| width = 30%|
| width = 30%| <math display ="block"> \bold{D} = f_{xx} \cdot f_{yy} - f_{xy} </math>
| style="text-align:right" width = 30%|(2)
|}

If the point is a global or local minimum, then the value <math display = "in line"> \bold{D}</math> and <math display = "in line"> f_{xx}(x_0, y_0) > 0</math>. The saddle point is confirmed if <math display = "in line"> \bold{D} = 0</math>, which correlates to the transition state. By observing Figure 2a, a distinctive saddle point is revealed.

{{fontcolor1|gray| Where did you get this from? I think you're trying to use the [second partial derivative test](https://www.wikiwand.com/en/Second_partial_derivative_test), but I'm not sure since you're not defining what your notation means. When using maths, it's important to let the reader know what each symbol means. I assume that D is the determinant of the Hessian matrix. In such case, I'm afraid you got the definition of it, and the sign of the test, wrong. D is actually equal to <math>f_{xx} \cdot f_{yy} - f_{xy}^{2}</math> and D<0 for a saddle point. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 15:20, 31 May 2017 (BST)}}

=== Trajectories from '''r1''' = '''r2''': locating the transition state ===

*Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.

In the reaction of H1–H2 with H3 atom, this involves the bond breaking of the H1–H2, as well as the bond formation of –H3. At the transition state, the potential energy of the system is maximum along the reaction trajectory, which occurs at the point where the separation of H1, H2 and H3 atoms are equidistant since the H + H2 surface is symmetric. At the transition state, it is also important to consider how the velocity associated with any atom affects the reaction coordinate. If p1 or p2 is not zero, a slight displacement from the saddle point would result in the H atoms sliding towards either the reactants’ or products’ valley.

As such, by modifying the values such that r1 = r2, as well as p1 = p2 = 0, it is possible to locate the transition state geometry, since there is no resultant force that drives the reaction either towards the reactants or the products.

Figure 4 illustrates the internuclear distance of the atoms against time. [[File:Stella Transition State for Exercise 1.png|centre|upright=2.5|thumb|Figure 4. Graph of internuclear distance against time t when r1 = r2 = 0.908.]]

The straight line graphs are representative of the motion of the atoms, indicating that there is no vibration between the atoms. Another prominent feature is that both r1 and r2 gradients are 0, which means that the internuclear forces of attraction between H1 and H2, as well as those between H1 and H3 balance, such that the resultant force on the atoms (given in Eq. 1) = 0.

Since r1 = r2, the graphs are identical and directly overlay each other, as evident in Figure 4. The transition state position rts is determined to be 0.908.

=== Trajectories from r1 = '''rts+δ''', '''r2''' = rts ===
*Comment on how the mep and the trajectory you just calculated differ.

The minimum energy path (mep) searches for transition states and intermediates between the reactants and products, by following along the reaction trajectory, until the saddle point is reached. This is attributed to the fact that the reaction trajectory passes through the local minima of the potential energy surface, and thus the height of the barrier is the lowest maximum of the potential between the reactants’ and products’ valleys.

One of the established methods of the mep is the utilisation of many geometric configurations of the system to describe the reaction pathway. At each time step, the velocity of the atoms are reset to zero. It can, therefore, present information about the total gradient of the minimum energy pathway, and a description of the reaction trajectory in a qualitative manner (the direction of the reaction progress). This is because all subsequent geometric configurations in each time step are dependent on the gradient of the potential surface at that point, '''''F''''', without any contribution from the translational or vibrational modes of the atoms. In contrast, a molecular dynamics calculation does not constraint the velocity of the atoms, and the trajectory traced includes contributions of '''''F''''' at that instant and the instantaneous motion of the atoms.

{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;width:90%;text-align:center"
|+ Table 1. Key differences between the results of molecular dynamics and mep calculations.
|-
! scope="Col" style = "width: 50%;"| Molecular Dynamics
! scope="Col" style = "width: 50%;"| mep
|- style="background: silver"
| The contour plot exhibits an oscillatory behaviour due to molecular vibrations.
| The contour plot is a smooth curve since any vibrational motion of the molecules are ignored.
|-
| The potential energy graph exhibits a noticeable oscillatory behaviour when t is approximately 0.5 from the H1–H2 vibration, and a regular oscillatory pattern at approximately t = 0.95, upon the complete bond formation between H2–H3, and the separation of H1 from the diatomic molecule.
| The potential energy curve smoothly decreases over time and does not plateau within t = 3.
|}

{{fontcolor1|gray| Good. Be careful: the ''mep'' isn't really searching for TS geometries -- it simply follows the path along where the energy is minimal. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 15:24, 31 May 2017 (BST)}}

*What would change if we used the initial conditions r1 = '''rts''' and r2 = '''rts'''+0.01 instead?

If the above conditions were used, H1–H2 + H3 are formed instead of H2–H3 and H1. Figure 5a. illustrates the subsequent motion of the atoms.

{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none; width:60%"
|-
| | [[File:Syl815 Internuclear distance AB 0.918.png|center|400px]]
| [[File:Syl815 Internuclear momenta AB 0.918.png|center|400px]]
|-
| style = "text-align:center;"| Figure 5a. Graph of internuclear distance against time t when r1 = 0.908 and r2 = 0.918.
| style = "text-align:center;"| Figure 5b. Graph of internuclear momenta against time t when r1 = 0.908 and r2 = 0.918.
|}

A slight oscillatory behaviour of the <math>f(\mathbf{r_1})</math> in Figure 5a. implies the presence of the H1–H2 vibrations. This also affects the behaviour of the <math>f(\mathbf{r_2})</math> plot, which is expected to be linear. This is ascribed to the periodic increase and decrease in separation between the H1 and H2 atoms, which corresponds to the value of r1, and between the H3 and H2 atoms (r2). Figure 5b elucidates the relationships between the atoms, where the <math>f(\mathbf{r_1})</math> displays sinusoidal behaviour and the <math>f(\mathbf{r_2})</math> graph is linear, verifying the presence of H1–H2 + H3.

=== Reactive and unreactive trajectories ===
*Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.

{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+ Table 1. Prediction of reactivity of trajectories with given p1 and p2 values.
|-
! scope="Col"| p1, p2
! scope = "Col" colspan="2"| Screenshot
! Reactivity
|- style="background: silver"
| rowspan="2" style="width:15%;"| -1.25, -2.5
| style="width:50%;"| [[File:Stella Internuclear Distance p1 -1.25 p2 -2.5.png|thumb|center|upright=1|Figure 6a. A graph of the internuclear distances r1 and r2 against time t]]
| [[File:Stella Surface plot p1 -1.25 p2 -2.5.png|thumb|center|upright=1|Figure 6b. A contour plot of the trajectory of the reaction r1 and r2]]
| rowspan = "2" style="width:20%;" align-text:"Center" | Reactive. The collision between H3 and H1–H2 results in a bond between H2 and H3 with the r2 value oscillating about a fixed value of approximately 0.741 Å.
|- style="background: silver"
| align-text:"Center"| The intersection of the r1 and r2 graphs denotes the transition state of the reaction since r1 = r2 at this point. The successful reaction of H1–H2 with H3 to form H2–H3 is evinced by intersection of r1 and r2 at only one point, which implies that the atoms have successfully slid into the products’ valley.
| align-text:"Center"| The reaction trajectory of H1–H2 crossed the transition state at a single point. Another noteworthy feature of the reaction trajectory is the linearity of the motion of H1–H2 and H3 at the entrance channel, which implies that there was no vibrational energy present in the H1–H2 molecule, and translational energy was sufficient for the reaction to proceed.
|-
| rowspan="2" style="width:15%;"| -1.5, -2.0
| style="width:50%;"| [[File:Syl815_Internuclear_Momentum_p1_-1.5_p2_-2.0.png|thumb|upright=1|Figure 7a. A graph of the internuclear distances r1 and r2 against time t]]
| [[File:Stella Surface plot p1 -1.5 p2 -2.0.png|thumb|upright=1|center|Figure 7b. A contour plot of the trajectory of the reaction r1 and r2]]
| rowspan = "2" style="width:20%;" align-text:"Center" | Not reactive. H3 did not approach H1–H2 with sufficient kinetic energy to effect a collision. The reaction path did not lead to the atoms sliding into the products valley, as the path was reversed.
|-
| align-text:"Center"| The sinusoidal curve of <math>f(\mathbf{p_1})</math> is indicative of the vibrational mode of H1–H2. The negative values of p1 and p2 indicate the initial approach of H1–H2 and H3, and a positive value of p2 indicates the increasing separation of H1–H2 and H3. <math>f(\mathbf{p_2})</math> increases to a horizontal limit of approximately 2.1, and the horizontal graph at that value indicates that H3 is moving at a constant velocity away from H1–H2.
| align-text:"Center"| H1–H2 is vibrationally excited as the bond was initially stretched. It is evident that vibrational modes are not as effective as translational modes in promoting the reaction.
|- style="background: silver"
| rowspan="2" style="width:15%;"| -1.5, -2.5
| [[File:Stella Internuclear Distance p1 -1.5 p2 -2.5.png|thumb|upright=1|center|Figure 8a. A graph of the internuclear distances r1 and r2 against time t]]
| [[File:Stella Surface plot p1 -1.5 p2 -2.5.png|thumb|upright=1|center|Figure 8b. A contour plot of the trajectory of the reaction r1 and r2]]
| rowspan = "2"| Reactive, since the reaction trajectory crossed the transition state at a single point.
|- style="background: silver"
| align-text:"Center"| The intersection of the r1 and r2 graphs denotes the transition state of the reaction since r1 = r2 at this point. The successful reaction of H1–H2 with H3 to form H2–H3 is evinced by intersection of r1 and r2 at only one point. H2–H3 formed has vibrational energy, and H1 moves away from the molecule at a constant velocity, as suggested by the approximately straight line function of <math>f(\mathbf{r_1})</math>. The slight variations are due to the vibrational motion of H2–H3 which results in the periodic increase and decrease in separation between H2 and H3.
| align-text:"Center"| The slight sinusoidal pattern of r1 implicates the vibrational energy of H1–H2 at the entrance channel. After the crossing of the transition state, a more pronounced oscillatory trajectory at r2 ≈ 0.741 Å suggests an increased vibrational mode of H2–H3 compared to H1–H2.
|-
| rowspan = "2"| -2.5, -5.0
|[[File:Stella Internuclear Distance p1 -2.5 p2 -5.png|thumb|upright=1|center|Figure 9a. A graph of the internuclear distances r1 and r2 against time t]]
|[[File:Stella Surface plot p1 -2.5 p2 -5.png|thumb|upright=1|center|Figure 9b. A contour plot of the trajectory of the reaction r1 and r2]]
| rowspan = "2" align-text:"Center"| Unreactive. A reaction occurs, leading to a transient formation of H2–H3. However, the subsequent collision between the H2–H3 molecule and H1 atom results in the formation of H1–H2 once more, as the energetic barrier has been recrossed.
|-
|The successful reaction to form H2–H3 molecule is demonstrated, through the intersection of the bond distance r1 and r2 as shown in the graph at t = 0.165. A second intersection indicates the recrossing of the reaction trajectory, and the subsequent sinusoidal curve of <math>f(\mathbf{r_2})</math> indicates the presence of H1–H2 vibrations.
| The initial system possesses large translational energy and no vibrational energy as suggested by linear reaction path at the entrance channel. Following collision, the reaction trajectory easily surmounts the energetic barrier, forming H1–H2 and H3 once more. The reaction trajectory exhibits a diagonal oscillatory pattern. Physical interpretation suggests that both vibrational and translational modes are present in the final system, i.e. some translational energy is converted to vibrational energy.
|- style="background: silver"
| rowspan = "2"| -2.5, -5.2
| [[File:Stella Internuclear Distance p1 -2.5 p2 -5.2.png|thumb|upright=1|center|Figure 10a. A graph of the internuclear distances r1 and r2 against time t]]
| [[File:Stella Surface plot p1 -2.5 p2 -5.2.png|thumb|upright=1|center|Figure 10b. A contour plot of the trajectory of the reaction r1 and r2]]
| rowspan = "2"| Reactive. 2 recrossings of the reaction trajectory are observed, indicating the formation of the products.
|- style="background: silver"
| Two intersections can be clearly observed. The first intersection implies the crossing of the energetic barrier from the products to the reactants, while the second will result in the reactant to product trajectory. The oscillatory function of r2 indicates the presence of vibrationally excited modes. The sinusoidal pattern of <math>f(\mathbf{r_1})</math> present is a result of the vibrations of H2–H3. The otherwise linear graph indicates the presence of translational energy as H1 separates from H2–H3.
| The initial system possesses large translational energy and insignificant vibrational energy as suggested by linear reaction path at the entrance channel. Following collision recrosses the energetic barrier to result in the products. The reaction trajectory exhibits a diagonal oscillatory pattern. Physical interpretation suggests that both vibrational and translational modes are present in the final system, i.e. some translational energy is converted to vibrational energy.
|}

=== Limitations of Transition State Theory ===
*State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition state theory is established on the principles of classical mechanics, where the reactants must collide with sufficient energy to form the transition state in order for the reaction to occur. However, this does not sufficiently describe the dynamics of a reaction.

A fundamental assumption of transition state theory is that classical trajectories do not recross the transition dividing surface, such that, starting from the reactants, if the system reaches the transition state configuration, it will necessarily proceed to form products.<ref name="(1)">Garrett, B.; Truhlar, D. The Journal of Physical Chemistry 1979, 83, 1079-1112." </ref><ref name="(2)"> Levine, R. Molecular reaction dynamics; 1st ed.; Cambridge University Press: Cambridge, 2009."</ref> If such a configuration can be identified then it must be crossed by any reactive trajectory and transition state theory assumes that it is only crossed once.

This is incorrect, however, and the breakdown of this fundamental approximation is illustrated in '''Figures 9a''', '''9b''', '''10a''' and '''10b'''. This effect is significant under high temperature conditions due to the large kinetic energy associated with the system, and is corroborated by literature review.<ref name="(1)" />

{{fontcolor1|gray| Good. This is the main limitation that you do observe in your simulations. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 15:28, 31 May 2017 (BST)}}

Another essential assumption of transition state theory is the concept of quasi-equilibrium. It states for a system at complete equilibrium, the activated complexes are at equilibrium. The concentration of the activated complexes is independent of changes in concentration of the reactants and products, and therefore, the concentration can be quantitatively determined based on principles of equilibrium theory. This could be a plausible approach, except an accurate position of the transition state cannot be ascertained. The accuracy of the calculated transition state is limited by the sophistication and complexity of the computational method, and to obtain a highly accurate picture is computationally inefficient and costly.

{{fontcolor1|gray| Not really applicable here, as this is a statistical treatment of TST. In this lab, you're just looking at a simple triatomic collision in issolation -- not an an ensemble of particles. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 15:28, 31 May 2017 (BST)}}

In such a case, a quantum mechanical descriptor might be more appropriate. There is a non-zero probability for H to tunnel through the potential barrier and form the products. Past research has suggested that for small atoms like hydrogen, such effects are significant.<ref name="(3)">Lauhon, L.; Ho, W. Physical Review Letters 2000, 85, 4566-4569.</ref>

Literature review has indicated that recrossing would result in the overestimation of thermal rate constants, whereas the neglect of quantum effects on the reaction coordinate motion would lead to underestimations of reaction rates.<ref name="(4)">Laidler, K.; King, M. The Journal of Physical Chemistry 1983, 87, 2657-2664.</ref> From this investigation, it seems more probable that reaction rate values are overestimated due to the large effects of recrossing observed, but this cannot be definitively concluded.

{{fontcolor1|gray| Good thinking. But bear in mind that the QM limitation isn't applicable here since the simulations are treating the atoms classically. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 15:28, 31 May 2017 (BST)}}

==EXERCISE 2: F–H–H system ==
=== PES inspection ===
*Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

The reaction between F and H2 is exothermic, since the minimum point energy of H2 is less exothermic than the minimum energy of the HF bond.

The reaction between F and H2 is illustrated in the following equation: F + H2 <math>\rightarrow</math> HF + H. The reaction between H + HF is simply the reverse of the reaction as displayed above.

To determine which reaction is exothermic, the potential energy of the reactants and the products for both scenarios have to be considered. {{fontcolor1|gray| you just did, though [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 15:33, 31 May 2017 (BST)}} Figure 11 presents the potential energy surface of the reaction between H2 and F.

[[File:Syl815 Potential Energy Surface of H 2.PNG|centre|upright=1.5|thumb|Figure 11. Graph of Potential Energy Surface of H2]]

Figure 11 has been adjusted to juxtapose the stability of H2 and HF molecules. The potential energy of the H2 molecule at equilibrium is as indicated in the figure. It is evident that this potential energy is a local minima, whereas the formation of the HF molecule is the global minima since the potential energy value is comparably more exothermic.

The reaction between F and H2 is therefore exothermic, while the inverse reaction (HF + H) is endothermic, and this is verified by the reaction trajectory of the H2 + F reaction in Figure 11.

To evaluate the validity of this conclusion, literature values of the H2 and HF bond dissociation energies were compared, to determine the standard enthalpy change of both reactions. The values found were 436.002 kJ/mol and 568.6 kJ/mol respectively. The enthalpy change of the reaction between H2 and F was calculated to be -132.6 kJ/mol, and conversely, +132.6 kJ/mol for the reaction between H and HF. <ref name="(1)">Cottrell, T. The strengths of chemical bonds; 1st ed.; Academic Press: New York, 1961." </ref> This is in agreement with the interpretation of the molecular dynamics calculation.

The exothermic reaction between H2 and F implies that the HF bond is stronger than that of H2.

=== Locating transition state ===
*Locate the approximate position of the transition state.

As elucidated above in Exercise 1, the transition state for the system can be determined by finding the distances r(HH) and r(HF) by fixing the values of pHH and pHF = 0. At the saddle point, the atoms would roll neither towards the reactants nor the products.

Hammond’s postulate predicts that in an exothermic reaction, the transition state is closer in energy to the reactants than the products. The highly exothermic reaction between F and H2 indicates that the energy level of the transition state is extremely close in energy to H2 than F.

As such, an initial molecular dynamics test was performed with a test value of r(HH) = 0.74, which is close to the equilibrium bond length of H2, and r(HF) = 2.00. Figure 12 shows the trajectory obtained.

[[File:Syl 815Initial Guess of Transition State of H2 and F.PNG|centre|upright=1.5|thumb|Figure 12. Reaction trajectory of initial guess.]]

An inspection of the trajectory reveals that this initial guess is slightly before the approach of the transition state, since H2 is formed, and F drifts away from H2. This implies that the rHF is larger than that of the actual rHF (transition) value.

After further modifications of the rHF values, an approximation of rHF (transition) and rHH (transition) was evaluated, which are 1.81 Å and 0.745 Å respectively. Figure 13 illustrates the reaction trajectory calculated, where it is observed that the reaction trajectory is merely a point.

[[File:Syl815 Transition State Energy Value Contour Plot.PNG|centre|upright=1.5|thumb|Figure 13. Surface plot of reaction trajectory at the transition state.]]

==== Activation Energy Calculation ====
*Report the activation energy for both reactions.

To determine the potential energy of the transition state, the potential energy against time graph was obtained at rHF (transition) = 1.79 Angstroms and rHH (transition) = 0.745. This was achieved by running a molecular dynamics calculation at 10 000 steps. The potential energy was determined to be -103.752 kcal/mol.

==== H2 + F <math>\rightarrow</math> HF ====

An initial mep calculation was performed at rHF (transition) = 1.79 Å and rHH (transition) = 0.745 Å with 25 000 steps, and the contour plot was obtained to observe the reaction trajectory.

[[File:Syl815 Contour Plot Convergence of HF.PNG|centre|upright=1.5|thumb|Figure 14. Reaction trajectory of H2 and F]]

Figure 14 confirms the formation of the HF bond, and the vertical line at rHF (transition) = 0.9199 Å and rHH (transition) = 2.427 Å indicates that it is probable the energy minimised structure of HF has been obtained. To test this, a dynamics calculation with 10 000 steps at rHF (transition) = 0.9199 Å and rHH (transition) = 2.427 Å was performed. The results are displayed in Figure 15a and Figure 15b.

{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none; width:60%"
|-
| | [[File:Syl815 Potential Energy Convergence of HF.PNG|center|400px]]
| [[File:Syl815 Kinetic Energy Convergence of H2 and F.PNG|center|400px]]
|-
| style = "text-align:center;"| Figure 15a. Potential energy plot.
| style = "text-align:center;"| Figure 15b. Kinetic energy plot.
|}

From Figure 15a, it is observed that the potential energy of the system initially decreases, but approaches a horizontal limit of -134.025 kcal/mol. This convergence implies that the potential energy of HF at equilibrium is -134.025 kcal/mol. The Figure 15b also validates this inference, as the value increases and approaches a horizontal limit as well.

The activation energy of the reaction is therefore given in Eq. 3.

{| align = "center" style="width:90%;background-color: white; border-spacing: 0px;"
| width = 30%|
| width = 30%|<math> \Delta E_a = E_{TS} - E_{HF} </math>
| style="text-align:right" width = 30%|(3)
|}

This value is calculated to be +30.273 kcal/mol.

==== HF + H <math>\rightarrow</math> H2 + F ====
Following a similar procedure as given above, an initial mep calculation was performed at rHF (transition) = 1.82 Å and rHH (transition) = 0.740 with 35 000 steps, and the contour plot was obtained to observe the reaction trajectory. At the end terminal, the values of rHF (transition) and rHH (transition) are 2.029 Å and 0.7416 Å respectively.

[[File:Syl815 Transition State of HF and H.PNG|centre|upright=1.5|thumb|Figure 16. Reaction trajectory of HF + H.]]

To determine whether the equilibrium structure of H2 was attained, a further dynamics calculation with 10 000 steps was performed at rHF (transition) = 2.029 Å and rHH (transition) = 0.7416 Å. The results are displayed in Figure 17a and Figure 17b.

{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none; width:60%"
|-
| | [[File:Syl815 Potential Energy Graph of HF and H.PNG|center|400px]]
| [[File:Syl815 Kinetic Energy Convergence of H and HF.PNG|center|400px]]
|-
| style = "text-align:center;"| Figure 17a. Graph of the potential energy of the system against t.
| style = "text-align:center;"| Figure 17b. Graph of the kinetic energy of the system against t.
|}

From the potential energy against time graph, it is observed that the potential energy of the system initially decreases, but approaches a horizontal limit of -104.020 kcal/mol. This convergence implies that the potential energy of HF at equilibrium is -104.020 kcal/mol. The kinetic energy against time graph also validates this inference, as the value increases and approaches a horizontal limit as well.

{| align = "center" style="width:90%;background-color: white; border-spacing: 0px;"
| width = 30%|
| width = 30%|<math> \Delta E_a = E_{TS} - E_{HH} </math>
| style="text-align:right" width = 30%|(4)
|}

The activation energy calculated is therefore +0.268 kcal/mol.

{{fontcolor1|gray| Very accurate results. GJ. [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 15:33, 31 May 2017 (BST)}}

=== Reaction Dynamics ===

To understand the dynamics of the reaction between F + H2, it may be useful to temporarily neglect any effects associated with the motion of the atoms, and scrutinise how the potential energy surface affects the reaction path taken.

As such, a preliminary dynamics calculation was performed with F + H2 with the initial conditions of rHF (transition) = 1.80 Angstroms, rHH (transition) = 0.745 Angstroms, pHF (transition) = pHH (transition) = 0. The trajectory computed is illustrated in Figure. The corresponding internuclear distance against time graph is provided in Figure.

[[FIGURES + TABLE]] {{fontcolor1|gray| <-- woops [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 15:33, 31 May 2017 (BST)}}

In this reaction, the release of energy following the initial bond formation of HF at t ≈ 2.6 is converted to translational and hot vibrational energy. The reaction does not proceed via the minimum energy path. This subsequent trajectory of the atoms is thus difficult to explicate based on qualitative analyses of molecular interactions.

===== Mechanism of release of reaction energy =====
*In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

As the atoms roll down to the products’ valley from the transition state, potential energy is converted to translational and vibrational energies.

The reaction energy can be evaluated experimentally through IR spectroscopy. H2 has a high degree of symmetry and as such, is IR inactive, whereas HF has a bond dipole associated with it, and is thus IR active. The intensity of the IR spectrum correlates to the vibrational energy that is released from the reaction.

Since H2 is IR inactive, Raman spectroscopy is a possible method for comparison as well.

=== Polanyi's empirical rules ===
*Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

Polanyi’s rules state that the vibrational energy is more efficient in promoting a late-barrier reaction than translational energy, and the reverse is true for an early barrier reaction. These empirically determined principles serve as a guide in understanding the mode/bond selectivity and the energy disposal in chemical processes.

The applicability of Polanyi’s rules are evaluated in the reaction between H and HF.

{| align = "center" style="width:90%;background-color: white; border-spacing: 0px;"
| width = 30%|
| width = 30%|<math>H + HF \rightarrow H_2 + F </math>
| style="text-align:right" width = 30%|
|}

The reaction between H and HF is highly endothermic, and as such, the reaction is an early barrier reaction. Polanyi’s rules therefore predict a highly efficient reaction between HF and H if a larger the vibrational energy is supplied to the system.

A test trajectory, A was calculated using rHF = 0.9 Å, rHH = 2.3 Å, pHF = 3 and pHH = -4. This is a representation of a HF + H system, where the HF bond is stretched and thus HF possesses vibrational energy. This is depicted in Figure 18a. The corresponding internuclear distance against time graph is provided as well in Figure 18a.

{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none; width:60%"
|-
| [[File:Syl815 System A Trajectory.PNG|center|400px]]
| [[File:Syl815 System A Internuclear Distance vs Time.PNG|center|400px]]
|-
| style = "text-align:center;"| Figure 18a. The trajectory of system A
| style = "text-align:center;"| Figure 18b. Internuclear Distance vs. t graph.
|}

An analysis based on the transition state theory would postulate that the formation of H2 molecule, since H and HF collide with sufficient kinetic energy to overcome the energetic barrier. Nonetheless, this is not observed.

A second system, B was configured as a contrast to system A, whereby the values of rHF and rHH is the same as those in A, while configuring the values of pHF and pHH to 7 and -4 respectively as illustrated in Figure 19a.

{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none; width:60%"
|-
| [[File:Syl815 System B Trajectory.PNG|center|400px]]
| [[File:Syl815 System B Internuclear Distance vs Time.PNG|center|400px]]
|-
| style = "text-align:center;"| Figure 19a. The trajectory of system B
| style = "text-align:center;"| Figure 19. Internuclear Distance vs. t graph.
|}

Here, the graph of function rHH indicates that the formation of H2 + F was successful.

A salient feature of both systems A and B is the diagonal relationship between rHH and rHF. The vibrations of HF should result in horizontal oscillations of the trajectory along the minimum energy path. However, following the recrossing of HF + H right arrow H2 + F as shown in system A, the trajectory of HF and H exhibit a diagonal relationship. Figures elucidate the variations in rHF and rHH more clearly, where both functions exhibit sinusoidal behaviour. This is expected for the function rHF, but a linear plot is expected for the function rHH.

This is attributable to the vibrational mode of HF, which results in the contraction and elongation of rHF. Since the quantity rHH is defined as the separation between the two H atoms, this would result in a periodic increase and decrease in the rHH separation. Therefore, both functions demonstrate a similar oscillatory behaviour.

A juxtaposition of Figures underscore the mode selectivity of reaction paths. The calculation performed in this investigation is in good agreement with Polanyi’s rules, where vibrational energy proves to be more efficacious in promoting the endothermic reaction of H and HF.

The converse is valid for the H2 + F exothermic reaction as well. Systems C and D were configured with the following values as shown in below, and dynamics was performed with a step of 500.

{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none; width:60%"
|-
| [[File:Syl815 System C Trajectory.PNG|center|400px]]
| [[File:Syl815 System C Internuclear Distance vs Time.PNG|center|400px]]
|-
| style = "text-align:center;"| Figure 20a. The trajectory of system C
| style = "text-align:center;"| Figure 20b. Internuclear Distance vs. t graph.
|}

{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none; width:60%"
|-
| [[File:Syl815 System D Trajectory.PNG|center|400px]]
| [[File:Syl815 System D Internuclear Distance vs Time.PNG|center|400px]]
|-
| style = "text-align:center;"| Figure 20a. The trajectory of system D
| style = "text-align:center;"| Figure 20b. Internuclear Distance vs. t graph.
|}

The reaction trajectories depicted in Figures substantiate Polanyi’s rules. Figure illustrates the outcome where a large initial vibrational energy is supplied to H2. The trajectory simply recrosses and H2 is regenerated. On the contrary, demonstrates a successful reaction in the formation of HF due to the large translational energy supplied to the system. This therefore establishes that exothermic reactions with an early barrier selectively favour large translational modes between the reactants.

{{fontcolor1|gray| Good discussion [[User:Je714|Je714]] ([[User talk:Je714|talk]]) 15:33, 31 May 2017 (BST)}}

== References ==
<references>
</references>