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		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805899</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805899"/>
		<updated>2020-05-15T22:00:16Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: /* Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as it is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum. The momentum graphs illustrate that dynamic calculation involves the vibration motions of molecules.&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_dy_d.png|300px|center|thumb|intermolecular distance on dynamics calculation]] ||[[File:01499685_dy_m.png|300px|center|thumb|momentum on dynamics calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier then form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. It across the col to form product without bouncing back. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they pass the transition state and then back to reform reactants. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| reactive || It is able to across the transition state, then encounter a potential energy wall and bounce back. After that, it across the transition state for the second time to form the product. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
From the example above, it illustrates that sometimes even the system has enough energy to across the col, it is still possible to reform reactants. The transition state theory assumption is not valid for this situation. This reveals the transition state theory overestimate the experiment values (which will be interpreted below)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants. Besides, CTST calculation only use standard expressions. The vibrational expression is a quantum expression, but it assumes the vibrations are harmonic. However, this is not true at high temperature.Moreover, the transmission may lower than unity due to an alternative reaction path. Additionally, tunnelling is probable for a thin energy barrier. Also, tunnelling is most important for the transfer of particles of small mass, but this effect is negligible for heavier species.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; /&amp;gt; &lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
F+H2 reaction is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond to reach the transition state. Overall, it is an exothermic reaction. Nevertheless, H+HF reaction is a backward reaction. Therefore, this reaction has to be an endothermic reaction.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the reactant. This type of transition state is defined as early transition state. In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74.5 pm. Then set the momentum zero and test the different value of the distance between F and H, which gives the approximate position of the transition is around 181.8 pm.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
Activation energy is the energy difference between transition state and reactants. By &#039;&#039; rolling down &#039;&#039; the molecule from the highest energy level with zero momentum, the energy level of reactant is able to be found. The total energy of the transition state is -433.981 kJ/mol. When the distance of AB increases to 600 pm, the total  energy is -435.057 kJ/mol. The total energy keeps unchanged if the distance of AB keeps increase. Therefore, the activation energy for F+H2 reaction is 1.076 kJ/mol. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
The types of released energy are electronic, vibrational, rotational, and translational energy. However, the excited electronic levels are too high for standard temperature. The partition function for translational motion in three degrees of freedom involves only the mass of the molecule. The rotational partition function for a linear molecule involves the moment of inertia I. In a non-linear molecule which contains N number of atoms, the three degrees of freedom relates to rotational energy, and 3N-6 relates to vibrational energy. This can be examined by IR spectrum. &amp;lt;ref name=&amp;quot;energy type&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
For an exothermic reaction, the transition state resembles the reactant. This early transition state has more efficient translational energy because the motion corresponds to the separation of products. Besides, the late transition state appears in endothermic reaction, which makes vibrational energy more efficient. &amp;lt;ref name=&amp;quot;energy modes&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===references:===&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; &amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; &amp;gt;K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy type&amp;quot; &amp;gt;D. M. Hirst Potential Energy Surfaces: Molecular Structure and Reaction Dynamics Taylor and Francis, 1985.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy modes&amp;quot; &amp;gt;I. N. Levine, Physical Chemistry, McGraw-Hill.&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805895</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805895"/>
		<updated>2020-05-15T21:57:42Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as it is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum. The momentum graphs illustrate that dynamic calculation involves the vibration motions of molecules.&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_dy_d.png|300px|center|thumb|intermolecular distance on dynamics calculation]] ||[[File:01499685_dy_m.png|300px|center|thumb|momentum on dynamics calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier then form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. It across the col to form product without bouncing back. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they pass the transition state and then back to reform reactants. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| reactive || It is able to across the transition state, then encounter a potential energy wall and bounce back. After that, it across the transition state for the second time to form the product. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
From the example above, it illustrates that sometimes even the system has enough energy to across the col, it is still possible to reform reactants. The transition state theory assumption is not valid for this situation. This reveals the transition state theory overestimate the experiment values (which will be interpreted below)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants. Besides, CTST calculation only use standard expressions. The vibrational expression is a quantum expression, but it assumes the vibrations are harmonic. However, this is not true at high temperature.Moreover, the transmission may lower than unity due to an alternative reaction path. Additionally, tunnelling is probable for a thin energy barrier. Also, tunnelling is most important for the transfer of particles of small mass, but this effect is negligible for heavier species.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; /&amp;gt; &lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
F+H2 reaction is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond to reach the transition state. Overall, it is an exothermic reaction. Nevertheless, H+HF reaction is a backward reaction. Therefore, this reaction has to be an endothermic reaction.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the reactant. This type of transition state is defined as early transition state. In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74.5 pm. Then set the momentum zero and test the different value of the distance between F and H, which gives the approximate position of the transition is around 181.8 pm.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
Activation energy is the energy difference between transition state and reactants. By &#039;&#039; rolling down &#039;&#039; the molecule from the highest energy level with zero momentum, the energy level of reactant is able to be found. The total energy of the transition state is -433.981 kJ/mol. When the distance of AB increases to 600 pm, the total  energy is -435.057 kJ/mol. The total energy keeps unchanged if the distance of AB keeps increase. Therefore, the activation energy for F+H2 reaction is 1.076 kJ/mol. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
The types of released energy are electronic, vibrational, rotational, and translational energy. However, the excited electronic levels are too high for standard temperature. The partition function for translational motion in three degrees of freedom involves only the mass of the molecule. The rotational partition function for a linear molecule involves the moment of inertia I. In a non-linear molecule which contains N number of atoms, the three degrees of freedom relates to rotational energy, and 3N-6 relates to vibrational energy. &amp;lt;ref name=&amp;quot;energy type&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
For an exothermic reaction, the transition state resembles the reactant. This early transition state has more efficient translational energy because the motion corresponds to the separation of products. Besides, the late transition state appears in endothermic reaction, which makes vibrational energy more efficient. &amp;lt;ref name=&amp;quot;energy modes&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===references:===&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; &amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; &amp;gt;K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy type&amp;quot; &amp;gt;D. M. Hirst Potential Energy Surfaces: Molecular Structure and Reaction Dynamics Taylor and Francis, 1985.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy modes&amp;quot; &amp;gt;I. N. Levine, Physical Chemistry, McGraw-Hill.&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805873</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805873"/>
		<updated>2020-05-15T21:50:13Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as it is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum. The momentum graphs illustrate that dynamic calculation involves the vibration motions of molecules.&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_dy_d.png|300px|center|thumb|intermolecular distance on dynamics calculation]] ||[[File:01499685_dy_m.png|300px|center|thumb|momentum on dynamics calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier then form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. It across the col to form product without bouncing back. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they pass the transition state and then back to reform reactants. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| reactive || It is able to across the transition state, then encounter a potential energy wall and bounce back. After that, it across the transition state for the second time to form the product. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
From the example above, it illustrates that sometimes even the system has enough energy to across the col, it is still possible to reform reactants. The transition state theory assumption is not valid for this situation. This reveals the transition state theory overestimate the experiment values (which will be interpreted below)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants. Besides, CTST calculation only use standard expressions. The vibrational expression is a quantum expression, but it assumes the vibrations are harmonic. However, this is not true at high temperature.Moreover, the transmission may lower than unity due to an alternative reaction path. Additionally, tunnelling is probable for a thin energy barrier. Also, tunnelling is most important for the transfer of particles of small mass, but this effect is negligible for heavier species.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; /&amp;gt; &lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
F+H2 reaction is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond to reach the transition state. Overall, it is an exothermic reaction. Nevertheless, H+HF reaction is a backward reaction. Therefore, this reaction has to be endothermic reaction.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the reactant. This type of transition state is defined as early transition state. In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74.5 pm. Then set the momentum zero, by testing the different value of the distance between F and H, the approximate position of the transition is around 181.8 pm.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
Activation energy is the energy difference between transition state and reactants. By &#039;&#039; rolling down &#039;&#039; the molecule from the highest energy with zero momentum, the energy level of reactant is able to be found. The total energy of the transition state is -433.981 kJ/mol. When the distance of AB increases to 600 pm, the total  energy is -435.057 kJ/mol. The total energy keeps unchanged if the distance of AB keeps increase. Therefore, the activation energy for F+H2 reaction is 1.076 kJ/mol. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
The types of released energy are electronic, vibrational, rotational, and translational energy. However, the excited electronic levels are too high for standard temperature. The partition function for translational motion in three degrees of freedom involves only the mass of the molecule. The rotational partition function for a linear molecule involves the moment of inertia I. In a non-linear molecule which contains N number of atoms, the three degrees of freedom relates to rotational energy, and 3N-6 relates to vibrational energy. &amp;lt;ref name=&amp;quot;energy type&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
For an exothermic reaction, the transition state resembles the reactant. This early transition state has more efficient translational energy because the motion corresponds to the separation of products. Besides, the late transition state appears in endothermic reaction, which makes vibrational energy more efficient. &amp;lt;ref name=&amp;quot;energy modes&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===references:===&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; &amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; &amp;gt;K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy type&amp;quot; &amp;gt;D. M. Hirst Potential Energy Surfaces: Molecular Structure and Reaction Dynamics Taylor and Francis, 1985.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy modes&amp;quot; &amp;gt;I. N. Levine, Physical Chemistry, McGraw-Hill.&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805869</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805869"/>
		<updated>2020-05-15T21:48:55Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as it is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum. The momentum graphs illustrate that dynamic calculation involves the vibration motions of molecules.&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_dy_d.png|300px|center|thumb|intermolecular distance on dynamics calculation]] ||[[File:01499685_dy_m.png|300px|center|thumb|momentum on dynamics calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier then form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. It across the col to form product without bouncing back. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they pass the transition state and then back to reform reactants. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| reactive || It is able to across the transition state, then encounter a potential energy wall and bounce back. After that, it across the transition state for the second time to form the product. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
From the example above, it illustrates that sometimes even the system has enough energy to across the col, it is still possible to reform reactants. The transition state theory assumption is not valid for this situation. This reveals the transition state theory overestimate the experiment values (which will be interpreted below)&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants. Besides, CTST calculation only use standard expressions. The vibrational expression is a quantum expression, but it assumes the vibrations are harmonic. However, this is not true at high temperature.Moreover, the transmission may lower than unity due to an alternative reaction path. Additionally, tunnelling is probable for a thin energy barrier. Also, tunnelling is most important for the transfer of particles of small mass, but this effect is negligible for heavier species.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; /&amp;gt; &lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
F+H2 reaction is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond to reach the transition state. Overall, it is an exothermic reaction. Nevertheless, H+HF reaction is a backward reaction. Therefore, this reaction has to be endothermic reaction.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the reactant. This type of transition state is defined as early transition state. In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74.5 pm. Then set the momentum zero, by testing the different value of the distance between F and H, the approximate position of the transition is around 181.8 pm.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
Activation energy is the energy difference between transition state and reactants. By &#039;&#039; rolling down &#039;&#039; the molecule from the highest energy with zero momentum, the energy level of reactant is able to be found. The total energy of the transition state is -433.981 kJ/mol. When the distance of AB increases to 600 pm, the total  energy is -435.057 kJ/mol. The total energy keeps unchanged if the distance of AB keeps increase. Therefore, the activation energy for F+H2 reaction is 1.076 kJ/mol. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
The types of released energy are electronic, vibrational, rotational, and translational energy. However, the excited electronic levels are too high for standard temperature. The partition function for translational motion in three degrees of freedom involves only the mass of the molecule. The rotational partition function for a linear molecule involves the moment of inertia I. In a non-linear molecule which contains N number of atoms, the three degrees of freedom relates to rotational energy, and 3N-6 relates to vibrational energy. &amp;lt;ref name=&amp;quot;energy type&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
For an exothermic reaction, the transition state resembles the reactant. This early transition state has more efficient translational energy because the motion corresponds to the separation of products. Besides, the late transition state appears in endothermic reaction, which makes vibrational energy more efficient. &amp;lt;ref name=&amp;quot;energy modes&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===references:===&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; &amp;gt;K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987.&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy type&amp;quot;&amp;gt;D. M. Hirst Potential Energy Surfaces: Molecular Structure and Reaction Dynamics Taylor and Francis, 1985.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy modes&amp;quot;&amp;gt;I. N. Levine, Physical Chemistry, McGraw-Hill.&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805858</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805858"/>
		<updated>2020-05-15T21:44:45Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum. The momentum graphs illustrate that dynamic calculation involves the vibration motions of molecules.&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_dy_d.png|300px|center|thumb|intermolecular distance on dynamics calculation]] ||[[File:01499685_dy_m.png|300px|center|thumb|momentum on dynamics calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier then form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. It across the col to form product without bouncing back. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they pass the transition state and then back to reform reactants. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| reactive || It is able to across the transition state, then encounter a potential energy wall and bounce back. After that, it across the transition state for the second time to form the product. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
From the example above, it illustrates that sometimes even the system has enough energy to across the col, it is still possible to reform reactants. The transition state theory assumption is not valid for this situation.&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants. Besides, CTST calculation only use standard expressions. The vibrational expression is a quantum expression, but it assumes the vibrations are harmonic. However, this is not true at high temperature.Moreover, the transmission may lower than unity due to an alternative reaction path. Additionally, tunnelling is probable for a thin energy barrier. Also, tunnelling is most important for the transfer of particles of small mass, but this effect is negligible for heavier species.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; /&amp;gt; &lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
F+H2 reaction is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond to reach the transition state. Overall, it is an exothermic reaction. Nevertheless, H+HF reaction is a backward reaction. Therefore, this reaction has to be endothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the reactant. This type of transition state is defined as early transition state. In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74.5 pm. Then set the momentum zero, by testing the different value of the distance between F and H, the approximate position of the transition is around 181.8 pm.&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
Activation energy is the energy difference between transition state and reactants. By &#039;&#039; rolling down &#039;&#039; the molecule from the highest energy with zero momentum, the energy level of reactant is able to be found. The total energy of the transition state is -433.981 kJ/mol. When the distance of AB increases to 600 pm, the total  energy is -435.057 kJ/mol. The total energy keeps unchanged if the distance of AB keeps increase. Therefore, the activation energy for F+H2 reaction is 1.076 kJ/mol. &lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
The types of released energy are electronic, vibrational, rotational, and translational energy. However, the excited electronic levels are too high for standard temperature. The partition function for translational motion in three degrees of freedom involves only the mass of the molecule. The rotational partition function for a linear molecule involves the moment of inertia I. In a non-linear molecule which contains N number of atoms, the three degrees of freedom relates to rotational energy, and 3N-6 relates to vibrational energy. &amp;lt;ref name=&amp;quot;energy type&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
For an exothermic reaction, the transition state resembles the reactant. This early transition state has more efficient translational energy because the motion corresponds to the separation of products. Besides, the late transition state appears in endothermic reaction, which makes vibrational energy more efficient. &amp;lt;ref name=&amp;quot;energy modes&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===references:===&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; &amp;gt;K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987.&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy type&amp;quot;&amp;gt;D. M. Hirst Potential Energy Surfaces: Molecular Structure and Reaction Dynamics Taylor and Francis, 1985.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy modes&amp;quot;&amp;gt;I. N. Levine, Physical Chemistry, McGraw-Hill.&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805845</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805845"/>
		<updated>2020-05-15T21:41:43Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum. The momentum graphs illustrate that dynamic calculation involves the vibrations of molecules.&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_dy_d.png|300px|center|thumb|intermolecular distance on dynamics calculation]] ||[[File:01499685_dy_m.png|300px|center|thumb|momentum on dynamics calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier then form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. It across the col to form product without bouncing back. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they pass the transition state and then back to reform reactants. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| reactive || It is able to across the transition state, then encounter a potential energy wall and bounce back. After that, it across the transition state for the second time to form the product. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
From the example above, it illustrates that sometimes even the system has enough energy to across the col, it is still possible to reform reactants. The transition state theory assumption is not valid for this situation.&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants. Besides, CTST calculation only use standard expressions. The vibrational expression is a quantum expression, but it assumes the vibrations are harmonic. However, this is not true at high temperature.Moreover, the transmission may lower than unity due to an alternative reaction path. Additionally, tunnelling is probable for a thin energy barrier. Also, tunnelling is most important for the transfer of particles of small mass, but this effect is negligible for heavier species.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; /&amp;gt; &lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
F+H2 reaction is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond to reach the transition state. Overall, it is an exothermic reaction. Nevertheless, H+HF reaction is a backward reaction. Therefore, this reaction has to be endothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the reactant. This type of transition state is defined as early transition state. In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74.5 pm. Then set the momentum zero, by testing the different value of the distance between F and H, the approximate position of the transition is around 181.8 pm.&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
Activation energy is the energy difference between transition state and reactants. By &#039;&#039; rolling down &#039;&#039; the molecule from the highest energy with zero momentum, the energy level of reactant is able to be found. The total energy of the transition state is -433.981 kJ/mol. When the distance of AB increases to 600 pm, the total  energy is -435.057 kJ/mol. The total energy keeps unchanged if the distance of AB keeps increase. Therefore, the activation energy for F+H2 reaction is 1.076 kJ/mol. &lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
The types of released energy are electronic, vibrational, rotational, and translational energy. However, the excited electronic levels are too high for standard temperature. The partition function for translational motion in three degrees of freedom involves only the mass of the molecule. The rotational partition function for a linear molecule involves the moment of inertia I. In a non-linear molecule which contains N number of atoms, the three degrees of freedom relates to rotational energy, and 3N-6 relates to vibrational energy. &amp;lt;ref name=&amp;quot;energy type&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
For an exothermic reaction, the transition state resembles the reactant. This early transition state has more efficient translational energy because the motion corresponds to the separation of products. Besides, the late transition state appears in endothermic reaction, which makes vibrational energy more efficient. &amp;lt;ref name=&amp;quot;energy modes&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===references:===&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; &amp;gt;K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987.&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy type&amp;quot;&amp;gt;D. M. Hirst Potential Energy Surfaces: Molecular Structure and Reaction Dynamics Taylor and Francis, 1985.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy modes&amp;quot;&amp;gt;I. N. Levine, Physical Chemistry, McGraw-Hill.&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805842</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805842"/>
		<updated>2020-05-15T21:41:17Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum. The momentum graphs illustrate that dynamic calculation involves the vibrations of molecules.&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_dy_d.png|300px|center|thumb|intermolecular distance on dynamics calculation]] ||[[File:01499685_dy_m.png|300px|center|thumb|momentum on dynamics calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier then form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. It across the col to form product without bouncing back. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they pass the transition state and then back to reform reactants. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| reactive || It is able to across the transition state, then encounter a potential energy wall and bounce back. After that, it across the transition state for the second time to form the product. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
From the example above, it illustrates that sometimes even the system has enough energy to across the col, it is still possible to reform reactants. The transition state theory assumption is not valid for this situation.&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants. Besides, CTST calculation only use standard expressions. The vibrational expression is a quantum expression, but it assumes the vibrations are harmonic. However, this is not true at high temperature.Moreover, the transmission may lower than unity due to an alternative reaction path. Additionally, tunnelling is probable for a thin energy barrier. Also, tunnelling is most important for the transfer of particles of small mass, but this effect is negligible for heavier species.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; /&amp;gt; &lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
F+H2 reaction is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond to reach the transition state. Overall, it is an exothermic reaction. Nevertheless, H+HF reaction is a backward reaction. Therefore, this reaction has to be endothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the reactant. This type of transition state is defined as early transition state. In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74.5 pm. Then set the momentum zero, by testing the different value of the distance between F and H, the approximate position of the transition is around 181.8 pm.&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
Activation energy is the energy difference between transition state and reactants. By &#039;&#039; rolling down &#039;&#039; the molecule from the highest energy with zero momentum, the energy level of reactant is able to be found. The total energy of the transition state is -433.981 kJ/mol. When the distance of AB increases to 600 pm, the total  energy is -435.057 kJ/mol. The total energy keeps unchanged if the distance of AB keeps increase. Therefore, the activation energy for F+H2 reaction is 1.076 kJ/mol. &lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
The types of released energy are electronic, vibrational, rotational, and translational energy. However, the excited electronic levels are too high for standard temperature. The partition function for translational motion in three degrees of freedom involves only the mass of the molecule. The rotational partition function for a linear molecule involves the moment of inertia I. In a non-linear molecule which contains N number of atoms, the three degrees of freedom relates to rotational energy, and 3N-6 relates to vibrational energy. &amp;lt;ref name=&amp;quot;energy type&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
For an exothermic reaction, the transition state resembles the reactant. This early transition state has more efficient translational energy because the motion corresponds to the separation of products. Besides, the late transition state appears in endothermic reaction, which makes vibrational energy more efficient. &amp;lt;ref name=&amp;quot;energy modes&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===references:===&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; &amp;gt;K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy type&amp;quot;&amp;gt;D. M. Hirst Potential Energy Surfaces: Molecular Structure and Reaction Dynamics Taylor and Francis, 1985.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy modes&amp;quot;&amp;gt;I. N. Levine, Physical Chemistry, McGraw-Hill.&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805841</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805841"/>
		<updated>2020-05-15T21:40:53Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum. The momentum graphs illustrate that dynamic calculation involves the vibrations of molecules.&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_dy_d.png|300px|center|thumb|intermolecular distance on dynamics calculation]] ||[[File:01499685_dy_m.png|300px|center|thumb|momentum on dynamics calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier then form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. It across the col to form product without bouncing back. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they pass the transition state and then back to reform reactants. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| reactive || It is able to across the transition state, then encounter a potential energy wall and bounce back. After that, it across the transition state for the second time to form the product. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
From the example above, it illustrates that sometimes even the system has enough energy to across the col, it is still possible to reform reactants. The transition state theory assumption is not valid for this situation.&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants. Besides, CTST calculation only use standard expressions. The vibrational expression is a quantum expression, but it assumes the vibrations are harmonic. However, this is not true at high temperature.Moreover, the transmission may lower than unity due to an alternative reaction path. Additionally, tunnelling is probable for a thin energy barrier. Also, tunnelling is most important for the transfer of particles of small mass, but this effect is negligible for heavier species.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; /&amp;gt; &lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
F+H2 reaction is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond to reach the transition state. Overall, it is an exothermic reaction. Nevertheless, H+HF reaction is a backward reaction. Therefore, this reaction has to be endothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the reactant. This type of transition state is defined as early transition state. In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74.5 pm. Then set the momentum zero, by testing the different value of the distance between F and H, the approximate position of the transition is around 181.8 pm.&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
Activation energy is the energy difference between transition state and reactants. By &#039;&#039; rolling down &#039;&#039; the molecule from the highest energy with zero momentum, the energy level of reactant is able to be found. The total energy of the transition state is -433.981 kJ/mol. When the distance of AB increases to 600 pm, the total  energy is -435.057 kJ/mol. The total energy keeps unchanged if the distance of AB keeps increase. Therefore, the activation energy for F+H2 reaction is 1.076 kJ/mol. &lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
The types of released energy are electronic, vibrational, rotational, and translational energy. However, the excited electronic levels are too high for standard temperature. The partition function for translational motion in three degrees of freedom involves only the mass of the molecule. The rotational partition function for a linear molecule involves the moment of inertia I. In a non-linear molecule which contains N number of atoms, the three degrees of freedom relates to rotational energy, and 3N-6 relates to vibrational energy. &amp;lt;ref name=&amp;quot;energy type&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
For an exothermic reaction, the transition state resembles the reactant. This early transition state has more efficient translational energy because the motion corresponds to the separation of products. Besides, the late transition state appears in endothermic reaction, which makes vibrational energy more efficient. &amp;lt;ref name=&amp;quot;energy modes&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===references:===&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; &amp;gt;K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy type&amp;quot;&amp;gt;D. M. Hirst Potential Energy Surfaces: Molecular Structure and Reaction Dynamics Taylor and Francis, 1985.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy modes&amp;quot;&amp;gt;I. N. Levine, Physical Chemistry, McGraw-Hill.&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805818</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805818"/>
		<updated>2020-05-15T21:37:12Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum (vibrations).&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_dy_d.png|300px|center|thumb|intermolecular distance on dynamics calculation]] ||[[File:01499685_dy_m.png|300px|center|thumb|momentum on dynamics calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier then form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. It across the col to form product without bouncing back. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they pass the transition state and then back to reform reactants. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| reactive || It is able to across the transition state, then encounter a potential energy wall and bounce back. After that, it across the transition state for the second time to form the product. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
From the example above, it illustrates that sometimes even the system has enough energy to across the col, it is still possible to reform reactants. The transition state theory assumption is not valid for this situation.&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants. Besides, CTST calculation only use standard expressions. The vibrational expression is a quantum expression, but it assumes the vibrations are harmonic. However, this is not true at high temperature.Moreover, the transmission may lower than unity due to an alternative reaction path. Additionally, tunnelling is probable for a thin energy barrier. Also, tunnelling is most important for the transfer of particles of small mass, but this effect is negligible for heavier species.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; /&amp;gt; &lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
F+H2 reaction is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond to reach the transition state. Overall, it is an exothermic reaction. Nevertheless, H+HF reaction is a backward reaction. Therefore, this reaction has to be endothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the reactant. This type of transition state is defined as early transition state. In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74.5 pm. Then set the momentum zero, by testing the different value of the distance between F and H, the approximate position of the transition is around 181.8 pm.&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
Activation energy is the energy difference between transition state and reactants. By &#039;&#039; rolling down &#039;&#039; the molecule from the highest energy with zero momentum, the energy level of reactant is able to be found. The total energy of the transition state is -433.981 kJ/mol. When the distance of AB increases to 600 pm, the total  energy is -435.057 kJ/mol. The total energy keeps unchanged if the distance of AB keeps increase. Therefore, the activation energy for F+H2 reaction is 1.076 kJ/mol. &lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
The types of released energy are electronic, vibrational, rotational, and translational energy. However, the excited electronic levels are too high for standard temperature. The partition function for translational motion in three degrees of freedom involves only the mass of the molecule. The rotational partition function for a linear molecule involves the moment of inertia I. In a non-linear molecule which contains N number of atoms, the three degrees of freedom relates to rotational energy, and 3N-6 relates to vibrational energy. &amp;lt;ref name=&amp;quot;energy type&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
For an exothermic reaction, the transition state resembles the reactant. This early transition state has more efficient translational energy because the motion corresponds to the separation of products. Besides, the late transition state appears in endothermic reaction, which makes vibrational energy more efficient. &amp;lt;ref name=&amp;quot;energy modes&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===references:===&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; &amp;gt;K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987.&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy type&amp;quot;&amp;gt;D. M. Hirst Potential Energy Surfaces: Molecular Structure and Reaction Dynamics Taylor and Francis, 1985.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy modes&amp;quot;&amp;gt;I. N. Levine, Physical Chemistry, McGraw-Hill.&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805815</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805815"/>
		<updated>2020-05-15T21:36:43Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum (vibrations).&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_dy_d.png|300px|center|thumb|intermolecular distance on dynamics calculation]] ||[[File:01499685_dy_m.png|300px|center|thumb|momentum on dynamics calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier then form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. It across the col to form product without bouncing back. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they pass the transition state and then back to reform reactants. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| reactive || It is able to across the transition state, then encounter a potential energy wall and bounce back. After that, it across the transition state for the second time to form the product. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
From the example above, it illustrates that sometimes even the system has enough energy to across the col, it is still possible to reform reactants. The transition state theory assumption is not valid for this situation.&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants. Besides, CTST calculation only use standard expressions. The vibrational expression is a quantum expression, but it assumes the vibrations are harmonic. However, this is not true at high temperature.Moreover, the transmission may lower than unity due to an alternative reaction path. Additionally, tunnelling is probable for a thin energy barrier. Also, tunnelling is most important for the transfer of particles of small mass, but this effect is negligible for heavier species.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; /&amp;gt; &lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
F+H2 reaction is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond to reach the transition state. Overall, it is an exothermic reaction. Nevertheless, H+HF reaction is a backward reaction. Therefore, this reaction has to be endothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the reactant. This type of transition state is defined as early transition state. In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74.5 pm. Then set the momentum zero, by testing the different value of the distance between F and H, the approximate position of the transition is around 181.8 pm.&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
Activation energy is the energy difference between transition state and reactants. By &#039;&#039; rolling down &#039;&#039; the molecule from the highest energy with zero momentum, the energy level of reactant is able to be found. The total energy of the transition state is -433.981 kJ/mol. When the distance of AB increases to 600 pm, the total  energy is -435.057 kJ/mol. The total energy keeps unchanged if the distance of AB keeps increase. Therefore, the activation energy for F+H2 reaction is 1.076 kJ/mol. &lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
The types of released energy are electronic, vibrational, rotational, and translational energy. However, the excited electronic levels are too high for standard temperature. The partition function for translational motion in three degrees of freedom involves only the mass of the molecule. The rotational partition function for a linear molecule involves the moment of inertia I. In a non-linear molecule which contains N number of atoms, the three degrees of freedom relates to rotational energy, and 3N-6 relates to vibrational energy. &amp;lt;ref name=&amp;quot;energy type&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
For an exothermic reaction, the transition state resembles the reactant. This early transition state has more efficient translational energy because the motion corresponds to the separation of products. Besides, the late transition state appears in endothermic reaction, which makes vibrational energy more efficient. &amp;lt;ref name=&amp;quot;energy modes&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===references:===&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; &amp;gt;K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987.&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy type&amp;quot;&amp;gt;D. M. Hirst Potential Energy Surfaces: Molecular Structure and Reaction Dynamics Taylor and Francis, 1985.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy modes&amp;quot;&amp;gt;I. N. Levine, Physical Chemistry, McGraw-Hill.&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805810</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805810"/>
		<updated>2020-05-15T21:35:11Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum (vibrations).&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_dy_d.png|300px|center|thumb|intermolecular distance on dynamics calculation]] ||[[File:01499685_dy_m.png|300px|center|thumb|momentum on dynamics calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier then form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. It across the col to form product without bouncing back. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they pass the transition state and then back to reform reactants. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| reactive || It is able to across the transition state, then encounter a potential energy wall and bounce back. After that, it across the transition state for the second time to form the product. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
From the example above, it illustrates that sometimes even the system has enough energy to across the col, it is still possible to reform reactants. The transition state theory assumption is not valid for this situation.&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants. Besides, CTST calculation only use standard expressions. The vibrational expression is a quantum expression, but it assumes the vibrations are harmonic. However, this is not true at high temperature.Moreover, the transmission may lower than unity due to an alternative reaction path. Additionally, tunnelling is probable for a thin energy barrier. Also, tunnelling is most important for the transfer of particles of small mass, but this effect is negligible for heavier species.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; /&amp;gt; &lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
F+H2 reaction is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond to reach the transition state. Overall, it is an exothermic reaction. Nevertheless, H+HF reaction is a backward reaction. Therefore, this reaction has to be endothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the reactant. This type of transition state is defined as early transition state. In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74.5 pm. Then set the momentum zero, by testing the different value of the distance between F and H, the approximate position of the transition is around 181.8 pm.&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
Activation energy is the energy difference between transition state and reactants. By &#039;&#039; rolling down &#039;&#039; the molecule from the highest energy with zero momentum, the energy level of reactant is able to be found. The total energy of the transition state is -433.981 kJ/mol. When the distance of AB increases to 600 pm, the total  energy is -435.057 kJ/mol. The total energy keeps unchanged if the distance of AB keeps increase. Therefore, the activation energy for F+H2 reaction is 1.076 kJ/mol. &lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
The types of released energy are electronic, vibrational, rotational, and translational energy. However, the excited electronic levels are too high for standard temperature. The partition function for translational motion in three degrees of freedom involves only the mass of the molecule. The rotational partition function for a linear molecule involves the moment of inertia I. In a non-linear molecule which contains N number of atoms, the three degrees of freedom relates to rotational energy, and 3N-6 relates to vibrational energy. &amp;lt;ref name=&amp;quot;energy type&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
For an exothermic reaction, the transition state resembles the reactant. This early transition state has more efficient translational energy because the motion corresponds to the separation of products. Besides, the late transition state appears in endothermic reaction, which makes vibrational energy more efficient. &amp;lt;ref name=&amp;quot;energy modes&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
references:&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; &amp;gt;K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987.&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy type&amp;quot;&amp;gt;D. M. Hirst Potential Energy Surfaces: Molecular Structure and Reaction Dynamics Taylor and Francis, 1985.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy modes&amp;quot;&amp;gt;I. N. Levine, Physical Chemistry, McGraw-Hill.&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805807</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805807"/>
		<updated>2020-05-15T21:34:44Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum (vibrations).&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_dy_d.png|300px|center|thumb|intermolecular distance on dynamics calculation]] ||[[File:01499685_dy_m.png|300px|center|thumb|momentum on dynamics calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier then form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. It across the col to form product without bouncing back. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they pass the transition state and then back to reform reactants. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| reactive || It is able to across the transition state, then encounter a potential energy wall and bounce back. After that, it across the transition state for the second time to form the product. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
From the example above, it illustrates that sometimes even the system has enough energy to across the col, it is still possible to reform reactants. The transition state theory assumption is not valid for this situation.&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants. Besides, CTST calculation only use standard expressions. The vibrational expression is a quantum expression, but it assumes the vibrations are harmonic. However, this is not true at high temperature.Moreover, the transmission may lower than unity due to an alternative reaction path. Additionally, tunnelling is probable for a thin energy barrier. Also, tunnelling is most important for the transfer of particles of small mass, but this effect is negligible for heavier species.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; /&amp;gt; &lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
F+H2 reaction is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond to reach the transition state. Overall, it is an exothermic reaction. Nevertheless, H+HF reaction is a backward reaction. Therefore, this reaction has to be endothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the reactant. This type of transition state is defined as early transition state. In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74.5 pm. Then set the momentum zero, by testing the different value of the distance between F and H, the approximate position of the transition is around 181.8 pm.&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
Activation energy is the energy difference between transition state and reactants. By &#039;&#039; rolling down &#039;&#039; the molecule from the highest energy with zero momentum, the energy level of reactant is able to be found. The total energy of the transition state is -433.981 kJ/mol. When the distance of AB increases to 600 pm, the total  energy is -435.057 kJ/mol. The total energy keeps unchanged if the distance of AB keeps increase. Therefore, the activation energy for F+H2 reaction is 1.076 kJ/mol. &lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
The types of released energy are electronic, vibrational, rotational, and translational energy. However, the excited electronic levels are too high for standard temperature. The partition function for translational motion in three degrees of freedom involves only the mass of the molecule. The rotational partition function for a linear molecule involves the moment of inertia I. In a non-linear molecule which contains N number of atoms, the three degrees of freedom relates to rotational energy, and 3N-6 relates to vibrational energy. &amp;lt;ref name=&amp;quot;energy type&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
For an exothermic reaction, the transition state resembles the reactant. This early transition state has more efficient translational energy because the motion corresponds to the separation of products. Besides, the late transition state appears in endothermic reaction, which makes vibrational energy more efficient. &amp;lt;ref name=&amp;quot;energy modes&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
references:&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; &amp;gt;K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987.&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy type&amp;quot;&amp;gt;D. M. Hirst Potential Energy Surfaces: Molecular Structure and Reaction Dynamics Taylor and Francis, 1985.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy modes&amp;quot;&amp;gt;I. N. Levine, Physical Chemistry, McGraw-Hill.&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805803</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805803"/>
		<updated>2020-05-15T21:33:51Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum (vibrations).&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_dy_d.png|300px|center|thumb|intermolecular distance on dynamics calculation]] ||[[File:01499685_dy_m.png|300px|center|thumb|momentum on dynamics calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier then form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. It across the col to form product without bouncing back. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they pass the transition state and then back to reform reactants. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| reactive || It is able to across the transition state, then encounter a potential energy wall and bounce back. After that, it across the transition state for the second time to form the product. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
From the example above, it illustrates that sometimes even the system has enough energy to across the col, it is still possible to reform reactants. The transition state theory assumption is not valid for this situation.&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants. Besides, CTST calculation only use standard expressions. The vibrational expression is a quantum expression, but it assumes the vibrations are harmonic. However, this is not true at high temperature.Moreover, the transmission may lower than unity due to an alternative reaction path. Additionally, tunnelling is probable for a thin energy barrier. Also, tunnelling is most important for the transfer of particles of small mass, but this effect is negligible for heavier species.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; /&amp;gt; &lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
F+H2 reaction is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond to reach the transition state. Overall, it is an exothermic reaction. Nevertheless, H+HF reaction is a backward reaction. Therefore, this reaction has to be endothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the reactant. This type of transition state is defined as early transition state. In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74.5 pm. Then set the momentum zero, by testing the different value of the distance between F and H, the approximate position of the transition is around 181.8 pm.&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
Activation energy is the energy difference between transition state and reactants. By &#039;&#039; rolling down &#039;&#039; the molecule from the highest energy with zero momentum, the energy level of reactant is able to be found. The total energy of the transition state is -433.981 kJ/mol. When the distance of AB increases to 600 pm, the total  energy is -435.057 kJ/mol. The total energy keeps unchanged if the distance of AB keeps increase. Therefore, the activation energy for F+H2 reaction is 1.076 kJ/mol. &lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
The types of released energy are electronic, vibrational, rotational, and translational energy. However, the excited electronic levels are too high for standard temperature. The partition function for translational motion in three degrees of freedom involves only the mass of the molecule. The rotational partition function for a linear molecule involves the moment of inertia I. In a non-linear molecule which contains N number of atoms, the three degrees of freedom relates to rotational energy, and 3N-6 relates to vibrational energy. &amp;lt;ref name=&amp;quot;energy type&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
For an exothermic reaction, the transition state resembles the reactant. This early transition state has more efficient translational energy because the motion corresponds to the separation of products. Besides, the late transition state appears in endothermic reaction, which makes vibrational energy more efficient. &amp;lt;ref name=&amp;quot;energy modes&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; &amp;gt;K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy type&amp;quot;&amp;gt;D. M. Hirst Potential Energy Surfaces: Molecular Structure and Reaction Dynamics Taylor and Francis, 1985.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy modes&amp;quot;&amp;gt;I. N. Levine, Physical Chemistry, McGraw-Hill.&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805799</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805799"/>
		<updated>2020-05-15T21:32:38Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum (vibrations).&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_dy_d.png|300px|center|thumb|intermolecular distance on dynamics calculation]] ||[[File:01499685_dy_m.png|300px|center|thumb|momentum on dynamics calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier then form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. It across the col to form product without bouncing back. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they pass the transition state and then back to reform reactants. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| reactive || It is able to across the transition state, then encounter a potential energy wall and bounce back. After that, it across the transition state for the second time to form the product. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
From the example above, it illustrates that sometimes even the system has enough energy to across the col, it is still possible to reform reactants. The transition state theory assumption is not valid for this situation.&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants. Besides, CTST calculation only use standard expressions. The vibrational expression is a quantum expression, but it assumes the vibrations are harmonic. However, this is not true at high temperature.Moreover, the transmission may lower than unity due to an alternative reaction path. Additionally, tunnelling is probable for a thin energy barrier. Also, tunnelling is most important for the transfer of particles of small mass, but this effect is negligible for heavier species.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; /&amp;gt; &lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
F+H2 reaction is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond to reach the transition state. Overall, it is an exothermic reaction. Nevertheless, H+HF reaction is a backward reaction. Therefore, this reaction has to be endothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the reactant. This type of transition state is defined as early transition state. In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74.5 pm. Then set the momentum zero, by testing the different value of the distance between F and H, the approximate position of the transition is around 181.8 pm.&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
Activation energy is the energy difference between transition state and reactants. By &#039;&#039; rolling down &#039;&#039; the molecule from the highest energy with zero momentum, the energy level of reactant is able to be found. The total energy of the transition state is -433.981 kJ/mol. When the distance of AB increases to 600 pm, the total  energy is -435.057 kJ/mol. The total energy keeps unchanged if the distance of AB keeps increase. Therefore, the activation energy for F+H2 reaction is 1.076 kJ/mol. &lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
The types of released energy are electronic, vibrational, rotational, and translational energy. However, the excited electronic levels are too high for standard temperature. The partition function for translational motion in three degrees of freedom involves only the mass of the molecule. The rotational partition function for a linear molecule involves the moment of inertia I. In a non-linear molecule which contains N number of atoms, the three degrees of freedom relates to rotational energy, and 3N-6 relates to vibrational energy. &amp;lt;ref name=&amp;quot;energy type&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
For an exothermic reaction, the transition state resembles the reactant. This early transition state has more efficient translational energy because the motion corresponds to the separation of products. Besides, the late transition state appears in endothermic reaction, which makes vibrational energy more efficient. &amp;lt;ref name=&amp;quot;energy modes&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; &amp;gt;K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy type&amp;quot;&amp;gt;D. M. Hirst Potential Energy Surfaces: Molecular Structure and Reaction Dynamics Taylor and Francis, 1985.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy modes&amp;quot;&amp;gt;I. N. Levine, Physical Chemistry, McGraw-Hill.&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805748</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805748"/>
		<updated>2020-05-15T21:17:21Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum (vibrations).&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_dy_d.png|300px|center|thumb|intermolecular distance on dynamics calculation]] ||[[File:01499685_dy_m.png|300px|center|thumb|momentum on dynamics calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier then form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. It across the col to form product without bouncing back. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they pass the transition state and then back to reform reactants. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| reactive || It is able to across the transition state, then encounter a potential energy wall and bounce back. After that, it across the transition state for the second time to form the product. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
From the example above, it illustrates that sometimes even the system has enough energy to across the col, it is still possible to reform reactants. The transition state theory assumption is not valid for this situation.&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants. Besides, CTST calculation only use standard expressions. The vibrational expression is a quantum expression, but it assumes the vibrations are harmonic. However, this is not true at high temperature.Moreover, the transmission may lower than unity due to an alternative reaction path. Additionally, tunnelling is probable for a thin energy barrier. Also, tunnelling is most important for the transfer of particles of small mass, but this effect is negligible for heavier species.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; /&amp;gt; &lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
F+H2 reaction is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond to reach the transition state. Overall, it is an exothermic reaction. Nevertheless, H+HF reaction is a backward reaction. Therefore, this reaction has to be endothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the reactant. This type of transition state is defined as early transition state. In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74.5 pm. Then set the momentum zero, by testing the different value of the distance between F and H, the approximate position of the transition is around 181.8 pm.&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
Activation energy is the energy difference between transition state and reactants. By &#039;&#039; rolling down &#039;&#039; the molecule from the highest energy with zero momentum, the energy level of reactant is able to be found. The total energy of the transition state is -433.981 kJ/mol. When the distance of AB increases to 600 pm, the total  energy is -435.057 kJ/mol. The total energy keeps unchanged if the distance of AB keeps increase. Therefore, the activation energy for F+H2 reaction is 1.076 kJ/mol. &lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
In this case, early barrier and translational energy are more efficient. &lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
For an exothermic reaction, the transition state resembles the reactant. This early transition state has more efficient translational energy because the motion corresponds to the separation of products. Besides, the late transition state appears in endothermic reaction, which makes vibrational energy more efficient. &amp;lt;ref name=&amp;quot;energy modes&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; &amp;gt;K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy modes&amp;quot;&amp;gt;I. N. Levine, Physical Chemistry, McGraw-Hill.&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805746</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805746"/>
		<updated>2020-05-15T21:16:55Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum (vibrations).&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_dy_d.png|300px|center|thumb|intermolecular distance on dynamics calculation]] ||[[File:01499685_dy_m.png|300px|center|thumb|momentum on dynamics calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier then form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. It across the col to form product without bouncing back. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they pass the transition state and then back to reform reactants. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| reactive || It is able to across the transition state, then encounter a potential energy wall and bounce back. After that, it across the transition state for the second time to form the product. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
From the example above, it illustrates that sometimes even the system has enough energy to across the col, it is still possible to reform reactants. The transition state theory assumption is not valid for this situation.&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants. Besides, CTST calculation only use standard expressions. The vibrational expression is a quantum expression, but it assumes the vibrations are harmonic. However, this is not true at high temperature.Moreover, the transmission may lower than unity due to an alternative reaction path. Additionally, tunnelling is probable for a thin energy barrier. Also, tunnelling is most important for the transfer of particles of small mass, but this effect is negligible for heavier species.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; /&amp;gt; &lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
F+H2 reaction is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond to reach the transition state. Overall, it is an exothermic reaction. Nevertheless, H+HF reaction is a backward reaction. Therefore, this reaction has to be endothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the reactant. This type of transition state is defined as early transition state. In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74.5 pm. Then set the momentum zero, by testing the different value of the distance between F and H, the approximate position of the transition is around 181.8 pm.&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
Activation energy is the energy difference between transition state and reactants. By &#039;&#039; rolling down &#039;&#039; the molecule from the highest energy with zero momentum, the energy level of reactant is able to be found. The total energy of the transition state is -433.981 kJ/mol. When the distance of AB increases to 600 pm, the total  energy is -435.057 kJ/mol. The total energy keeps unchanged if the distance of AB keeps increase. Therefore, the activation energy for F+H2 reaction is 1.076 kJ/mol. &lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
In this case, early barrier and translational energy are more efficient. &lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
For an exothermic reaction, the transition state resembles the reactant. This early transition state has more efficient translational energy because the motion corresponds to the separation of products. Besides, the late transition state appears in endothermic reaction, which makes vibrational energy more efficient. &amp;lt;ref name=&amp;quot;energy modes&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; &amp;gt;K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy modes&amp;quot;&amp;gt;I. N. Levine, Physical Chemistry, McGraw-Hill.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805742</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805742"/>
		<updated>2020-05-15T21:14:51Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum (vibrations).&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_dy_d.png|300px|center|thumb|intermolecular distance on dynamics calculation]] ||[[File:01499685_dy_m.png|300px|center|thumb|momentum on dynamics calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier then form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. It across the col to form product without bouncing back. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they pass the transition state and then back to reform reactants. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| reactive || It is able to across the transition state, then encounter a potential energy wall and bounce back. After that, it across the transition state for the second time to form the product. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
From the example above, it illustrates that sometimes even the system has enough energy to across the col, it is still possible to reform reactants. The transition state theory assumption is not valid for this situation.&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants. Besides, CTST calculation only use standard expressions. The vibrational expression is a quantum expression, but it assumes the vibrations are harmonic. However, this is not true at high temperature.Moreover, the transmission may lower than unity due to an alternative reaction path. Additionally, tunnelling is probable for a thin energy barrier. Also, tunnelling is most important for the transfer of particles of small mass, but this effect is negligible for heavier species.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; /&amp;gt; &lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
F+H2 reaction is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond to reach the transition state. Overall, it is an exothermic reaction. Nevertheless, H+HF reaction is a backward reaction. Therefore, this reaction has to be endothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the reactant. This type of transition state is defined as early transition state. In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74.5 pm. Then set the momentum zero, by testing the different value of the distance between F and H, the approximate position of the transition is around 181.8 pm.&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
Activation energy is the energy difference between transition state and reactants. By &#039;&#039; rolling down &#039;&#039; the molecule from the highest energy with zero momentum, the energy level of reactant is able to be found. The total energy of the transition state is -433.981 kJ/mol. When the distance of AB increases to 600 pm, the total  energy is -435.057 kJ/mol. The total energy keeps unchanged if the distance of AB keeps increase. Therefore, the activation energy for F+H2 reaction is 1.076 kJ/mol. &lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
In this case, early barrier and translational energy are more efficient. &lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
For an exothermic reaction, the transition state resembles the reactant. This early transition state has more efficient translational energy because the motion corresponds to the separation of products. Besides, the late transition state appears in endothermic reaction, which makes vibrational energy more efficient. &amp;lt;ref name=&amp;quot;energy modes&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; &amp;gt;K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987.&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy modes&amp;quot;&amp;gt;I. N. Levine, Physical Chemistry, McGraw-Hill.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805739</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805739"/>
		<updated>2020-05-15T21:14:09Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum (vibrations).&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_dy_d.png|300px|center|thumb|intermolecular distance on dynamics calculation]] ||[[File:01499685_dy_m.png|300px|center|thumb|momentum on dynamics calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier then form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. It across the col to form product without bouncing back. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they pass the transition state and then back to reform reactants. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| reactive || It is able to across the transition state, then encounter a potential energy wall and bounce back. After that, it across the transition state for the second time to form the product. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
From the example above, it illustrates that sometimes even the system has enough energy to across the col, it is still possible to reform reactants. The transition state theory assumption is not valid for this situation.&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants. Besides, CTST calculation only use standard expressions. The vibrational expression is a quantum expression, but it assumes the vibrations are harmonic. However, this is not true at high temperature.Moreover, the transmission may lower than unity due to an alternative reaction path. Additionally, tunnelling is probable for a thin energy barrier. Also, tunnelling is most important for the transfer of particles of small mass, but this effect is negligible for heavier species.&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; /&amp;gt; &lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
F+H2 reaction is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond to reach the transition state. Overall, it is an exothermic reaction. Nevertheless, H+HF reaction is a backward reaction. Therefore, this reaction has to be endothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the reactant. This type of transition state is defined as early transition state. In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74.5 pm. Then set the momentum zero, by testing the different value of the distance between F and H, the approximate position of the transition is around 181.8 pm.&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
Activation energy is the energy difference between transition state and reactants. By &#039;&#039; rolling down &#039;&#039; the molecule from the highest energy with zero momentum, the energy level of reactant is able to be found. The total energy of the transition state is -433.981 kJ/mol. When the distance of AB increases to 600 pm, the total  energy is -435.057 kJ/mol. The total energy keeps unchanged if the distance of AB keeps increase. Therefore, the activation energy for F+H2 reaction is 1.076 kJ/mol. &lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
In this case, early barrier and translational energy are more efficient. &lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
For an exothermic reaction, the transition state resembles the reactant. This early transition state has more efficient translational energy because the motion corresponds to the separation of products. Besides, the late transition state appears in endothermic reaction, which makes vibrational energy more efficient. &amp;lt;ref name=&amp;quot;energy modes&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; &amp;gt;K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;energy modes&amp;quot;&amp;gt;I. N. Levine, Physical Chemistry, McGraw-Hill.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805697</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805697"/>
		<updated>2020-05-15T21:04:49Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum (vibrations).&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_dy_d.png|300px|center|thumb|intermolecular distance on dynamics calculation]] ||[[File:01499685_dy_m.png|300px|center|thumb|momentum on dynamics calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier then form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. It across the col to form product without bouncing back. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they pass the transition state and then back to reform reactants. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| reactive || It is able to across the transition state, then encounter a potential energy wall and bounce back. After that, it across the transition state for the second time to form the product. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
From the example above, it illustrates that sometimes even the system has enough energy to across the col, it is still possible to reform reactants. The transition state theory assumption is not valid for this situation.&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants. Besides, CTST calculation only use standard expressions. The vibrational expression is a quantum expression, but it assumes the vibrations are harmonic. However, this is not true at high temperature.Moreover, the transmission may lower than unity due to an alternative reaction path. Additionally, tunnelling is probable for a thin energy barrier. Also, tunnelling is most important for the transfer of particles of small mass, but this effect is negligible for heavier species.&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; /&amp;gt; &lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
F+H2 reaction is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond to reach the transition state. Overall, it is an exothermic reaction. Nevertheless, H+HF reaction is a backward reaction. Therefore, this reaction has to be endothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the reactant. This type of transition state is defined as early transition state. In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74.5 pm. Then set the momentum zero, by testing the different value of the distance between F and H, the approximate position of the transition is around 181.8 pm.&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
Activation energy is the energy difference between transition state and reactants. By &#039;&#039; rolling down &#039;&#039; the molecule from the highest energy with zero momentum, the energy level of reactant is able to be found. The total energy of the transition state is -433.981 kJ/mol. When the distance of AB increases to 600 pm, the total  energy is -435.057 kJ/mol. The total energy keeps unchanged if the distance of AB keeps increase. Therefore, the activation energy for F+H2 reaction is 1.076 kJ/mol. &lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
In this case, early barrier and translational energy are more efficient. &lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
For an exothermic reaction, the transition state resembles &lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=problems of Transition State Theory&amp;gt;K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805694</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805694"/>
		<updated>2020-05-15T21:04:19Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum (vibrations).&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_dy_d.png|300px|center|thumb|intermolecular distance on dynamics calculation]] ||[[File:01499685_dy_m.png|300px|center|thumb|momentum on dynamics calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier then form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. It across the col to form product without bouncing back. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they pass the transition state and then back to reform reactants. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| reactive || It is able to across the transition state, then encounter a potential energy wall and bounce back. After that, it across the transition state for the second time to form the product. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
From the example above, it illustrates that sometimes even the system has enough energy to across the col, it is still possible to reform reactants. The transition state theory assumption is not valid for this situation.&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants. Besides, CTST calculation only use standard expressions. The vibrational expression is a quantum expression, but it assumes the vibrations are harmonic. However, this is not true at high temperature.Moreover, the transmission may lower than unity due to an alternative reaction path. Additionally, tunnelling is probable for a thin energy barrier. Also, tunnelling is most important for the transfer of particles of small mass, but this effect is negligible for heavier species.&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; /&amp;gt; &lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
F+H2 reaction is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond to reach the transition state. Overall, it is an exothermic reaction. Nevertheless, H+HF reaction is a backward reaction. Therefore, this reaction has to be endothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the reactant. This type of transition state is defined as early transition state. In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74.5 pm. Then set the momentum zero, by testing the different value of the distance between F and H, the approximate position of the transition is around 181.8 pm.&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
Activation energy is the energy difference between transition state and reactants. By &#039;&#039; rolling down &#039;&#039; the molecule from the highest energy with zero momentum, the energy level of reactant is able to be found. The total energy of the transition state is -433.981 kJ/mol. When the distance of AB increases to 600 pm, the total  energy is -435.057 kJ/mol. The total energy keeps unchanged if the distance of AB keeps increase. Therefore, the activation energy for F+H2 reaction is 1.076 kJ/mol. &lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
In this case, early barrier and translational energy are more efficient. &lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
For an exothermic reaction, the transition state resembles &lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot;&amp;gt;K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805690</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805690"/>
		<updated>2020-05-15T21:03:21Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum (vibrations).&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_dy_d.png|300px|center|thumb|intermolecular distance on dynamics calculation]] ||[[File:01499685_dy_m.png|300px|center|thumb|momentum on dynamics calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier then form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. It across the col to form product without bouncing back. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they pass the transition state and then back to reform reactants. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| reactive || It is able to across the transition state, then encounter a potential energy wall and bounce back. After that, it across the transition state for the second time to form the product. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
From the example above, it illustrates that sometimes even the system has enough energy to across the col, it is still possible to reform reactants. The transition state theory assumption is not valid for this situation.&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants. Besides, CTST calculation only use standard expressions. The vibrational expression is a quantum expression, but it assumes the vibrations are harmonic. However, this is not true at high temperature.Moreover, the transmission may lower than unity due to an alternative reaction path. Additionally, tunnelling is probable for a thin energy barrier. Also, tunnelling is most important for the transfer of particles of small mass, but this effect is negligible for heavier species.&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; /&amp;gt; &lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
F+H2 reaction is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond to reach the transition state. Overall, it is an exothermic reaction. Nevertheless, H+HF reaction is a backward reaction. Therefore, this reaction has to be endothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the reactant. This type of transition state is defined as early transition state. In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74.5 pm. Then set the momentum zero, by testing the different value of the distance between F and H, the approximate position of the transition is around 181.8 pm.&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
Activation energy is the energy difference between transition state and reactants. By &#039;&#039; rolling down &#039;&#039; the molecule from the highest energy with zero momentum, the energy level of reactant is able to be found. The total energy of the transition state is -433.981 kJ/mol. When the distance of AB increases to 600 pm, the total  energy is -435.057 kJ/mol. The total energy keeps unchanged if the distance of AB keeps increase. Therefore, the activation energy for F+H2 reaction is 1.076 kJ/mol. &lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
In this case, early barrier and translational energy are more efficient. &lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
For an exothermic reaction, the transition state resembles &lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&#039;&#039;problems of Transition State Theory&amp;quot;&amp;gt;K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805647</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805647"/>
		<updated>2020-05-15T20:49:10Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum (vibrations).&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_dy_d.png|300px|center|thumb|intermolecular distance on dynamics calculation]] ||[[File:01499685_dy_m.png|300px|center|thumb|momentum on dynamics calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier then form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. It across the col to form product without bouncing back. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they pass the transition state and then back to reform reactants. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| reactive || It is able to across the transition state, then encounter a potential energy wall and bounce back. After that, it across the transition state for the second time to form the product. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
From the example above, it illustrates that sometimes even the system has enough energy to across the col, it is still possible to reform reactants. The transition state theory assumption is not valid for this situation.&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants. Besides, CTST calculation only use standard expressions. The vibrational expression is a quantum expression, but it assumes the vibrations are harmonic. However, this is not true at high temperature.Moreover, the transmission may lower than unity due to an alternative reaction path. Additionally, tunnelling is probable for a thin energy barrier. Also, tunnelling is most important for the transfer of particles of small mass, but this effect is negligible for heavier species.&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; /&amp;gt; &lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
F+H2 reaction is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond to reach the transition state. Overall, it is an exothermic reaction. Nevertheless, H+HF reaction is a backward reaction. Therefore, this reaction has to be endothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the reactant. This type of transition state is defined as early transition state. In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74.5 pm. Then set the momentum zero, by testing the different value of the distance between F and H, the approximate position of the transition is around 181.8 pm.&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
Activation energy is the energy difference between transition state and reactants. By &#039;&#039; rolling down &#039;&#039; the molecule from the highest energy with zero momentum, the energy level of reactant is able to be found. The total energy of the transition state is -433.981 kJ/mol. When the distance of AB increases to 600 pm, the total  energy is -435.057 kJ/mol. The total energy keeps unchanged if the distance of AB keeps increase. Therefore, the activation energy for F+H2 reaction is 1.076 kJ/mol. &lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
In this case, early barrier and translational energy are more efficient.&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&#039;&#039;problems of Transition State Theory&amp;quot;&amp;gt;K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805643</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805643"/>
		<updated>2020-05-15T20:47:58Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum (vibrations).&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_dy_d.png|300px|center|thumb|intermolecular distance on dynamics calculation]] ||[[File:01499685_dy_m.png|300px|center|thumb|momentum on dynamics calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier then form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. It across the col to form product without bouncing back. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they pass the transition state and then back to reform reactants. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| reactive || It is able to across the transition state, then encounter a potential energy wall and bounce back. After that, it across the transition state for the second time to form the product. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
From the example above, it illustrates that sometimes even the system has enough energy to across the col, it is still possible to reform reactants. The transition state theory assumption is not valid for this situation.&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants. Besides, CTST calculation only use standard expressions. The vibrational expression is a quantum expression, but it assumes the vibrations are harmonic. However, this is not true at high temperature.&amp;lt;ref name=&amp;quot;problems of Transition State Theory&amp;quot; /&amp;gt; Moreover, the transmission may lower than unity due to an alternative reaction path. Additionally, tunnelling is probable for a thin energy barrier. Also, tunnelling is most important for the transfer of particles of small mass, but this effect is negligible for heavier species.&lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
F+H2 reaction is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond to reach the transition state. Overall, it is an exothermic reaction. Nevertheless, H+HF reaction is a backward reaction. Therefore, this reaction has to be endothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the reactant. This type of transition state is defined as early transition state. In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74.5 pm. Then set the momentum zero, by testing the different value of the distance between F and H, the approximate position of the transition is around 181.8 pm.&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
Activation energy is the energy difference between transition state and reactants. By &#039;&#039; rolling down &#039;&#039; the molecule from the highest energy with zero momentum, the energy level of reactant is able to be found. The total energy of the transition state is -433.981 kJ/mol. When the distance of AB increases to 600 pm, the total  energy is -435.057 kJ/mol. The total energy keeps unchanged if the distance of AB keeps increase. Therefore, the activation energy for F+H2 reaction is 1.076 kJ/mol. &lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
In this case, early barrier and translational energy are more efficient.&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&#039;&#039;problems of Transition State Theory&amp;quot;&amp;gt;K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805635</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805635"/>
		<updated>2020-05-15T20:46:28Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum (vibrations).&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_dy_d.png|300px|center|thumb|intermolecular distance on dynamics calculation]] ||[[File:01499685_dy_m.png|300px|center|thumb|momentum on dynamics calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier then form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. It across the col to form product without bouncing back. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they pass the transition state and then back to reform reactants. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| reactive || It is able to across the transition state, then encounter a potential energy wall and bounce back. After that, it across the transition state for the second time to form the product. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
From the example above, it illustrates that sometimes even the system has enough energy to across the col, it is still possible to reform reactants. The transition state theory assumption is not valid for this situation.&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants. Besides, CTST calculation only use standard expressions. The vibrational expression is a quantum expression, but it assumes the vibrations are harmonic. However, this is not true at high temperature.&amp;lt;ref name=&amp;quot;Transition State Theory_1&amp;quot; /&amp;gt; Moreover, the transmission may lower than unity due to an alternative reaction path. Additionally, tunnelling is probable for a thin energy barrier. Also, tunnelling is most important for the transfer of particles of small mass, but this effect is negligible for heavier species.&lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
F+H2 reaction is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond to reach the transition state. Overall, it is an exothermic reaction. Nevertheless, H+HF reaction is a backward reaction. Therefore, this reaction has to be endothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the reactant. This type of transition state is defined as early transition state. In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74.5 pm. Then set the momentum zero, by testing the different value of the distance between F and H, the approximate position of the transition is around 181.8 pm.&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
Activation energy is the energy difference between transition state and reactants. By &#039;&#039; rolling down &#039;&#039; the molecule from the highest energy with zero momentum, the energy level of reactant is able to be found. The total energy of the transition state is -433.981 kJ/mol. When the distance of AB increases to 600 pm, the total  energy is -435.057 kJ/mol. The total energy keeps unchanged if the distance of AB keeps increase. Therefore, the activation energy for F+H2 reaction is 1.076 kJ/mol. &lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
In this case, early barrier and translational energy are more efficient.&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805630</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805630"/>
		<updated>2020-05-15T20:45:10Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum (vibrations).&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_dy_d.png|300px|center|thumb|intermolecular distance on dynamics calculation]] ||[[File:01499685_dy_m.png|300px|center|thumb|momentum on dynamics calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier then form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. It across the col to form product without bouncing back. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they pass the transition state and then back to reform reactants. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| reactive || It is able to across the transition state, then encounter a potential energy wall and bounce back. After that, it across the transition state for the second time to form the product. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
From the example above, it illustrates that sometimes even the system has enough energy to across the col, it is still possible to reform reactants. The transition state theory assumption is not valid for this situation.&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants. Besides, CTST calculation only use standard expressions. The vibrational expression is a quantum expression, but it assumes the vibrations are harmonic. However, this is not true at high temperature.&amp;lt;ref name=&amp;quot;Transition State Theory_1&amp;quot; /&amp;gt; Moreover, the transmission may lower than unity due to an alternative reaction path. Additionally, tunnelling is probable for a thin energy barrier. Also, tunnelling is most important for the transfer of particles of small mass, but this effect is negligible for heavier species.&lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
F+H2 reaction is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond to reach the transition state. Overall, it is an exothermic reaction. Nevertheless, H+HF reaction is a backward reaction. Therefore, this reaction has to be endothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the reactant. This type of transition state is defined as early transition state. In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74.5 pm. Then set the momentum zero, by testing the different value of the distance between F and H, the approximate position of the transition is around 181.8 pm.&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
Activation energy is the energy difference between transition state and reactants. By &#039;&#039; rolling down &#039;&#039; the molecule from the highest energy with zero momentum, the energy level of reactant is able to be found. The total energy of the transition state is -433.981 kJ/mol. When the distance of AB increases to 600 pm, the total  energy is -435.057 kJ/mol. The total energy keeps unchanged if the distance of AB keeps increase. Therefore, the activation energy for F+H2 reaction is 1.076 kJ/mol. &lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
In this case, early barrier and translational energy are more efficient.&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805393</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805393"/>
		<updated>2020-05-15T19:27:53Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta. However, the dynamics calculation includes the change in momentum (vibrations).&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|300px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep_d.png|300px|center|thumb|intermolecular distance on mep calculation]] ||[[File:01499685_mep_m.png|300px|center|thumb|momentum on mep calculation]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier and form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they don&#039;t have enough energy to across the energy barrier. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| unreactive || They collide with each other and unable to form the transition state. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants.&lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
It is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond and reach the transition state. Overall, it is an exothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the structure of reactant. This type of transition state is defined as early transition state. In this case, early barrier and translational energy are more efficient.In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74 pm. Then set the momentum zero, and by testing the different value of the distance between F and H, the approximate position of the transition is around 182 pm.&lt;br /&gt;
&lt;br /&gt;
activation energy&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:01499685_mep_d.png&amp;diff=805316</id>
		<title>File:01499685 mep d.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:01499685_mep_d.png&amp;diff=805316"/>
		<updated>2020-05-15T19:00:49Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:01499685_mep_m.png&amp;diff=805313</id>
		<title>File:01499685 mep m.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:01499685_mep_m.png&amp;diff=805313"/>
		<updated>2020-05-15T19:00:18Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:01499685_dy_m.png&amp;diff=805312</id>
		<title>File:01499685 dy m.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:01499685_dy_m.png&amp;diff=805312"/>
		<updated>2020-05-15T18:59:44Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:01499685_dy_d.png&amp;diff=805307</id>
		<title>File:01499685 dy d.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:01499685_dy_d.png&amp;diff=805307"/>
		<updated>2020-05-15T18:58:58Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805302</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805302"/>
		<updated>2020-05-15T18:57:46Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy. While on mep calculation, there is no change in momenta.,&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
|-&lt;br /&gt;
|[[File:01499685_mep.png|400px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]] ||[[File:01499685_dy.png|400px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on dynamics calculation]]&lt;br /&gt;
|-&lt;br /&gt;
|&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier and form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they don&#039;t have enough energy to across the energy barrier. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| unreactive || They collide with each other and unable to form the transition state. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants.&lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
It is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond and reach the transition state. Overall, it is an exothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the structure of reactant. This type of transition state is defined as early transition state. In this case, early barrier and translational energy are more efficient.In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74 pm. Then set the momentum zero, and by testing the different value of the distance between F and H, the approximate position of the transition is around 182 pm.&lt;br /&gt;
&lt;br /&gt;
activation energy&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805245</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805245"/>
		<updated>2020-05-15T18:40:17Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+δ, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy, where the kinetic energy keeps zero all the time and there is no vibrational energy on mep calculation, and the total energy is not constant in mep. However, on the dynamics calculation, the atoms vibrate, and total energy conserves.&lt;br /&gt;
&lt;br /&gt;
[[File:01499685_mep.png|01499685_dy.png|400px|center|thumb|r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+5, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt; on mep calculation]]&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier and form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they don&#039;t have enough energy to across the energy barrier. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| unreactive || They collide with each other and unable to form the transition state. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants.&lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
It is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond and reach the transition state. Overall, it is an exothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the structure of reactant. This type of transition state is defined as early transition state. In this case, early barrier and translational energy are more efficient.In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74 pm. Then set the momentum zero, and by testing the different value of the distance between F and H, the approximate position of the transition is around 182 pm.&lt;br /&gt;
&lt;br /&gt;
activation energy&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:01499685_dy.png&amp;diff=805237</id>
		<title>File:01499685 dy.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:01499685_dy.png&amp;diff=805237"/>
		<updated>2020-05-15T18:36:22Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:01499685_mep.png&amp;diff=805235</id>
		<title>File:01499685 mep.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:01499685_mep.png&amp;diff=805235"/>
		<updated>2020-05-15T18:35:40Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805233</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805233"/>
		<updated>2020-05-15T18:34:44Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center|thumb|Internuclear Distances vs Time plot for transition state]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+δ, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy, where the kinetic energy keeps zero all the time and there is no vibrational energy on mep calculation, and the total energy is not constant in mep. However, on the dynamics calculation, the atoms vibrate, and total energy conserves.&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier and form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they don&#039;t have enough energy to across the energy barrier. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| unreactive || They collide with each other and unable to form the transition state. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants.&lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
It is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond and reach the transition state. Overall, it is an exothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the structure of reactant. This type of transition state is defined as early transition state. In this case, early barrier and translational energy are more efficient.In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74 pm. Then set the momentum zero, and by testing the different value of the distance between F and H, the approximate position of the transition is around 182 pm.&lt;br /&gt;
&lt;br /&gt;
activation energy&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805215</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805215"/>
		<updated>2020-05-15T18:29:37Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|400px|center]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+δ, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy, where the kinetic energy keeps zero all the time and there is no vibrational energy on mep calculation, and the total energy is not constant in mep. However, on the dynamics calculation, the atoms vibrate, and total energy conserves.&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier and form the product. || [[File:0149685_t1.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. || [[File:01499685_t3.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they don&#039;t have enough energy to across the energy barrier. ||[[File:01499685_t4.png|200px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| unreactive || They collide with each other and unable to form the transition state. ||[[File:01499685_t5.png|200px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants.&lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
It is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond and reach the transition state. Overall, it is an exothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the structure of reactant. This type of transition state is defined as early transition state. In this case, early barrier and translational energy are more efficient.In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74 pm. Then set the momentum zero, and by testing the different value of the distance between F and H, the approximate position of the transition is around 182 pm.&lt;br /&gt;
&lt;br /&gt;
activation energy&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805207</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805207"/>
		<updated>2020-05-15T18:28:30Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
[[File:01499685_dt.png|180px|center]]&lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+δ, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy, where the kinetic energy keeps zero all the time and there is no vibrational energy on mep calculation, and the total energy is not constant in mep. However, on the dynamics calculation, the atoms vibrate, and total energy conserves.&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier and form the product. || [[File:0149685_t1.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. || [[File:01499685_t3.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they don&#039;t have enough energy to across the energy barrier. ||[[File:01499685_t4.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| unreactive || They collide with each other and unable to form the transition state. ||[[File:01499685_t5.png|180px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants.&lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
It is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond and reach the transition state. Overall, it is an exothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the structure of reactant. This type of transition state is defined as early transition state. In this case, early barrier and translational energy are more efficient.In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74 pm. Then set the momentum zero, and by testing the different value of the distance between F and H, the approximate position of the transition is around 182 pm.&lt;br /&gt;
&lt;br /&gt;
activation energy&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:01499685_dt.png&amp;diff=805203</id>
		<title>File:01499685 dt.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:01499685_dt.png&amp;diff=805203"/>
		<updated>2020-05-15T18:25:44Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805201</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805201"/>
		<updated>2020-05-15T18:25:05Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. A saddle point is produced on the graph. The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Furthermore, local minimum of the potential energy surface usually represents products or intermediates. Mathematically, the transition state can be defined using first and second derivatives. The first derivative equals to zero, and the second derivative should be a negative value. Second derivatives can be used to distinguish the saddle point or the local minimum of the potential energy surface.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. In the H-H-H system, the transition state should be symmetrical, so AB=BC, and no force acts on AB or BC. The best estimate of the transition state position is 90.775 pm. The internuclear distance vs time plot demonstrates that at this specific bond length, line AB overlaps line BC and remains constant. &lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+δ, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy, where the kinetic energy keeps zero all the time and there is no vibrational energy on mep calculation, and the total energy is not constant in mep. However, on the dynamics calculation, the atoms vibrate, and total energy conserves.&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier and form the product. || [[File:0149685_t1.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. || [[File:01499685_t3.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they don&#039;t have enough energy to across the energy barrier. ||[[File:01499685_t4.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| unreactive || They collide with each other and unable to form the transition state. ||[[File:01499685_t5.png|180px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants.&lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
It is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond and reach the transition state. Overall, it is an exothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the structure of reactant. This type of transition state is defined as early transition state. In this case, early barrier and translational energy are more efficient.In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74 pm. Then set the momentum zero, and by testing the different value of the distance between F and H, the approximate position of the transition is around 182 pm.&lt;br /&gt;
&lt;br /&gt;
activation energy&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805068</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805068"/>
		<updated>2020-05-15T17:45:14Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Mathematically, the maximum can be defined using first and second derivative. The first derivative equals to zero, and the second derivative should be a negative value. Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. Local minimum of the potential &lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. So if there are two horizontal straight lines on the internuclear distance vs time plot, that means the atoms are not move at all. The best estimate of the transition state position is 90.775 pm. &lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+δ, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy, where the kinetic energy keeps zero all the time and there is no vibrational energy on mep calculation, and the total energy is not constant in mep. However, on the dynamics calculation, the atoms vibrate, and total energy conserves.&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier and form the product. || [[File:0149685_t1.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. || [[File:01499685_t3.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they don&#039;t have enough energy to across the energy barrier. ||[[File:01499685_t4.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| unreactive || They collide with each other and unable to form the transition state. ||[[File:01499685_t5.png|180px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions:&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants.&lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
It is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond and reach the transition state. Overall, it is an exothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the structure of reactant. This type of transition state is defined as early transition state. In this case, early barrier and translational energy are more efficient.In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74 pm. Then set the momentum zero, and by testing the different value of the distance between F and H, the approximate position of the transition is around 182 pm.&lt;br /&gt;
&lt;br /&gt;
activation energy&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
&amp;lt;references&amp;gt;&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805058</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805058"/>
		<updated>2020-05-15T17:43:47Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Mathematically, the maximum can be defined using first and second derivative. The first derivative equals to zero, and the second derivative should be a negative value. Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. Local minimum of the potential &lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. So if there are two horizontal straight lines on the internuclear distance vs time plot, that means the atoms are not move at all. The best estimate of the transition state position is 90.775 pm. &lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+δ, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy, where the kinetic energy keeps zero all the time and there is no vibrational energy on mep calculation, and the total energy is not constant in mep. However, on the dynamics calculation, the atoms vibrate, and total energy conserves.&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier and form the product. || [[File:0149685_t1.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. || [[File:01499685_t3.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they don&#039;t have enough energy to across the energy barrier. ||[[File:01499685_t4.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| unreactive || They collide with each other and unable to form the transition state. ||[[File:01499685_t5.png|180px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&amp;lt;references&amp;gt;:&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants.&lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
It is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond and reach the transition state. Overall, it is an exothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the structure of reactant. This type of transition state is defined as early transition state. In this case, early barrier and translational energy are more efficient.In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74 pm. Then set the momentum zero, and by testing the different value of the distance between F and H, the approximate position of the transition is around 182 pm.&lt;br /&gt;
&lt;br /&gt;
activation energy&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805057</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805057"/>
		<updated>2020-05-15T17:42:51Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Mathematically, the maximum can be defined using first and second derivative. The first derivative equals to zero, and the second derivative should be a negative value. Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. Local minimum of the potential &lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. So if there are two horizontal straight lines on the internuclear distance vs time plot, that means the atoms are not move at all. The best estimate of the transition state position is 90.775 pm. &lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+δ, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy, where the kinetic energy keeps zero all the time and there is no vibrational energy on mep calculation, and the total energy is not constant in mep. However, on the dynamics calculation, the atoms vibrate, and total energy conserves.&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier and form the product. || [[File:0149685_t1.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. || [[File:01499685_t3.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they don&#039;t have enough energy to across the energy barrier. ||[[File:01499685_t4.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| unreactive || They collide with each other and unable to form the transition state. ||[[File:01499685_t5.png|180px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&amp;lt;references&amp;gt;:&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants.&lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
It is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond and reach the transition state. Overall, it is an exothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the structure of reactant. This type of transition state is defined as early transition state. In this case, early barrier and translational energy are more efficient.In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74 pm. Then set the momentum zero, and by testing the different value of the distance between F and H, the approximate position of the transition is around 182 pm.&lt;br /&gt;
&lt;br /&gt;
activation energy&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805055</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805055"/>
		<updated>2020-05-15T17:41:54Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Mathematically, the maximum can be defined using first and second derivative. The first derivative equals to zero, and the second derivative should be a negative value. Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. Local minimum of the potential &lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. So if there are two horizontal straight lines on the internuclear distance vs time plot, that means the atoms are not move at all. The best estimate of the transition state position is 90.775 pm. &lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+δ, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy, where the kinetic energy keeps zero all the time and there is no vibrational energy on mep calculation, and the total energy is not constant in mep. However, on the dynamics calculation, the atoms vibrate, and total energy conserves.&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier and form the product. || [[File:0149685_t1.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. || [[File:01499685_t3.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they don&#039;t have enough energy to across the energy barrier. ||[[File:01499685_t4.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| unreactive || They collide with each other and unable to form the transition state. ||[[File:01499685_t5.png|180px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot; /&amp;gt;&lt;br /&gt;
&amp;lt;references&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants.&lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
It is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond and reach the transition state. Overall, it is an exothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the structure of reactant. This type of transition state is defined as early transition state. In this case, early barrier and translational energy are more efficient.In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74 pm. Then set the momentum zero, and by testing the different value of the distance between F and H, the approximate position of the transition is around 182 pm.&lt;br /&gt;
&lt;br /&gt;
activation energy&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;ref name=&amp;quot;Transition State Theory&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805053</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=805053"/>
		<updated>2020-05-15T17:40:51Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Mathematically, the maximum can be defined using first and second derivative. The first derivative equals to zero, and the second derivative should be a negative value. Potential energy surface showing the minimal energy path, and the dividing surface through the transition state is at right angles to the minimum-energy path, as is the alternative dividing surface. Local minimum of the potential &lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. So if there are two horizontal straight lines on the internuclear distance vs time plot, that means the atoms are not move at all. The best estimate of the transition state position is 90.775 pm. &lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+δ, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy, where the kinetic energy keeps zero all the time and there is no vibrational energy on mep calculation, and the total energy is not constant in mep. However, on the dynamics calculation, the atoms vibrate, and total energy conserves.&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier and form the product. || [[File:0149685_t1.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. || [[File:01499685_t3.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they don&#039;t have enough energy to across the energy barrier. ||[[File:01499685_t4.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| unreactive || They collide with each other and unable to form the transition state. ||[[File:01499685_t5.png|180px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
Conventional Transition-State Theory has some assumptions&amp;lt;ref name=&amp;quot;1&amp;quot; /&amp;gt;&lt;br /&gt;
&amp;lt;references&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
1. Once the molecular systems surmounted the transition state, it is not able to turn back and form reactant again.&lt;br /&gt;
&lt;br /&gt;
2. The energy distribution of the reactant is determined by Maxwell-Boltzmann distribution. Moreover, the con-concentration of those activated complexes that are becoming products can also be calculated using equilibrium theory.&lt;br /&gt;
&lt;br /&gt;
3. The motion along the transition state can be separated from other motions associated with the activated complex.&lt;br /&gt;
&lt;br /&gt;
4. The chemical reactions can be treated using classical motion and ignore quantum effects.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants.&lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
It is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond and reach the transition state. Overall, it is an exothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the structure of reactant. This type of transition state is defined as early transition state. In this case, early barrier and translational energy are more efficient.In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74 pm. Then set the momentum zero, and by testing the different value of the distance between F and H, the approximate position of the transition is around 182 pm.&lt;br /&gt;
&lt;br /&gt;
activation energy&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&amp;lt;ref name=&amp;quot;1&amp;quot;&amp;gt;J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998.&amp;lt;/ref&amp;gt;&lt;br /&gt;
&amp;lt;/references&amp;gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=804538</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=804538"/>
		<updated>2020-05-15T14:22:15Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Basic Theory ==&lt;br /&gt;
&lt;br /&gt;
===Transition State Theory ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Mathematically, the maximum can be defined using first and second derivative. The first derivative equals to zero, and the second derivative should be a negative value. &lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. So if there are two horizontal straight lines on the internuclear distance vs time plot, that means the atoms are not move at all. The best estimate of the transition state position is 90.775 pm. &lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+δ, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy, where the kinetic energy keeps zero all the time and there is no vibrational energy on mep calculation, and the total energy is not constant in mep. However, on the dynamics calculation, the atoms vibrate, and total energy conserves.&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier and form the product. || [[File:0149685_t1.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. || [[File:01499685_t3.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they don&#039;t have enough energy to across the energy barrier. ||[[File:01499685_t4.png|180px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| unreactive || They collide with each other and unable to form the transition state. ||[[File:01499685_t5.png|180px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants.&lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
It is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond and reach the transition state. Overall, it is an exothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the structure of reactant. This type of transition state is defined as early transition state. In this case, early barrier and translational energy are more efficient.In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74 pm. Then set the momentum zero, and by testing the different value of the distance between F and H, the approximate position of the transition is around 182 pm.&lt;br /&gt;
&lt;br /&gt;
activation energy&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=804474</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=804474"/>
		<updated>2020-05-15T14:07:06Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Basic Theory ==&lt;br /&gt;
&lt;br /&gt;
===Transition State Theory ==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Mathematically, the maximum can be defined using first and second derivative. The first derivative equals to zero, and the second derivative should be a negative value. &lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. So if there are two horizontal straight lines on the internuclear distance vs time plot, that means the atoms are not move at all. The best estimate of the transition state position is 90.775 pm. &lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+δ, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy, where the kinetic energy keeps zero all the time and there is no vibrational energy on mep calculation, and the total energy is not constant in mep. However, on the dynamics calculation, the atoms vibrate, and total energy conserves.&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier and form the product. || [[File:0149685_t1.png|150px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|150px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. || [[File:01499685_t3.png|150px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they don&#039;t have enough energy to across the energy barrier. ||[[File:01499685_t4.png|150px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| unreactive || They collide with each other and unable to form the transition state. ||[[File:01499685_t5.png|150px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants.&lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
It is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond and reach the transition state. Overall, it is an exothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the structure of reactant. This type of transition state is defined as early transition state. In this case, early barrier and translational energy are more efficient.In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74 pm. Then set the momentum zero, and by testing the different value of the distance between F and H, the approximate position of the transition is around 182 pm.&lt;br /&gt;
&lt;br /&gt;
activation energy&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=804262</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=804262"/>
		<updated>2020-05-15T12:36:36Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Mathematically, the maximum can be defined using first and second derivative. The first derivative equals to zero, and the second derivative should be a negative value. &lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. So if there are two horizontal straight lines on the internuclear distance vs time plot, that means the atoms are not move at all. The best estimate of the transition state position is 90.775 pm. &lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+δ, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy, where the kinetic energy keeps zero all the time and there is no vibrational energy on mep calculation, and the total energy is not constant in mep. However, on the dynamics calculation, the atoms vibrate, and total energy conserves.&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier and form the product. || [[File:0149685_t1.png|150px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|150px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. || [[File:01499685_t3.png|150px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they don&#039;t have enough energy to across the energy barrier. ||[[File:01499685_t4.png|150px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| unreactive || They collide with each other and unable to form the transition state. ||[[File:01499685_t5.png|150px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants.&lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
It is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond and reach the transition state. Overall, it is an exothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, the transition state has a energy closer to the reactant. Based on the Hammond&#039;s Postulate, the structure of the transition state will resemble the structure of reactant. This type of transition state is defined as early transition state. In this case, early barrier and translational energy are more efficient.In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74 pm. Then set the momentum zero, and by testing the different value of the distance between F and H, the approximate position of the transition is around 182 pm.&lt;br /&gt;
&lt;br /&gt;
activation energy&lt;br /&gt;
&lt;br /&gt;
=== Q3. Report the activation energy for both reactions.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Q4. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
=== Q5. 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.===&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=804198</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=804198"/>
		<updated>2020-05-15T12:15:06Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Mathematically, the maximum can be defined using first and second derivative. The first derivative equals to zero, and the second derivative should be a negative value. &lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. So if there are two horizontal straight lines on the internuclear distance vs time plot, that means the atoms are not move at all. The best estimate of the transition state position is 90.775 pm. &lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+δ, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy, where the kinetic energy keeps zero all the time and there is no vibrational energy on mep calculation, and the total energy is not constant in mep. However, on the dynamics calculation, the atoms vibrate, and total energy conserves.&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier and form the product. || [[File:0149685_t1.png|150px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|150px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. || [[File:01499685_t3.png|150px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they don&#039;t have enough energy to across the energy barrier. ||[[File:01499685_t4.png|150px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| unreactive || They collide with each other and unable to form the transition state. ||[[File:01499685_t5.png|150px]]&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
=== Q5.Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===&lt;br /&gt;
&lt;br /&gt;
The Transition State Theory prediction underestimate the reaction rate values compare with experimental values. Because the transition state theory assumes that once the reactant across the energy barrier and it cannot back to reactant. However, in the actual experiment, the reactants can go across the transition state and form products, then reform the reactants.&lt;br /&gt;
&lt;br /&gt;
== EXERCISE 2: F - H - H system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===&lt;br /&gt;
&lt;br /&gt;
It is an exothermic reaction. Since H can form a stronger bond with F, it will release more energy when forming the new H-F bond. It only needs a little energy to break the H-H bond and reach the transition state. Overall, it is an exothermic reaction.&lt;br /&gt;
&lt;br /&gt;
=== Q2. Locate the approximate position of the transition state. ===&lt;br /&gt;
&lt;br /&gt;
Since this reaction of F + H2 is an exothermic reaction, In the F-H-H system,  the H-H bond will not be elongated too much, so the bond length is 74 pm. Then set the momentum zero, and by testing the different value of the distance between F and H, the approximate position of the transition is around 182 pm.&lt;br /&gt;
&lt;br /&gt;
activation energy&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=File:01499685_t5.png&amp;diff=802797</id>
		<title>File:01499685 t5.png</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=File:01499685_t5.png&amp;diff=802797"/>
		<updated>2020-05-14T12:15:26Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
	<entry>
		<id>https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=802793</id>
		<title>MRD:01499685</title>
		<link rel="alternate" type="text/html" href="https://chemwiki.ch.ic.ac.uk/index.php?title=MRD:01499685&amp;diff=802793"/>
		<updated>2020-05-14T12:12:55Z</updated>

		<summary type="html">&lt;p&gt;Yw21318: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== EXERCISE 1: H + H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; system ==&lt;br /&gt;
&lt;br /&gt;
=== Q1. On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===&lt;br /&gt;
&lt;br /&gt;
The  transition state is the maximum in energy (saddle point) along the reaction pathway. On the potential energy surface diagram, the saddle point is minimum when viewing potential against r1/r2. However, it is maximum for the angle orthogonal to it. Mathematically, the maximum can be defined using first and second derivative. The first derivative equals to zero, and the second derivative should be a negative value. &lt;br /&gt;
&lt;br /&gt;
=== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===&lt;br /&gt;
&lt;br /&gt;
If the system starts at the transition state with no momentum, there will be no force acting on the atom, so the atom will not oscillate. So if there are two horizontal straight lines on the internuclear distance vs time plot, that means the atoms are not move at all. The best estimate of the transition state position is 90.775 pm. &lt;br /&gt;
&lt;br /&gt;
=== Q3. Comment on how the mep and the trajectory you just calculated differ. ===&lt;br /&gt;
&lt;br /&gt;
Under the same condition ( r&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;+δ, r&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; = r&amp;lt;sub&amp;gt;ts&amp;lt;/sub&amp;gt;), the potential energy equals to the total energy, where the kinetic energy keeps zero all the time and there is no vibrational energy on mep calculation, and the total energy is not constant in mep. However, on the dynamics calculation, the atoms vibrate, and total energy conserves.&lt;br /&gt;
&lt;br /&gt;
=== Q4. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;wikitable&amp;quot; border=1&lt;br /&gt;
! p&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! p&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt;/&amp;amp;nbsp;g.mol&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt;.pm.fs&amp;lt;sup&amp;gt;-1&amp;lt;/sup&amp;gt; !! E&amp;lt;sub&amp;gt;tot&amp;lt;/sub&amp;gt; !! Reactive? !! Description of the dynamics !! Illustration of the trajectory&lt;br /&gt;
|-&lt;br /&gt;
| -2.56 || -5.1  || -414.280 || reactive  || The H atom and H&amp;lt;sub&amp;gt;2&amp;lt;/sub&amp;gt; molecule approach to each other, and the system has enough energy to across the energy barrier and form the product. || [[File:0149685_t1.png|150px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -4.1  || -420.077 || unreactive || The atom is approaching to the molecule, then they separate again since they don&#039;t have enough energy to overcome the activation energy. || [[File:01499685_t2.png|150px]]&lt;br /&gt;
|-&lt;br /&gt;
| -3.1  || -5.1  || -413.977|| reactive ||The atom is going to react with the molecule and the gives the product. || [[File:01499685_t3.png|150px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.1 || -357.277 || unreactive || The reactants just collide with each other, they don&#039;t have enough energy to across the energy barrier. ||[[File:01499685_t4.png|150px]]&lt;br /&gt;
|-&lt;br /&gt;
| -5.1  || -10.6 || -349.477|| unreactive || They collide with each other and unable to form the transition state. ||&lt;br /&gt;
|}&lt;/div&gt;</summary>
		<author><name>Yw21318</name></author>
	</entry>
</feed>