1. Chemistry 2402 - Thermodynamics
Lecture 14 : Chemical Equilibria
Lecture 15: Kinetic Coefficients & the Transition State

2. The Dependence of Reaction Rates on Temperature
The idea that we can speed a process up by raising the temperature is commonplace enough to be regarded as intuitive but how does it come about?
The average velocity of a particle is proportional to T1/2 so all processes are expected to speed up on heating by at least this much.
1889 Arrhenius proposed, based on experimental data, that the temperature dependence of a rate constant k went like
k = A exp(-ΔE/RT)
This is a much stronger dependence on T than that of the average particle velocity (i.e. T1/2) and is now referred to as an Arrhenius temperature dependence.
Svante Arrhenius

3. Chemical Reactions seen as Trajectories
To understand what physical properties influence the size of a rate coefficient we must start by considering the microscopic dynamics of particles that we avoided in the second lecture with the assumption of equal a priori probabilities of microstates.
In classical mechanics, a particle continues at its initial velocity unless acted on by a force. For our purposes, the force on a particle along a coordinate is equal to the derivative of the potential energy with respect to that coordinate and directed ‘downhill’ in energy.
Example:
A particle moving along a coordinate x under the influence of a potential U(x)

4. Chemical Reactions seen as Trajectories
To understand what physical properties influence the size of a rate coefficient we must start by considering the microscopic dynamics of particles that we avoided in the second lecture with the assumption of equal a priori probabilities of microstates.
In classical mechanics, a particle continues at its initial velocity unless acted on by a force. For our purposes, the force on a particle along a coordinate is equal to the derivative of the potential energy with respect to that coordinate and directed ‘downhill’ in energy.
denotes the magnitude and direction of the force
Example:
A particle moving along a coordinate x under the influence of a potential U(x)

5. Introducing the Potential Energy Surface
What happens when there is movement along more than one coordinate?
In the case of two coordinates, the potential energy is now a surface.
Even though it is hard to visualize for more than two coordinates, we talk about a potential energy surface for any number of coordinated.
The potential energy surface can be pictured using a contour plot where each line connects points of equal potential energy (i.e. ‘height’).

6. Introducing the Potential Energy Surface
What happens when there is movement along more than one coordinate?
In the case of two coordinates, the potential energy is now a surface.
Even though it is hard to visualize for more than two coordinates, we talk about a potential energy surface for any number of coordinated.
The potential energy surface can be pictured using a contour plot where each line connects points of equal potential energy (i.e. ‘height’).
Note how the saddle looks in the contour plot.

7. The Simplest Reaction: Hydrogen Exchange
A simple reaction – involving only 3 protons and 3 electrons – is the exchange of a hydrogen atom in a hydrogen molecule, i.e.
Ha-Hb + Hc → Ha + Hb-Hc
We consider only collisions along the bond axis (i.e. collinear collisions).
Studied by Eyring and Polanyi in 1931

8. The Simplest Reaction: Hydrogen Exchange
A simple reaction – involving only 3 protons and 3 electrons – is the exchange of a hydrogen atom in a hydrogen molecule, i.e.
Ha-Hb + Hc → Ha + Hb-Hc
The red curve is an example of a non-reactive trajectory.
Note the vibrations of the molecular bond (r2 in this case) as the molecule approaches the atom (along r1).

9. The Simplest Reaction: Hydrogen Exchange
A simple reaction – involving only 3 protons and 3 electrons – is the exchange of a hydrogen atom in a hydrogen molecule, i.e.
Ha-Hb + Hc → Ha + Hb-Hc
The blue curve represents a reactive trajectory.
Note that reactive trajectories are forced to pass close to the saddle point (the ‘mountain pass’) between the reactant and product ‘valleys’.

10. The Transition State
The transition state is the configuration of reactants corresponding to the saddle point of the potential energy surface.
Examples
For the H2 +H reaction, the transition state is the configuration H—H—H where both bond length are equal.
The transition state can also be characterised as that point along the reaction coordinate at which the forward and backward rates are equal.

11. The Transition State Assumption
The transition state assumption is that the transition state is in a ‘quasi-equilibrium’ with the reactants. This means that the details of how the transition state was reached do not matter.
This idea was introduced simultaneously by Eyring and Polanyi in 1935.
Consider the following description of the A → B isomerization
We shall resolve the observed kinetics
At the transition state k-1 = kp Typically this is found by choosing the saddle point. For surfaces without a sharply defined saddle, the transition state is varied until the desired equality is achieved.
A → B k k
A → B
k-1 kp
into this 2-step process involving the transition state
A ⇋ Aǂ → B
A ⇋ Aǂ → B k1 k1
So that we have dAǂ/dt = k1A-(k-1+kp)Aǂ
dB/dt = kpAǂ = kA

12. The Steady State Approximation
The Transition State Assumption is a particular case of the more general Steady State Assumption.
We have dAǂ/dt = k1A-(k-1+kp)Aǂ
If we can assume that Aǂ is always very small (i.e. k-1+kp >> k1)
then dAǂ/dt ≈ 0 (the steady state approximation)
Aǂ = A k1/(k-1+kp) = A Kǂ/2
where Kǂ = k1/k-1, the transition state equilibrium constant.
(We have used the fact that k-1 = kp..)
Finally dB/dt = kpAǂ = kp(Kǂ/2)A = kA so that the overall rate constant k = kpKǂ/2

13. The Gibbs Free Energy of the Transition State
The transition state equilibrium constant Kǂ can be expressed in terms of the difference in the Gibbs free energies between the transition state and the reactants.
Kǂ = exp(-ΔGǂ/RT) = exp(-ΔHǂ/RT) exp(ΔSǂ/R)
Note that ΔGǂ here refers to the molar Gibbs free energy difference.
The rate coefficient kp describes the microscopic dynamics of moving off the saddle point and is typically set to kBT/h, where h is Planck’s constant.
This leaves us with our final expression for the rate constant k
k = kBT/2h exp(ΔSǂ) exp(-ΔHǂ/RT)
We have recovered the temperature dependence described by Arrhenius in 1889.
ΔHǂ is often written as Ea, the activation energy.

14. Catalysis
Catalysis refers to the increase in the rate coefficient for a reaction without altering the equilibrium constant. This is accomplished by introducing a completely new mechanism by which the reaction proceeds.
Catalysis is essential to life and industry by allowing reactions to occur at a sufficiently high rate so as to be of use.
Example: the hydrogenation of ethylene on a nickel surface 1 2 3 4 5

15. Transition States and Catalysis
While the specific mechanism by which a catalyst acts may be complicated, the overall explanation for the increase in the rate constant is simply that the activation energy Ea (or, more correctly, the Gibbs free energy of activation ΔGǂ) has been decreased.

16. Summary
You should now
Understand the representation of a reaction by a potential energy surface
Be able to define and identify the transition state of a reaction
Understand the role of the Transition State Assumption in the microscopic calculations of reaction rate coefficients
Understand the connection between the properties of the transition state, and the reaction rate coefficient
Explain the concept of catalysis, and provide a microscopic description of how it can enhance reaction rates

17. Representative equations you should know: statistical thermodynamics
Pr(n0, n1, n2,…) = W(n0, n1, n2,…)/Wtot
emolecule = etranslation + erotation + evibration + eelectronic

18. Representative equations you should know (or derive): classical thermodynamics
dS = q/T
G = U -TS +PV
= N
γi = ai/xi

19. Equations you don’t have to remember but need to be able to use: statistical thermodynamics

20. Equations you don’t have to remember but need to be able to use: classical thermodynamics
Clapeyron Equation (differential)
Clapeyron Equation (integrated)
Van’t Hoff Equation (differential)
Van’t Hoff Equation (integrated)
Osmotic Pressure
Pi(gas)= xi(solution)Pi* Raoult’s Law
i = io + kBTln(Pi/atm) chemical potential for an ideal gas mixture
i(solution) = io(liq) + kBTln(xi(solution)) chemical potential for an ideal solution