Activation barriers, transition state theory and catalysis

Most collisions fail, because the molecules have to climb a barrier first

Intuition

Think of pushing a ball over a hill to get it into the next valley. Rolling it gently at the hill does nothing. It needs enough energy to reach the top, and it has to go over at the right place. Most reactions are like this. Molecules bump into each other all the time, but only a few bumps carry enough energy and the right orientation. Heat puts more bumps over the top.

Most reactions do not happen at every collision. The reactants need the energy and the correct orientation to pass through a transition state, sometimes called an activated complex. A molecule’s energy is very sensitive to temperature. So the diffusion-controlled rate needs a temperature-sensitive factor. Svante Arrhenius proposed in 1889 that rate constants follow

$k = A\exp(-E_a/RT)$

$E_a$ is the activation energy and $A$ is the pre-exponential factor. The slide 7 energy diagram plots Gibbs free energy against reaction progress. The curve rises from the reactants to a maximum marked $\Delta G^\ddagger$ above them, then falls to the products at a level slightly below the reactants, marked $\Delta G$.

A reaction is a trip across an energy landscape, and the transition state is the mountain pass

In the classical picture, a reaction is movement along a potential energy curve. Dissociation is a bond-stretching motion, drawn as potential energy against bond length. Dimerisation starts as simple travel, but at short distances intermolecular forces switch on and a bond forms. The force on an atom is the derivative of the potential energy along the reaction coordinate. Internal rotations of organic molecules are curves of energy against dihedral angle. Slide 8 shows barriers of 16 kJ/mol (eclipsed), 3.8 kJ/mol (gauche) and 19 kJ/mol (eclipsed methyls) for one such rotation.

A reaction may involve more than one coordinate. For two coordinates the energy is a surface, and the lecture keeps the phrase potential energy surface even for more coordinates. The transition state is a particular configuration along the reaction trajectory. It sits at the saddle point of the surface. The lecture’s example is HO⁻ + bromoalkane going through [HO···C···Br]‡ to alcohol + Br⁻. On a 2-D contour plot, each line joins points of equal energy, like a map’s height lines. Reactive trajectories are forced to pass close to the saddle point, the mountain pass between the reactant and product valleys.

The simplest example is hydrogen exchange, studied by Eyring and Polanyi in 1931. Any bimolecular process is a variation on it.

$\mathrm{H_a + H_b\text{-}H_c \rightarrow H_a\text{-}H_b + H_c}$

Exam detail

Know that the transition state is the saddle point of the potential energy surface, and that reactive trajectories pass near it. Know the example reactions on slides 9 and 10.

Transition state theory links the barrier height to the rate

Intuition

Suppose a few walkers are always standing at the very top of the pass at any moment. The rate at which people cross depends on two things: what fraction are up there, and how fast they tumble down the far side. Transition state theory multiplies those two.

Transition state theory (TST) assumes reactants and the transition state are in equilibrium. The equilibrium constant $K^\ddagger$ depends on the Gibbs free energy difference $\Delta G^\ddagger$ between the transition state and the reactants. That splits into an enthalpy part $\Delta H^\ddagger$ and an entropy part $\Delta S^\ddagger$.

$K^\ddagger = \exp(-\Delta G^\ddagger/RT) = \exp(-\Delta H^\ddagger/RT)\exp(\Delta S^\ddagger/R)$

Once at the transition state, a very fast rate coefficient $k_p$ describes the dynamics of moving off it:

$k_p = \frac{k_B T}{h}$

with $h$ Planck’s constant. Combining them gives the TST rate constant.

$k_{\mathrm{TST}} = \frac{k_p K^\ddagger}{2} = \frac{k_B T}{2h}\exp(\Delta S^\ddagger/R)\exp(-\Delta H^\ddagger/RT)$

This has the temperature dependence Arrhenius described. $\Delta H^\ddagger$ is often written as $E_a$, so

$k_{\mathrm{TST}} = A\exp(-E_a/RT) \qquad A = \frac{k_B T}{2h}\exp(\Delta S^\ddagger/R)$

The lecture adds that quantum mechanics with statistical thermodynamics can be used, via computer, to calculate the properties of the transition state and so the rate constant, just as for any molecular structure.

A catalyst speeds things up by lowering the barrier

The way a catalyst works may be complicated. But the general reason the rate constant rises is simply that the Gibbs free energy of activation $\Delta G^\ddagger$ has been lowered. That can happen through a change in $\Delta H^\ddagger$ (sometimes written $E_a$), or through a change in $\Delta S^\ddagger$.

Compare

Uncatalysed path

A single hump with a large barrier, $\Delta G^\ddagger$ (no catalyst).

Catalysed path

Several smaller intermediate humps and a lower barrier, $\Delta G^\ddagger$ (catalyst). The products end at the same level.

Pitfall

A catalyst does not change where the reaction ends up. On the slide 12 diagram, both paths finish at the same product level $\Delta G$. It changes only the barrier on the way there.

Optional aside

Slide 11 shows a factor of 2 in $k_{\mathrm{TST}} = k_pK^\ddagger/2$ and in $A$. The slide does not explain where it comes from, so it is reported as given.

Recall

Why does a lower activation free energy speed up a reaction?

$K^\ddagger = \exp(-\Delta G^\ddagger/RT)$, so a smaller $\Delta G^\ddagger$ means more molecules sit at the transition state at any moment. The rate coefficient is proportional to $K^\ddagger$, so it rises.