Activation energy and catalysis

MDCAT Chemistry · Chemistry · Reaction kinetics. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

Particles react only when they collide with enough energy and in the right orientation. The minimum energy a collision needs is the activation energy, E_a. On an energy profile it is the height of the hill from the reactants up to the peak.

At the peak is the activated complex (transition state): a short-lived, unstable arrangement in which old bonds are partly broken and new bonds are partly formed. It has the highest energy on the path and cannot be isolated.

Anything that gives more effective collisions per second raises the rate. A catalyst does this by offering a new path with a lower E_a. It lowers the forward and reverse barriers by the same amount, so ΔH does not change.

Rules to remember

ΔH = E_a(forward) − E_a(reverse). Exothermic: the forward barrier is the smaller one. Endothermic: the forward barrier is the larger one.
Higher concentration (or gas pressure) and larger surface area: more collisions per second, so a faster rate.
Higher temperature: a much larger fraction of molecules have energy of at least E_a. The rate roughly doubles for each 10 °C rise.
Lower E_a means a faster reaction at the same temperature. ΔH does not decide the rate.
Arrhenius equation: k = Ae^−E_a/RT. A graph of ln k against 1/T is a straight line of slope −E_a/R.
Homogeneous catalyst: same phase as the reactants. Heterogeneous catalyst: different phase. Autocatalysis: a product catalyses the reaction (Mn²⁺ in the KMnO₄ and oxalic acid reaction).
An inhibitor (negative catalyst) slows a reaction. A catalyst never changes ΔH or the position of equilibrium.

Common mistake

Students say a catalyst lowers ΔH or gives the molecules more energy. It only lowers the activation energy, for the forward and reverse reactions equally.

Worked example

For a reaction, E_a(forward) = 75 kJ mol⁻¹ and E_a(reverse) = 120 kJ mol⁻¹. A catalyst lowers the forward value to 50 kJ mol⁻¹. Find ΔH and the reverse activation energy with the catalyst.

ΔH = E_a(forward) − E_a(reverse) = 75 − 120 = −45 kJ mol⁻¹, so the reaction is exothermic.
The catalyst lowers the forward barrier by 75 − 50 = 25 kJ mol⁻¹.
The reverse barrier falls by the same amount: 120 − 25 = 95 kJ mol⁻¹.
Check: 50 − 95 = −45 kJ mol⁻¹, so ΔH is unchanged.

Answer: ΔH = −45 kJ mol⁻¹; E_a(reverse) with the catalyst = 95 kJ mol⁻¹.

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