Activation energy, factors affecting rate and catalysis

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. The energy of activation, E_a, is the minimum energy, above the average energy, that colliding particles need to react. At the top of the energy profile is the unstable activated complex.

A higher concentration or a larger surface area gives more collisions. A higher temperature greatly increases the fraction of molecules with energy of at least E_a. The Arrhenius equation links the rate constant to temperature and E_a.

A catalyst gives another pathway with a lower E_a. It is not used up, and it does not change ΔH or the position of equilibrium.

Rules to remember

ΔH = E_a(forward) − E_a(reverse). For an exothermic reaction, E_a(reverse) is the larger one.
Arrhenius equation: k = Ae^−E_a/RT, so ln k = ln A − E_a ÷ (RT), with R = 8.314 J K⁻¹ mol⁻¹ and T in kelvin.
Graph of ln k against 1/T: straight line, slope = −E_a ÷ R. Graph of log k against 1/T: slope = −E_a ÷ (2.303R).
A rise of 10 K roughly doubles the rate of many reactions.
Homogeneous catalysis: catalyst and reactants in the same phase. Heterogeneous catalysis: different phases, e.g. nickel in the hydrogenation of vegetable oils, V₂O₅ in the Contact process.
Autocatalysis: a product catalyses the reaction, e.g. Mn²⁺ in the reaction of KMnO₄ with oxalic acid. A poison destroys the activity of a catalyst (arsenic compounds poison platinum); a promoter increases it.
Enzymes are protein catalysts: highly specific, very efficient, and most active at an optimum temperature (about 37 °C) and pH.

Common mistake

Students say a catalyst gives the molecules more energy, or that it changes ΔH. It does neither. It lowers the activation energy, by the same amount for the forward and reverse reactions.

Worked example

For a reaction, a graph of log k against 1/T is a straight line of slope −2500 K. What is the activation energy? (R = 8.314 J K⁻¹ mol⁻¹)

Slope = −E_a ÷ (2.303R), so E_a = −slope × 2.303 × R
2.303 × 8.314 = 19.15 J K⁻¹ mol⁻¹
E_a = 2500 × 19.15 = 47 900 J mol⁻¹ (to 3 significant figures)

Answer: E_a = 47.9 kJ mol⁻¹

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