Arrhenius, Reaction Coordinate Diagrams & Catalysis

You’ll be able to

Use k = A·e^(−Eₐ/RT) to reason about how temperature and activation energy control the rate constant
Read a reaction-coordinate diagram for Eₐ (forward and reverse), ΔH, the transition state, and any intermediate
Explain how a catalyst lowers Eₐ without changing ΔH or the position of equilibrium

The Arrhenius equation, piece by piece

The rate constant is not truly constant — it depends on temperature through the Arrhenius equation, k = A·e^(−Eₐ/RT). Here A is the frequency factor (how often collisions occur with the right orientation), and e^(−Eₐ/RT) is the fraction of collisions with enough energy to clear the barrier. Because Eₐ sits in a negative exponent, a bigger Eₐ makes that fraction smaller and shrinks k; a bigger RT makes the exponent less negative and grows k.

Arrhenius equation

k = A·e^(−Eₐ/RT)

R = 8.314 J·mol⁻¹·K⁻¹ (use Eₐ in J/mol). Two-temperature form: ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂). A plot of ln k vs 1/T is linear with slope = −Eₐ/R.

How T and Eₐ move the rate

Two levers change k. Raising temperature raises k (the exponent −Eₐ/RT becomes less negative, so the energetic fraction grows) — this is the same high-energy-tail effect from the Maxwell–Boltzmann picture. Lowering Eₐ also raises k, because it makes the barrier easier to clear. Reactions with a large Eₐ are the most temperature-sensitive: their k climbs by the biggest factor for a given temperature rise, because the exponent depends on Eₐ.

Reading a reaction-coordinate diagram

A reaction-coordinate (energy) diagram plots potential energy along the reaction path. The peak is the transition state (activated complex). The forward activation energy is the climb from reactants up to the peak; the reverse Eₐ is the climb from products up to the same peak. ΔH is the reactants-to-products difference — downhill (ΔH < 0) is exothermic, uphill (ΔH > 0) is endothermic. A multi-step mechanism shows two or more humps, and the valley between them is an intermediate (a real species that is made then consumed).

Energy relationships on the diagram

Eₐ(forward) = E(transition state) − E(reactants) Eₐ(reverse) = E(transition state) − E(products) ΔH = Eₐ(forward) − Eₐ(reverse)

ΔH also equals E(products) − E(reactants). If Eₐ(forward) < Eₐ(reverse) the reaction is exothermic (ΔH < 0); if Eₐ(forward) > Eₐ(reverse) it is endothermic (ΔH > 0).

Worked example

Two reactions share the same frequency factor A and are run at the same temperature. Reaction X has Eₐ = 40 kJ/mol; Reaction Y has Eₐ = 80 kJ/mol. Using k = A·e^(−Eₐ/RT), decide which has the larger rate constant, and describe how raising the temperature affects each.

1.Compare the exponential factors. Since A and T are identical, the only difference is e^(−Eₐ/RT). Eₐ appears in a negative exponent, so a larger Eₐ gives a smaller e^(−Eₐ/RT) and therefore a smaller k.
2.Reaction X has the smaller Eₐ (40 vs 80 kJ/mol), so its exponent −Eₐ/RT is less negative, e^(−Eₐ/RT) is larger, and k(X) > k(Y). Reaction X is faster at this temperature.
3.Raising T increases RT, which makes −Eₐ/RT less negative for both reactions, so e^(−Eₐ/RT) increases and k increases for both.
4.The effect is not equal: because the exponent scales with Eₐ, the larger-barrier reaction Y is more temperature-sensitive — its k rises by the greater factor when T goes up.

Answer: X has the larger k (smaller Eₐ). Raising T increases k for both, but Y (the larger Eₐ) speeds up more steeply because a bigger barrier makes k more temperature-sensitive.

What a catalyst does to the diagram

A catalyst provides an alternative pathway with a lower Eₐ — on the diagram, a lower hump (or a set of smaller humps via a new intermediate). Because the reactant and product energies are untouched, ΔH is unchanged, and because both the forward and reverse barriers drop by exactly the same amount, the catalyst speeds the forward and reverse reactions equally. It therefore lets equilibrium arrive faster but does not shift its position or change K.

Watch out

A catalyst changes kinetics, not thermodynamics. It lowers Eₐ (both directions) and raises k, but it leaves ΔH, K, and the equilibrium position unchanged. If a choice says a catalyst makes a reaction more exothermic or shifts equilibrium toward product, it is wrong.

Checkpoint

On a reaction-coordinate diagram, the reactants sit at 20 kJ, the transition state at 90 kJ, and the products at 60 kJ. What is the activation energy of the reverse reaction?

Checkpoint

A catalyst is added to a reaction. According to k = A·e^(−Eₐ/RT), what happens to the rate constant k and to ΔH for the reaction?

On the exam

To annotate a diagram fast: mark the peak as the transition state, arrow up to it from reactants (Eₐ forward) and up to it from products (Eₐ reverse), and read ΔH from reactant level to product level. Check with ΔH = Eₐ(forward) − Eₐ(reverse). Count humps: two humps means a two-step mechanism with an intermediate in the valley.

Answer the 2 checkpoints as you read.