Part 3: Rate-determining steps and the Arrhenius equation

Use a rate law as evidence about a mechanism, then calculate how temperature changes the rate constant without confusing kinetics with equilibrium.

A mechanism must explain more than the overall equation

A multistep mechanism is a sequence of elementary events. The rate-determining step limits the overall rate under the stated conditions. For a proposed elementary slow step A + B → intermediate, the simplest model gives rate = k[A][B]. A later fast reaction of that intermediate with C does not necessarily put [C] into the observed rate law.

Add the steps and cancel intermediates to recover the overall equation. An intermediate is made in one step and consumed in another; a catalyst is consumed and later regenerated. Check atom and charge balance in each step, not only the sum. Agreement with the rate law supports a mechanism but does not prove it is unique.

Avoid the shortcut “anything absent from the rate law reacts after the slow step” as a universal rule. A preceding equilibrium can connect an intermediate concentration to reactant concentrations. At this level, use the supplied steps and data to test consistency rather than inventing an unseen mechanism.

Worked example: distinguish two proposals

Consider original hypothetical gas-phase chemistry with overall equation 2NO + O₂ → 2NO₂ and observed rate = k[NO]²[O₂]. A single elementary slow step NO + O₂ → NO₃ would predict first order in NO and is inconsistent with that observed law if there are no additional assumptions.

A possible alternative has a rapid pre-equilibrium 2NO ⇌ N₂O₂ followed by slow N₂O₂ + O₂ → 2NO₂. The slow-step rate is proportional to [N₂O₂][O₂]. If the preceding equilibrium gives [N₂O₂] proportional to [NO]², the observed dependence follows. This is an illustrative reasoning extension; a full treatment of pre-equilibrium kinetics is not a required derivation.

The lesson is to distinguish a rate equation for an elementary step from the experimentally measured overall rate equation. In a simpler exam proposal where the slow step contains the original reactants directly, compare its reacting particles to the observed powers and then check the sum of steps.

Why a higher temperature changes k

At the same concentrations, warming a reacting mixture usually increases rate because a larger fraction of collisions has sufficient energy to react. Since the concentration terms have not changed, the increase appears in k. For a fixed pathway with positive activation energy, the Arrhenius equation describes the strong temperature dependence.

T must be in kelvin, Ea in J mol⁻¹ when R = 8.314 J mol⁻¹ K⁻¹, and ln is the natural logarithm. The pre-exponential factor A is the intercept-related constant; OCR does not require a detailed interpretation of A. Do not confuse this constant with a reactant labelled A in a different equation.

Worked example: read the straight-line form

Plot y = ln k against x = 1/T. The gradient is −Ea/R and the intercept is ln A. For an original illustrative best-fit line with gradient −7200 K and intercept 24.0, Ea = −(−7200) × 8.314 = 59 860.8 J mol⁻¹ = 59.9 kJ mol⁻¹. A = e²⁴ = 2.65 × 10¹⁰, with the same units as k.

At 300 K this line gives ln k = −7200/300 + 24.0 = 0, so k = 1.00 in the chosen rate-constant units. At 320 K, ln k = 1.50 and k = 4.48. This check shows increasing temperature increases k even though the graph slopes down: warming makes 1/T smaller, so you move left on the graph.

Use a large triangle on a best-fit line, not two noisy adjacent measurements. If the horizontal axis is 1000/T rather than 1/T, account for the factor of 1000: its gradient is −Ea/(1000R). A negative gradient does not mean a negative activation energy.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official OCR A mark allocations.

Sources