Part 3: Arrhenius calculations and reaction mechanisms

Calculate activation energy and use experimental orders to assess a proposed sequence of steps.

Temperature changes the rate constant

The Arrhenius model relates k to absolute temperature. A is the pre-exponential or Arrhenius factor and has the same units as k. For a positive activation energy, increasing T makes −Ea/RT less negative, so the exponential and k increase. Use natural logarithms for the linear form, kelvin for temperature and joules for Ea when R is in J K⁻¹ mol⁻¹.

Worked Arrhenius substitution

For an illustrative first-order reaction, A = 2.00 × 10⁸ s⁻¹, Ea = 60.0 kJ mol⁻¹ and T = 320 K. With R = 8.31 J K⁻¹ mol⁻¹, the exponent is −60000/(8.31 × 320) = −22.563. Thus k = 0.0318 s⁻¹ (3 s.f.). Enter the full exponent in parentheses.

To find A, rearrange A = k exp(Ea/RT). To find Ea, rearrange Ea = RT(ln A − ln k). Do not confuse eˣ with 10ˣ, or ln with log₁₀. If only the temperature changes and A is treated as constant, the two-temperature equation follows by subtracting two linear forms.

Use the axis scale when finding Ea

A graph of ln k against 1/T is linear under the Arrhenius approximation, with gradient −Ea/R and intercept ln A. Use two widely separated points on the best-fit line. A slope of −7200 K gives Ea = 7200 × 8.31 = 59832 J mol⁻¹ = 59.8 kJ mol⁻¹.

If the horizontal axis is 1000/T instead, its numerical gradient is one thousandth as large: Ea = −gradient × 1000R. A graph of ln(initial rate) or ln(1/t) may replace ln k only when concentrations and the clock threshold are controlled so the rate measure remains proportional to k.

Use the slow step and any preceding equilibria

A mechanism is a set of elementary steps whose sum gives the overall reaction. Intermediates are produced then consumed; catalysts are consumed then regenerated. In the simple rate-determining-step model, the slow step limits the overall rate. For an elementary slow first step, its reactant molecularity gives the expected rate-law powers.

Consider NO₂ + NO₂ → NO + NO₃ as a slow step followed by NO₃ + CO → NO₂ + CO₂ fast. Cancelling the intermediate and one NO₂ gives NO₂ + CO → NO + CO₂. The model predicts rate = k[NO₂]², with zero order in CO in this regime because CO reacts after the bottleneck. This is consistent evidence, not proof that no alternative mechanism exists.

Do not leave an unmeasured intermediate in the final law

If a slow step uses an intermediate, connect its concentration to earlier steps using an appropriate stated assumption. For a hypothetical fast equilibrium A + B ⇌ I followed by slow I + B → P, [I] = K[A][B] and rate = k₂[I][B] = k₂K[A][B]². Here the measured third-order law reflects both the pre-equilibrium and the slow step.

Simply counting everything appearing before the slow step is not a general kinetic derivation. A fast irreversible step is not automatically an equilibrium, and an overall rate law alone does not uniquely identify a mechanism. Use the model and conditions supplied in the question.

A catalyst provides an alternative mechanism with a lower activation barrier and changes the kinetic behaviour. It does not change the reaction’s enthalpy, Gibbs energy or equilibrium constant at the same temperature.

Quick checks

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

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