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2.2.11 (HL)—Rate constant (k)

Syllabus
First assessment 2025
Objective
2.2.11
Level
HL

The Rate Constant k

HL only

k=rate/([A]x[B]y)k = rate / ([A]^x[B]^y)

The units of k depend on the overall order and must cancel the concentration and time units in the rate equation. For a particular reaction, k changes with temperature.

Derive rather than memorize the units: first-order k has units s⁻¹, while an overall second-order law written with mol dm⁻³ and seconds gives dm³ mol⁻¹ s⁻¹. Concentration changes rate but does not change k at fixed temperature.

Worked kk example: for v=k[FeX3+][IX−]2v=k[\ce{Fe^{3+}}][\ce{I^-}]^2, use v=1.62×10−5 mol dm−3 s−1v=1.62\times10^{-5}\,\mathrm{mol\,dm^{-3}\,s^{-1}} and both concentrations 1.00×10−2 mol dm−31.00\times10^{-2}\,\mathrm{mol\,dm^{-3}}. k=1.62×10−5(1.00×10−2)(1.00×10−2)2=16.2 dm6 mol−2 s−1k=\frac{1.62\times10^{-5}}{(1.00\times10^{-2})(1.00\times10^{-2})^2}=16.2\,\mathrm{dm^6\,mol^{-2}\,s^{-1}}. Substitution returns the measured rate, and the units cancel those of a third-order concentration term.

Calculating k and Its Units

HL only

Assessment in practice

Representative question

Question 1

[Maximum number: 2]

Calculate the value of the rate constant stating its units.

Rate and Mechanisms Summary

Retrieve the route: measure a tangent rate, explain effective collisions, map rate factors, read Ea and energy profiles, evaluate mechanisms, determine molecularity and orders, calculate k, then use Arrhenius gradient and intercept for Ea and A.

Check tangent versus average slope, energy versus orientation, barrier labels, intermediate versus transition state, one-variable trial comparisons, order-dependent units, kelvin temperature and the signs of gradient and Ea.