2.2.13 (HL)—Arrhenius factor (A)

Syllabus
First assessment 2025
Objective
2.2.13
Level
HL

The Arrhenius Factor A

HL only

A is the frequency factor: it represents the frequency of collisions with proper orientation. In the linear Arrhenius form, the y-intercept is ln A.

Read the intercept, find A by exponentiating ln A, and keep A distinct from Ea: Ea describes the energy barrier, while A describes collision frequency and orientation.

From a fitted line, intercept b gives A = eᵇ, while gradient m gives Ea = −mR. Check the pair against k = Ae^(−Ea/RT) at one data temperature. A has the same units as k for the stated rate law; it is not an activation energy or a universal constant.

Worked intercept example: for the same first-order Arrhenius line, use lnk=6.02\ln k=-6.02, 1/T=0.00296K11/T=0.00296\,\mathrm{K^{-1}} and gradient Ea/R=12500K-E_a/R=-12500\,\mathrm{K}. From lnk=(Ea/R)(1/T)+lnA\ln k=(-E_a/R)(1/T)+\ln A, 6.02=(12500)(0.00296)+lnA-6.02=(-12500)(0.00296)+\ln A, so lnA=30.98\ln A=30.98 and A=e30.98=2.85×1013s1A=e^{30.98}=2.85\times10^{13}\,\mathrm{s^{-1}}. The unit is s1\mathrm{s^{-1}} because this reaction is first order and AA has the same units as kk.

Finding A from the Intercept

HL only

Assessment in practice

Representative question

Question 1

[Maximum number: 2]

Calculate the numerical value of A.

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.