A level Chemistry · Edexcel 9CH0 · Topic 16C

The Arrhenius equation

Key points

Arrhenius equation: $k = Ae^{-E_a/RT}$ , where $A$ is the pre-exponential factor (collision frequency and orientation).
Log form: $\ln k = -\frac{E_a}{R} \times \frac{1}{T} + \ln A$ .
Plot ln k against 1/T: a straight line with gradient $-\frac{E_a}{R}$ , so $E_a = -\text{gradient} \times R$ (R = 8.31 J mol⁻¹ K⁻¹).
You can use $\ln(\text{rate})$ or $\ln(1/t)$ in place of ln k when other concentrations are fixed.
A small temperature rise causes a large rate increase because many more molecules exceed $E_a$ .
Catalysts lower $E_a$ , so $k$ increases at a given temperature.
Remember to use $T$ in kelvin and give $E_a$ in kJ mol⁻¹ (divide J by 1000).

Topic notesPremium

Temperature has a dramatic effect on rate: a 10 K rise often roughly doubles it. The Arrhenius equation explains this through the rate constant, and turns a set of rates at different temperatures into a value for the activation energy.

Unlock the full notes with Premium

Premium includes full topic notes, the paper generator, self-quizzes and progress tracking.

See Premium plans Already Premium? Sign in

Practise this topicPremium

Premium members can generate an exam-style paper with a mark scheme, or take a quick quiz, on The Arrhenius equation.

Back to all Chemistry topics