📘 IB Understanding
The Arrhenius equation uses the temperature dependence of the rate constant to determine the activation energy. Analyse graphical representations including its linear form.
The Exponential Form
\[k = Ae^{-\frac{E_a}{RT}}\]
| Symbol | Meaning | Units |
|---|---|---|
| \(k\) | Rate constant | Depends on order |
| \(A\) | Pre-exponential (frequency) factor | Same as k |
| \(E_a\) | Activation energy | J mol⁻¹ |
| \(R\) | Gas constant | 8.314 J K⁻¹ mol⁻¹ |
| \(T\) | Absolute temperature | K (Kelvin) |
The Linear Form
Taking natural logs of both sides gives:
\[\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A\]
This is in the form y = mx + c, which gives a straight line when you plot \(\ln k\) vs \(\frac{1}{T}\):
Arrhenius Plot: ln k vs 1/T
The Two-Point Form
If you have rate constants at two different temperatures:
\[\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\]
Worked Example
Problem: From an Arrhenius plot, the gradient of the line is −12,500 K. Find Eₐ.
\(E_a = -\text{gradient} \times R = -(-12500) \times 8.314\)
\(E_a = 103,925 \text{ J mol}^{-1} = \textbf{104 kJ mol}^{-1}\)
📋 Exam Tip
R uses Joules (8.314 J K⁻¹ mol⁻¹), so your calculated Eₐ will be in J mol⁻¹. Always divide by 1000 to convert to kJ mol⁻¹ for your final answer.