The Linearised Arrhenius Form
The linearised Arrhenius equation \(\ln k = -\dfrac{E_{\text{a}}}{R}\left(\dfrac{1}{T}\right) + \ln A\) maps to \(y = mx + c\). A plot of \(\ln k\) against \(\dfrac{1}{T}\) yields a straight line with gradient \(m = -\dfrac{E_{\text{a}}}{R}\) and y-intercept \(c = \ln A\).
The Exponential Form
| 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:
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:
Worked Example
Calculating Activation Energy from Arrhenius Gradient
Problem: From an Arrhenius plot of \(\ln k\) vs \(1/T\), the gradient of the line is \(-12,500\text{ K}\). Calculate the activation energy \(E_{\text{a}}\) in \(\text{kJ mol}^{-1}\).
Step 1: Calculate \(E_{\text{a}}\) in Joules
\(\text{Gradient} = -\dfrac{E_{\text{a}}}{R} \implies E_{\text{a}} = -\text{gradient} \times R = -(-12500) \times 8.314 = 103,925\text{ J mol}^{-1}\)
Step 2: Convert to kiloJoules
\(E_{\text{a}} = \dfrac{103,925}{1000} = \mathbf{104\text{ kJ mol}^{-1}}\)
Gas Constant Units in Arrhenius Calculations
Unit Trap: Because \(R = 8.314 ext{ J K}^{-1} ext{mol}^{-1}\) is in Joules, calculating \(E_{ ext{a}} = - ext{gradient} imes R\) produces an answer in \( ext{J mol}^{-1}\). Always divide by 1000 to state your final activation energy in \( ext{kJ mol}^{-1}\).
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