Arrhenius equation

CIE A-Level Chemistry· 15 min read

1. Forms of the Arrhenius Equation and Key Terms★★☆☆☆⏱ 15 min

The Arrhenius equation describes how the rate constant (k) of a reaction changes with absolute temperature (T), and depends on the reaction's activation energy (E_a).

📘 Definition

Arrhenius Equation

A mathematical relationship that connects the rate of a reaction to temperature and activation energy, based on experimental observations.

Example:

For the hydrolysis of a primary haloalkane, (k) doubles for approximately every 10 K temperature increase, matching the equation's prediction.

The original (exponential) form of the equation is:

k = A e^{-E_a / RT}

Taking the natural logarithm of both sides gives the linear form, which is much more useful for calculations and graphical interpretation:

\ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A

(k): rate constant, units match the reaction order
(A): pre-exponential (frequency) factor, same units as (k)
(E_a): activation energy, typically reported in kJ mol⁻¹
(R = 8.31) J K⁻¹ mol⁻¹: universal gas constant
(T): absolute temperature, always in Kelvin

📐 Worked Example

Derive the logarithmic form of the Arrhenius equation from the exponential form.

1
Start with the exponential form, then take the natural logarithm of both sides:
2
$k = A e^{-E_a / RT} \implies \ln k = \ln\left(A e^{-E_a / RT}\right)$
3
Use the logarithm product rule (\ln(ab) = \ln a + \ln b) to split the right-hand side:
4
$\ln k = \ln A + \ln\left(e^{-E_a / RT}\right)$
5
Simplify using the rule (\ln(e^x) = x) to get the final linear form:
6
$\ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A$

2. Arrhenius Plots and Activation Energy Calculations★★★☆☆⏱ 20 min

The linear form of the Arrhenius equation matches the equation of a straight line (y = mx + c), so we can plot experimental data to find (E_a) and (A). This is called an Arrhenius plot.

y-axis = (\ln k)
x-axis = (1/T) (units: K⁻¹)
Gradient (m = -E_a / R)
y-intercept (c = \ln A)

If you only have two sets of (k) and (T) data, you can use a two-point calculation instead of plotting. This is the most common exam question on this topic.

📐 Worked Example

Calculate the activation energy for a reaction with the following data: (k_1 = 1.2 \times 10^{-3} \text{ dm}^3 \text{ mol}^{-1} \text{ s}^{-1}) at (T_1 = 20^\circ \text{C}), (k_2 = 6.5 \times 10^{-3} \text{ dm}^3 \text{ mol}^{-1} \text{ s}^{-1}) at (T_2 = 40^\circ \text{C}). Give your answer in kJ mol⁻¹.

1
First convert temperatures from Celsius to Kelvin:
2
$T_1 = 20 + 273 = 293 \text{ K}, \quad T_2 = 40 + 273 = 313 \text{ K}$
3
Subtract the Arrhenius equation for (T_1) from the equation for (T_2) and simplify:
4
$\ln k_2 - \ln k_1 = \left(-\frac{E_a}{R T_2} + \ln A\right) - \left(-\frac{E_a}{R T_1} + \ln A\right) \\ \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)$
5
Substitute all known values into the rearranged equation:
6
$\ln\left(\frac{6.5 \times 10^{-3}}{1.2 \times 10^{-3}}\right) = \frac{E_a}{8.31} \left(\frac{1}{293} - \frac{1}{313}\right)$
7
Calculate each term: (\ln(5.42) ≈ 1.69), and (1/293 - 1/313 ≈ 2.18 \times 10^{-4} \text{ K}^{-1}):
8
$1.69 = \frac{E_a \times 2.18 \times 10^{-4}}{8.31}$
9
Solve for (E_a) and convert to kJ mol⁻¹:
10
$E_a = \frac{1.69 \times 8.31}{2.18 \times 10^{-4}} ≈ 64400 \text{ J mol}^{-1} = 64.4 \text{ kJ mol}^{-1}$

3. Qualitative Interpretation and Common Exam Questions★★★☆☆⏱ 15 min

The pre-exponential factor (A) is related to how often successful collisions occur. It depends on the frequency of collisions between reactants and the fraction of collisions that have the correct orientation to react. It has the same units as the rate constant (k).

A common qualitative exam question asks how changing temperature affects (k) for reactions with different activation energies. The key takeaway is:

✓ Quick check

Test your understanding of core concepts:

What happens to the rate constant (k) when temperature increases for most reactions?
- It decreases
- It stays the same
- It increases
- It changes randomly
Reveal answer
2 —
Increasing temperature means more particles have energy equal to or greater than the activation energy, so the rate of reaction and rate constant always increase.
What is the gradient of an Arrhenius plot of (\ln k) against (1/T)?
- (E_a/R)
- (-E_a/R)
- (E_a)
- (\ln A)
Reveal answer
1 —
From the linear Arrhenius equation (\ln k = (-E_a/R)(1/T) + \ln A), the coefficient of (1/T) is the gradient, which is (-E_a/R).

4. Common Pitfalls

Wrong move:

Using temperature in Celsius instead of Kelvin in calculations

Why:

The Arrhenius equation requires absolute temperature, so using Celsius values will give a drastically incorrect activation energy

Correct move:

Always add 273 to any temperature given in °C before substituting it into the equation

Wrong move:

Forgetting to cancel the negative sign from the gradient, resulting in negative activation energy

Why:

The gradient of the Arrhenius plot is negative, but activation energy is always a positive value

Correct move:

Use the relationship (E_a = -m \times R), where (m) is the measured gradient of the plot, to get a positive final value

Wrong move:

Mixing natural logarithms and base 10 logarithms

Why:

The Arrhenius equation uses natural logarithms, so mixing bases with the standard value of (R = 8.31) J K⁻¹ mol⁻¹ gives wrong results

Correct move:

Always use natural logarithms ((\ln)) unless the question explicitly tells you to use base 10 logarithms

Wrong move:

Mismatching units for activation energy and the gas constant

Why:

(R) is given in J K⁻¹ mol⁻¹, so calculating (E_a) gives a value in J mol⁻¹, but exams usually ask for kJ mol⁻¹

Correct move:

Always check the required units for the final answer, divide by 1000 to convert J mol⁻¹ to kJ mol⁻¹

5. Quick Reference Cheatsheet

Concept	Formula/Value	Key Notes
Exponential form	$k = A e^{- E_{a} / R T}$	Use for calculating $k$ from known $E_{a}$
Linear (log) form	$ln k = - \frac{E _{a}}{R} \cdot \frac{1}{T} + ln A$	Use for plots and calculations
Arrhenius plot	y = $ln k$ , x = $1/ T$	Gradient = $- E_{a} / R$ , y-intercept = $ln A$
Two-point calculation	$ln (\frac{k _{2}}{k _{1}}) = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$	Use for two (k,T) data pairs
Gas constant	$R = 8.31$ J K⁻¹ mol⁻¹	Always convert $T$ to Kelvin

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 22
Calculate Ea from two (k,T) pairs
2023 · 12
Interpret gradient of Arrhenius plot
2021 · 33
Find pre-exponential factor from plot

What's Next

The Arrhenius equation is a core foundation for further topics in reaction kinetics, connecting experimental rate data to underlying reaction mechanisms and collision theory. Calculations of activation energy are regularly tested in multiple-choice, structured and practical questions across all CIE A-Level Chemistry papers, so mastering this skill is critical for scoring well in kinetics sections. This topic builds on your understanding of rate equations and rate constants, and leads into more advanced concepts like rate-determining step and activation energy profiles.