# Arrhenius equation

> CIE A-Level Chemistry · 9701
> Source: https://www.owlsprep.com/study/cie-9701-u20-arrhenius-equation/

This sub-topic covers the temperature dependence of reaction rate constants, linking activation energy, temperature and the pre-exponential factor. You will learn to use both forms of the equation and calculate activation energy for exam questions.

**Prerequisites:** [Rate equations and rate constants](https://www.owlsprep.com/study/cie-9701-u19-rate-equations/); Logarithmic transformations

## Learning objectives

- State the Arrhenius equation in exponential and logarithmic forms
- Calculate activation energy from rate constant and temperature data
- Interpret Arrhenius plots and relate gradient to activation energy
- Explain the meaning of the pre-exponential factor

## Forms of the Arrhenius Equation and Key Terms

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 \(E_a\).

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

## Arrhenius Plots and Activation Energy Calculations

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. $$E_a = \frac{1.69 \times 8.31}{2.18 \times 10^{-4}} ≈ 64400 \text{ J mol}^{-1} = 64.4 \text{ kJ mol}^{-1}$$

> **tip**
>
> Always check that your final activation energy value is positive. If you get a negative value, you probably messed up the sign of the gradient or temperature order.

## Qualitative Interpretation and Common Exam Questions

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:

> **note**
>
> For the same increase in temperature, a reaction with higher activation energy will show a larger percentage increase in \(k\) than a reaction with lower activation energy.

**Check your understanding**

Test your understanding of core concepts:

1. What happens to the rate constant \(k\) when temperature increases for most reactions?

   - It decreases
   - It stays the same
   - It increases
   - It changes randomly

   *Answer:* It increases

   *Why:* 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.

2. What is the gradient of an Arrhenius plot of \(\ln k\) against \(1/T\)?

   - \(E_a/R\)
   - \(-E_a/R\)
   - \(E_a\)
   - \(\ln A\)

   *Answer:* \(-E_a/R\)

   *Why:* From the linear Arrhenius equation \(\ln k = (-E_a/R)(1/T) + \ln A\), the coefficient of \(1/T\) is the gradient, which is \(-E_a/R\).

## Common pitfalls

- **Wrong:** Using temperature in Celsius instead of Kelvin in calculations
  - Why it fails: The Arrhenius equation requires absolute temperature, so using Celsius values will give a drastically incorrect activation energy
  - Correct: Always add 273 to any temperature given in °C before substituting it into the equation
- **Wrong:** Forgetting to cancel the negative sign from the gradient, resulting in negative activation energy
  - Why it fails: The gradient of the Arrhenius plot is negative, but activation energy is always a positive value
  - Correct: Use the relationship \(E_a = -m \times R\), where \(m\) is the measured gradient of the plot, to get a positive final value
- **Wrong:** Mixing natural logarithms and base 10 logarithms
  - Why it fails: 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: Always use natural logarithms (\(\ln\)) unless the question explicitly tells you to use base 10 logarithms
- **Wrong:** Mismatching units for activation energy and the gas constant
  - Why it fails: \(R\) 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: Always check the required units for the final answer, divide by 1000 to convert J mol⁻¹ to kJ mol⁻¹

## Cheatsheet

| Concept | Formula/Value | Key Notes |
| --- | --- | --- |
| Exponential form | $k = A e^{-E_a/RT}$ | 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 | $\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$ | Use for two (k,T) data pairs |
| Gas constant | $R = 8.31$ J K⁻¹ mol⁻¹ | Always convert $T$ to Kelvin |

## 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.

- [Rate-determining step](https://www.owlsprep.com/study/cie-9701-u20-rate-determining-step/)
- [Rate constant calculations](https://www.owlsprep.com/study/cie-9701-u20-rate-constant-calculations/)
- [Transition elements](https://www.owlsprep.com/study/cie-9701-u21-overview/)

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