# AHL: Activation energy and Arrhenius equation

> IB Chemistry HL · IB Chemistry HL R2: How much / how fast / how far?
> Source: https://www.owlsprep.com/study/ib-chemistry-hl-u5-ahl-activation-energy-and-arrhenius/

This advanced topic explores the temperature dependence of reaction rate, introducing the Arrhenius equation that links rate constant, temperature, and activation energy. You will learn calculation and plotting skills commonly assessed in IB exams.

**Prerequisites:** [Rate of reaction and collision theory](https://www.owlsprep.com/study/ib-chemistry-hl-u5-collision-theory/)

## Learning objectives

- Define activation energy and relate it to reaction rate via collision theory
- Use both the exponential and linear forms of the Arrhenius equation
- Calculate activation energy from experimental rate constant data
- Interpret Arrhenius plots to find Ea and the pre-exponential factor

## Activation Energy and Temperature Dependence

Reaction rate increases with temperature because a greater proportion of reactant molecules have kinetic energy equal to or greater than the activation energy, leading to more successful collisions per second.

**Activation energy** — The minimum total kinetic energy that reactant particles must possess for a successful collision that results in chemical reaction

*Notation:* $E_a$

*Example:* The decomposition of hydrogen peroxide at 298 K has $E_a \approx 76$ kJ mol⁻¹

The fraction of molecules with energy $\geq E_a$ is given by the Boltzmann factor $f = e^{-E_a/RT}$, which increases exponentially with temperature.

**Worked example:** A reaction has an activation energy of 50 kJ mol⁻¹. By what factor does the fraction of collisions with energy $\geq E_a$ increase when temperature rises from 298 K to 308 K? Use $R = 8.31$ J K⁻¹ mol⁻¹.

1. Convert activation energy to matching units:
2. $$E_a = 50 \times 1000 = 50000 \text{ J mol}^{-1}$$
3. Calculate the exponent at 298 K:
4. $$-\frac{E_a}{RT_1} = -\frac{50000}{8.31 \times 298} \approx -20.16$$
5. Calculate the fraction at 308 K:
6. $$-\frac{E_a}{RT_2} = -\frac{50000}{8.31 \times 308} \approx -19.51$$
7. Find the ratio of fractions:
8. $$\frac{f_2}{f_1} = \frac{e^{-19.51}}{e^{-20.16}} = e^{0.65} \approx 1.92$$
9. Conclusion: The fraction of sufficiently energetic collisions almost doubles over this 10 K temperature rise.

> **Exam tip:** Always convert $E_a$ to J mol⁻¹ to match the units of $R = 8.31$ J K⁻¹ mol⁻¹, or use $R = 0.00831$ kJ K⁻¹ mol⁻¹.

## Arrhenius Equation: Forms and Calculations

The Arrhenius equation quantifies the relationship between the rate constant $k$, absolute temperature $T$, and activation energy $E_a$. It has two commonly used forms for problem solving.

**Arrhenius equation** — A mathematical model for the temperature dependence of rate constants

*Example:* Exponential form: $k = A e^{-E_a/RT}$; Linear form: $\ln k = -\frac{E_a}{RT} + \ln A$

The term $A$ (the pre-exponential factor) accounts for the frequency of collisions between reactants with the correct orientation to react. For two sets of $(k, T)$ data, we use the two-point rearranged form:

$$\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$

**Worked example:** A first-order reaction has $k_1 = 2.5 \times 10^{-3}$ s⁻¹ at 25°C and $k_2 = 8.5 \times 10^{-3}$ s⁻¹ at 45°C. Calculate the activation energy.

1. Convert temperatures to Kelvin:
2. $$T_1 = 25 + 273 = 298 \text{ K}, \quad T_2 = 45 + 273 = 318 \text{ K}$$
3. Calculate the ratio of rate constants:
4. $$\ln\left(\frac{8.5 \times 10^{-3}}{2.5 \times 10^{-3}}\right) = \ln(3.4) \approx 1.224$$
5. Calculate the temperature term:
6. $$\frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{318} - \frac{1}{298} \approx -2.11 \times 10^{-4} \text{ K}^{-1}$$
7. Rearrange to solve for $E_a$:
8. $$E_a = -R \times \frac{\ln(k_2/k_1)}{(1/T_2 - 1/T_1)} = -8.31 \times \frac{1.224}{-2.11 \times 10^{-4}} \approx 48000 \text{ J mol}^{-1}$$
9. Final answer: $E_a = 48$ kJ mol⁻¹

## Arrhenius Plots

When you have multiple measurements of $k$ at different temperatures, you can construct a linear Arrhenius plot to find $E_a$ and $A$ graphically. From the linear form of the Arrhenius equation, plotting $y = \ln k$ on the vertical axis against $x = 1/T$ (in K⁻¹) on the horizontal axis gives a straight line.

- Slope of the line = $-E_a/R$
- Y-intercept of the line = $\ln A$

**Worked example:** An Arrhenius plot of $\ln k$ against $1/T$ gives a best-fit line with a slope of $-6250$ K. Calculate the activation energy.

1. Relate slope to activation energy:
2. $$m = -\frac{E_a}{R} \implies E_a = -mR$$
3. Substitute values:
4. $$E_a = -(-6250 \text{ K}) \times 8.31 \text{ J K}^{-1} \text{mol}^{-1} = 51937.5 \text{ J mol}^{-1}$$
5. Convert to standard units: $E_a = 52$ kJ mol⁻¹ (2 significant figures)

> **Exam tip:** Always check that temperature is in Kelvin before calculating 1/T for an Arrhenius plot. Using Celsius will give an incorrect value for Ea.

## Common pitfalls

- **Wrong:** Using Ea in kJ mol⁻¹ directly with R = 8.31 J K⁻¹ mol⁻¹
  - Why it fails: Units do not match, leading to a calculated Ea 1000 times smaller than the correct value
  - Correct: Convert Ea from kJ mol⁻¹ to J mol⁻¹ before substitution, or use R = 0.00831 kJ K⁻¹ mol⁻¹
- **Wrong:** Using Celsius temperature directly in the Arrhenius equation
  - Why it fails: The Arrhenius equation requires absolute temperature, so Celsius values give incorrect proportionality
  - Correct: Always add 273 to Celsius temperature to get Kelvin before substitution
- **Wrong:** Mixing up the order of k1, k2, T1, T2 in the two-point equation
  - Why it fails: This results in a negative activation energy, which is physically impossible
  - Correct: Label k2 as the rate constant at the higher temperature T2, and confirm your final Ea is positive
- **Wrong:** Taking the slope of an Arrhenius plot as equal to Ea
  - Why it fails: The slope equals -Ea/R, so the negative sign and gas constant are ignored
  - Correct: Calculate Ea as Ea = -(slope) × R to get the correct value

## Cheatsheet

| Quantity/Form | Symbol/Expression | Notes |
| --- | --- | --- |
| Activation Energy | $E_a$ | Convert to J mol⁻¹ for calculations |
| Absolute Temperature | $T$ | $T (K) = T (^\circ C) + 273$ |
| Pre-exponential Factor | $A$ | Same units as $k$ |
| Exponential Form | $k = A e^{-E_a/RT}$ | Used for calculating $k$ from $E_a$ |
| Linear Form | $\ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A$ | Used for Arrhenius plots |
| Two-Point Form | $\ln(k_2/k_1) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)$ | Used for two data sets |
| Arrhenius Plot Slope | Slope | Slope = $-E_a/R$ |
| Arrhenius Plot Intercept | Y-intercept | Intercept = $\ln A$ |

## What's next

Understanding activation energy and the Arrhenius equation is foundational for further study of reaction mechanisms and catalysis in IB Chemistry HL. This topic directly explains how catalysts lower activation energy to speed up reactions without changing the enthalpy of reaction, a core concept assessed in both Paper 1 and Paper 2. You will also use the relationships introduced here when exploring more advanced activation parameters in optional topics, and it is often combined with rate law calculations in extended response questions. Mastery of these calculation skills is essential for achieving high marks on kinetics exam questions.

- [AHL: Extended Le Chatelier's principle](https://www.owlsprep.com/study/ib-chemistry-hl-u5-ahl-extended-le-chatelier-s/)
- [AHL: Reaction quotient](https://www.owlsprep.com/study/ib-chemistry-hl-u5-ahl-reaction-quotient/)
- [R3: What are the mechanisms of chemical change?](https://www.owlsprep.com/study/ib-chemistry-hl-u6-overview/)

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