AHL: Activation energy and Arrhenius equation

IB Chemistry HL· 6 min read

1. Activation Energy and Temperature Dependence★★☆☆☆HL only⏱ 15 min

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.

📘 Definition

Activation energy

$E_{a}$

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

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 $f = e^{- E_{a} / R T}$ , 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.

2. Arrhenius Equation: Forms and Calculations★★★☆☆HL only⏱ 20 min

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.

📘 Definition

Arrhenius equation

A mathematical model for the temperature dependence of rate constants

Example:

Exponential form: $k = A e^{- E_{a} / R T}$ ; Linear form: $ln k = - \frac{E _{a}}{R T} + ln A$

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 $k_{1} = 2.5 \times 1 0^{- 3}$ s⁻¹ at 25°C and $k_{2} = 8.5 \times 1 0^{- 3}$ 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

3. Arrhenius Plots★★★☆☆HL only⏱ 15 min

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.

4. Common Pitfalls

Wrong move:

Using Ea in kJ mol⁻¹ directly with R = 8.31 J K⁻¹ mol⁻¹

Why:

Units do not match, leading to a calculated Ea 1000 times smaller than the correct value

Correct move:

Convert Ea from kJ mol⁻¹ to J mol⁻¹ before substitution, or use R = 0.00831 kJ K⁻¹ mol⁻¹

Wrong move:

Using Celsius temperature directly in the Arrhenius equation

Why:

The Arrhenius equation requires absolute temperature, so Celsius values give incorrect proportionality

Correct move:

Always add 273 to Celsius temperature to get Kelvin before substitution

Wrong move:

Mixing up the order of k1, k2, T1, T2 in the two-point equation

Why:

This results in a negative activation energy, which is physically impossible

Correct move:

Label k2 as the rate constant at the higher temperature T2, and confirm your final Ea is positive

Wrong move:

Taking the slope of an Arrhenius plot as equal to Ea

Why:

The slope equals -Ea/R, so the negative sign and gas constant are ignored

Correct move:

Calculate Ea as Ea = -(slope) × R to get the correct value

5. Quick Reference 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	$k = A e^{- E_{a} / R T}$	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	$ln (k_{2} / k_{1}) = - \frac{E _{a}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})$	Used for two data sets
Arrhenius Plot Slope	Slope	Slope = $- E_{a} / R$
Arrhenius Plot Intercept	Y-intercept	Intercept = $ln A$

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.

2025 · 2
Two-point Ea calculation
2024 · 1
Arrhenius plot slope interpretation
2023 · 2
Plot Ea from experimental data

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.