Arrhenius Equation & Activation Energy Cheat Sheet

A printable reference covering the Arrhenius equation, activation energy, rate constants, temperature effects, and Arrhenius plots for grades 11-12.

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The Arrhenius equation explains how temperature affects the rate of a chemical reaction. This cheat sheet helps students connect particle collisions, activation energy, and rate constants in one clear reference. It is especially useful for interpreting lab data, comparing reactions, and solving chemistry problems involving temperature changes. Students need it because small temperature changes can cause large changes in reaction rate. The core equation is $k = Ae^{-\frac{E_a}{RT}}$ , where $k$ is the rate constant, $A$ is the frequency factor, $E_a$ is activation energy, $R$ is the gas constant, and $T$ is temperature in kelvins. The linear form, $\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A$ , allows students to find activation energy from a graph. A two-temperature form, $\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$ , compares rate constants at two temperatures. Higher activation energy usually means a reaction is more sensitive to temperature changes.

Key Facts

The Arrhenius equation is $k = Ae^{-\frac{E_a}{RT}}$ , where $k$ is the rate constant and $T$ must be measured in kelvins.
Activation energy $E_a$ is the minimum energy particles must have for a successful reaction to occur.
The gas constant is commonly used as $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ when $E_a$ is measured in joules per mole.
The linear Arrhenius form is $\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A$ .
On a graph of $\ln k$ versus $\frac{1}{T}$ , the slope is $m = -\frac{E_a}{R}$ , so $E_a = -mR$ .
The two-point Arrhenius equation is $\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$ .
Temperature must be converted using $T_{K} = T_{^{\circ}\text{C}} + 273.15$ before using any Arrhenius equation.
A catalyst lowers $E_a$ , which increases $k$ at the same temperature without changing the overall reaction energy difference.

Vocabulary

Arrhenius Equation: An equation, $k = Ae^{-\frac{E_a}{RT}}$ , that relates a reaction rate constant to temperature and activation energy.
Activation Energy: The minimum energy, $E_a$ , that reacting particles must have to form products successfully.
Rate Constant: The value $k$ that connects reactant concentration to reaction rate for a specific reaction at a specific temperature.
Frequency Factor: The value $A$ that represents how often particles collide with the proper orientation for reaction.
Arrhenius Plot: A graph of $\ln k$ versus $\frac{1}{T}$ used to determine activation energy from the slope.
Catalyst: A substance that increases reaction rate by providing a lower-energy pathway and reducing $E_a$ .

Common Mistakes to Avoid

Using Celsius instead of kelvins is wrong because the Arrhenius equation requires absolute temperature, so always convert with $T_{K} = T_{^{\circ}\text{C}} + 273.15$ .
Mixing joules and kilojoules is wrong because $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ requires $E_a$ in $\text{J mol}^{-1}$ , not $\text{kJ mol}^{-1}$ .
Forgetting the negative slope is wrong because an Arrhenius plot has slope $m = -\frac{E_a}{R}$ , so $E_a$ must be calculated as $E_a = -mR$ .
Using $\log$ instead of $\ln$ without conversion is wrong because the standard Arrhenius forms use natural logarithms, not base-10 logarithms.
Assuming a catalyst changes the products is wrong because a catalyst lowers $E_a$ and speeds the reaction without changing the balanced equation or overall energy change.

Practice Questions

1 A reaction has $E_a = 75.0\ \text{kJ mol}^{-1}$ and $A = 2.5 \times 10^{12}\ \text{s}^{-1}$ . Calculate $k$ at $T = 298\ \text{K}$ using $k = Ae^{-\frac{E_a}{RT}}$ .
2 For an Arrhenius plot of $\ln k$ versus $\frac{1}{T}$ , the slope is $-9500\ \text{K}$ . Calculate $E_a$ in $\text{kJ mol}^{-1}$ using $E_a = -mR$ .
3 A reaction has $k_1 = 0.015\ \text{s}^{-1}$ at $T_1 = 300\ \text{K}$ and $E_a = 48.0\ \text{kJ mol}^{-1}$ . Use $\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$ to find $k_2$ at $T_2 = 330\ \text{K}$ .
4 Explain why a reaction with a larger $E_a$ usually shows a greater increase in rate when temperature rises.

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