# Step-by-Step Solution Given: - **Activation energy of forward reaction** (\( E_a^{\text{forward}} \)) = **60 kJ/mol** - **Heat of reaction** (\( \Delta H \)) = **20 kJ/mol** We are to find: - **Activation energy of reverse reaction** (\( E_a^{\text{reverse}} \)) --- ## Step 1: Understanding the Relationship The relationship between the activation energies and the heat of reaction is: \[ E_a^{\text{reverse}} = E_a^{\text{forward}} - \Delta H \] - If \( \Delta H \) is **positive**, the reaction is **endothermic** (products have higher energy). - If \( \Delta H \) is **negative**, the reaction is **exothermic** (products have lower energy). --- ## Step 2: Substitute Values Given that \( \Delta H = +20 \) kJ/mol (endothermic): \[ E_a^{\text{reverse}} = 60\ \text{kJ/mol} - 20\ \text{kJ/mol} \] --- ## Step 3: Calculate \[ E_a^{\text{reverse}} = 40\ \text{kJ/mol} \] --- ## Final Answer **The activation energy of the reverse reaction is:** \[ \boxed{40\ \text{kJ/mol}} \]