Question

The rate of a chemical reaction doubles when the temperature is raised from 298 K to 308 K. Calculate the activation energy (\(E_a\)) for this reaction assuming it does not change with temperature. (Given: \( R = 8.314 \, \text{J mol}^{-1}\text{K}^{-1} \), \( \log 2 = 0.30 \))

Hint:

If rate doubles with 10 K rise → \(E_a\) usually ~50–60 kJ/mol. Use Arrhenius logarithmic form for temperature change problems.

Answer 1

Concept: Temperature dependence of reaction rate is given by the
Arrhenius equation: \[ k = A e^{-E_a/RT} \] For two temperatures: \[ \log \left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R} \left(\frac{T_2 - T_1}{T_1 T_2}\right) \]
Step 1: Given data. \[ \frac{k_2}{k_1} = 2 \quad (\text{rate doubles}) \] \[ T_1 = 298\,K, \quad T_2 = 308\,K \]
Step 2: Substitute into Arrhenius form. \[ \log 2 = \frac{E_a}{2.303 \times 8.314} \left(\frac{308 - 298}{298 \times 308}\right) \]
Step 3: Simplify temperature term. \[ \frac{T_2 - T_1}{T_1 T_2} = \frac{10}{298 \times 308} \] \[ 298 \times 308 = 91784 \] \[ \Rightarrow \frac{10}{91784} \approx 1.089 \times 10^{-4} \]
Step 4: Substitute values. \[ 0.30 = \frac{E_a}{2.303 \times 8.314} \times 1.089 \times 10^{-4} \] First calculate denominator: \[ 2.303 \times 8.314 \approx 19.15 \] \[ 0.30 = \frac{E_a}{19.15} \times 1.089 \times 10^{-4} \]
Step 5: Solve for \(E_a\). \[ E_a = \frac{0.30 \times 19.15}{1.089 \times 10^{-4}} \] \[ 0.30 \times 19.15 = 5.745 \] \[ E_a = \frac{5.745}{1.089 \times 10^{-4}} \] \[ E_a \approx 5.28 \times 10^4 \, \text{J mol}^{-1} \]
Step 6: Convert to kJ mol\(^{-1}\). \[ E_a \approx 52.8 \, \text{kJ mol}^{-1} \]
Final Answer: \[ \therefore E_a \approx 53 \, \text{kJ mol}^{-1} \]