The rate constant of a reaction increases by five times on increase in temperature from 27 C to 52 C — Chemical Kinetics Chemistry Question
Question
The rate constant of a reaction increases by five times on increase in temperature from 27 C to 52 C. The value of activation energy in kJ mol is _______. (Rounded-off to the nearest integer) [R = 8.314 J K mol ] o o -1 -1 -1
💡 Solution & Explanation
**Step 1: Identify the given information** - k₂/k₁ = 5 (rate constant increases 5 times) - T₁ = 27°C = 300 K - T₂ = 52°C = 325 K - R = 8.314 J K⁻¹ mol⁻¹ **Step 2: Apply the Arrhenius equation** Use the two-temperature form: $$\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$ **Step 3: Substitute values** $$\ln(5) = \frac{E_a}{8.314}\left(\frac{1}{300} - \frac{1}{325}\right)$$ $$1.609 = \frac{E_a}{8.314}\left(\frac{325-300}{300 \times 325}\right)$$ $$1.609 = \frac{E_a}{8.314}\left(\frac{25}{97500}\right)$$ **Step 4: Calculate the temperature term** $$\frac{25}{97500} = 2.564 \times 10^{-4} \text{ K}^{-1}$$ **Step 5: Solve for Eₐ** $$E_a = \frac{1.609 \times 8.314}{2.564 \times 10^{-4}}$$ $$E_a = \frac{13.37}{2.564 \times 10^{-4}} = 52,124 \text{ J/mol}$$ $$E_a = 52.1 \text{ kJ/mol}$$ Therefore, the answer is 52.00.