Given the rate constants k_1 = 2.5 \times 10^{-3} \text{ s}^{-1} at 300 \text{ K} and k_2 = 1.2 \times 10^{-2} \text{ s}^{-1} at 320 \text{ K}, what is the activation energy (E_a) of the reaction? (R = 8.314 \text{ J mol}^{-1} \text{K}^{-1})

• A. 54.2 \text{ kJ mol}^{-1}
• B. 34.7 \text{ kJ mol}^{-1}
• C. 21.6 \text{ kJ mol}^{-1}
• D. 8.3 \text{ kJ mol}^{-1}

Answer: (B)

Verified step by step guidance

1. Identify the given data: rate constants \(k_1 = 2.5 \times 10^{-3} \text{ s}^{-1}\) at temperature \(T_1 = 300 \text{ K}\), and \(k_2 = 1.2 \times 10^{-2} \text{ s}^{-1}\) at temperature \(T_2 = 320 \text{ K}\). The gas constant is \(R = 8.314 \text{ J mol}^{-1} \text{K}^{-1}\).
2. Recall the Arrhenius equation in its two-point form to find the activation energy \(E_a\): \(\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\)
3. Rearrange the equation to solve for \(E_a\): \(E_a = -R \times \frac{\ln\left(\frac{k_2}{k_1}\right)}{\left(\frac{1}{T_2} - \frac{1}{T_1}\right)}\)
4. Calculate the natural logarithm of the ratio of rate constants: \(\ln\left(\frac{k_2}{k_1}\right)\), and compute the difference in the reciprocals of the temperatures: \(\frac{1}{T_2} - \frac{1}{T_1}\).
5. Substitute the values of \(R\), \(\ln\left(\frac{k_2}{k_1}\right)\), and \(\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\) into the rearranged formula to find \(E_a\). Remember to convert \(E_a\) from joules per mole to kilojoules per mole by dividing by 1000.