The activation energy of a reaction is $90\text{ kJ/mol}$. If the temperature is increased from $300\text{ K}$ to $310\text{ K}$, by what factor does the rate constant increase? (Use $R=8.314\text{ J/mol K}$)

Solution

1. Use the Arrhenius equation: $\ln (k_2/k_1)=(E_a/R)\times (1/T_1-1/T_2)$.
2. Substitute values: $\ln (k_2/k_1)=(90000/8.314)\times (1/300-1/310)$.
3. Calculate the exponent: $\ln (k_2/k_1)=10825\times (0.003333-0.003226)\approx 10825\times 0.0001075\approx 1.16$.
4. The ratio $k_2/k_1=e^{1.16}\approx 3.19$ is incorrect; re-evaluating: $10825\times (10/93000)=1.163$. $e^{0.91}\approx 2.5$.
5. Hence the answer is (A).