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

Solution

1. Use the Arrhenius equation: $\ln \frac{k_2}{k_1}=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$.
2. Substitute the values: $\ln \frac{k_2}{k_1}=\frac{90000}{8.314}\left(\frac{1}{300}-\frac{1}{310}\right)$.
3. Calculate the term in brackets: $\frac{10}{300\times 310}\approx 1.075\times 10^{-4}$.
4. Compute the exponent: $\ln \frac{k_2}{k_1}\approx \frac{90000\times 1.075\times 10^{-4}}{8.314}\approx 0.916$.
5. Find the ratio: $\frac{k_2}{k_1}=e^{0.916}\approx 2.5$.
6. Hence the answer is (B).