Question

Two reactions $R_1$ and $R_2$ have identical pre-exponential factors. Activation energy of $R_1$ exceeds that of $R_2$ by $10 \, kJ \, mol^{-1}$. If $k_1$ and $k_2$ are rate constants for reactions $R_1$ and $R_2$ respectively at $300\, K$, then ln $(k_2/k_1)$ is equal to : $(R=8.314 \, J \, mol^{-1} K^{-1})$

• A. 6
• B. 4
• C. 8
• D. 12

Answer 1

Answer: (B)

$k_{1}=Ae^{-E_{a_1}/RT}$

$k_{2}=Ae^{-E_{a_2}/RT}$

$\frac{k_{2}}{k_{1}}=e^{\frac{1}{RT}\left(E_{a_1-E_{a_2}}\right)}$

In$\frac{k_{2}}{k_{1}}= \frac{E_{a_1}-E_{a_2}}{RT}$

$=\frac{10 \times 10^{3}}{8.314 \times 300} \approx 4$