JEE Main 2017 · 2 Apr · Shift 1, Chemistry question 39

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$^{-1}$)