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 $ln(k2/k1)$ is equal to : $(R=8.314 \;J \;mol^{−1} K^{−1} )$

Question

$\begin{array}{1 1} (1)6 \\ (2) 4 \\ (3) 8 \\ (4) 12 \end{array}$