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 : $(R = 8.314 \;J mol^{–1} K ^{–1} )$

Question

( A ) 8
( B ) 6
( C ) 4
( D ) 12