Two reactions ${\mathrm{A}}_1$ and ${\mathrm{A}}_2$ have identical pre-exponential factors. The activation energy of ${\mathrm{A}}_1$ is more than ${\mathrm{A}}_2$ by $10 \mathrm{kJ} {\mathrm{mol}}^{-1}$ . If ${\mathrm{k}}_1$ and ${\mathrm{k}}_2$ are the rate constants for reactions ${\mathrm{A}}_1$ and ${\mathrm{A}}_2$, respectively at $300 \mathrm{K}$, then $\ln \left(\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}\right)$ is equal to

$\left(\mathrm{R}=8.314\mathrm{ }\mathrm{J}\mathrm{ }{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}\right)$

• A

  $12$
• B

  $6$
• Correct answer
  C

  $4$
• D

  $8$

Hence, the correct answer is option: 'C'.