Question

Two reactions $R_1$ and $R_2$ have identical pre-exponential factors. Activation energy of $R_1$ exceeds that of $R_2$ by $20KJmo{l}^{-1}$ . If $K_1$ and $K_2$ are rate constants for reactions $R_1$ and $R_2$ respectively at $300K$ , then $ln\left(\frac{K_2}{K_1}\right)$ is equal to $(R=8.314J{\left(mol\right)}^{-1}{K}^{-1})$ :

• A. $12$
• B. $6$
• C. $4$
• D. $8$

Answer 1

Answer: (D)

$E{a}_1=E{a}_2+20$
$\ell n{K}_1=cnA-\frac{E_1}{R\times 300}$
$\ell n{K}_2=lnA-\frac{E_2}{R\times 300}$
eq.(2) $-$ eq.(1)
$\ell n\left(\frac{K_2}{K_1}\right)=\frac{E_1-E_2}{R\times 300}$
$\ell n\left(\frac{K_2}{K_1}\right)=\frac{20\times {\left(10\right)}^3}{8.314\times 300}\approx 8$