Question

The energy change accompanying the equilibrium reaction $A\rightleftharpoons B$ is $-33.0kJmo{l}^{-1}$

Assuming that pre-exponential factor is same for forward and backward reaction answer the following
The equilibrium constant $k$ for the reaction at $300K$

• A. $5.55\times 10^5$
• B. $5.67\times 10^3$
• C. $5.55\times 10^6$
• D. $5.67\times 10^2$

Answer 1

Answer: (A)

$k_1=A{e}^{-E_f/RT};k_b=A{e}^{-E_b/RT}$
$\therefore k=\frac{k_f}{k_b}=\frac{A{e}^{-E_f/RT}}{A{e}^{-E_b/RT}}=e^{\left(E_b-E_f\right)RT}$
or $\log k=\frac{E_b-E_f}{2.303RT}$
$=\frac{33000Jmo{l}^{-1}}{2.303\times 8.314J{K}^{-1}mo{l}^{-1}\times 300K}=5.745$
$\therefore k=$ Antilog $(5.745)=5.55\times 10^5$