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. 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)

$5.55\times {10}^5$

$k_f=A{e}^{-E_f/RT};k_b=A{e}^{-E_b/RT}$

$\therefore K=\frac{k_f}{k_b}=\frac{A{e}^{-E_f/RT}}{A{e}^{-E_b/RT}}=e^{(E_b-E_f)/RT}$

or $logK=\frac{E_b-E_f}{2.303RT}$

$logK=\frac{33000Jmo{l}^{-1}}{2.303\times 8.314J{K}^{-1}mo{l}^{-1}\times 300K}=5.745$

$\therefore K=antilog\left(5.745\right)=5.55\times {10}^5$

Hence, option A is correct.