Question

The energy change accompanying the equilibrium reaction $A \rightleftharpoons B$ is $-33.0 \,kJ\, mol ^{-1}$

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

• 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)

$k_1=A e^{-E_f / R T} ; k_b=A e^{-E_b / R T}$
$\therefore k=\frac{k_{ f }}{k_{ b }}=\frac{A e^{-E_{ f } / R T}}{A e^{-E_b / R T}}=e^{\left(E_{ b }-E_{ f }\right) R T}$
or $\log k=\frac{E_{ b }-E_{ f }}{2.303 R T}$
$=\frac{33000 \,J \,mol ^{-1}}{2.303 \times 8.314 \,J \,K ^{-1} mol ^{-1} \times 300 \,K }=5.745$
$\therefore k=$ Antilog $(5.745)=5.55 \times 10^5$