Question

For a reaction $A_{(g)} \rightleftharpoons B_{(g)}$ at equilibrium, the partial pressure of $B$ is found to be one fourth of the partial pressure of $A$. The value of $\Delta G^{\circ}$ for the reaction $A \rightarrow B$ is

• A. $R T \ln 4$
• B. $-R T \ln 4$
• C. $R T \log 4$
• D. $-R T \log 4$

Answer 1

Answer: (A)

Equilibrium constant, $K=\frac{p_{B}}{p_{A}}=\frac{1}{4}$

$\Delta G^{\circ}=-R T \ln K=-R T \ln \frac{1}{4}=R T \ln 4$