Question

For a reaction $A_{\left(g\right)}\rightleftharpoons B_{\left(g\right)}$ at equilibrium the partial pressure of $B$ is found to be one-fourth of the partial pressure of $A$. The value of $\Delta G^0$ of the reaction $A\to B$ is:

• A. $+RT$ $ln\left(4\right)$
• B. $-RTlog\frac{1}{4}$
  
• C. $+RTlog4$
• D. $-RTlog4$

Answer 1

The correct option is B $-RTlog\frac{1}{4}$

$A_{\left(g\right)}\rightleftharpoons B_{\left(g\right)}$

Formula for standard Gibbs free energy-

$\Delta G^o=-RT\times lnK$

For $K=\frac{P_B}{P_A}=\frac{P}{4P}=\frac{1}{4}$

$\Delta G^o=-R\times T\times ln\frac{1}{4}$