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 $\Delta G^{\circ }$ of the reaction A $\to$ B is:

• A. $RTln4$
• B. $-RTln4$
• C. $RT\text{log}4$
• D. $-RT\text{log}4$

Answer 1

The correct option is A $RTln4$

$A\left(g\right)\rightleftharpoons B\left(g\right)$
According to the question,
$P_B=\frac{1}{4}{P}_A$
$P_A=4P_B$

Equilibirum constant,
$K_p=\frac{P_B}{P_A}=\frac{\frac{P_A}{4}}{P_A}=\frac{1}{4}$

We know that, $\Delta G=-RTlnK$
Hence, $\Delta G=-RTln\frac{1}{4}=RTln4$