Question

For the reaction X(g) $\rightarrow$ Y(g) at equilibrium, the partial pressure of Y is one-third that of X. What is the standard Gibbs energy change ($\Delta G^\circ$) for this reaction?

• A. $-RT \ln 3$
• B. $RT \ln 3$
• C. $RT \log 3$
• D. $-RT \log 3$

Hint:

N/A

Answer 1

The Correct Option is B

The reaction is X(g) → Y(g). At equilibrium, \(K_p = \frac{P_Y}{P_X}\).
Given \(P_Y = \frac{1}{3} P_X\), so \(K_p = \frac{P_Y}{P_X} = \frac{1}{3}\).
The standard Gibbs energy change is given by \(\Delta G^\circ = -RT \ln K_p\).
Substitute \(K_p\): \(\Delta G^\circ = -RT \ln \left(\frac{1}{3}\right) = -RT (-\ln 3) = RT \ln 3\).