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?
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?
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Updated On: Jun 3, 2025
- $-RT \ln 3$
- $RT \ln 3$
- $RT \log 3$
- $-RT \log 3$
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The Correct Option is B
Solution and Explanation
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 \(K_p\): \(\Delta G^\circ = -RT \ln \left(\frac{1}{3}\right) = -RT (-\ln 3) = RT \ln 3\).
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 \(K_p\): \(\Delta G^\circ = -RT \ln \left(\frac{1}{3}\right) = -RT (-\ln 3) = RT \ln 3\).
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