Question

Which of the following expression is correct

• A. \(\Delta G = -RT \ln K\)
• B. \(\Delta G = \dfrac{1}{RT^2\ln K}\)
• C. \(\Delta G^\circ = -RT \ln K\)
• D. \(\Delta G^\circ = -\dfrac{1}{RT^2\ln K}\)

Hint:

Always remember the important equilibrium relation:
\[ \Delta G^\circ = -RT\ln K \] A larger equilibrium constant implies more negative \(\Delta G^\circ\), indicating greater spontaneity.

Answer 1

The Correct Option is C

Step 1: Recall the relationship between Gibbs free energy and equilibrium constant.
In chemical thermodynamics, the standard Gibbs free energy change \(\Delta G^\circ\) is related to the equilibrium constant \(K\) by the equation:
\[ \Delta G^\circ = -RT \ln K \] where,
\[ R = \text{Universal gas constant} \] \[ T = \text{Absolute temperature in Kelvin} \] \[ K = \text{Equilibrium constant} \] This equation is one of the most important thermodynamic relations for chemical equilibrium.

Step 2: Understand the meaning of the equation.
The equation shows the connection between spontaneity and equilibrium.
If:
\[ K\gt 1 \] then:
\[ \ln K \gt 0 \] and therefore:
\[ \Delta G^\circ \lt 0 \] which indicates a spontaneous forward reaction.
If:
\[ K\lt 1 \] then:
\[ \ln K \lt 0 \] and thus:
\[ \Delta G^\circ \gt 0 \] which indicates the reaction is non-spontaneous in the forward direction.

Step 3: Analyze each option carefully.
\[ \textbf{Option (1): } \Delta G=-RT\ln K \] This is incorrect because the correct relation involves standard Gibbs free energy change \(\Delta G^\circ\), not \(\Delta G\).
\[ \textbf{Option (2): } \Delta G=\dfrac{1}{RT^2\ln K} \] This expression is dimensionally incorrect and does not represent any standard thermodynamic relation.
\[ \textbf{Option (3): } \Delta G^\circ=-RT\ln K \] This is the correct thermodynamic equation relating equilibrium constant and standard Gibbs free energy change.
\[ \textbf{Option (4): } \Delta G^\circ=-\dfrac{1}{RT^2\ln K} \] This expression is also incorrect and has no thermodynamic significance.

Step 4: Select the correct option.
Hence, the correct expression is:
\[ \Delta G^\circ = -RT\ln K \] which corresponds to option (3).

Step 5: Final conclusion.
Therefore, the correct answer is:
\[ \boxed{\Delta G^\circ = -RT\ln K} \]