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 link between free energy and equilibrium.
Thermodynamics connects the standard Gibbs free energy change to the equilibrium constant. We must pick the correct expression among the four options.
Step 2: Write the standard relation.
The correct and standard relation is \[ \Delta G^\circ = -RT \ln K \] where $R$ is the gas constant, $T$ is the absolute temperature, and $K$ is the equilibrium constant.
Step 3: Note the importance of the standard symbol.
The relation uses $\Delta G^\circ$, the standard free energy change, not the general $\Delta G$. So any option using plain $\Delta G$ with $-RT \ln K$ is not the standard form.
Step 4: Check the logarithm placement.
The natural logarithm of $K$ appears linearly, multiplied by $-RT$. Options that place $\ln K$ in a denominator or use $\tfrac{1}{RT^2 \ln K}$ are dimensionally and physically wrong.
Step 5: Eliminate the incorrect options.
The forms $\Delta G = \tfrac{1}{RT^2 \ln K}$ and $\Delta G^\circ = -\tfrac{1}{RT^2 \ln K}$ are incorrect, and $\Delta G = -RT \ln K$ uses the wrong symbol.
Step 6: State the answer.
The correct expression is $\Delta G^\circ = -RT \ln K$, matching the key.
\[ \boxed{\Delta G^\circ = -RT \ln K} \]