Question

If $\Delta H^\circ$ and $\Delta S^\circ$ for the reaction $N_2O_4(g) \rightarrow 2NO_2(g)$ is $57.24\ \text{kJ}$ and $175.8\ \text{J K}^{-1}\text{mol}^{-1}$ respectively, what is the value of $\Delta G^\circ$ for this reaction at $298\ \text{K}$?

• A. $57.24\ \text{kJ}$
• B. $-17.58\ \text{kJ}$
• C. $-4.85\ \text{kJ}$
• D. $4.85\ \text{kJ}$

Hint:

Always convert entropy into kJ before substituting values in the Gibbs free energy equation to avoid calculation errors.

Answer 1

The Correct Option is D

Step 1: Write the Gibbs free energy relation.
The Gibbs free energy change is calculated using the formula:
\[ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \]
Step 2: Convert entropy into consistent units.
Given $\Delta S^\circ = 175.8\ \text{J K}^{-1}\text{mol}^{-1}$
Convert into kJ:
\[ \Delta S^\circ = 0.1758\ \text{kJ K}^{-1}\text{mol}^{-1} \]
Step 3: Substitute the values.
\[ \Delta G^\circ = 57.24 - (298 \times 0.1758) \]
\[ \Delta G^\circ = 57.24 - 52.39 \]
Step 4: Final calculation.
\[ \Delta G^\circ = 4.85\ \text{kJ} \]
Step 5: Conclusion.
The Gibbs free energy change for the reaction at $298\ \text{K}$ is $4.85\ \text{kJ}$.