Question

At 298 K, the entropy of rhombic sulphur is 32.04 J mol\(^{-1}\) K\(^{-1}\) and that of monoclinic sulphur is 32.68 J mol\(^{-1}\) K\(^{-1}\). The heats of combustion are respectively \(-298246\) and \(-297948\) J mol\(^{-1}\). \( \Delta G \) for the reaction: S\(_{\text{rhombic}}\) \( \rightarrow \) S\(_{\text{monoclinic}}\) will be

• A. 107.28 J
• B. 10.728 J
• C. 107.28 kJ
• D. 10728 J

Hint:

Always keep units consistent while using $\Delta G = \Delta H - T\Delta S$.

Answer 1

The Correct Option is A

Concept: \[ \Delta G = \Delta H - T\Delta S \]

Step 1: Calculate entropy change:
\[ \Delta S = 32.68 - 32.04 = 0.64 \, \text{J mol}^{-1}\text{K}^{-1} \]

Step 2: Calculate enthalpy change using heats of combustion:
\[ \Delta H = (-297948) - (-298246) = 298 \, \text{J mol}^{-1} \]

Step 3: Substitute in Gibbs equation at $T = 298K$:
\[ \Delta G = 298 - (298 \times 0.64) \] \[ = 298 - 190.72 = 107.28 \, \text{J} \] Conclusion:
$\Delta G = 107.28$ J