Question

The molar entropy of which of the following reaction is negative?

• A. C(gr) + O$_2MATH_5f6b369e40b64d9d83ba106f80ba4273_{(g)}$
• B. C$_6$H$_6MATH_b49d39854f1b4bbda05d6d0de5746b7c_{(g)} \to$ 6CO$_2$$_{(g)}$ + 3H$_2$O\(_{(l)}\)
• C. CaCO$_3MATH_e2e93751f70841eba7deb7efd88634e9_{(g)}$
• D. PCl$_5MATH_9977ea4cae5f4df293ead7d4d825bf09_{(g)}$ + Cl$_2$$_{(g)}$

Hint:

Quick check for \(\Delta S\): Count only gas molecules.
Reactant side: 7.5 gas. Product side: 6 gas.
Fewer gas molecules \(\to\) Less disorder \(\to\) Negative \(\Delta S\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Entropy change (\(\Delta S\)) represents the change in randomness or disorder. For reactions involving gases, the sign of \(\Delta S\) is primarily determined by the change in the number of moles of gas (\(\Delta n_g\)).
- If \(\Delta n_g < 0\), the reaction likely has \(\Delta S < 0\) (negative).
- If \(\Delta n_g > 0\), the reaction likely has \(\Delta S > 0\) (positive).

Step 2: Detailed Explanation:
1. Option A: \(C(s) + O_2(g) \to CO_2(g)\). \(\Delta n_g = 1 - 1 = 0\). Entropy change is near zero or slightly positive.
2. Option B: \(C_6H_6(l) + 7.5O_2(g) \to 6CO_2(g) + 3H_2O(l)\).
\(\Delta n_g = 6 - 7.5 = -1.5\).
Since the number of moles of gas decreases, the system becomes more ordered. Thus, \(\Delta S\) is negative.
3. Options C, D, E: All show an increase in the number of moles of gas (\(\Delta n_g > 0\)), indicating a positive entropy change.

Step 3: Final Answer:
The reaction in Option B has a negative molar entropy.