The entropy $(S^{\circ})$ of the following substances are:
$\begin{array}{l} {CH}_{4} (g) 186.2 {JK}^{- 1} {mol}^{- 1} {CO}_{2} (g) 213.6 {JK}^{- 1} {mol}^{- 1} \\ O_{2} (g) 205.0 {JK}^{- 1} {mol}^{- 1} H_{2} O (l) 69.9 {JK}^{- 1} {mol}^{- 1} \end{array}$
The entropy change $(Δ S^{\circ})$ for the reaction ${CH}_{4} (g) + 2 O_{2} (g) ⟶ {CO}_{2} (g) + 2 H_{2} O (l)$ is

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$- 242.8 {JK}^{- 1} {mol}^{- 1}$

$- 108.1 {JK}^{- 1} {mol}^{- 1}$

$- 312.5 {JK}^{- 1} {mol}^{- 1}$

$- 37.6 {JK}^{- 1} {mol}^{- 1}$

answer is B.

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Detailed Solution

We have ${CH}_{4} (g) + 2 O_{2} (g) ⟶ {CO}_{2} (g) + 2 H_{2} O (l)$

$\begin{array}{l} Δ S^{\circ} = S^{\circ} ({CO}_{2}, g) + 25^{\circ} (H_{2} O, l) - S^{\circ} ({CH}_{4}, g) - 2 S^{\circ} (O_{2}, g) \\ = (213.6 + 2 \times 69.9 - 186.2 - 2 \times 205.0) {JK}^{- 1} {mol}^{- 1} \\ = - 242.8 {JK}^{- 1} {mol}^{- 1} \end{array}$