At 300 K, the entropy change for the reaction given below is2H2(g)+O2(g) → 2H2O(l)Standard entropies of H2(g), O2(g) and H2O(l) are 126.6, 201.20 and 68.0 J K-1 mol-1 respectively.

∆SReaction=∑Sproduct-∑Sreactant =2 × SH2O- 2 × SH2 + SO2 =2 × 68- 2 × 126.6 + 201.20=-318.4 JK-1 mol-1

At 300 K, the entropy change for the reaction given below is2H2(g)+O2(g) → 2H2O(l)Standard entropies of H2(g), O2(g) and H2O(l) are 126.6, 201.20 and 68.0 J K-1 mol-1 respectively.

∆SReaction=∑Sproduct-∑Sreactant =2 × SH2O- 2 × SH2 + SO2 =2 × 68- 2 × 126.6 + 201.20=-318.4 JK-1 mol-1

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At 300 K, the entropy change for the reaction given below is
$2 H_{2} (g) + O_{2} (g) \overset{}{\to} 2 H_{2} O (l)$
Standard entropies of $H_{2} (g), O_{2} (g) and H_{2} O (l)$ are 126.6, 201.20 and 68.0 $J K^{- 1} {mol}^{- 1}$ respectively.

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$- 318.4 J K^{- 1} {mol}^{- 1}$

$+ 318.4 J K^{- 1} {mol}^{- 1}$

$+ 31.84 J K^{- 1} {mol}^{- 1}$

$- 159.2 J K^{- 1} {mol}^{- 1}$

answer is A.

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$∆ S_{Reaction} = \sum S_{product} - \sum S_{reactant} = 2 \times S_{H_{2} O} - [2 \times S_{H_{2}} + S_{O_{2}}] = 2 \times 68 - [2 \times 126.6 + 201.20] = - 318.4 {JK}^{- 1} {mol}^{- 1}$

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At 300 K, the entropy change for the reaction given below is2H2(g)+O2(g) → 2H2O(l)Standard entropies of H2(g), O2(g) and H2O(l) are 126.6, 201.20 and 68.0 J K-1 mol-1 respectively.

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At 300 K, the entropy change for the reaction given below is
$2 H_{2} (g) + O_{2} (g) \overset{}{\to} 2 H_{2} O (l)$
Standard entropies of $H_{2} (g), O_{2} (g) and H_{2} O (l)$ are 126.6, 201.20 and 68.0 $J K^{- 1} {mol}^{- 1}$ respectively.