Calculate standard entropy change in the reaction
${Fe}_{2} O_{3} (s) + 3 H_{2} (g) ⟶ 2 Fe (s) + 3 H_{2} O (l)$
Given : $S_{m}^{\circ} ({Fe}_{2} O_{3}, s) = 87.4, S_{m}^{\circ} (Fe, s) = 27.3$
$S_{m}^{\circ} (H_{2}, g) = 130.7, S_{m}^{\circ} (H_{2} O, l) = 69.9 {JK}^{- 1} {mol}^{- 1}$

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$- 215.2 {JK}^{- 1} {mol}^{- 1}$

$- 212.5 {JK}^{- 1} {mol}^{- 1}$

$- 120.9 {JK}^{- 1} {mol}^{- 1}$

None of these

answer is B.

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Detailed Solution

Given reaction,

${Fe}_{2} O_{3} (s) + 3 H_{2} (g) ⟶ 2 Fe (s) + 3 H_{2} O (l)$

The standard entropy change in the reaction is found as,
$\begin{array}{l} Δ S_{R x^{n}}^{\circ} = 2 \times 27.3 + 3 \times 69.9 - 87.4 - 3 \times 130.7 \\ = - 215.2 {JK}^{- 1} {mol}^{- 1} \end{array}$

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