Calculate standard entropy change in the reaction Fe2O3 s - 3H2 g - 2Fe s - 3H2OlGiven - S0m Fe2 O3- S - 87-4- S0m Fe- S - 27-3- S0m H2- g - 130-7- S0m H2 O- l - 69-9 JK-1 mol-1

Fe _ 2 O _ 3 (s)+3 H _ 2 ⟶ 2 Fe _ (s) +3 H _ 2 O(l) Δ S^ o #160; _ reaction =ΣΔ S ^ o #160; (products) ΣΔ S ^ o #160; (reactants) = 2 S _ ...

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Byju's Answer

Standard XII

Chemistry

Entropy

Calculate sta...

Question

Calculate standard entropy change in the reaction
$F e_{2} O_{3} (s) + 3 H_{2} (g) \to 2 F e (s) + 3 H_{2} O (l)$
Given : $S_{m}^{0} (F e_{2} O_{3}, S) = 87.4, S_{m}^{0} (F e, S) = 27.3, S_{m}^{0} (H_{2}, g) = 130.7, S_{m}^{0} (H_{2} O, l) = 69.9 J K^{- 1} m o l^{- 1}$

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Solution

${F e}_{2} O_{3} (s) + 3 H_{2} ⟶ 2 {F e}_{(s)} + 3 H_{2} O (l)$
$Δ S_{r e a c t i o n}^{o} = Σ Δ S^{o} (p r o d u c t s) - Σ Δ S^{o} (r e a c t a n t s)$
= $2 S_{m}^{o} (F e, S) + 3 S_{m}^{o} (H_{2} O, l) - [S_{m}^{o} ({F e}_{2} O_{3}, S) + 3 S_{m}^{o} (H_{2}, g)]$
= $(2 \times 27.3) + (3 \times 69.9) - 87.4 - (3 \times 130.7)$
$Δ S_{r e a c t i o n}^{o}$ =- $215.2 J k^{- 1}$

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Calucalculate standard entropy change in the reaction

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