Giveni 2Fe2O3s- 4Fes-3O2g- -rGo-1487-0 kJ mol-1ii 2COg-O2g- 2CO2g- -rGo-514-4kJ mol-1Free energy change- -rGo for the reaction- 2Fe2O3s-6COg- 4Fes-6CO2g will be-

The correct option is B 56.2 kJ mol^ 1 (i) 2Fe_2O_3(s)→ 4Fe(s)+3O_2(g); Δ_rG^o=+1487.0 kJ mol^ 1 (ii) 2CO(g)+O_2(g)→ 2CO_2(g); ...

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Byju's Answer

Standard XII

Chemistry

Third Law of Thermodynamics

Giveni 2Fe2...

Question

Given
(i) $2 F e_{2} O_{3} (s) \to 4 F e (s) + 3 O_{2} (g);$
$Δ_{r} G^{o} = + 1487.0$ kJ $m o l^{- 1}$
(ii) $2 C O (g) + O_{2} (g) \to 2 C O_{2} (g);$
$Δ_{r} G^{o} = - 514.4$ kJ $m o l^{- 1}$
Free energy change, $Δ_{r} G^{o}$ for the reaction,
$2 F e_{2} O_{3} (s) + 6 C O (g) \to 4 F e (s) + 6 C O_{2} (g)$ will be?

- 112.4

m o l^{- 1}

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- 56.2

m o l^{- 1}

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- 208.0

m o l^{- 1}

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- 168.2

m o l^{- 1}

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Solution

The correct option is B $- 56.2$ kJ $m o l^{- 1}$
(i) $2 F e_{2} O_{3} (s) \to 4 F e (s) + 3 O_{2} (g);$ $Δ_{r} G^{o} = + 1487.0$ kJ $m o l^{- 1}$
(ii) $2 C O (g) + O_{2} (g) \to 2 C O_{2} (g);$ $Δ_{r} G^{o} = - 514.4$ kJ $m o l^{- 1}$
Multiply above reaction with 3
(iii) $6 C O (g) + 3 O_{2} (g) \to 6 C O_{2} (g);$ $Δ_{r} G^{o} = 3 \times - 514.4 = - 1543.2$ kJ $m o l^{- 1}$
When we add reaction (i) and reaction (iii), we get reaction (iv)
(iv) $2 F e_{2} O_{3} (s) + 6 C O (g) \to 4 F e (s) + 6 C O_{2} (g)$
Free energy change, $Δ_{r} G^{o}$ for the reaction will be,
$1487.0 - 1543.2 = - 56.2$ kJ $m o l^{- 1}$

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Given(i) 2Fe2O3(s)→4Fe(s)+3O2(g);ΔrGo=+1487.0 kJ mol−1(ii) 2CO(g)+O2(g)→2CO2(g);ΔrGo=−514.4kJ mol−1Free energy change, ΔrGo for the reaction,2Fe2O3(s)+6CO(g)→4Fe(s)+6CO2(g) will be?