Given 2Fe2O3s - 4Fes -3O2g- G-1487 KJ mol-1 2COg-O2g - 2CO2g- G-514-4 KJ mol-1Free energy change - Go for the reaction 2Fe2O3s - 6COg - 4Fes - 6CO2g will be

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Gibbs Free Energy & Spontaneity

Given 2Fe2O3...

Question

Given
$2 F e_{2} O_{3} (s) \to 4 F e (s) + 3 O_{2} (g); Δ G^{\circ} = - 1487 K J m o l^{- 1}$

$2 C O (g) + O_{2} (g) \to 2 C O_{2} (g); Δ G^{\circ} = - 514.4 K J m o l^{- 1}$

Free energy change $Δ G^{o}$ 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

- 168.2 k J m o l^{- 1}

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- 208.0 k J m o l^{- 1}

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- 56.2 k J m o l^{- 1}

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- 112.4 k J m o l^{- 1}

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Solution

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Similar questions

2 F e_{2} O_{3} (s) \to 4 F e (s) + 3 O_{2} (g);

Δ_{r} G^{o} = + 1487.0

m o l^{- 1}

2 C O (g) + O_{2} (g) \to 2 C O_{2} (g);

Δ_{r} G^{o} = - 514.4

m o l^{- 1}

Δ_{r} G^{o}

2 F e_{2} O_{3} (s) + 6 C O (g) \to 4 F e (s) + 6 C O_{2} (g)

F e_{2} O_{3} (s)

m o l^{- 1}

4 F e (s) + 3 O_{2} (g) \to 2 F e_{2} O_{3} (s)

2 {Fe}_{2} O_{3} (s) + 3 C (s) \to 4 Fe (s) + 3 {CO}_{2} (g)

1773 K

C (s) + O_{2} (g) ⟶ C O_{2} (g); Δ G^{o} = - 380 k J {m o l}^{- 1}

C (s) + O_{2} (g) ⇌ 2 C O (g); Δ G^{o} = - 500 k J {m o l}^{- 1}

{A l}_{2} O_{3}

P b O

4 A l + 3 O_{2} (g) ⟶ 2 {A l}_{2} O_{3} (s); Δ G^{o} = - 22500 k J {m o l}^{- 1}

2 P b + O_{2} (g) ⟶ 2 P b O (s); Δ G^{o} = - 120 k J {m o l}^{- 1}

{A g}_{2} O (s) ⇌ 2 A g (s) + \frac{1}{2} O_{2} (g)

Δ H, Δ S, Δ T

40.63 k J {m o l}^{- 1}, 1.808 J K^{- 1} {m o l}^{- 1}

373.4 K

Δ G

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