If - G-177Kcal for 1 2Fe s - 3 - 2 O 2 g - Fe 2 O 3 s and - G-19Kcal for 2 4Fe 2 O 3 s - 3Fe 3 O 4 s What is the Gibbs free energy of formation of Fe 3 O 4-

The correct option is A 242.3 kcal / mol According to question, 3×Δ H_f 4× 177= 19 Δ H_f = (19 708)/3 = 242kcal/mol ...

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Nernst Equation

If Δ G=-177...

Question

If $Δ G = - 177 K c a l$ for $(1) 2 F e (s) + \frac{3}{2} O_{2} (g) ⟶ {F e}_{2} O_{3} (s)$
and $Δ G = - 19 K c a l f o r (2) {4 F e}_{2} O_{3} (s) ⟶ {3 F e}_{3} O_{4} (s)$
What is the Gibbs free energy of formation of ${F e}_{3} O_{4}$ ?

+ 229.6 \frac{k c a l}{m o l}

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- 242.3 \frac{k c a l}{m o l}

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- 727 \frac{k c a l}{m o l}

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- 229.6 \frac{k c a l}{m o l}

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Similar questions

$Δ G = - 177 K c a l$ for $2 F e (s) + \frac{3}{2} O_{2} (g) \to F e_{2} O_{3} (s)$ and

What is the Gibbs free energy of formation of $F e_{3} O_{4}$ ?;

Δ G = - 177 k cal

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

Δ G = - 19 k cal

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

F e_{3} O_{4} (s)

Δ G = - 177 K c a l

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

Δ G = - 19 K

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

F e_{3} O_{4}

25^{o} C

{F e}_{2} O_{3} (s) + 3 C_{g r a p h i t e} ⟶ 2 F e (s) + 3 C O (g); Δ H^{o} = 492 k J / m o l

F e O (s) + C_{g r a p h i t e} ⟶ F e (s) + C O (g); Δ H^{o} = 155.8 k J / m o l

C_{g r a p h i t e} + O_{2} (g) ⟶ C O_{2} (g); Δ H^{o} = - 393.51 k J / m o l

C O (g) + \frac{1}{2} O_{2} (g) ⟶ C O_{2} (g); Δ H^{o} = - 282.98 k J / m o l

F e O (s)

{F e}_{2} O_{3} (s)

F e O (s) + {F e}_{2} O_{3} (s) ⟶ {F e}_{3} O_{4} (s)

2 F e (s) + O_{2} (g) ⟶ 2 F e O (s); Δ H^{o} = - 544 k J

4 F e (s) + 3 O_{2} (g) ⟶ 2 {F e}_{2} O_{; 3} (s); Δ H^{o} = - 1648.4 k J

{F e}_{3} O_{4} (s) ⟶ 3 F e (s) + 2 O_{2} (g); Δ H^{o} = + 1118.4 k J

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