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Q.

If $ΔG = - 177 k$ cal for
(1) $2 Fe (s) + \frac{3}{2} O_{2} ⟶ {Fe}_{2} O_{3} (s)$ and $Δ G = - 19 k$ cal for
(2) $4 {Fe}_{2} O_{3} (s) + Fe (s) ⟶ 3 {Fe}_{3} O_{4} (s)$
What is the Gibbs free energy of formation of ${Fe}_{3} O_{4}$ ?

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a

$- 727 kcal / mol$

b

$-242.3 kcal/mol$

c

$- 229.6 kcal / mol$

d

$+ 229.6 kcal / mol$

answer is B.

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Detailed Solution

Gibbs free energy of formation,

$Δ H_{f} = \frac{1}{3} [4 \times (- 177) - 19] = - 242.33 kcal / mol$

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