- G -177 Kcal for 2 Fe s -3-2 O 2 g - F e 2 O 3 s andWhat is the Gibbs free energy of formation of F e 3 O 4 -A- -229-6 kcal - molB- -242-4 kcal - molC- -727 kcal - molD- -229-6 kcal - mol

The correct option is B (i) 2Fe(s) + 3/2O_2(g) → Fe_2O_3(s) Δ G = 177 kcal (ii) 4Fe_2O_3(s) + Fe(s) → 3Fe_3O_4(s) Δ G = 19 kcal Δ G_r = 3Δ G_F(F_3O_4) 4Δ_ ...

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Byju's Answer

Standard XII

Chemistry

Nitrogen

Δ G =-177 Kca...

Question

$Δ G = - 177 K c a l$ 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 $F e_{3} O_{4}$ ?;

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Solution

The correct option is B

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

$Δ G = - 177 k c a l$

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

$Δ G = - 19 k c a l$

$Δ G_{r} = 3 Δ G_{F} (F_{3} O_{4}) - 4 Δ_{F} (F e_{2} O_{3}) - Δ G_{F} (F e_{2} S)$

$3 D e l t a_{F} (F e_{3} O_{4}, S) = - 19 - 4 \times 177$ .

$Δ G_{F} (F e_{3} O_{4}, S) = \frac{777}{3} = - 242.4 k c a l / m o l e$

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Similar questions

Q. If

Δ G = - 177 k cal

for
(1)

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

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Δ G = - 19 k cal

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

What is the Gibbs free energy of formation of

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

Q. If

Δ G = - 177 K c a l

for
(1)

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

and

Δ G = - 19 K

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

What is the Gibbs free energy of formation of

F e_{3} O_{4}

Q. If

Δ G = - 177

Kcal for

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

and

Δ G = - 19 K c a l

for

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

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

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Δ G = - 177

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

and

Δ G = - 19

K cal for

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

What is the Gibbs free energy of formation of

F e_{3} O_{4}

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Δ G = - 177 k cal

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(1)

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

and

Δ G = - 19 k cal

for
(2)

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

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ΔG = −177Kcal for 2Fe(s) + 32O2(g) → Fe2O3(s) and What is the Gibbs free energy of formation of Fe3O4 ?;

The correct option is B (i) 2Fe(s) + 32O2(g) → Fe2O3(s) ΔG = −177 kcal (ii) 4Fe2O3(s) + Fe(s) → 3Fe3O4(s) ΔG = −19kcal ΔGr = 3ΔGF(F3O4) − 4ΔF(Fe2O3) − ΔGF(Fe2S) 3DeltaF(Fe3O4,S) = −19 − 4 × 177. ΔGF(Fe3O4,S) = 7773 = −242.4kcal/mole

$Δ G = - 177 K c a l$ 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 $F e_{3} O_{4}$ ?;