Question

If $\Delta$G = – 177 K cal for (1) 2 Fe(s) +$\frac{3}{2}$O2 (g) $\longrightarrow$ Fe2O3 (s)

and $\Delta$G = – 19 K cal for (2) 4 Fe2O3 (s) + Fe(s) $\longrightarrow$ 3 Fe3O4 (s)

What is the Gibbs free energy of formation of Fe3O4(s)?

• A.
  • 229.6 kcal/mol
• B. – 242.3 kcal/mol
• C. – 727 kcal/mol
• D. – 229.6 kcal/mol

Answer 1

Answer: (B)

– 242.3 kcal/mol

Answer 2

$\Delta$G for 3Fe(s) + 2O2 (g) $\longrightarrow$ Fe3O4 (s) can be obtained by taking

[(2) + 4 × (1)] × $\frac{1}{3}$

Hence, we get $\Delta$Gf = [– 19 + 4 × (– 177)] × $\frac{1}{3}$ = – 242.3 k cal for 1 mole Fe3O4