Question

If $\Delta G=-177\text{k cal}$ for
(1) $2Fe\left(s\right)+\frac{3}{2}{O}_2\left(g\right)\to F{e}_2{O}_3\left(s\right)$
and $\Delta G=-19\text{k cal}$ for
(2) $4F{e}_2{O}_3\left(s\right)+Fe\left(s\right)\to 3F{e}_3{O}_4\left(s\right)$
What is the Gibbs free energy of formation of $F{e}_3{O}_4\left(s\right)$ ?

• A. $+229.6\text{kcal/mol}$
• B. $-242.3\text{kcal/mol}$
• C. $-727\text{kcal/mol}$
• D. $-229.6\text{kcal/mol}$

Answer 1

The correct option is B $-242.3\text{kcal/mol}$

$\left(1\right)2Fe\left(s\right)+\frac{3}{2}{O}_2\left(g\right)\to F{e}_2{O}_3\left(s\right)\Delta G=-177\text{k cal}$

$\left(2\right)4F{e}_2{O}_3\left(s\right)+Fe\left(s\right)\to 3F{e}_3{O}_4\left(s\right)\Delta G=-19\text{k cal}$

$\Delta G$ for $3Fe\left(s\right)+2O_2\left(g\right)\to F{e}_3{O}_4\left(s\right)$ can be obtained by taking
$\left[\right(2)+4\times (1\left)\right]\times \frac{1}{3}$
Hence, we get $\Delta G_f=[19+4\times (-177\left)\right]\times \frac{1}{3}=-242.3\text{k cal}$ for 1 mole $F{e}_3{O}_4$