Question

The standard enthalpies of formation of $FeO$ and $Fe _{2} O _{3}$ are $-65\, kcal\, mol ^{-1}$ and $-197 \,kcal\, mol ^{-1}$ respectively. A mixture of two oxides contains $FeO$ and $Fe _{2} O _{3}$ in the mole ratio $2 : 1$. If, by oxidation, the mixture is changed into $1: 2$ mole ratio, how much thermal energy (in kcal) is released per mole from the initial mixture?

Answer 1

$Fe +\frac{1}{2} O _{2} \longrightarrow FeO , \Delta H =-65\, kcal$...(1)
$2 Fe +\frac{3}{2} O _{2} \longrightarrow Fe _{2} O _{3}, \Delta H =-197 \,kcal$ ...(2)
$\therefore $ equation $(2)-2 \times$ eq $(1)$ we get
$2 FeO +\frac{1}{2} O _{2} \longrightarrow Fe _{2} O _{3} \Delta H =-67 \,kcal$...(3)
Let initially the total moles of $FeO$ and $Fe _{2} O _{3}$ be ' $a$ '
Then $FeO =\frac{2 a}{3}$ and $Fe _{2} O _{3}=\frac{a}{3}$
Let ' $b$ ' mole of $FeO$ are converted in to $Fe _{2} O _{3}$,
then mole of $FeO$ left $=\frac{2 a}{3}-b$,
mole of $Fe _{2} O _{3}$ formed $=\frac{b}{2}$
$\therefore Fe _{2} O _{3}$ total $=\frac{a}{3}+\frac{b}{2}$
given $\frac{ FeO }{ Fe _{2} O _{3}}$ after oxidation $=\frac{1}{2}$
$ \frac{2 a}{\frac{3}{a}-b}=\frac{1}{2}$ or $a=\frac{5 b}{2}, b=\frac{2}{3} a$
$\therefore 2$ mole of $FeO$ released heat $=-67 \,kcal$
$\therefore $ heat produced by ' $b$ '
mole of $FeO =\frac{-67 \times b}{2}$
$=\frac{-67 \times 2 \times a}{2 \times 5}=-13.4 a$
$\therefore $ Heat liberated per mole of mixt $=\frac{-13.4 a}{a}$
$=-13.4 \,kcal / mol .$