Question

The standard enthalpies of formation of $FeO$ and $F{e}_2{O}_3$ are $-65kcalmo{l}^{-1}$ and $-197kcalmo{l}^{-1}$ respectively. A mixture of two oxides contains $FeO$ and $F{e}_2{O}_3$ 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

Answer: -13.4

$Fe+\frac{1}{2}{O}_2⟶FeO,\Delta H=-65kcal$...(1)
$2Fe+\frac{3}{2}{O}_2⟶F{e}_2{O}_3,\Delta H=-197kcal$ ...(2)
$\therefore$ equation $(2)-2\times$ eq $(1)$ we get
$2FeO+\frac{1}{2}{O}_2⟶F{e}_2{O}_3\Delta H=-67kcal$...(3)
Let initially the total moles of $FeO$ and $F{e}_2{O}_3$ be ' $a$ '
Then $FeO=\frac{2a}{3}$ and $F{e}_2{O}_3=\frac{a}{3}$
Let ' $b$ ' mole of $FeO$ are converted in to $F{e}_2{O}_3$,
then mole of $FeO$ left $=\frac{2a}{3}-b$,
mole of $F{e}_2{O}_3$ formed $=\frac{b}{2}$
$\therefore F{e}_2{O}_3$ total $=\frac{a}{3}+\frac{b}{2}$
given $\frac{FeO}{F{e}_2{O}_3}$ after oxidation $=\frac{1}{2}$
$\frac{2a}{\frac{3}{a}-b}=\frac{1}{2}$ or $a=\frac{5b}{2},b=\frac{2}{3}a$
$\therefore 2$ mole of $FeO$ released heat $=-67kcal$
$\therefore$ heat produced by ' $b$ '
mole of $FeO=\frac{-67\times b}{2}$
$=\frac{-67\times 2\times a}{2\times 5}=-13.4a$
$\therefore$ Heat liberated per mole of mixt $=\frac{-13.4a}{a}$
$=-13.4kcal/mol.$