Question

The standard enthalpy of formation of $FeO$ and $F{e}_2{O}_3$ is $-65kcalmo{l}^{-1}$ and $-197kcalmo{l}^{-1}$ respectively. If a mixture containing $FeO$ and $Fe{O}_3$ in $2:1$ mole ratio on oxidation is changed into $1:2$ mole ratio, thermal energy in kcal released per $6$ mole of mixture will be (nearest integer)____________.

Answer 1

$Fe+\frac{1}{2}{O}_2⟶FeO;\Delta H=-65kcalmo{l}^{-1}$ . ..(i)
$2Fe+\frac{3}{2}{O}_2⟶F{e}_2{O}_3;\Delta H=-197kcalmo{l}^{-1}$ ...(ii)
By eq, $\left[\right(ii)-2\times (i\left)\right]$
Also, $2FeO+\frac{1}{2}{O}_2⟶F{e}_2{O}_3;\Delta H=-67kcalmo{l}^{-1}$ ...(iii)
Let $a$ mole of FeO and $b$ mole of $F{e}_2{O}_3$, are present such that
$a+b=1and\frac{a}{b}=2$
$a=\frac{2}{3}andb=\frac{1}{3}$
$2FeO+\frac{1}{2}{O}_2⟶F{e}_2{O}_3$
Initial $a$ $b$
After oxidation $(a-2a^{\prime })$ $b+a^{\prime }$
Given, $\frac{a}{b}=2and\frac{a-2a^{\prime }}{b+a^{\prime }}=\frac{1}{2}$
$\therefore \frac{a-2a^{\prime }}{\frac{a}{2}+a^{\prime }}=\frac{1}{2}$
$\therefore 2a-4a^{\prime }=\frac{a}{2}+a^{\prime }$
$\therefore a^{\prime }=\frac{3a}{10}=\frac{3\times 2}{10\times 3}=\frac{1}{5}$
$\therefore FeOused=2a^{\prime }=2/5mol$
$\because 2moleFeOgives67kcal$.
$\therefore \frac{2}{5}moleFeOgives\frac{67\times 2}{2\times 5}=13.4kcalheat$.

Answer $=\frac{13.4}{6}\approx 2$