Question

The standard enthalpy of formation of $FeO$ and $F{E}_2{O}_3$ is $-65kcalmo{l}^{-1}$ and $-197kcalmo{l}^{-1}$ respectively mixture of two oxides contains $FeO$ and $F{e}_2{O}_3$ in the mole ratio $2:1$. If by oxidation, it is changed into a $1:2$ mole ratio mixture, how much of thermal energy will be released per mole of the initial mixture?

• A. $13.4kcal/mole$
• B. $14.6kcal/mole$
• C. $15.7kcal/mole$
• D. $16.8kcal/mole$

Answer 1

The correct option is A $13.4kcal/mole$

Solution:- (A) $13.4kcal/mol$

$Fe+\frac{1}{2}{O}_2⟶FeO\Delta H=-65kcal/mol$

$2\times [Fe+\frac{1}{2}{O}_2⟶FeO\Delta H=-65kcal/mol]$

$2Fe+O_2⟶2FeO\Delta H=-130kcal/mol.....\left(1\right)$

$2Fe+\frac{3}{2}{O}_2⟶{Fe}_2{O}_3\Delta H=-197kcal/mol.....\left(2\right)$

Subtracting equation $\left(1\right)$ from $\left(2\right)$, we get

$2FeO+\frac{1}{2}{O}_2⟶{Fe}_2{O}_3\Delta H=-67kcal/mol$

Let $a$ and $b$ be the no. of moles of $FeO$ and ${Fe}_2{O}_3$ respectively such that,

$a+b=1.....\left(3\right)$

$\frac{a}{b}=2\Rightarrow a=2b$

Substituting the value of $a$ in equation $\left(3\right)$, we get

$b=\frac{1}{3}$

$\therefore a=\frac{2}{3}$

$\underset{\text{After oxidation}:}{\underset{\text{Initially}:}{}}\underset{a-2\alpha }{\underset{a}{2FeO}}+\frac{1}{2}{O}_2⟶\underset{b+\alpha }{\underset{b}{{Fe}_2{O}_3}}$

Given:-

$\frac{a}{b}=2\Rightarrow a=2b$

$\frac{a-2\alpha }{b+\alpha }=\frac{1}{2}$

$\frac{2b-2\alpha }{b+\alpha }=\frac{1}{2}$

$4b-4\alpha =b+\alpha$

$3b=5\alpha$

$\Rightarrow \alpha =\frac{3}{5}b=\frac{3}{5}\times \frac{1}{3}=\frac{1}{5}$

Therefore,

No. of moles of $FeO$ used $=2\alpha =\frac{2}{5}\text{mol}$

Heat released by $2$ mole of $FeO=67kcal$

Therefore,

Heat released by $\frac{2}{5}$ mole of $FeO=\frac{2}{5}\times \frac{67}{2}=13.4kcal$