Question

The standard enthalpy of formation of FeO & Fe2O3 is -65 kcal mol-1 and -197kcalmol-1 respectively. A mixture of two oxides contains FeO & Fe2O3 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. 13.4 kcal/mole
• B. 14.6 kcal/mole
• C. 15.7 kcal/mole
• D. 16.8 kcal/mole

Answer 1

Answer: (A)

13.4 kcal/mole

Answer 2

FeO + Fe2O3

2x x

Fe + $\frac{1}{2}$O2 $\longrightarrow$ FeO $\Delta H_{f}^{0}$ = – 65 Kcal/mole

2Fe + $\frac{3}{2}$O2 $\longrightarrow$ Fe2O3 $\Delta H_{f}^{0}$= – 197 Kcal/mole

2FeO +$\frac{1}{2}$ O2 $\longrightarrow$ Fe2O3

$\Delta$H = –197 + 65 × 2 $\Rightarrow$ $\Delta$H = – 67 Kcal/mole

2FeO +$\frac{1}{2}$ O2 $\longrightarrow$ Fe2O3

$\frac{2}{3}$ $\frac{1}{3}$ $\frac{2}{3}$ – 2x$\frac{1}{3}$+ x

$\frac{\frac{2}{3} - 2x}{\frac{1}{3} + x} = \frac{1}{2}$

$\Rightarrow$ x =$\frac{1}{5}$

So, energy released = $\frac{1}{5}$× 67 = 13.4 kcal/mole