Question

A gas mixture $3.67L$ in volume contains $C_2{H}_4$ and $C{H}_4$ is proportion of $2:1$ by moles and is at ${25}^{\circ }C$ and $1atm$. If the $\Delta H_C$ ($C_2{H}_4)$ and $\Delta H_C$ $\left(C{H}_4\right)$ are $-1400$ and $-900kJ/mol$ find heat evolved on burning this mixture

• A. 20.91 kJ
• B. 50.88 kJ
• C. 185 kJ
• D. 160 kJ

Answer 1

The correct option is C

185 kJ

Volume of $C_2{H}_4$ = $\frac{3.67\times 2}{3}=2.45Lit$

Volume of $C{H}_4$ = $1.22Lit$

Mole of $C_2{H}_5$ = $PV$ = $nRTn$ = $\frac{PV}{RT}$

= $\frac{1\times 2.45}{0.082\times 298}=n=0.1mole$

Mole of $C{H}_4$ = ${}^{n}C{H}_4$ = $\frac{1\times 1.22}{0.082\times 298}=0.05mole$

$\because$ Energy released due to combusting 1 mole $C_2{H}_4=1400KJ$

$\therefore$ Energy released due to combusting 1 mole $C_2{H}_4$ = $1400\times 0.1=140kJ$

Similar energy released in combusting 0.05 mole

$C{H}_4=-900\times 0.05=45$

So that heat realized = $140+45$ = $185kJ/mole$