Question

A gas mixture of $3.67L$ of ethylene and methane on complete combustion at ${25}^{o}C$ produces $6.11L$ of $C{O}_{2-}$. Find out the heat evolved on burning $1L$ of the gas mixture. The heats of combustion of ethylene and methane are $-1423$ and $-891kJmo{l}^{-1}$, respectively, at ${25}^{o}C$.

Answer 1

Combustion reaction of ethylene and methane are
$C_2{H}_4+3O_2\to 2C{O}_2+2H_{2}O,\Delta H=-1423kJ$

$C{H}_4+2O_2\to C{O}_2+2H_{2}O,\Delta H=-891kJ$

The volume of $C{H}_4$ in the mixture $=(3.67-x)$ litres
From the above reaction
$1$ litre of $C_2{H}_4$ gives $C{O}_2=2$ litres
$\therefore x$ litres of $C_2{H}_4$ gives $C{O}_2=2x$ litres
$1$ litre of $C{H}_4$ gives $C{O}_2=1$ litres
$\therefore (3.67-x)$ litres of $C{H}_4$ gives $C{O}_2$
$=(3.67-x)$ litres
Total $C{O}_2$ produced $=2x+(3.67-x)$
$=(3.67+x)$ litres
$\therefore 3.67+x=6.11$
or $x=2.44$
$\therefore 1$ litres of the mixture will contain $C_2{H}_4$
$=\frac{2.44}{3.67}=0.66$ litre
and $C{H}_4=1-0.66=0.34$ litre
$22.4$ litres of $C_2{H}_4$ give heat $=1423kJ$
$\therefore 0.66$ litres of $C_2{H}_4$ will gives heat
$=\frac{1423}{22.4}\times 0.66kJ=41.93kJ$
$22.4$ litre of $C{H}_4$ give heat $=891kJ$
$\therefore 0.34$ litre of $C{H}_4$ will give heat
$=\frac{891}{22.4}\times 0.34kJ=13.52kJ$
$\therefore$ Total heat produced $=41.93+13.52$
$=55.45kJ$