Question

A gas mixture of $3.67$ litre of ethylene and methane on complete combustion at ${25}^{o}C$ produces $6.11$ litre of ${CO}_2$. Find out the heat evolved on burning $1$ litre of the gas mixture. The heats of combustion of ethylene and methane are $-1423$ and $-891kJ$ ${mol}^{-1}$ at ${25}^{o}C$.

• A. 60kJ
• B. -50.90kJ
• C. 30kJ
• D. 100kJ

Answer 1

The correct option is B -50.90kJ

Solution:- (B) $-50.90KJ$

Combustion of $C{H}_4$-

$\underset{\text{1 vol.}}{C{H}_4}+2O_2⟶\underset{\text{1 vol.}}{C{O}_2}+2H_{2}O$

Combustion of $C_2{H}_4$-

$\underset{\text{1 vol.}}{C_2{H}_4}+3O_2⟶2\underset{\text{2 vol.}}{C{O}_2}+2H_{2}O$

Let $V$ litre be the volume of $C{H}_4$ in the mixture.

Total volume of mixture $=3.67L\left(\text{Given}\right)$

$\therefore$ Volume of $C_2{H}_4$ in the mixture $=(3.67-V)L$

$\therefore$ Volume of $C{O}_2$ formed during combustion of $C_2{H}_4=V+2(3.67-V)$

But given that the volume of $C{O}_2$ formed during combustion of $C_2{H}_4$ is $6.11L$

$\therefore V+2(3.67-V)=6.11$

$\Rightarrow V=1.23L$

Therefore,

Volume of $C{H}_4$ in the mixture $=1.23L$

Volume of $C_2{H}_4$ in the mixture $=3.67-1.23=2.44L$

Now,

Volume of $C{H}_4$ in $1L$ of mixture $=\frac{1.23}{3.67}$

Volume of $C_2{H}_4$ in $1L$ of mixture $=\frac{2.44}{3.67}$

As we know that volue of $1$ mole of any gas at $25℃$ and $1atm=\frac{22.4}{273}\times 298=24.45L\left[\text{By using charle's law}\right]$

Given that:-

heat of combustion of methane $=-891KJ/mol$

Heat of combustion of ethylene $=-1423KJ/mol$

Heat evolved due to the combustion of $C{H}_4=\frac{1.23\times (-891)}{3.67\times 24.45}=-12.213KJ$

Heat evolved due to the combustion of $C_2{H}_4=\frac{2.44\times (-1423)}{3.67\times 24.45}=-38.694KJ$

$\therefore$ Total heat evolved $=(-12.213)+(-38.694)=-50.907KJ$

Hence the correct answer is $-50.90KJ$.