Thermodynamics and Thermochemistry - Result Question 72
72. A gas mixture of $3.67$ $ L$ of ethylene and methane on complete combustion at $25^{\circ} C$ produces $6.11 $ $L$ of $CO _2$. Find out the amount of heat evolved on burning $1$ $ L$ of the gas mixture. The heat of combustion of ethylene and methane are $-1423$ and $-891$ $ kJ $ $mol^{-1}$ at $25^{\circ} C$.
(1991, 5M)
Show Answer
Answer:
Correct Answer: 72. ($49.82 $ $kJ$)
Solution:
- Let the mixture contain $x$ litre of $\mathrm{CH}_4$ and $3.67-x$ litre of ethylene.
$ \underset{x}{\mathrm{CH}_4}+\mathrm{O}_2 \longrightarrow \underset{x}{\mathrm{CO}_2} $
$ \underset{3.67-x}{\mathrm{C}_2 \mathrm{H}_4}+\mathrm{O}_2 \longrightarrow \underset{2(3.67-x)}{2 \mathrm{CO}_2} $
$ \text { Given : } x+2(3.67-x)=6.11 \mathrm{~L} $
$ \Rightarrow \quad x=1.23 \mathrm{~L} $
Volume of ethylene $=2.44 \mathrm{~L}$
Total moles of gases in $1$ litre $=\dfrac{p V}{R T}=\dfrac{1 \times 1}{0.082 \times 298}=0.04$
Also, $\mathrm{CH}_4$ and ethylene are in $1: 2$ volume (or mole) ratio, moles of $\mathrm{CH}_4=\dfrac{0.04}{3}$ and moles of ethylene $=\dfrac{2 \times 0.04}{3}$
$\Rightarrow \quad$ Heat evolved due to methane $=\dfrac{0.04}{3} \times 891=11.88 \mathrm{~kJ}$
Heat evolved due to ethylene $=\dfrac{2 \times 0.04}{3} \times 1423=37.94 \mathrm{~kJ}$
$\Rightarrow$ Total heat evolved on combustion of $1.0$ L gaseous mixture at $25^{\circ} \mathrm{C}$ is $11.88+37.94=49.82 \mathrm{~kJ}$
BETA
