Question

A gas mixture of $3.67$ lit of ethylene & methane on complete combustion at ${25}^o$C produces $6.11L$ of $C{O}_2$. Find out the amount of heat evolved on burning one litre of the gas mixture. The heat for combustion of ethylene and methane are $-1423$ & $-891kJmo{l}^{-1}$.

Answer 1

Reaction involved (At ${25}^{\circ }C=298K$)

(I)$C_2{H}_4+3O_2\to 2C{O}_2+2H_{2}O$

1 mole (1 litre) 2 moles (2 litre)

(II) $C{H}_4+2O_2\to C{O}_2+2H_{2}O$

1 litre 1 litre

Let $V_{C{H}_4}=x$ L then $V_{C_2{H}_2}=(3.67-x)L$

Then, vol. of $C{O}_2$ produced by xL of $C{H}_4=xL$

Vol. of $C{O}_2$ produced by $(3.67-x)L$ of $C_2{H}_4=2\times (3.67-x)L$

Given:- Total vol. of $C{O}_2$ produced = 6.11 L

$\Rightarrow x+2(3.67-x)=6.11L$

$\Rightarrow x=1.23L$

$\therefore V_{C{H}_4}=1.23L$ and $V_{C_2{H}_4}=2.44L$

Now, Vol. of 1 mol of a gas at 298 K = $\frac{22.4}{273}\times 298=24.45L$

No. of moles of $C{H}_4={}^{n}C{H}_4=\frac{1.23}{24.45}=0.0503$ moles

No. of moles of $C_2{H}_4={}^{n}C{H}_4=\frac{2.44}{24.45}=0.0997$ moles

Amount of heat evolved on burning 3.67 L of gas

$=0.0997\times 1423+891\times 0.0503=141.87+44.84=186.69KJ$

Thus, Amount of heat evolved on burning 1L of gas mixture

$=\frac{186.69}{3.67}=50.86KJ$

Ans 50.86 KJ