Question

$11.6g$ of an organic compound having formula $C_n{H}_{2n+2}$ is burnt in excess of $O_2\left(g\right)$ initially taken in a $22.41$ litre steel vessel. Before reaction, the gaseous mixture was at $273K$ with pressure reading $2atm$. After complete combustion and loss of considerable amount of heat, the mixture of product and excess of $O_2$ had a temperature of $546K$ and $4.6atm$ pressure. The formula of organic compound is:

• A. $C_2{H}_6$
• B. $C_3{H}_8$
• C. $C_5{H}_{12}$
• D. $C_4{H}_{10}$

Answer 1

The correct option is D $C_4{H}_{10}$

$C_n{H}_{2n+2}+\left(\frac{3n+1}{2}\right)O_2\to nC{O}_2+(n+1)H_{2}O$
Let initial pressure of $C_n{H}_{2n+2}$ is $P$, then increase in pressure :
$=P\left[\right(2n+1)-1-\left(\frac{3n+1}{2}\right)]$
$=\left(\frac{n-1}{2}\right)P$
For a constant volume of the steel vessel, $T\propto P$
So,
At $546K$ pressure is $4.6atm$
$\therefore$ at $\frac{546}{2}=273K$ pressure is $\frac{4.6}{2}=2.3atm$;
Increase in pressure= $(2.3-2)atm=0.3atmR=\text{ideal gas constant}=0.0821L.atmmo{l}^{-1}{K}^{-1}V=\text{volume of steel vessel}=22.41L$
$P=\frac{nRT}{V}=\frac{11.6}{M}\times \left(\frac{0.0821\times 273}{22.41}\right)$
$P=\frac{11.6}{M}$
where $M$ = Molecular weight of organic compound
$M=n\times 12+(2n+2)\times 1M=14n+2$
Since,
$\left(\frac{n-1}{2}\right)P=0.3$
Putting $P=\frac{11.6}{14n+2}$
$\Rightarrow \left(\frac{n-1}{2}\right)\times \frac{11.6}{(14n+2)}=0.3$
$\Rightarrow \frac{n-1}{14n+2}=\frac{0.6}{11.6}$
$\Rightarrow n=4$
$\therefore$ Compound is $C_4{H}_{10}$