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 $22.41L$ steel vessel. Before the 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 the 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 $=P\left[\right(2n+1)-1-\left(\frac{3n+1}{2}\right)]$ $=\left(\frac{n-1}{2}\right)P$
Initial Pressure = $2$ atm and Final Pressure = $4.6$ atm
Initial temperature = $273$ K and Final temperature = $546$ K
Increase in pressure$=0.3atm$
$P=\frac{nRT}{V}=\frac{11.6}{M}\times \left(\frac{0.0821\times 273}{22.41}\right)$
$\left(\frac{n-1}{2}\right)\times \frac{11.6}{(14n+2)}=0.3$
$\left(\frac{n-1}{14n+2}\right)=\frac{0.6}{11.6}=n=4$
$\therefore$ Compound is $C_4{H}_{10}$.