Question

A mixture of ethylene and excess of $H_2$ had a pressure of $600$ mm of Hg. The mixture was passed over nickel catalyst to convert ethylene to ethane. The pressure of the resultant mixture at the similar conditions of temperature and volume dropped to $400$ mm of Hg. The fraction of $C_2{H}_4$ by volume in the original mixture is :

• A. $\frac{1}{3}$rd of the total volume
• B. $\frac{1}{4}$th of the total volume
• C. $\frac{2}{3}$rd of the total volume
• D. $\frac{1}{2}$nd of the total volume

Answer 1

The correct option is A $\frac{1}{3}$rd of the total volume

Let $n$ mol of $(C_2{H}_4+H_2)$ be present.
Let $x$ mol of $C_2{H}_4$ be present.
Therefore, $H_2=(n-x)$ moles
$\underset{x}{C_2{H}_4}+\underset{x}{H_2}\to \underset{xmol}{C_2{H}_6}$
After reaction $\underset{x}{(C_2{H}_6}\underset{+}{+}\underset{n-x-x=n-2x}{H_{2}left)}$
[Total $H_2=(n-x),H_2$ reacted $=x$]
$H_{2}left=(n-x-x)$
$n=600;n-x=400$
$\frac{n}{n-x}=\frac{600}{400};x=\frac{n}{3}$ volume of $C_2{H}_4$ $=\frac{1}{3}$ rd of total volume