A mixture of ethylene and excess of `H_(2)` had a pressure of `600 mmHg` 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 Hg` The fraction of `C_(2) H_(4)` by volume in the original mixture is

Options

A. 1/3rd of the total volume
B. 1/4th of the total volume
C. 2/3rd of the total volume
D. 1/2 of the total volume

Answer

A

Explanation

Let `n` mol of `(C_2H_4+H_2)` and `x` mol of `C_2H_4`
`H_2=(n-x)mol`
`underset(x)(C_2H_4)+underset(x)(H_2)rarrunderset(x mol)(C_2H_4)`
After reaction `(C_2H_6+H_2` left)
`x+n-x-x=n-x`
[Total `H_2=(n-x),H_2` reacted `=x]`
`H_2` left `=(n-x-x)`
`n=600,n-x=400`
`(n)/(n-x)=(600)/(400)`
`x=(n)/(3)` volume of `C_2H_4`
`=(1)/(3)` rd of total volume