A mixture of ethylene and excess of `H_2` has a pressure of 600 mm Hg. The mixture was passed over nickel catalyst to convert ethylene to ethane. The pressure of the resultant mixture at the similar condition of temperature and volume dropped to 400 mm Hg. the fraction of `C_2H_4` by volume dropped to 400 mm Hg. The fraction of `C_2H_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