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A mixture of ethylene and excess of H(2)...

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

A

`1//3` rd of the total volume

B

`1//4` th of the total volume

C

`2//3` rd of the total volume

D

`1//2` nd of the total volume

Text Solution

Verified by Experts

The correct Answer is:
A
Let `n` mol of `(C_(2) H_(3) + H_(2))`
`X "mol of" C_(2)H_(4)`
`H_(2) =(n - x)` mole
`underset(x)(C_(2))H_(4) + underset(x)(H_(2)) rarr underset(x "mol")(C_(2) H_(6))`
After reaction `(C_(2) H_(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 = (4)/(4)` volume of `C_(2) H_(4)`
`= (1)/(3)` rd of total volume
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