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A mixture of ethylene and excess of H2 ...

A mixture of ethylene and excess of `H_2` had 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 conditions of temperature and volume dropped to 400 mm Hg. The fraction of `C_2H_4` by volume in the original mixture is:

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

Text Solution

Verified by Experts

The correct Answer is:
A
Let n mol of `(C_2H_4 +H_2)` and x mol of `C_2H_4 " " H_2 = (n-x)` mol
`{:(C_2H_4,+H_2 rarr,C_2H_6),(x,x,"x mol"):}`
After reaction ( `C_2H_6 +H_2 ` left )` x+n - x - x = m - 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
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