Mole ConceptMedium
Question
A mixture of ethylene and excess of H2 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 C2H4 by volume in the original mixture is
Options
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 of the total volume
Solution
Let there be n mole of (C2H4 + H2)
Mole of C2H4 = x
H2 = (n - x) mole
C2H4 + H2 → C2H6
Mole of C2H4 = x
H2 = (n - x) mole
C2H4 + H2 → C2H6
x n-x
0 n - 2x x mol
After reaction (C2H6 + H2 left)
x + n - x - x = n - x
[Total H2 = (n - x), H2 reacted = x]
H2 left = (n - x - x)
n ∝ 600; n - x ∝ 400
;
0 n - 2x x mol
After reaction (C2H6 + H2 left)
x + n - x - x = n - x
[Total H2 = (n - x), H2 reacted = x]
H2 left = (n - x - x)
n ∝ 600; n - x ∝ 400
; x/n = 1/3 = fraction of C2H4 by volume
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