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 $C_{2} H_{4}$ by volume in the original mixture is:

\frac{1}{3} r d

of total volume

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\frac{1}{4} t h

of total volume

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\frac{2}{3} r d

of total volume

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\frac{1}{2}

of total volume

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Solution

The correct option is A $\frac{1}{3} r d$ of total volume
Let $n$ mol of ( $C_{2} H_{4} + H_{2})$ and $x$ mol of $C_{2} H_{4}$
$H_{2} = (n - x) m o l$

$C_{2} H_{4} + H_{2} ⟶ C_{2} H_{6}$

$x$ $x$ $x m o l$

After reaction $(C_{2} H_{6} + H_{2} l e f t)$

$x + n - x - x = n - x$

[Total $H_{2} = (n - x), H_{2}$ reacted $= x$ ]

$H_{2} l e f t = (n - x - x)$

$n = 600, n - x = 400$

$\frac{n}{n - x} = \frac{600}{400}$ ;

$x = \frac{n}{3}$ volume of $C_{2} H_{4}$

$= \frac{1}{3}$ rd of total volume

Hence, the correct option is $A$

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Similar questions

Q. A mixture of ethylene and excess of

H_{2}

had a pressure of

600

mm of 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 of Hg. The fraction of

C_{2} H_{4}

by volume in the original mixture is :

Gaseous benzene reacts with hydrogen gas in presence of a nickel catalyst to form gaseous cyclohexane according to the reaction: $C_{6} H_{6} (g) + 3 H_{2} (g) \to C_{6} H_{12} (g) .$ A mixture of $C_{6} H_{6}$ and excess $H_{2}$ has a pressure of 60 mm of Hg in an unknown volume. After the gas has been passed over a nickel catalyst and all benzene converted to cyclohexane, the pressure of the gas was 30 mm of Hg at same volume and temperature. The fraction of $C_{6} H_{6}$ (by volume) present in the original volume is___.