Gaseous benzene reacts with hydrogen gas in the presence of a nickel catalyst to form gaseous cyclohexane according to the reaction,

$C_{6} H_{6} (g) + 3 H_{2} (g) ⟶ C_{6} H_{12} (g)$

A mixture of $C_{6} H_{6}$ and excess $H_{2}$ has a pressure of $60 m m$ of $H g$ in an unknown volume. After the gas had been passed over a nickel catalyst and all the benzene converted to cyclohexane, the pressure of the gas was $30 m m$ of $H g$ in the same volume at the same temperature. The fraction of $C_{6} H_{6}$ (by volume) present in the original volume is:

1 / 3

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1 / 6

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Solution

The correct option is C $1 / 6$
Let initially, pressure of $C_{6} H_{6} (g)$ is $p_{1} m m$ and for $H_{2} (g)$ is $p_{2} m m$

$∴$ $p_{1} + p_{2} = 60 m m$ ....(i)

After the reaction pressure of

$C_{6} H_{6} (g) = 0$ (as all $C_{6} H_{6}$ has reacted)

$H_{2} (g) = p_{2} - 3 p_{1}$

So, total pressure $= p_{2} - 3 p_{1} + p_{1} = 30 m m$

$p_{2} - 2 p_{1} = 30 m m$ ....(ii)

On solving equation (i) and (ii)

$p_{1} = 10 m m$ , $p_{2} = 50 m m$

Fraction of $C_{6} H_{6}$ by volume $=$ moles fraction

Fraction of pressure $= \frac{10}{60} = \frac{1}{6}$

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