Question

Gaseous benzene reacts with hydrogen gas in the presence of nickel catalyst to give gaseous cyclohexane. A mixture of benzene vapour and hydrogen had a pressure of 60 mm Hg in vessel. After all benzene converted to cyclohexane, the pressure of the gas was 30 mm Hg in the same volume and at the same temperature. What fraction (by mole) of the original mixture was benzene?

• A. 0.167
• B. 0.333
• C. 0.666
• D. 1

Answer 1

Answer: (A)

$C_{6}H_{6}\left(g\right)+3H_{2}\left(g\right) \rightarrow C_{6}H_{12}\left(g\right)$

Let x and y be the partial pressures (initial) of gaseous benzene and hydrogen in the vessel.

Then, $\text{x} + \text{y} = 6 0$ ......... (i)

After the reaction, whole of benzene is converted to cyclohexane.

Partial pressure of cyclohexane $= \, x mm Hg$

Partial pressure of hydrogen left unreacted $= \text{y} - 3 \text{x}$

Total pressure of the system after the reaction,

$\text{x} + \text{y} - 3 \text{x} = \text{y} - 2 \text{x} = 3 0$ .......... (ii)

Solving (i) and (ii), $\text{x} = 10 mm Hg$

Number of moles $ \propto $ pressure ; Volume $ \propto $ number of moles

Hence, fraction of benzene by volume in the original mixture $= \frac{1 0}{6 0} = \frac{1}{6} = \text{0.167}$