Question

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_6H_{12} (g) $
A mixture of $C_6H_6 $ and excess $H_2$ has a pressure of $60 \, mm$ of $Hg$ 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 $30 \, mm$ of $Hg$ in the same volume at the same temperature. The fraction of $C_6H_6$ (by volume) present in the original volume is

• A. 1/3
• B. 1/4
• C. 1/5
• D. 1/6

Answer 1

Answer: (D)

Suppose initially, pressure of $C _{6} H _{6}(g)=p_{1}$ mmand that of $H _{2}(g)= p _{2} mm$
$\therefore p_{1}+p_{2}=60$ mm ..... (i)
After the reaction,
Pressure of $C _{6} H _{6}(g)=0 $ (as all has reacted)
Pressure of $H _{2}(g)=p_{2}-3 p_{1}$
Pressure of $C _{6} H _{12}(g)=p_{1}$
Total pressure $=p_{2}-3 p_{1}+p_{1}=30\, mm$
or, $p_{2}-2 p_{1}=30\, mm$
Solving (i) and (ii), we get
$p_{1}=10\, mm , p_{2}=50\,mm$
Fraction of $C _{6} H _{6}$ by volume $=$ fraction of moles
fraction of pressure $=\frac{10}{60}=\frac{1}{6}$