Question

Gaseous benzene reacts with hydrogen gas in the presence of a nickel catalyst to form gaseous cyclohexane according to the reaction,
\[{{C}_{6}}{{H}_{6}}(g)+3{{H}_{2}}(g)\to {{C}_{6}}{{H}_{12}}(g)\]
A mixture of \[{{C}_{6}}{{H}_{6}}\] and excess \[{{H}_{2}}\] has a pressure of 60mm 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 30mm of Hg in the same volume at the same temperature. The fraction of \[{{C}_{6}}{{H}_{6}}\] (by volume) present in the original volume is:
(a)- \[\dfrac{1}{3}\]
(b)- \[\dfrac{1}{4}\]
(c)- \[\dfrac{1}{5}\]
(d)- \[\dfrac{1}{6}\]

Answer 1

Hint: The fraction can be calculated by taking 2 equations, with initial and final process pressure. Add the 2 equations and divide it with original pressure.

Complete step by step answer:
Let us first write down the equation given in the question:
 \[{{C}_{6}}{{H}_{6}}(g)+3{{H}_{2}}(g)\to {{C}_{6}}{{H}_{12}}(g)\]
For the first condition,
Let the initial pressure of \[{{C}_{6}}{{H}_{6}}(g)\] is \[{{p}_{1}}mm\] and for \[{{H}_{2}}(g)\] is \[{{p}_{2}}mm\],
In the question, it is given that the mixture has a pressure of 60mm of Hg.
Therefore, the equation is-
\[{{p}_{1}}+{{p}_{2}}=60mm\text{ }of\text{ }Hg\] - Equation 1
For the second condition,
After heating the final pressure of \[{{C}_{6}}{{H}_{6}}(g)=0\] (because all the benzene has reacted during heating)
For \[{{H}_{2}}(g)={{p}_{2}}-3{{p}_{1}}\]
Because the initial pressure of benzene is \[{{p}_{1}}\] , hydrogen is \[{{p}_{2}}\] , and cyclohexane is 0.
Final pressure of benzene is 0, hydrogen is\[{{p}_{2}}-3{{p}_{1}}\] , and cyclohexane is \[{{p}_{1}}\]
So, the total pressure is-
\[{{p}_{2}}-3{{p}_{1}}+{{p}_{1}}=30mm\text{ }of\text{ }Hg\]
\[{{p}_{2}}-2{{p}_{1}}=30mm\text{ }of\text{ }Hg\]--Equation 2
On solving Equation 1 and 2, we get \[{{p}_{1}}=10mm\text{ and }{{p}_{2}}=50mm\]
So, the fraction of \[{{C}_{6}}{{H}_{6}}\] by volume is = mole fraction,
Hence, the fraction of pressure = \[\dfrac{{{p}_{1}}}{{{p}_{1}}+{{p}_{2}}}=\dfrac{10}{60}=\dfrac{1}{6}\]

So, the correct answer is option (d) \[\dfrac{1}{6}\].

Note: The mole fraction of the initial and final pressure should be taken, and not the fraction of \[{{p}_{1}}\text{ }and\text{ }{{p}_{2}}\].
So, you may get confused between option (c) and option (d).