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) + 3H_(2) (h) rArr 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 the benzene converted to cyclohexane, the pressure of the gas was 30 mm of Hg in the same volume and temperature. The fraction of `C_(6)H_(6)` (by volume) present in the original mixture is :

Options

A. `½`
B. `¼`
C. `.^(1)//_(5)`
D. `.^(1)//_(6)`

Explanation

To solve the problem, we need to determine the fraction of benzene (C₆H₆) by volume in the original mixture based on the pressures before and after the reaction. Here’s a step-by-step solution:

### Step 1: Define the Variables
Let:
- \( P_1 \) = pressure of benzene (C₆H₆) in mm Hg
- \( P_2 \) = pressure of hydrogen (H₂) in mm Hg

From the problem, we know the total initial pressure of the mixture is:
\[ P_1 + P_2 = 60 \, \text{mm Hg} \]
(Equation 1)

### Step 2: Analyze the Reaction
The reaction is:
\[ C_6H_6(g) + 3H_2(g) \rightarrow C_6H_{12}(g) \]

According to the stoichiometry of the reaction, for every 1 mole of benzene that reacts, 3 moles of hydrogen are consumed, and 1 mole of cyclohexane is produced.

### Step 3: Determine the Final Pressure
After the reaction, all the benzene is converted to cyclohexane, and the total pressure is given as:
\[ P_2 - 3P_1 + P_1 = 30 \, \text{mm Hg} \]
This simplifies to:
\[ P_2 - 2P_1 = 30 \, \text{mm Hg} \]
(Equation 2)

### Step 4: Solve the Equations
Now we have two equations:
1. \( P_1 + P_2 = 60 \)
2. \( P_2 - 2P_1 = 30 \)

From Equation 1, we can express \( P_2 \) in terms of \( P_1 \):
\[ P_2 = 60 - P_1 \]

Substituting \( P_2 \) into Equation 2:
\[ (60 - P_1) - 2P_1 = 30 \]
\[ 60 - 3P_1 = 30 \]
\[ 3P_1 = 30 \]
\[ P_1 = 10 \, \text{mm Hg} \]

Now substituting \( P_1 \) back into Equation 1 to find \( P_2 \):
\[ P_2 = 60 - 10 = 50 \, \text{mm Hg} \]

### Step 5: Calculate the Fraction of C₆H₆
The fraction of benzene by volume can be calculated using the pressures:
\[ \text{Fraction of } C_6H_6 = \frac{P_1}{P_1 + P_2} = \frac{10}{10 + 50} = \frac{10}{60} = \frac{1}{6} \]

### Conclusion
The fraction of C₆H₆ by volume present in the original mixture is:
\[ \frac{1}{6} \]