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?

Options

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

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Reaction
The reaction between benzene (C6H6) and hydrogen (H2) in the presence of a nickel catalyst can be represented as:
\[ \text{C}_6\text{H}_6 (g) + 3 \text{H}_2 (g) \rightarrow \text{C}_6\text{H}_{12} (g) \]
This indicates that one mole of benzene reacts with three moles of hydrogen to produce one mole of cyclohexane.

### Step 2: Set Up Initial Conditions
We know that the total pressure of the mixture of benzene and hydrogen is 60 mm Hg. Let:
- \( P_1 \) = pressure of benzene
- \( P_2 \) = pressure of hydrogen

From the problem, we have:
\[ P_1 + P_2 = 60 \, \text{mm Hg} \quad \text{(1)} \]

### Step 3: Analyze Final Conditions
After the reaction, all benzene is converted to cyclohexane, and the pressure of the gas mixture is 30 mm Hg. The pressure of hydrogen after the reaction can be expressed as:
\[ P_2 - 3P_1 \]
The total pressure after the reaction is then:
\[ P_2 - 3P_1 + 0 = 30 \, \text{mm Hg} \quad \text{(2)} \]

### Step 4: Solve the Equations
Now we have two equations:
1. \( P_1 + P_2 = 60 \)
2. \( P_2 - 3P_1 = 30 \)

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

Substituting this into equation (2):
\[ (60 - P_1) - 3P_1 = 30 \]
\[ 60 - 4P_1 = 30 \]
\[ 4P_1 = 60 - 30 \]
\[ 4P_1 = 30 \]
\[ P_1 = \frac{30}{4} = 7.5 \, \text{mm Hg} \]

Now substituting \( P_1 \) back into equation (1) to find \( P_2 \):
\[ P_2 = 60 - 7.5 = 52.5 \, \text{mm Hg} \]

### Step 5: Calculate the Mole Fraction of Benzene
The mole fraction of benzene in the original mixture can be calculated using the formula:
\[ \text{Mole fraction of benzene} = \frac{P_1}{P_{\text{total}}} \]
Where \( P_{\text{total}} = 60 \, \text{mm Hg} \):
\[ \text{Mole fraction of benzene} = \frac{7.5}{60} = \frac{1}{8} \]

### Final Answer
The fraction (by mole) of the original mixture that was benzene is:
\[ \frac{1}{8} \]