A mixture of ethylene and excess of `H_(2)` had a pressure of `600 mmHg` the mixture was passed over nickel catalyst to convert ethylene to ethane.The pressure of the resultant mixture at the similar conditions of temperature and volume dropped to `400 mm Hg` The fraction of `C_(2) H_(4)` by volume in the original mixture is

Options

A. `1//3` rd of the total volume
B. `1//4` th of the total volume
C. `2//3` rd of the total volume
D. `1//2` nd of the total volume

Explanation

To solve the problem, we need to find the fraction of ethylene (C₂H₄) by volume in the original mixture of ethylene and hydrogen (H₂). Here’s a step-by-step breakdown of the solution:

### Step 1: Define the Variables
Let:
- \( x \) = moles of C₂H₄ in the original mixture
- \( n \) = total moles of the gas mixture (C₂H₄ + H₂)

Since the mixture contains excess H₂, the moles of H₂ can be expressed as:
\[ \text{Moles of H₂} = n - x \]

### Step 2: Write the Reaction
The reaction that occurs is:
\[ \text{C₂H₄} + \text{H₂} \rightarrow \text{C₂H₆} \]
From the stoichiometry of the reaction, \( x \) moles of C₂H₄ will react with \( x \) moles of H₂ to produce \( x \) moles of C₂H₆.

### Step 3: Determine Moles After Reaction
After the reaction, the moles of each component will be:
- Moles of C₂H₆ = \( x \)
- Moles of H₂ left = \( (n - x) - x = n - 2x \)

### Step 4: Total Moles After Reaction
The total moles after the reaction will be:
\[ \text{Total moles after reaction} = \text{Moles of C₂H₆} + \text{Moles of H₂ left} = x + (n - 2x) = n - x \]

### Step 5: Relate Pressures to Moles
According to the problem:
- Initial pressure (before reaction) = 600 mmHg
- Final pressure (after reaction) = 400 mmHg

Using the ideal gas law, we can relate the pressure to the number of moles:
\[ \frac{P_{\text{initial}}}{P_{\text{final}}} = \frac{n}{n - x} \]
Substituting the pressures:
\[ \frac{600}{400} = \frac{n}{n - x} \]

### Step 6: Solve for \( n - x \)
Simplifying the fraction:
\[ \frac{3}{2} = \frac{n}{n - x} \]
Cross-multiplying gives:
\[ 3(n - x) = 2n \]
Expanding and rearranging:
\[ 3n - 3x = 2n \implies n = 3x \]

### Step 7: Find the Fraction of C₂H₄
The fraction of C₂H₄ by volume in the original mixture is given by:
\[ \text{Fraction of C₂H₄} = \frac{x}{n} = \frac{x}{3x} = \frac{1}{3} \]

### Final Answer
Thus, the fraction of C₂H₄ by volume in the original mixture is:
\[ \frac{1}{3} \]