A mixture of a mol of `C_(3) H_(8)` and `b` mol of `C_(2) H_(4)` was kept is a container of `V L` exerts a pressure of 4.93 atm at temperature `T`. Mixture was burnt in presence of `O_(2)` to convert `C_(3) H_(8)` and `C_(2) H_(4)` into `CO_(2)` in the container at the same temperature. The pressure of gases after the reaction and attaining the thermal equilirium with atomsphere at temperature `T` was found to be 11.08 atm.
The mole fraction of `C_(2)H_(4)` in the mixture is

Options

A. 0.25
B. 0.75
C. 0.45
D. 0.55

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Initial Conditions
We have a mixture of 1 mole of \( C_3H_8 \) (propane) and \( b \) moles of \( C_2H_4 \) (ethene) in a container of volume \( V \) at temperature \( T \). The initial pressure of the mixture is given as 4.93 atm.

### Step 2: Apply the Ideal Gas Law
Using the ideal gas law \( PV = nRT \), we can express the initial conditions:
\[ P_1 = 4.93 \text{ atm}, \quad n_1 = 1 + b \text{ moles} \]
Thus,
\[ 4.93V = (1 + b)RT \quad \text{(1)} \]

### Step 3: Write the Combustion Reactions
The combustion reactions for propane and ethene are:
1. For \( C_3H_8 \):
\[ C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O \]
2. For \( C_2H_4 \):
\[ C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O \]

### Step 4: Calculate Moles of \( CO_2 \) Produced
From the combustion reactions:
- 1 mole of \( C_3H_8 \) produces 3 moles of \( CO_2 \).
- \( b \) moles of \( C_2H_4 \) produce \( 2b \) moles of \( CO_2 \).

Total moles of \( CO_2 \) after combustion:
\[ n_2 = 3 \cdot 1 + 2b = 3 + 2b \]

### Step 5: Apply the Ideal Gas Law After Combustion
After combustion, the pressure of the gases is 11.08 atm. Thus, we can write:
\[ P_2 = 11.08 \text{ atm}, \quad n_2 = 3 + 2b \]
Using the ideal gas law again:
\[ 11.08V = (3 + 2b)RT \quad \text{(2)} \]

### Step 6: Set Up the Equations
Now we have two equations:
1. From the initial state:
\[ 4.93V = (1 + b)RT \quad \text{(1)} \]
2. From the final state:
\[ 11.08V = (3 + 2b)RT \quad \text{(2)} \]

### Step 7: Divide Equation (2) by Equation (1)
To eliminate \( V \) and \( RT \):
\[ \frac{11.08V}{4.93V} = \frac{3 + 2b}{1 + b} \]
This simplifies to:
\[ \frac{11.08}{4.93} = \frac{3 + 2b}{1 + b} \]
Calculating the left side:
\[ \frac{11.08}{4.93} \approx 2.25 \]
Thus, we have:
\[ 2.25 = \frac{3 + 2b}{1 + b} \]

### Step 8: Cross-Multiply and Solve for \( b \)
Cross-multiplying gives:
\[ 2.25(1 + b) = 3 + 2b \]
Expanding and rearranging:
\[ 2.25 + 2.25b = 3 + 2b \]
\[ 2.25b - 2b = 3 - 2.25 \]
\[ 0.25b = 0.75 \]
\[ b = 3 \]

### Step 9: Calculate the Mole Fraction of \( C_2H_4 \)
The total moles in the mixture are:
\[ 1 + b = 1 + 3 = 4 \]
The mole fraction of \( C_2H_4 \) is:
\[ \text{Mole fraction of } C_2H_4 = \frac{b}{1 + b} = \frac{3}{4} = 0.75 \]

### Final Answer
The mole fraction of \( C_2H_4 \) in the mixture is \( 0.75 \).

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