In the following reaction started only with `A_(8), 2A_(8)(g) hArr3A_(2)(g)+A_(4)(g)` mole fraction of `A_(2)` is found to `0.36` at a total pressure of `100 atm` at equilibrium. The mole fraction of `A_(8)(g)` at equlibrium is `:`

Options

A. `0.28`
B. `0.72`
C. `0.18`
D. None of these

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Understand the Reaction
The reaction given is:
\[ 2A_8(g) \rightleftharpoons 3A_2(g) + A_4(g) \]

### Step 2: Use the Given Information
We are given that the mole fraction of \( A_2 \) at equilibrium is \( 0.36 \) and the total pressure is \( 100 \, \text{atm} \).

### Step 3: Calculate the Partial Pressure of \( A_2 \)
The partial pressure of \( A_2 \) can be calculated using its mole fraction and the total pressure:
\[ P_{A_2} = \text{Mole Fraction of } A_2 \times \text{Total Pressure} \]
\[ P_{A_2} = 0.36 \times 100 \, \text{atm} = 36 \, \text{atm} \]

### Step 4: Determine the Relationship Between Partial Pressures
From the stoichiometry of the reaction, we know:
\[ \frac{P_{A_2}}{P_{A_8}} = \frac{3}{2} \]
This means:
\[ P_{A_8} = \frac{2}{3} P_{A_2} \]
Substituting the value of \( P_{A_2} \):
\[ P_{A_8} = \frac{2}{3} \times 36 \, \text{atm} = 24 \, \text{atm} \]

### Step 5: Calculate the Mole Fraction of \( A_8 \)
The mole fraction of \( A_8 \) can be calculated using its partial pressure:
\[ \text{Mole Fraction of } A_8 = \frac{P_{A_8}}{\text{Total Pressure}} \]
\[ \text{Mole Fraction of } A_8 = \frac{24 \, \text{atm}}{100 \, \text{atm}} = 0.24 \]

### Step 6: Conclusion
The mole fraction of \( A_8 \) at equilibrium is \( 0.24 \).

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