`PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g)`, In above reaction, at equilibrium condition mole fraction of `PCl_(5)` is 0.4 and mole fraction of `Cl_(2)` is 0.3. Then find out mole fraction of `PCl_(3)`

Options

A. 0.3
B. 0.7
C. 0.4
D. 0.6

Answer

To find the mole fraction of \( PCl_3 \) in the equilibrium reaction

\[ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \]

we are given the mole fractions of \( PCl_5 \) and \( Cl_2 \). Let's denote the mole fractions as follows:

- \( \chi_{PCl_5} = 0.4 \)
- \( \chi_{Cl_2} = 0.3 \)

We need to find the mole fraction of \( PCl_3 \), denoted as \( \chi_{PCl_3} \).

### Step-by-Step Solution:

1. **Understand the Concept of Mole Fraction**:
The mole fraction of a component in a mixture is defined as the ratio of the number of moles of that component to the total number of moles of all components in the mixture. The sum of the mole fractions of all components in a system must equal 1.

2. **Set Up the Equation**:
Since we have three components in the reaction, we can write the equation for the total mole fractions:
\[ \chi_{PCl_5} + \chi_{PCl_3} + \chi_{Cl_2} = 1 \]

3. **Substitute Known Values**:
Substitute the known mole fractions into the equation:
\[ 0.4 + \chi_{PCl_3} + 0.3 = 1 \]

4. **Combine the Known Values**:
Combine the known mole fractions:
\[ 0.4 + 0.3 = 0.7 \]
So the equation now becomes:
\[ 0.7 + \chi_{PCl_3} = 1 \]

5. **Solve for \( \chi_{PCl_3} \)**:
To find \( \chi_{PCl_3} \), subtract 0.7 from both sides:
\[ \chi_{PCl_3} = 1 - 0.7 = 0.3 \]

6. **Conclusion**:
The mole fraction of \( PCl_3 \) is:
\[ \chi_{PCl_3} = 0.3 \]

### Final Answer:
The mole fraction of \( PCl_3 \) is \( 0.3 \).