Vapour density of equilibrium `PCI_(5)(g)hArrPCI_(3)(g)+CI_(2)(g)` is decreased by

Options

A. increasing temperature
B. decreasing pressure
C. increasing pressure
D. decreasing temperature

Explanation

To solve the problem regarding the vapor density of the equilibrium reaction \( \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \), we will analyze how changes in temperature and pressure affect the vapor density.

### Step-by-Step Solution:

1. **Write the Reaction**:
The equilibrium reaction is given as:
\[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \]

2. **Identify Initial Moles**:
Assume we start with \( A \) moles of \( \text{PCl}_5 \) and 0 moles of \( \text{PCl}_3 \) and \( \text{Cl}_2 \). At equilibrium, if \( \alpha \) is the degree of dissociation, we have:
- Moles of \( \text{PCl}_5 \) = \( A(1 - \alpha) \)
- Moles of \( \text{PCl}_3 \) = \( A\alpha \)
- Moles of \( \text{Cl}_2 \) = \( A\alpha \)

3. **Calculate Total Moles at Equilibrium**:
The total number of moles at equilibrium is:
\[ n = A(1 - \alpha) + A\alpha + A\alpha = A(1 + \alpha) \]

4. **Vapor Density Formula**:
The vapor density (VD) is given by:
\[ \text{VD} = \frac{\text{Molar Mass}}{2} \]
The molar mass can be expressed in terms of the total moles:
\[ \text{Molar Mass} = \frac{\text{Given Mass}}{n} = \frac{\text{Given Mass}}{A(1 + \alpha)} \]

5. **Substituting into Vapor Density**:
Therefore, the vapor density can be expressed as:
\[ \text{VD} = \frac{\text{Given Mass}}{2A(1 + \alpha)} \]
From this, we can see that as \( \alpha \) increases (more dissociation), the vapor density decreases.

6. **Effect of Temperature**:
Since the dissociation of \( \text{PCl}_5 \) is an endothermic reaction, increasing the temperature will shift the equilibrium to the right (forward direction), increasing \( \alpha \) and thus decreasing vapor density.

7. **Effect of Pressure**:
Decreasing the pressure will also shift the equilibrium to the right because there are more moles of gas on the product side (2 moles of products vs. 1 mole of reactant). This will also increase \( \alpha \) and decrease vapor density.

8. **Conclusion**:
The vapor density of the equilibrium decreases when:
- **Increasing temperature**
- **Decreasing pressure**

### Final Answer:
The vapor density of the equilibrium \( \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \) is decreased by increasing temperature and decreasing pressure.