The following two reactions:
i. `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`
(ii) `COCl_(2)(g) hArr CO(g)+Cl_(2)(g)`
are simultaneously in equilibrium in a container at constant volume. A few moles of `CO(g)` are later introduced into the vessel. After some time, the new equilibrium concentration of

Options

A. `PCl_(5)` will increases
B. `PCl_(5)` will decreases
C. `PCl_(5)` will remain unsffected
D. `Cl_(2)` will increases

Answer

To solve the problem, we need to analyze the two equilibrium reactions given and how the introduction of CO affects them.

### Step-by-Step Solution:

1. **Identify the Reactions**:
- Reaction (i): \( \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \)
- Reaction (ii): \( \text{COCl}_2(g) \rightleftharpoons \text{CO}(g) + \text{Cl}_2(g) \)

2. **Understand the Effect of Adding CO**:
- When CO is added to the system, the concentration of CO increases. According to Le Chatelier's principle, the system will respond to this change by shifting the equilibrium of the reaction that produces CO to the left (backward direction).

3. **Analyze Reaction (ii)**:
- The addition of CO will shift the equilibrium of reaction (ii) to the left:
\[ \text{CO}(g) + \text{Cl}_2(g) \rightleftharpoons \text{COCl}_2(g) \]
- This means that the concentration of Cl2 will decrease as it is being consumed to form COCl2.

4. **Effect on Reaction (i)**:
- As the concentration of Cl2 decreases due to the backward shift of reaction (ii), the equilibrium of reaction (i) will also be affected.
- The decrease in Cl2 concentration will shift the equilibrium of reaction (i) to the left:
\[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \]
- This means that PCl5 will dissociate less, leading to an increase in the concentration of PCl5.

5. **Conclusion**:
- After some time, the new equilibrium concentrations will show that the concentration of PCl5 will increase, while the concentrations of PCl3 and Cl2 will decrease.

### Final Answer:
- The new equilibrium concentration of PCl5 will **increase**.