Formation of `ClF_(3)` from `Cl_(2)and F_(2)` is an exothermic process . The equilibrium system can be represented as
`Cl_(2(g))+3F_(2(g))rArr2ClF_(3(g)),DeltaH=-329kJ` Which of the following will increase quantity of `ClF_(3)` in the equilibrium mixture ? i(A)ncrease in temperature, decrease in pressure, addition of C l 2 (B)Decrease in temperature and pressure, addition of C l F 3 (C)Increase in temperature and pressure, removal of C l 2 (D) Decrease in temperature, increase in pressure, addition of F 2

Options

A. Increase in temperature, decrease in pressure addition of `Cl_(2)`
B. Decrease in temperature and pressure, addition of `ClF_(3)`
C. Increase in temperature and pressure, removal of `Cl_(2)`
D. Decrease in temperature, increase in pressure, addition of `F_(2)`

Explanation

To determine which option will increase the quantity of \( \text{ClF}_3 \) in the equilibrium mixture, we can apply Le Chatelier's principle. This principle states that if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the equilibrium will shift in a direction that counteracts the change.

The given reaction is:
\[ \text{Cl}_2(g) + 3\text{F}_2(g) \rightleftharpoons 2\text{ClF}_3(g), \quad \Delta H = -329 \, \text{kJ} \]
This indicates that the formation of \( \text{ClF}_3 \) is an exothermic process.

### Step-by-Step Solution:

1. **Identify the nature of the reaction**:
- The reaction is exothermic, which means that heat is released when \( \text{ClF}_3 \) is formed.

2. **Analyze the effect of temperature**:
- For exothermic reactions, decreasing the temperature shifts the equilibrium to the right (towards the products) to produce more heat. Therefore, decreasing the temperature will favor the formation of \( \text{ClF}_3 \).

3. **Analyze the effect of pressure**:
- The reaction has 4 moles of gas on the reactant side (\( 1 \, \text{Cl}_2 + 3 \, \text{F}_2 \)) and 2 moles of gas on the product side (\( 2 \, \text{ClF}_3 \)). According to Le Chatelier's principle, increasing the pressure will shift the equilibrium towards the side with fewer moles of gas, which is the product side in this case. Thus, increasing the pressure will also favor the formation of \( \text{ClF}_3 \).

4. **Analyze the effect of concentration**:
- Adding more reactants (\( \text{Cl}_2 \) or \( \text{F}_2 \)) will increase the concentration of the reactants, shifting the equilibrium to the right and producing more \( \text{ClF}_3 \).

5. **Evaluate each option**:
- **(A)** Increase in temperature, decrease in pressure, addition of \( \text{Cl}_2 \):
- Increasing temperature is not favorable. This option is incorrect.
- **(B)** Decrease in temperature and pressure, addition of \( \text{ClF}_3 \):
- Decreasing temperature is favorable, but decreasing pressure is not. This option is incorrect.
- **(C)** Increase in temperature and pressure, removal of \( \text{Cl}_2 \):
- Increasing temperature is not favorable, and removing \( \text{Cl}_2 \) will shift the equilibrium left. This option is incorrect.
- **(D)** Decrease in temperature, increase in pressure, addition of \( \text{F}_2 \):
- Decreasing temperature is favorable, increasing pressure is favorable, and adding \( \text{F}_2 \) (a reactant) is favorable. This option is correct.

### Conclusion:
The correct answer is **(D)**: Decrease in temperature, increase in pressure, addition of \( \text{F}_2 \).