The exothermic formation of `ClF_(3)` is represented by thr equation:
`Cl_(2)(g)+3F_(2)(g) hArr 2ClF_(3)(g), DeltaH=-329 kJ`
Which of the following will increase the quantity of `ClF_(3)` in an equilibrium mixture of `Cl_(2), F_(2)`, and `ClF_(3)`?

Options

A. Increassing the temperature
B. Removing `Cl_(2)`
C. Increasing the volume of the container
D. Adding `F_(2)`

Answer

To solve the problem, we need to analyze the given equilibrium reaction and apply Le Chatelier's principle to determine which of the provided options will increase the quantity of \( \text{ClF}_3 \) in the equilibrium mixture.

### Step-by-Step Solution:

1. **Write the equilibrium reaction**:
\[ \text{Cl}_2(g) + 3\text{F}_2(g) \rightleftharpoons 2\text{ClF}_3(g), \quad \Delta H = -329 \, \text{kJ} \]
This indicates that the reaction is exothermic.

2. **Understand the implications of exothermic reactions**:
In exothermic reactions, heat is released. According to Le Chatelier's principle, if we decrease the temperature, the equilibrium will shift to the right to produce more products (in this case, \( \text{ClF}_3 \)). Conversely, increasing the temperature would shift the equilibrium to the left, favoring the reactants.

3. **Analyze the options**:
- **Option A: Increasing the temperature**: This will decrease the quantity of \( \text{ClF}_3 \) because it shifts the equilibrium to the left.
- **Option B: Removing chlorine**: Removing a reactant will shift the equilibrium to the left, thus decreasing the quantity of \( \text{ClF}_3 \).
- **Option C: Increasing the volume of the container**: Increasing the volume will shift the equilibrium towards the side with more moles of gas. Here, the reactants have 4 moles (1 mole of \( \text{Cl}_2 \) and 3 moles of \( \text{F}_2 \)) and the products have 2 moles (\( 2\text{ClF}_3 \)). Therefore, increasing the volume will also decrease the quantity of \( \text{ClF}_3 \).
- **Option D: Adding fluorine**: Adding a reactant will shift the equilibrium to the right, favoring the formation of \( \text{ClF}_3 \) and thus increasing its quantity.

4. **Conclusion**:
The only option that will increase the quantity of \( \text{ClF}_3 \) is **Option D: Adding fluorine**.

### Final Answer:
The correct answer is **D: Adding fluorine**.

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