The volume of the reaction vessel containing an equilibrium mixture is increased in the following reaction
`SO_(2)Cl_(2)(g)hArrSO_(2)(g)+Cl_(2)(g)`
When equilibrium is re-established :

Options

A. The amount of `Cl_(2)(g)` remains unchanged
B. the amount of `Cl_(2)(g)` increases
C. The amount of` SO_(@)Cl_(2)(g)`decreases
D. The amount of `SO_(@)(g)` decrsases

Answer

To solve the problem regarding the equilibrium reaction:

\[ \text{SO}_2\text{Cl}_2(g) \rightleftharpoons \text{SO}_2(g) + \text{Cl}_2(g) \]

we need to analyze the effects of increasing the volume of the reaction vessel on the equilibrium position.

### Step-by-Step Solution:

1. **Understand the Reaction**:
The reaction involves one mole of sulfur dichloride (\( \text{SO}_2\text{Cl}_2 \)) on the reactant side and produces one mole of sulfur dioxide (\( \text{SO}_2 \)) and one mole of chlorine gas (\( \text{Cl}_2 \)) on the product side.

- **Moles of Reactants**: 1 mole of \( \text{SO}_2\text{Cl}_2 \)
- **Moles of Products**: 1 mole of \( \text{SO}_2 \) + 1 mole of \( \text{Cl}_2 \) = 2 moles

2. **Effect of Volume Increase**:
When the volume of the reaction vessel is increased, the pressure of the system decreases. According to Le Chatelier's principle, the equilibrium will shift in the direction that counteracts this change, which means it will shift towards the side with more moles of gas.

3. **Determine Direction of Shift**:
In this case, the product side has 2 moles of gas (1 mole of \( \text{SO}_2 \) and 1 mole of \( \text{Cl}_2 \)), while the reactant side has only 1 mole of gas (\( \text{SO}_2\text{Cl}_2 \)). Therefore, the equilibrium will shift to the right (toward the products).

4. **Changes in Concentration**:
- As the equilibrium shifts to the right, the concentration of the products (\( \text{SO}_2 \) and \( \text{Cl}_2 \)) will increase.
- Conversely, the concentration of the reactant (\( \text{SO}_2\text{Cl}_2 \)) will decrease.

5. **Conclusion**:
- The amount of \( \text{SO}_2 \) increases.
- The amount of \( \text{Cl}_2 \) increases.
- The amount of \( \text{SO}_2\text{Cl}_2 \) decreases.

### Final Answer:
- The concentration of \( \text{SO}_2 \) increases.
- The concentration of \( \text{Cl}_2 \) increases.
- The concentration of \( \text{SO}_2\text{Cl}_2 \) decreases.