To solve the problem, we need to analyze the effect of reducing the volume of a system at equilibrium, specifically for the reaction:
\[ \text{SO}_2\text{Cl}_2(g) \rightleftharpoons \text{SO}_2(g) + \text{Cl}_2(g) \]
### Step-by-Step Solution:
1. **Understand the Reaction**:
The reaction involves one mole of sulfur dichloride (\( \text{SO}_2\text{Cl}_2 \)) decomposing into one mole of sulfur dioxide (\( \text{SO}_2 \)) and one mole of chlorine (\( \text{Cl}_2 \)). Therefore, the total number of moles of gas on the reactant side is 1, and on the product side, it is 2.
2. **Effect of Reducing Volume**:
According to the ideal gas law, reducing the volume of a gas increases its pressure. When the volume is reduced, the concentration of all gaseous species in the reaction increases because concentration is defined as the number of moles divided by the volume.
3. **Applying Le Chatelier's Principle**:
Le Chatelier's principle states that if a system at equilibrium is subjected to a change in pressure, temperature, or concentration, the system will adjust to counteract that change and establish a new equilibrium. In this case, since the pressure is effectively increased by reducing the volume, the equilibrium will shift in the direction that reduces the number of moles of gas.
4. **Direction of Shift**:
In this reaction:
- Reactants: 1 mole of \( \text{SO}_2\text{Cl}_2 \)
- Products: 1 mole of \( \text{SO}_2 \) + 1 mole of \( \text{Cl}_2 \) = 2 moles
Since there are more moles of gas on the product side (2 moles) compared to the reactant side (1 mole), the equilibrium will shift to the left (towards the reactants) to reduce the total number of moles of gas in response to the increased pressure.
5. **Conclusion**:
Therefore, when the volume is reduced (and thus pressure is increased), the equilibrium will shift to the left, favoring the formation of \( \text{SO}_2\text{Cl}_2 \) and increasing the concentration of the reactants. However, it is important to note that the concentrations of all species (reactants and products) will increase due to the reduction in volume.
### Final Answer:
The effect on equilibrium due to reducing the volume (increasing pressure) will be that the reaction shifts to the left, favoring the formation of \( \text{SO}_2\text{Cl}_2 \), while the concentrations of all gases will increase.
