In a vessel containing `SO_(3), SO_(2)` and `O_(2)` at equilibrium, some helium gas is introduced so that total pressure increases while temperature and volume and volume remain the same. According to Le Chatelier's principle, the dissociation of `SO_(3)`:

Options

A. Increases
B. Decreases
C. Remains unaltered
D. Changes unpredictably

Answer

To solve the problem, we need to analyze the situation using Le Chatelier's principle, which states that if a system at equilibrium is subjected to a change in conditions, the system will adjust to counteract that change and restore a new equilibrium.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- We have a vessel containing sulfur trioxide (SO₃), sulfur dioxide (SO₂), and oxygen (O₂) at equilibrium.

2. **Introduce a Change**:
- Helium gas is introduced into the vessel, which increases the total pressure. However, the temperature and volume of the system remain constant.

3. **Understand the Effect of Helium**:
- Helium is an inert gas and does not react with the components of the equilibrium system. Its introduction increases the total pressure but does not affect the number of moles of the reacting gases (SO₃, SO₂, O₂).

4. **Analyze the Partial Pressures**:
- Since the volume is constant and the number of moles of the reactive gases does not change, the partial pressures of SO₃, SO₂, and O₂ remain unchanged. The increase in total pressure is due solely to the addition of helium.

5. **Apply Le Chatelier's Principle**:
- According to Le Chatelier's principle, if there is no change in the partial pressures of the reactants or products, the equilibrium position will not shift. Therefore, the dissociation of SO₃ will not be affected.

6. **Conclusion**:
- The equilibrium remains unaltered despite the increase in total pressure due to the introduction of helium gas.

### Final Answer:
The dissociation of SO₃ remains unaltered.

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