`SO_(3)(g)+CO(g)hArrSO_(2)(g)+CO_(2)(g)`
Which of the following activity can reduce the equilibrium concentration of CO to half?

Explanation

To solve the problem of how to reduce the equilibrium concentration of CO to half in the reaction:

\[ \text{SO}_3(g) + \text{CO}(g) \rightleftharpoons \text{SO}_2(g) + \text{CO}_2(g) \]

we need to analyze the effect of different changes on the equilibrium state of the reaction.

### Step 1: Write the expression for the equilibrium constant (K).
The equilibrium constant (K) for the reaction can be expressed as:

\[ K = \frac{[\text{SO}_2][\text{CO}_2]}{[\text{SO}_3][\text{CO}]} \]

### Step 2: Analyze the options provided.
We need to determine which action can effectively reduce the concentration of CO to half.

#### Option A: Increasing the pressure twice.
- Increasing the pressure generally favors the side of the reaction with fewer moles of gas. However, in this case, both sides have the same number of moles (2 moles on each side). Therefore, increasing the pressure will have no effect on the equilibrium concentrations.
- **Conclusion:** This option can be eliminated.

#### Option B: Increasing the volume of the reaction container to twice its initial value.
- Increasing the volume of the container decreases the concentration of all gaseous reactants and products because concentration is inversely proportional to volume. If the volume is doubled, the concentration of CO will decrease to half.
- **Conclusion:** This option is correct.

#### Option C: Adding two moles of helium gas at constant pressure.
- Helium is an inert gas and does not participate in the reaction. Adding it at constant pressure will increase the total volume of the system, but it does not change the partial pressures or concentrations of the reactants and products involved in the reaction.
- **Conclusion:** This option can be eliminated.

### Final Answer:
The correct action to reduce the equilibrium concentration of CO to half is **Option B: Increasing the volume of the reaction container to twice its initial value.**

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