`SO_(2)+(1)/(2)O_(2)hArrSO_(3)`, for the above reaction, if 'a' and 'b' mole/1 are the initial concentrations of `SO_(2) and O_(2)` respectively and x mole `SO_(3)` is formed at equlibrium. The equilibrium concentration of `O_(2)` will be

Options

A. `b-x`
B. `b-2x`
C. `b-(x)/(2)`
D. `2b-x`

Answer

To solve the problem, we need to analyze the equilibrium reaction given:

\[ \text{SO}_2 + \frac{1}{2} \text{O}_2 \rightleftharpoons \text{SO}_3 \]

### Step-by-Step Solution:

1. **Identify Initial Concentrations**:
- Let the initial concentration of \( \text{SO}_2 \) be \( a \) moles/L.
- Let the initial concentration of \( \text{O}_2 \) be \( b \) moles/L.

2. **Define Change in Concentrations**:
- At equilibrium, let \( x \) moles/L of \( \text{SO}_3 \) be formed.
- According to the stoichiometry of the reaction:
- For every 1 mole of \( \text{SO}_2 \) that reacts, \( \frac{1}{2} \) mole of \( \text{O}_2 \) is consumed.
- Therefore, if \( x \) moles of \( \text{SO}_3 \) are formed, \( x \) moles of \( \text{SO}_2 \) will be consumed and \( \frac{x}{2} \) moles of \( \text{O}_2 \) will be consumed.

3. **Write the Equilibrium Concentrations**:
- The equilibrium concentration of \( \text{SO}_2 \) will be:
\[ [\text{SO}_2] = a - x \]
- The equilibrium concentration of \( \text{O}_2 \) will be:
\[ [\text{O}_2] = b - \frac{x}{2} \]
- The equilibrium concentration of \( \text{SO}_3 \) will be:
\[ [\text{SO}_3] = x \]

4. **Find the Equilibrium Concentration of \( \text{O}_2 \)**:
- From the previous step, we have already derived the expression for the equilibrium concentration of \( \text{O}_2 \):
\[ [\text{O}_2] = b - \frac{x}{2} \]

### Final Answer:
The equilibrium concentration of \( \text{O}_2 \) is:
\[ b - \frac{x}{2} \]