`2 SO _(2) (g)+ O _(2) (g) to 2 SO _(3) (g)`
The above reaction is carried out in a vessel starting with partial pressure `P _( SO _(2)) = 250 m` bar `P _( O _(2)) = 750 m` bar and `P _( SO _(3)) = 0` bar . When the reaction is complete, the total pressure in the reaction vessel is _____ m bar.
(Round off to the nearest integer)

Answer

To solve the problem, we need to analyze the reaction and calculate the total pressure in the vessel after the reaction is complete.

### Step-by-Step Solution:

1. **Write the balanced chemical equation:**
\[ 2 \text{SO}_2(g) + \text{O}_2(g) \rightarrow 2 \text{SO}_3(g) \]

2. **Identify initial pressures:**
- \( P_{\text{SO}_2} = 250 \, \text{mbar} \)
- \( P_{\text{O}_2} = 750 \, \text{mbar} \)
- \( P_{\text{SO}_3} = 0 \, \text{mbar} \)

3. **Determine the limiting reactant:**
- From the balanced equation, 2 moles of \(\text{SO}_2\) react with 1 mole of \(\text{O}_2\).
- The ratio of \(\text{SO}_2\) to \(\text{O}_2\) is 2:1.
- Calculate the amount of \(\text{O}_2\) needed for the available \(\text{SO}_2\):
\[ \text{Required } P_{\text{O}_2} = \frac{250 \, \text{mbar}}{2} = 125 \, \text{mbar} \]
- Since we have \(750 \, \text{mbar}\) of \(\text{O}_2\), \(\text{SO}_2\) is the limiting reactant.

4. **Calculate the change in pressure:**
- All \(250 \, \text{mbar}\) of \(\text{SO}_2\) will be consumed.
- The amount of \(\text{O}_2\) consumed:
\[ \text{Consumed } P_{\text{O}_2} = 125 \, \text{mbar} \]
- Remaining pressure of \(\text{O}_2\):
\[ P_{\text{O}_2 \text{ remaining}} = 750 \, \text{mbar} - 125 \, \text{mbar} = 625 \, \text{mbar} \]

5. **Calculate the pressure of \(\text{SO}_3\) produced:**
- According to the reaction, \(2 \, \text{moles of SO}_2\) produce \(2 \, \text{moles of SO}_3\).
- Therefore, \(250 \, \text{mbar}\) of \(\text{SO}_2\) will produce \(250 \, \text{mbar}\) of \(\text{SO}_3\).

6. **Calculate the total pressure in the vessel after the reaction is complete:**
\[ P_{\text{total}} = P_{\text{SO}_3} + P_{\text{O}_2 \text{ remaining}} = 250 \, \text{mbar} + 625 \, \text{mbar} = 875 \, \text{mbar} \]

### Final Answer:
The total pressure in the reaction vessel when the reaction is complete is **875 mbar**.

---