1. Given reaction

$2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)$

Initial partial pressures:

$P_{SO_2}=250\ \text{mbar},\quad P_{O_2}=750\ \text{mbar},\quad P_{SO_3}=0$

We assume constant temperature and volume, so partial pressure is proportional to moles.

2. Find the limiting reactant

From the reaction:

$2SO_2 : 1O_2$

For $250$ mbar of $SO_2$, required $O_2$ is:

$P_{O_2,\ required}=\frac{1}{2}\times 250=125\ \text{mbar}$

Available $O_2$ is $750$ mbar, which is much larger than $125$ mbar.

So, $SO_2$ is the limiting reactant.

3. Calculate pressure changes

If all $250$ mbar of $SO_2$ reacts:

• $SO_2$ consumed = $250$ mbar
• $O_2$ consumed = $\frac{250}{2}=125$ mbar
• $SO_3$ formed = $250$ mbar

Final partial pressures:

$P_{SO_2}=250-250=0$ $P_{O_2}=750-125=625\ \text{mbar}$ $P_{SO_3}=0+250=250\ \text{mbar}$

4. Total final pressure

$P_{\text{total, final}}=0+625+250=875\ \text{mbar}$

5. Final answer

$\boxed{875\ \text{mbar}}$