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

We are given the gas-phase reaction $$2\,SO_2(g)+O_2(g)\rightarrow 2\,SO_3(g)$$ which proceeds to completion in a rigid vessel at constant temperature. Because temperature and volume do not change, the partial pressure of each gas is directly proportional to the number of moles present. Hence, we can treat the given partial pressures exactly like the initial “mole” amounts for stoichiometric calculations.

Initially we have

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

According to the balanced equation, the stoichiometric ratio is $$2\;:\;1\;:\;2$$ for $$SO_2 : O_2 : SO_3$$. That is, $$2$$ units of $$SO_2$$ react with $$1$$ unit of $$O_2$$ to form $$2$$ units of $$SO_3$$.

To identify the limiting reactant, we compare the available amounts in the required ratio. The reaction needs half as much $$O_2$$ as $$SO_2$$. Specifically, the amount of $$O_2$$ required for the given $$SO_2$$ is

$$P_{O_2}^{\,\text{needed}} \;=\;\frac{1}{2}\,P_{SO_2}^{\,\text{initial}} \;=\;\frac{1}{2}\,(250\,\text{ mbar}) \;=\;125\,\text{ mbar}.$$

Because the vessel actually contains $$750\,\text{ mbar}$$ of $$O_2$$, which is much larger than the $$125\,\text{ mbar}$$ required, $$SO_2$$ is the limiting reactant. Therefore the reaction will consume all of the initial $$SO_2$$.

Now we calculate how much $$O_2$$ is consumed. From the stoichiometry $$2\,SO_2 \longrightarrow 1\,O_2,$$ every $$2$$ units of $$SO_2$$ use up $$1$$ unit of $$O_2$$, so

$$P_{O_2}^{\,\text{consumed}} \;=\;\frac{1}{2}\,P_{SO_2}^{\,\text{consumed}} \;=\;\frac{1}{2}\,(250\,\text{ mbar}) \;=\;125\,\text{ mbar}.$$

The remaining $$O_2$$ pressure after completion is therefore

$$P_{O_2}^{\,\text{final}} \;=\;P_{O_2}^{\,\text{initial}} \;-\; P_{O_2}^{\,\text{consumed}} \;=\;750\,\text{ mbar}\;-\;125\,\text{ mbar} \;=\;625\,\text{ mbar}.$$

Next, we determine the pressure of $$SO_3$$ formed. The stoichiometry gives $$2\,SO_2 \longrightarrow 2\,SO_3,$$ meaning the same number of units of $$SO_3$$ are produced as the number of $$SO_2$$ units consumed. Hence,

$$P_{SO_3}^{\,\text{formed}} \;=\;P_{SO_2}^{\,\text{consumed}} \;=\;250\,\text{ mbar}.$$

Because no $$SO_3$$ was present initially, the final $$SO_3$$ pressure is simply this amount:

$$P_{SO_3}^{\,\text{final}} = 250\,\text{ mbar}.$$

Finally, we add the partial pressures of all gases remaining in the vessel to obtain the total final pressure:

$$P_{\text{total}}^{\,\text{final}} \;=\;P_{SO_2}^{\,\text{final}} + P_{O_2}^{\,\text{final}} + P_{SO_3}^{\,\text{final}}.$$

The entire initial $$SO_2$$ has reacted, so $$P_{SO_2}^{\,\text{final}} = 0$$. Substituting the values, we get

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

Rounding to the nearest integer is unnecessary because the result is already an integer. Hence, the correct answer is Option A: $$875\,\text{mbar}$$.