Question

2SO$_2$(g) + O$_2$(g) $\rightarrow$ 2SO$_3$(g)
The above reaction is carried out in a vessel starting with partial pressures 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).

Hint:

For reactions with gases, partial pressures are directly proportional to the number of moles (Dalton's Law). You can use an ICE (Initial, Change, Equilibrium/End) table with pressures just as you would with moles to solve stoichiometry problems.

Answer 1

Correct Answer: 875

The balanced chemical equation is: $2\text{SO}_2(g) + \text{O}_2(g) \rightarrow 2\text{SO}_3(g)$.
The reaction stoichiometry shows that 2 moles (or units of pressure) of SO$_2$ react with 1 mole (or unit of pressure) of O$_2$.
Initial partial pressures:
P$_{SO_2}$ = 250 mbar
P$_{O_2}$ = 750 mbar
P$_{SO_3}$ = 0 mbar
First, we must identify the limiting reactant.
To completely react 250 mbar of SO$_2$, we would need $(1/2) \times 250 = 125$ mbar of O$_2$.
Since we have 750 mbar of O$_2$, which is more than 125 mbar, O$_2$ is in excess and SO$_2$ is the limiting reactant.
The reaction goes to completion, which means all of the limiting reactant (SO$_2$) is consumed.
Change in pressures based on stoichiometry:
Change in P$_{SO_2}$ = -250 mbar
Change in P$_{O_2}$ = -125 mbar (half of the change in SO$_2$)
Change in P$_{SO_3}$ = +250 mbar (same as the change in SO$_2$)
Final partial pressures:
Final P$_{SO_2}$ = 250 - 250 = 0 mbar
Final P$_{O_2}$ = 750 - 125 = 625 mbar
Final P$_{SO_3}$ = 0 + 250 = 250 mbar
The total pressure in the vessel after the reaction is the sum of the final partial pressures.
P$_{total}$ = Final P$_{SO_2}$ + Final P$_{O_2}$ + Final P$_{SO_3}$
P$_{total}$ = 0 + 625 + 250 = 875 mbar.