Final answer:
The partial pressure of the excess reactant, sulfur dioxide, remains at 0.84 atm, which equates to 638.4 torr, as oxygen was the limiting reactant and was completely consumed.
Step-by-step explanation:
The student is asking about calculating the partial pressure of the excess reactant in a chemical reaction between sulfur dioxide and oxygen to form sulfur trioxide. Given the total pressure and the partial pressure of one of the reactants, we can determine the partial pressure of the other reactant before the reaction. After the reaction, assuming 100% yield and no change in volume or temperature, the partial pressure of the excess reactant will be the same as it was initially if that reactant was in excess, or it will be zero if it was completely consumed.
The total initial pressure is given as 1.23 atm and the partial pressure of molecular oxygen is 0.39 atm. Assuming that only sulfur dioxide and oxygen are present in the mixture, we can calculate the initial partial pressure of sulfur dioxide by subtracting the partial pressure of oxygen from the total pressure:
Partial pressure of SO2 = Total pressure - Partial pressure of O2
= 1.23 atm - 0.39 atm
= 0.84 atm
Since the reaction stoichiometry is 2:1, it means that sulfur dioxide will be the limiting reactant if its partial pressure is exactly twice that of oxygen. However, in this case, the amount of sulfur dioxide is more than twice the partial pressure of oxygen, indicating that oxygen is the limiting reactant and sulfur dioxide is in excess. After the reaction, the partial pressure of oxygen will be zero, and the excess partial pressure of sulfur dioxide will remain the same, which is 0.84 atm. To express this in torr, we multiply by the conversion factor (1 atm = 760 torr).
Partial pressure of excess reactant (SO2) in torr = 0.84 atm × 760 torr / atm
= 638.4 torr