Consider the reaction:
2 SO2(g) + O2(g) → 2 SO3(g)
A reaction flask initially contains 0.10 atm of SO2 and 0.10 atm of O2. What is the total pressure in the flask once the limiting reactant is completely consumed? Assume a constant temperature and volume and a 100% reaction yield.
Here, we are going to calculate the total pressure in the given reaction flask once the limiting reactant is completely consumed.
The given reaction is:
2SO2(g) + O2(g) → 2SO3(g)
From the above equation, it is clear that 2 moles of SO2 requires 1 mole of O2 for complete reaction.
Now, at constant temperature and volume, the pressure(p) of a gas is directly proportional to the number of moles(n).
p ∝ n
Now, if p1, n1 are the pressure and number of moles of a gas and p2, n2 are the volume and number of moles of another gas , then,
p1 = kn1
p2 = kn2
Combining the two equations, we get, p1 / n1 = p2 / n2 -----(1)
Let, p1, n1 are the pressure and number of moles of SO2 and p2, n2 are the volume and number of moles of O2 , then, pressure of O2 required to react with 0.10 atm pressure of SO2 is given by:
p2 = (p1 x n2) / n1
= (0.10 atm x 1 mol) / 2 mol
= 0.05 atm
But we have oxygen at 0.10 atm pressure, i.e., the pressure of O2 is in excess. Thus, SO2 acts as a limiting reactant.
Pressure of O2 after the reaction is complete = initial pressure - pressure required in the reaction
= 0.10 atm - 0.05 atm
= 0.05 atm
Again, since 2 moles of SO2 produces 2 moles of SO3
Therefore, pressure of SO3 after reaction is complete = pressure of SO2 consumed = 0.10 atm.
Total pressure in the flask after the limiting reactant is completely consumed = Pressure of SO3 + pressure of remaining O2
= (0.10 + 0.05) atm
= 0.15 atm.
Total pressure in the flask once the limiting reactant is completely consumed is 0.15 atm.