Question

The formation of SO3 from SO2 and O2 is an intermediate step in the manufacture of sulfuric acid, and it is also responsible for the acid rain phenomenon. The equilibrium constant KP for the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g) is 0.13 at 803°C. In one experiment, 2.00 mol SO2 and 2.00 mol O2 were initially present in a flask. What must be the total pressure at equilibrium in order to have an 61.0% yield of SO3?

Answer 1

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

2.0          2.0               0

2-2x         2-x               2x

at equilibrium 61 % of SO3 obtained

moles of SO3 = 0.61 x 2 = 1.22 mol

at equilibrium :

moles of SO3 = 1.22

moles of SO2 = 0.78

moles of O2 = 1.39

total moles = 1.22 + 0.78 + 1.39 = 3.39

mole fraction of SO2 = 0.78 / 3.39 = 0.23

mole fraction of O2 = 1.39 / 3.39 = 0.41

mole fraction of SO3 = 1.22 / 3.39 = 0.36

let us assume :

total pressure = P atm

PSO2 = mole fraction x total pressure

PSO2 = 0.23 x P

PO2 = 0.41 x P

PSO3 = 0.36 x P

Kp = (PSO3)^2 / (PSO2)^2 x PO2

0.13 = (0.36 P)^2 / (0.23 P)^2 x (0.41P)

P = 45.96

total pressure = 46.0 atm