Question

For 2SO2(g)+O2(g)⇌2SO3(g),

Kp=3.0×104 at 700 K. In a 2.00-L vessel the equilibrium mixture contains 1.13 g of SO3 and 0.104 g of O2

How many grams of SO2 are in the vessel?

Answer 1

The equilibrium reaction:
2 SO₂(g) + O₂(g) <------> 2 SO₃(g)

Molar mass of SO₃ = 80 g/mol
Molar mass of O₂ = 32 g/mol

Mole of SO₃ = 1.13 g / 80 g/mol = 0.014 mol
Mole of O₂ = 0.104 g / 32 g/mol = 0.0032 mol

Molarity of SO₃ = n / V = 0.014 mol / 2 L = 0.007 M
Molarity of O₂ = n / V = 0.0032 mol / 2 L = 0.0016 M

You cannot put them into the equilibrium expression, because Kc (equilibrium constant in terms of concentrations) is not given. The given is Kp (equilibrium constant in terms of partial pressures). Therefore you must first convert Kp to Kc.

Kp = Kc(RT)^dn

R = 22.4 / 273
T = 700 K
dn = moles of products - moles of reactants = 2 - 3 = -1

Kp = Kc (RT)^-1 = Kc / RT
Kc = Kp x RT = 3.0x10⁴ x (22.4/273)x(700) = 172.3x10⁴
Kc = 1.7 x 10^6

Kc = [SO₃]² / [SO₂]₂ x [O₂] = 1.7 x 10⁶
(0.007)² / [SO₂]₂ x (0.0016) = 1.7 x 10⁶

[SO₂]₂ = 0.0306 / 1.7 x 10⁶ = 1.801 x 10^-8
Taking square root of both sides;
[SO₂] = 1.34x 10^-4 = 0.000134 M

Moles of SO₂ = M x V = (0.000134 mol/L) x (2.00 L)
Moles of SO₂ = 0.000268 mol

Molar mass of SO₂ = 64 g/mol
Mass of SO₂ = (0.000268 mol) x (64 g/mol) = 0.017 g SO₂ in the vessel.