Question

2H₂S (g) + 3O₂ (g) <---> 2SO₂ (g) + H₂O (g)

K_c= 1.35x10^-8 at 100K

If 30.0 g H₂S (g) is sealed in a 250.0mL flask with 50.0 g of oxygen, what is the concentration of SO₂ (g) when equilibrium is established?

Answer 1

we know that

moles = mass / molar mass

so

moles of H2S = 30 / 34 = 0.88235

moles of 02 = 50 / 32 = 1.5625

now

concentration = moles x 1000 / volume (ml)

so

initally

[H2S] = 0.88235 x 1000 / 250 = 3.529

[02] = 1.5625 x 1000 / 250 = 6.25

now

consider the given reaction

2 H2S + 3 02 ---> 2 S02 + 2H20

using ICE table

at equilibrium

[H2S] = 3.529 - 2x

[02] = 6.25 - 3x

[S02] = 2x

[H20] = 2x

now

2 H2S + 3 02 ---> 2 S02 + 2H20

the equilibrium constant is given by

Kc = [S02]^2 [H20]^2 / [H2S]^2 [02]^3

so

Kc = [2x]^2 [2x]^2 / [3.529 -2x]^2 [6.25 - 3x]^3

Kc = 16x^4 / [3.529 -2x]^2 [6.25 - 3x]^3

now

x is very small so

3.529 - 2x = 3.529

and

6.25 - 3x = 6.25

so

Kc = 16x^4 / [3.529]^2 [6.25]^3

1.35 * 10-8 = 16x^4 / [3.529]^2 [6.25]^3

x = 0.04

so

at equilibrium

[S02] = 2x = 2 * 0.04 = 0.08

so

the concentration of S02 at equilibrium is 0.08 M