Question

Hydrogen Sulfide decomposes according to the following reaction,for which Kc=9.3 x 10^-8 at 400 degrees Celsius. 2 H2S(g) ⇄ 2 H2(g) + S2(g)

If 0.47 moles of H2S are placed in a 3.0 L container, and the system is allowed to reach equilibrium. Calculate the concentration of H2(g) at equilibrium?

Answer 1

The concentration of H2 at equilibrium = 2X = 0.001659 M

Step-by-step explanation:

Step 1: The balanced equation

2 H2S(g) ⇔ 2 H2(g) + S2(g)

This means that for 2 moles of H2S consumed there is produced 2 moles of H2 and 1 mole S2

Step 2: Given data

0.47 moles of a gas H2S is placed in a 3L container.

Kc = 9.3 * 10^-8 (this at 400 °C)

Step 3: Calculate initial concentration of H2S

Molarity of H2S = moles of H2S / volume of H2S

Molarity of H2S = 0.47 moles / 3L

Molarity of H2S = 0.1567 M

Step 4: The initial concentration of H2S is 0.1567 M

The initial concentration of H2 and S2 is 0M

There will react 2X of H2S, that means there will be also produced 2X of H2 and X of S2 ( Since the ratio is 2:2:1).

The final concentration of H2S is 0.1567 - 2X

The final concentration of H2 is 2X

The final concentration of S2 is X

Step 5: Write the Kc

Kc =9.3 * 10^-8 = ([S2]*[H2]² )/ [H2S]²

Kc = 9.3 * 10^-8 = X *(2X)² / (0.1567 - 2X)²

4X³ = 0.1567²X * 9.3*10^-8

if we calculate X = 0.000829571

The concentration of S2 at equilibrium = X = 0.000829571 M

The concentration of H2 at equilibrium = 2X = 0.001659 M