Question

Hydrogen sulfide decomposes according to the following reaction,for which Kc = 9.30 multiplied by 10-8 at 700°C. 2 H2S(g) reverse reaction arrow 2H2(g) + S2(g) If 0.29 mol H2S is placed in a 3.0 L container, what is the equilibrium concentration of H2(g) at 700°C?

Answer 1

Solution :-

2H2S --- > 2 H2 + S2                kc = 9.30*10^-8

0.29 mol in 3.0 L container

What is the equilibrium concentration of H2

Lets first calculate the initial concentration of the H2S

Molarity = moles / liter

Molarity of H2S = 0.29 mol / 3.0 L = 0.0967 M

Now lets make the ICE table

           2H2S           --- >        2 H2         +         S2

0.0967 M                               0                             0

-2x                                        +2x                           +x

0.0967-2x                               2x                         x

Now lets write the Kc equation

Kc= [H2]^2[S2]/[H2S]^2

9.30*10^-8 = [2x]^2[x]/[0.0967-2x]^2

9.30*10^-8 *[0.0967-2x]^2 = [2x]^2[x]

9.30*10^-8 *[0.0967-2x]^2 = 4x^3

Solving this equation to get x

We get

X= 0.00060 M

Therefore the equilibrium concentration of the H2 = 2x = 2*0.0006 M = 1.20 M