Please help with chem!!?

Question

Hydrogen sulfide decomposes according to the following reaction, for which Kc = 9.30 10-8 at 700°C.

2 H2S(g) <~> 2 H2(g) + S2(g)

If 0.46 mol H2S is placed in a 2.0 L container, what is the equilibrium concentration of H2(g) at 700°C?
____ M

Answer 1

It is important to have the unit of Kc. Let us assume that the unit is mol/L.
Let the equilibrium concentration of S2(g) at 700°C be X.
............2 H2S(g) <==> 2 H2(g) + S2(g)
Initial: 0.23 (mol/L)......0 mol/L....0 mol/L
finally: 0.23 - 2X mol/L..2X mol/L..X mol/L
Hence Kc = 9.30x10^(-8) = X(2X)^2/(0.23 - 2X)^2
9.30x10^(-8)*(0.23 - 2X)^2 = 4X^3
let 2X = Y, we have:
18.60x10^(-8)*(0.23 - Y)^2 = Y^3
Solve this equation for Y and Y is the requested answer.