Question

Enter your answer in the provided box.Hydrogen sulfide decomposes according to the following reaction, for whichKc = 9.30 × 10−8 at 700°C:2 H2S(g) ⇌ 2 H2(g) + S2(g)If 0.55 mol of H2S is placed in a 3.0−L container, what is the equilibrium concentration of H2(g) at 700°C?? M

Answer 1

Explanation: The equation for the reaction is:

2 H2S(g) ⇌ 2 H2(g) + S2(g)

We can use the concentration of H2S to calculate the concentration of H2 at equilibrium. Let's call the initial concentration of H2S [H2S]initial and the concentration of H2 at equilibrium [H2]equilibrium.

At equilibrium, the rate of the forward reaction must equal the rate of the reverse reaction, which means that:

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

Since [H2S]initial = 0.55 mol / 3.0 L = 0.183 mol/L, we can substitute this value into the expression for Kc:

Kc = 9.30 × 10^−8 = [H2]^2 / (0.183)^2

Now we can solve for [H2]equilibrium:

[H2]equilibrium = √(Kc * [H2S]initial^2) = √(9.30 × 10^−8 * (0.183)^2) = 0.0276 mol/L

So the equilibrium concentration of H2 at 700°C is 0.0276 mol/L.