2 H2S(g) ⇄ 2 H2(g) + S2(g) Kc = 9.3× 10-8 at 400ºC 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 at equilibrium.

The concentration of hydrogen gas at equilibrium is

Explanation:

We are given:

Initial moles of hydrogen sulfide gas = 0.47 moles

Volume of the container = 3.0 L

The molarity of solution is calculated by using the equation:

So,

The given chemical equation follows:

Initial:                  0.1567

At eqllm:           0.1567-2x       2x         x

The expression of for above equation follows:

We are given:

Putting values in above equation, we get:

So, equilibrium concentration of hydrogen gas =

Hence, the concentration of hydrogen gas at equilibrium is

it would be a gas formation