Question

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.

Answer 1

Answer: The concentration of hydrogen gas at equilibrium is $1.648\times 10^{-3}M$

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:

$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of the solution}}$

So, $\text{Initial molarity of hydrogen sulfide gas}=\frac{0.47}{3}=0.1567M$

The given chemical equation follows:

                          $2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)$

Initial:                  0.1567

At eqllm:           0.1567-2x       2x         x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[H_2]^2[S_2]}{[H_2S]^2}$

We are given:

$K_c=9.3\times 10^{-8}$

Putting values in above equation, we get:

$9.3\times 10^{-8}=\frac{(2x)^2\times x}{(0.1567-2x)^2}\\\\x=8.24\times 10^{-4}$

So, equilibrium concentration of hydrogen gas = $2x=(2\times 8.24\times 10^{-4})=1.648\times 10^{-3}M$

Hence, the concentration of hydrogen gas at equilibrium is $1.648\times 10^{-3}M$