Respuesta :
Answer: The concentration of hydrogen gas at equilibrium is [tex]1.648\times 10^{-3}M[/tex]
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:
[tex]\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of the solution}}[/tex]
So, [tex]\text{Initial molarity of hydrogen sulfide gas}=\frac{0.47}{3}=0.1567M[/tex]
The given chemical equation follows:
[tex]2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)[/tex]
Initial: 0.1567
At eqllm: 0.1567-2x 2x x
The expression of [tex]K_c[/tex] for above equation follows:
[tex]K_c=\frac{[H_2]^2[S_2]}{[H_2S]^2}[/tex]
We are given:
[tex]K_c=9.3\times 10^{-8}[/tex]
Putting values in above equation, we get:
[tex]9.3\times 10^{-8}=\frac{(2x)^2\times x}{(0.1567-2x)^2}\\\\x=8.24\times 10^{-4}[/tex]
So, equilibrium concentration of hydrogen gas = [tex]2x=(2\times 8.24\times 10^{-4})=1.648\times 10^{-3}M[/tex]
Hence, the concentration of hydrogen gas at equilibrium is [tex]1.648\times 10^{-3}M[/tex]
