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2 H2(g) + S2(g) equilibrium reaction arrow 2 H2S(g) At a certain temperature, Kc = 1.30 ✕ 1010 for the reaction above. Calculate the equilibrium concentration of H2S if [H2] = 0.00400 M and [S2] = 0.00270 M at equilibrium in a container.

Respuesta :

Answer: The equilibrium concentration of hydrogen sulfide is 23.7 M

Explanation:

For the given chemical equation:

[tex]2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g)[/tex]

The expression of [tex]K_{c}[/tex] for above equation follows:

[tex]K_{c}=\frac{[H_2S]^2}{[H_2]^2[S_2]}[/tex]

We are given:

[tex]K_{c}=1.30\times 10^{10}[/tex]

[tex][H_2]_{eq}=0.00400M[/tex]

[tex][S_2]_{eq}=0.00270M[/tex]

Putting values in above expression, we get:

[tex]1.30\times 10^{10}=\frac{[H_2S]^2}{(0.00400)^2\times 0.00270}[/tex]

[tex][H_2S]_{eq}=23.7,-23.7[/tex]

Neglecting the negative value of 'x' because concentration cannot be negative

Hence, the equilibrium concentration of hydrogen sulfide is 23.7 M

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