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Calculate Kc for the reaction: 2 HI(g) ⇄ H2(g) + I2(g) given that the concentrations of each species at equilibrium are as follows: [HI] = 0.85 mol/L, [I2] = 0.60 mol/L, [H2] = 0.27 mol/L.

Respuesta :

Answer: The value of [tex]K_c[/tex] for the given reaction is 0.224

Explanation:

For the given chemical equation:

[tex]2HI(g)\rightleftharpoons H_2(g)+I_2(g)[/tex]

The expression of [tex]K_c[/tex] for given equation follows:

[tex]K_c=\frac{[H_2][I_2]}{[HI]^2}[/tex]

We are given:

[tex][HI]_{eq}=0.85M[/tex]

[tex][H_2]_{eq}=0.27M[/tex]

[tex][I_2]_{eq}=0.60M[/tex]

Putting values in above expression, we get:

[tex]K_c=\frac{0.27\times 0.60}{(0.85)^2}=0.224[/tex]

Hence, the value of [tex]K_c[/tex] for the given reaction is 0.224

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