Question

What is the value of the equilibrium constant for the following reaction? Assume 25 C, ideal behavior

CrOH2+ + H2O ⇌ Cr(OH)2+ + H+

Answer 1

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Answer 2

The equilibrium constant expression for the given reaction is:

K = [Cr(OH)2+][H+] / [CrOH2+][H2O]

Since the reaction is written with one molecule of water on the reactant side, we need to use the expression for the ion product of water (Kw) to substitute for [H2O]:

Kw = [H+][OH-] = 1.0 x 10^-14 at 25°C

Rearranging the expression, we get:

[H2O] = Kw / [H+]

Substituting [H2O] in the equilibrium constant expression, we get:

K = [Cr(OH)2+][H+] / [CrOH2+][Kw / [H+]]

Simplifying, we get:

K = (Kw[C r(OH)2+]) / [CrOH2+]

At 25°C, the ion product of water (Kw) is 1.0 x 10^-14, so:

K = (1.0 x 10^-14)([Cr(OH)2+]) / [CrOH2+]

The value of the equilibrium constant depends on the concentration of the species at equilibrium. Without knowing the initial concentrations and the equilibrium concentrations of the species involved, we cannot calculate the equilibrium constant.