Problem Page Fill in the left side of this equilibrium constant equation for the reaction of hypobromous acid with water.

Chemistry

1 answer:

jeyben [28]4 years ago

4 0

<u>Answer:</u> The equilibrium constant for the given reaction is $K_a=\frac{[BrO^-][H_3O^+]}{[HBrO]}$

<u>Explanation:</u>

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{eq}$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_{eq}$ is written as:

$K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

The concentration of pure liquids and pure solids are taken as 1.

The chemical equation for the reaction of hypobromous acid with water follows:

$HBrO(aq.)+H_2O(l)\rightarrow BrO^-(aq.)+H_3O^+(aq.)$

The expression of equilibrium constant for above equation follows:

$K_a=\frac{[BrO^-][H_3O^+]}{[HBrO]}$

Hence, the equilibrium constant for the given reaction is $K_a=\frac{[BrO^-][H_3O^+]}{[HBrO]}$

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