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HBrO(aq) + H2O(l) ⇄ H3O+(aq) + BrO−(aq)   Keq=2.8×10−9


The equilibrium reaction in 0.100M HBrO(aq) at equilibrium is represented by the equation above. Based on the magnitude of the equilibrium constant, which of the following correctly compares the equilibrium concentrations of substances involved in the reaction, and why?

answer choices
The equilibrium concentration of BrO− will be much smaller than the equilibrium concentration of H3O+, because H2O is the solvent and is present in the largest amount.

The equilibrium concentration of BrO− will be much smaller than the equilibrium concentration of HBrO, because Keq<<1

The equilibrium concentration of H3O+ will be much smaller than the equilibrium concentration of BrO−, because all the HBrO will react to produce BrO−

The equilibrium concentration of H3O+ will be much larger than the equilibrium concentration of HBrO, because Keq<<1.

Answer :

Answer: The equilibrium concentration of [tex]BrO^-[/tex] will be much smaller than the equilibrium concentration of [tex]HBrO[/tex], because Keq<<1

Explanation:

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 [tex]K_{eq}[/tex]

K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

For the given chemical reaction:

[tex]HBrO(aq)+H_2O(l)\rightarrow H_3O^+(aq)+BrO^-(aq)[/tex]

The expression for [tex]K_{eq}[/tex] is written as:

[tex]K=\frac{[H_3O^+]\times [BrO^-]}{[HBrO]}[/tex]

Concentration of pure solids and liquids is taken as 1.

[tex]K=2.8\times 10^{-9}[/tex]

Thus as [tex]K_{eq}<1[/tex] , That means the concentration of products is less as the reaction does not proceed much towards the forward direction.

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