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Fed [463]
3 years ago
13

Consider the following equilibrium: H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-1(aq). What is the correct equilibrium expression? Ka=[H3

O+][HCO1−3][H2CO3] Ka=[H3O+][H2CO3][HCO1−3] Ka=[H2CO3][H2O][H3O+][HCO1−3] Ka=[H3O+][HCO1−3][H2CO3][H2O]
Chemistry
2 answers:
Andrews [41]3 years ago
5 0

<u>Answer:</u> The correct answer is Ka=\frac{[H_3O^+][HCO_3^{-1}]}{[H_2CO_3]}

<u>Explanation:</u>

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power its stoichiometric coefficients.

For the general equation:

aA+bB\rightarrow cC+dD

The expression for equilibrium constant is given as:

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

For the given chemical equation:

H_2CO_3(aq.)+H_2O(l)\rightarrow H_3O^+(aq.)+HCO_3^{-1}(aq.)

The expression for equilibrium constant is:

K_a=\frac{[H_3O^+][HCO_3^{-1}]}{[H_2CO_3]}

The concentration of pure solids and liquids is taken as 1, therefore H_2O is not written in the given expression.

Hence, the correct expression is given above.

kupik [55]3 years ago
4 0
The activity of pure substance is taken as 1 because it has no effect on the reaction. Hence the equilibrium constant of the reaction is given by: <span>Ka=[H3O+][HCO1−3]/[H2CO3]. Hope this helps :)</span>
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