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Nitrogen dioxide and water react to form nitric acid and nitrogen monoxide, like this:

2HNO3 (aq) + NO (g) → 3NO2 (g) + H2O (l)

At a certain temperature, a chemist finds that a 9.5L reaction vessel containing a mixture of nitric acid, nitrogen monoxide, nitrogen dioxide, and water at equilibrium has the following composition:

Compound Amount
HNO3 15.5g
NO 16.6g
NO2 22.5g
H2O 189.0g

Required:
Calculate the value of the equilibrium constant Kc for this reaction.

Respuesta :

Answer: The value of the equilibrium constant Kc for this reaction is 3.72

Explanation:

Equilibrium concentration of [tex]HNO_3[/tex] = [tex]\frac{15.5g}{63g/mol\times 9.5L}=0.026M[/tex]

Equilibrium concentration of [tex]NO[/tex] = [tex]\frac{16.6g}{30g/mol\times 9.5L}=0.058M[/tex]

Equilibrium concentration of [tex]NO_2[/tex] = [tex]\frac{22.5g}{46g/mol\times 9.5L}=0.051M[/tex]

Equilibrium concentration of [tex]H_2O[/tex] = [tex]\frac{189.0g}{18g/mol\times 9.5L}=1.10M[/tex]

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_c[/tex]  

For the given chemical reaction:

[tex]2HNO_3(aq)+NO(g)\rightarrow 3NO_2(g)+H_2O(l)[/tex]

The expression for [tex]K_c[/tex] is written as:

[tex]K_c=\frac{[NO_2]^3\times [H_2O]^1}{[HNO_3]^2\times [NO]^1}[/tex]

[tex]K_c=\frac{(0.051)^3\times (1.10)^1}{(0.026)^2\times (0.058)^1}[/tex]

[tex]K_c=3.72[/tex]

Thus  the value of the equilibrium constant Kc for this reaction is 3.72

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