Respuesta :
Answer:
Equilibrium concentration of [tex]NO_{2}[/tex] is 0.088 M.
Explanation:
Chemical equilibrium: [tex]N_{2}O_{4}(g)\rightleftharpoons 2NO_{2}(g)[/tex]
Equilibrium constant in terms of concentration ( [tex]K_{c}[/tex])for the given reaction is expressed as:
[tex]K_{c}=\frac{[NO_{2}]_{eq}^{2}}{[N_{2}O_{4}]_{eq}}[/tex]
where, [tex][NO_{2}]_{eq}[/tex] and [tex][N_{2}O_{4}]_{eq}[/tex] refer equilibrium concentration of [tex]NO_{2}[/tex] and [tex]N_{2}O_{4}[/tex] respectively.
Here [tex]K_{c}[/tex] = 0.21 and [tex][N_{2}O_{4}]_{eq}[/tex] = 0.037 M
So, [tex][NO_{2}]_{eq}=\sqrt{K_{c}.[N_{2}O_{4}]_{eq}}[/tex] = [tex]\sqrt{(0.21)\times 0.037}[/tex] M =0.088 M
Hence equilibrium concentration of [tex]NO_{2}[/tex] is 0.088 M.
The equilibrium concentration of [tex]NO2(g)[/tex] would be:
[tex]0.088 M[/tex]
What is equilibrium concentration?
Equilibrium concentration is where the ratio remains constant among the reactant, as well as, the product.
Given the reaction,
[tex]N2O4(g)[/tex] ⇌ [tex]2 NO2(g)[/tex]
Equilibrium will be established as:
[tex]N_{2} O_{4}(g)[/tex] ⇄ [tex]2NO_{2}(g)[/tex]
The equilibrium constant can be determined through:
[tex]K_{c} = \frac{(NO_{2})^2_{eq.} }{(N_{2}O_{4})_{eq.} }[/tex]
So,
[tex]K_{c}[/tex] [tex]= 0.21[/tex]
[tex][N_{2} O_{4} ]_{eq.}[/tex] [tex]= 0.037 M[/tex]
So,
[tex]\sqrt{K_{c}.[N_{2} O_{4}]_{eq.} } \\= \sqrt{0.21 * 0.037}[/tex]
[tex]= 0.088 M[/tex]
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