What is the numerical value of Kc for the following reaction if the equilibrium mixture contains 0.035 M N2O4 and 0.19 M NO2?

N2O4(g)⇌2NO2(g)

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Respuesta :

Dejanras Dejanras · Answer 1 · 2017-12-21T12:47:17+01:00

Kc = c(NO₂)²/c(N₂O₄)
Kc= (0,19 mol/dm³)²/0,035 mol/dm³
Kc= 1,03 mol/dm³.

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Abdulazeez10 Abdulazeez10 · Answer 2 · 2021-12-03T14:00:34+01:00

The numerical value of the equilibrium constant, Kc for the given reaction is 1.0314 M

The given reaction is

N₂O₄(g) ⇌ 2NO₂(g)

The equilibrium constant, Kc for the given reaction is given by

[tex]K_{c}=\frac{[NO_{2}]^{2} }{[N_{2}O_{4}] }[/tex]

From the question

[N₂O₄] = 0.035 M

and

[NO₂] = 0.19 M

∴ [tex]K_{c} =\frac{0.19^{2} \ M^{2} }{0.035 \ M}[/tex]

[tex]K_{c} =\frac{0.0361 \ M}{0.035}[/tex]

Kc = 1.0314 M

Hence, the numerical value of the equilibrium constant, Kc for the given reaction is 1.0314 M

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