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