Respuesta :
Answer: Concentration of N₂ is 4.8.[tex]10^{9}[/tex] M.
Explanation: [tex]K_{c}[/tex] is a constant of equilibrium and it is dependent of the concentrations of the reactants and the products of a balanced reaction. For
N2(g) + 2 O2(g) ⇄ 2 NO2(g)
[tex]K_{c}[/tex] = [tex]\frac{[NO2]^{2} }{[N2][O2]^{2} }[/tex]
From the question concentration of NO2 is twice of O2:
[NO2] = 2[O2]
Substituting this into [tex]K_{c}[/tex]:
[tex]K_{c}[/tex] = [tex]\frac{[2O2]^{2} }{[N2][O2]^{2} }[/tex]
8.3.[tex]10^{-10}[/tex] = [tex]\frac{4O2^{2} }{[N2].O2^{2} }[/tex]
[N2] = [tex]\frac{4O2^{2} }{8.3.10^{-10}.O2^{2} }[/tex]
[N2] = [tex]\frac{4}{8.3.10^{-10} }[/tex]
[N2] = 4.8.[tex]10^{9}[/tex]
The concentration of N2 in the equilibrium is [N2] = 4.8.[tex]10^{9}[/tex]M.
The concentration of [tex]N_2[/tex] is [tex]4.8.10^9M[/tex]
Calculation of concentration:
Since k_e represent the equilibrium constant and based on the concentrations of the reactants and the products of a balanced reaction.
Also, the reaction: N2(g) + 2 O2(g) ⇌ 2 NO2(g), And, the concentration of NO2 is twice the concentration of O2 gas
Kc = 8.3 × 10-10 at 25°C.
NO2 is twice of O2.
Now
[tex]8.310^{-10} = \frac{4O_2^2}{N_2O_2^2} \\\\N_2 = \frac{4O_2^2}{8.3.10^{-10}O_2^2}\\\\ = 4\div 8.3.10^{-10}\\\\= 4.8.10^9[/tex]
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