Respuesta :
Answer:
Kc = 4.8x10^13
Explanation:
2 NO2(g) ⇌ N2(g) + 2 O2(g)
(1) 1/2 N2(g) + 1/2 O2(g) ⇌ NO(g) Kc = 4.8 × 10 −10
(2) 2 NO2(g) ⇌ 2NO(g) + O2(g) Kc = 1.1 × 10−5
Using the 2 given reactions you need to form the asked in the first place, in order to do that you multiply one of them by 2 and switch reactants with products and then add them together
(1): 1/2 N2(g) + 1/2 O2(g) ⇌ NO(g) / x2
N2(g) + O2(g) ⇌ 2 NO(g) / switch reactants with products (equivalent to multiply by -1)
2 NO(g) ⇌ N2(g) + O2(g)
(2): 2 NO2(g) ⇌ 2NO(g) + O2(g)
Adding them you get:
2 NO(g) + 2 NO2(g) ⇌ 2NO(g) + O2(g) + O2(g)
After eliminating NO in both sides:
2 NO2(g) ⇌ 2NO(g) + O2(g)
Now that the reaction is formed, everything you multiplied to the reactions you have to do the same in the constant but in the exponent and if you add reactions you multiply each other constants, then:
Kc = (4.8 × 10 −10)^(2*(-1)) * 1.1 × 10−5 = 4.8x10^13
The equilibrium constant Kc, for the reaction 2 NO₂(g) ⇌ N2(g) + 2O₂(g) is 4.77 × 10¹³
From the question,
We are to determine the equilibrium constant, Kc for the reaction
2 NO₂(g) ⇌ N2(g) + 2O₂(g)
The equilibrium constant, Kc for this reaction is given by
[tex]K_{c} =\frac{[N_{2}][O_{2}]^{2} }{[NO_{2}]^{2} }[/tex]
From the given data,
We have that
¹/₂ N₂(g) + ¹/₂ O₂(g) ⇌ NO(g) Kc₁ = 4.8 × 10⁻¹⁰
The equilibrium constant, Kc₁, for this reaction is given by
[tex]K_{c1}= \frac{[NO]}{[N_{2}]^{\frac{1}{2} }[O_{2}]^{\frac{1}{2} } }[/tex] --------------- (X)
Also,
2NO₂(g) ⇌ 2NO(g) + O₂(g) Kc₂ = 1.1 × 10⁻⁵
The equilibrium constant, Kc₂, for this reaction is given by
[tex]K_{c2} = \frac{[NO]^{2}[O_{2}] }{[NO_{2}]^{2} }[/tex] -------------- (Y)
Now,
From (X)
[tex]K_{c1}= \frac{[NO]}{[N_{2}]^{\frac{1}{2} }[O_{2}]^{\frac{1}{2} } }[/tex]
We can write that
[tex][NO] =K_{c1}[N_{2}]^{\frac{1}{2} }[O_{2}]^{\frac{1}{2} }[/tex]
Put this into (Y)
[tex]K_{c2} = \frac{[NO]^{2}[O_{2}] }{[NO_{2}]^{2} }[/tex]
We get
[tex]K_{c2} = \frac{(K_{c1}[N_{2}]^{\frac{1}{2} }[O_{2}]^{\frac{1}{2} })^{2}[O_{2}] }{[NO_{2}]^{2} }[/tex]
Then,
[tex]K_{c2} = \frac{K_{c1}^{2} [N_{2}][O_{2}][O_{2}] }{[NO_{2}]^{2} }[/tex]
[tex]K_{c2} = \frac{K_{c1}^{2} [N_{2}][O_{2}]^{2} }{[NO_{2}]^{2} }[/tex]
∴[tex]\frac{K_{c2} }{K_{c1}^{2} } = \frac{[N_{2}][O_{2}]^{2} }{[NO_{2}]^{2} }[/tex]
Recall that,
The equilibrium constant Kc, for the reaction 2NO₂(g) ⇌ N2(g) + 2O₂(g) is given by
[tex]K_{c} =\frac{[N_{2}][O_{2}]^{2} }{[NO_{2}]^{2} }[/tex]
∴ [tex]K_{c} =\frac{K_{c2} }{K_{c1}^{2} }[/tex]
From the question,
Kc₁ = 4.8 × 10⁻¹⁰
Kc₂ = 1.1 × 10⁻⁵
∴ [tex]K_{c} =\frac{1.1 \times 10^{-5} }{(4.8 \times 10^{-10} )^{2} }[/tex]
[tex]K_{c} =\frac{1.1 \times 10^{-5} }{2.304\times 10^-19 }[/tex]
∴ Kc = 4.77 × 10¹³
Hence, the equilibrium constant Kc, for the reaction 2NO₂(g) ⇌ N2(g) + 2O₂(g) is 4.77 × 10¹³
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