Respuesta :
Answer:
[tex]Kc=~1.49x10^3^4}[/tex]
Explanation:
We have the reactions:
A: [tex]N_2_(_g_) + O_2_(_g_) <-> 2NO_(_g_)~~~~~~Kc = 4.3x10^-^2^5[/tex]
B: [tex]2NO_(_g_)+~O_2_(_g_)<->~2NO_2_(_g_)~~~Kc = 6.4x10^9[/tex]
Our target reaction is:
[tex]4NO_(_g_) <-> N_2_(_g_) + 2NO_2_(_g_)[/tex]
We have [tex]NO_(_g_)[/tex] as a reactive in the target reaction and [tex]NO_(_g_)[/tex] is present in A reaction but in the products side. So we have to flip reaction A.
A: [tex]2NO_(_g_)<-> N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}[/tex]
Then if we add reactions A and B we can obtain the target reaction, so:
A: [tex]2NO_(_g_)<-> N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}[/tex]
B: [tex]2NO_(_g_)+~O_2_(_g_)<->~2NO_2_(_g_)~Kc=6.4x10^9[/tex]
For the final Kc value, we have to keep in mind that when we have to add chemical reactions the total Kc value would be the multiplication of the Kc values in the previous reactions.
[tex]4NO_(_g_) <-> N_2_(_g_) + 2NO_2_(_g_)~~~Kc=\frac{6.4x10^9}{4.3x10^-^2^5}[/tex]
[tex]Kc=~1.49x10^+^3^4}[/tex]
The value of Kc for the reaction given will be
[tex]Kc = 4.3*10^-^2^5 * 6.4*10^9 = 2.752*10^-^1^5[/tex]
Data;
- N2(g) + O2(g) ⇌ 2NO(g) Kc = 4.3 × 10^−25
- 2NO(g) + O2(g) ⇌ 2NO2(g) Kc = 6.4 × 10^9
The Equilibrium Constant
From the first equation;
[tex]N_2+O_2 \to 2NO\\K_c = 4.3*10^-^2^5[/tex]
Taking Kc1
[tex]Kc_1 = \frac{[NO]^2}{[N_2][O_2]}.. equation (i)[/tex]
From equation (ii)
[tex]2NO+ O_2 \to 2NO_2\\Kc_2 = 6.4*10^9\\Kc_2 = \frac{[NO_2]^2}{[NO]^2[O_2]}...equation (ii)[/tex]
Our given equation of reaction
[tex]N_2 + 2O_2 \to 2NO_2[/tex]
For this reaction
[tex]Kc = \frac{[NO_2]^2}{[N_2][O_2]^2} ... equation (iii)\\[/tex]
If we multiply both equation i and ii
[tex]Kc_i * Kc_i_i = \frac{[NO_2]^2}{[N_2][O_2]^2}\\Kc_i * Kc_i_i = Kc[/tex]
The value of Kc for the reaction given will be
[tex]Kc = 4.3*10^-^2^5 * 6.4*10^9 = 2.752*10^-^1^5[/tex]
Learn more on equilibrium constant here;
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