Determine the value of the equilibrium constant, Kgoal, for the reaction N2(g)+O2(g)+H2(g)⇌12N2H4(g)+NO2(g), Kgoal=? by making use of the following information: 1. N2(g)+O2(g)⇌2NO(g), K1 = 4.10×10−31 2. N2(g)+2H2(g)⇌N2H4(g), K2 = 7.40×10−26 3. 2NO(g)+O2(g)⇌2NO2(g), K3 = 6.00×10−13 Express your answer numerically.

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Alleei Alleei · Answer 1 · 2020-03-03T04:54:23+01:00

Answer : The value of the equilibrium constant is, [tex]K_{goal}=1.35\times 10^{-34}[/tex]

Explanation :

The given main reaction is:

[tex]N_2(g)+O_2(g)+H_2(g)\rightleftharpoons \frac{1}{2}N_2H_4(g)+NO_2(g)[/tex]

The intermediate reactions are:

[tex]N_2(g)+O_2(g)\rightleftharpoons 2NO(g)[/tex]; [tex]K_1=4.10\times 10^{-31}[/tex]

[tex]N_2(g)+2H_2(g)\rightleftharpoons N_2H_4(g)[/tex]; [tex]K_2=7.40\times 10^{-26}[/tex]

[tex]2NO(g)+O_2(g)\rightleftharpoons 2NO_2(g)[/tex]; [tex]K_3=6.00\times 10^{-13}[/tex]

Now we are adding all the equation, we get:

[tex]2N_2(g)+2O_2(g)+2NO(g)+2H_2(g)\rightleftharpoons 2NO(g)+N_2H_4(g)+2NO_2(g)[/tex]

[tex]2N_2(g)+2O_2(g)+2H_2(g)\rightleftharpoons N_2H_4(g)+2NO_2(g)[/tex] ...........(1)

The equilibrium constant expression will be:

[tex]K_{goal}=K_1\times K_2\times K_3[/tex]

Now we are dividing equation 1 by 2, we get:

[tex]N_2(g)+O_2(g)+H_2(g)\rightleftharpoons \frac{1}{2}N_2H_4(g)+NO_2(g)[/tex] ...........(1)

The equilibrium constant expression will be:

[tex]K_{goal}=(K_1\times K_2\times K_3)^{1/2}[/tex]

Now put all the given values in this expression, we get:

[tex]K_{goal}=[(4.10\times 10^{-31})\times (7.40\times 10^{-26})\times (6.00\times 10^{-13})]^{1/2}[/tex]

[tex]K_{goal}=1.35\times 10^{-34}[/tex]

Thus, the value of the equilibrium constant is, [tex]K_{goal}=1.35\times 10^{-34}[/tex]