Question

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.

Answer 1

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

Explanation :

The given main reaction is:

$N_2(g)+O_2(g)+H_2(g)\rightleftharpoons \frac{1}{2}N_2H_4(g)+NO_2(g)$

The intermediate reactions are:

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

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

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

Now we are adding all the equation, we get:

$2N_2(g)+2O_2(g)+2NO(g)+2H_2(g)\rightleftharpoons 2NO(g)+N_2H_4(g)+2NO_2(g)$

$2N_2(g)+2O_2(g)+2H_2(g)\rightleftharpoons N_2H_4(g)+2NO_2(g)$           ...........(1)

The equilibrium constant expression will be:

$K_{goal}=K_1\times K_2\times K_3$

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

$N_2(g)+O_2(g)+H_2(g)\rightleftharpoons \frac{1}{2}N_2H_4(g)+NO_2(g)$           ...........(1)

The equilibrium constant expression will be:

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

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

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

$K_{goal}=1.35\times 10^{-34}$

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