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* 10^(-34)`

Explanation :

The given main reaction is:

`N_2(g)+O_2(g)+H_2(g)\rightleftharpoons (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* 10^(-31)`

`N_2(g)+2H_2(g)\rightleftharpoons N_2H_4(g)`;
`K_2=7.40* 10^(-26)`

`2NO(g)+O_2(g)\rightleftharpoons 2NO_2(g)`;
`K_3=6.00* 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* K_2* K_3`

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

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

The equilibrium constant expression will be:

`K_(goal)=(K_1* K_2* K_3)^(1/2)`

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

`K_(goal)=[(4.10* 10^(-31))* (7.40* 10^(-26))* (6.00* 10^(-13))]^(1/2)`

`K_(goal)=1.35* 10^(-34)`

Thus, the value of the equilibrium constant is,
`K_(goal)=1.35* 10^(-34)`