Question

What is the value for the reaction: N2(g) + 2 O2(g) --> N2O4(g) in terms of K values from the reactions:

½ N2(g) + ½ O2(g) ---> NO(g)

K1

2 NO(g) + O2(g) ---> N2O4(g)

K2






K1^2 + K2
2 K1 x K2
K1^2 x K2
K1 + K2

Answer 1

Answer : The correct expression will be:

$K=(K_1)^2\times K_2$

Explanation :

The chemical reactions are :

(1) $\frac{1}{2}N_2(g)+\frac{1}{2}O_2(g)\rightleftharpoons NO(g)$ $K_1$

(2) $2NO(g)+O_2(g)\rightleftharpoons N_2O_4(g)$ $K_2$

The final chemical reaction is :

$N_2(g)+2O_2(g)\rightleftharpoons N_2O_4$ $K=?$

Now we have to calculate the value of $K$ for the final reaction.

Now equation 1 is multiply by 2 and then add both the reaction we get the value of 'K'.

If the equation is multiplied by a factor of '2', the equilibrium constant will be the square of the equilibrium constant of initial reaction.

If the two equations are added then equilibrium constant will be multiplied.

Thus, the value of 'K' will be:

$K=(K_1)^2\times K_2$