Question

For these chemical equilibriums:

$N_2+O_2\leftrightharpoons 2N{O}^{},K_1$
$\frac{1}{2}{N}_2+\frac{1}{2}{O}_2\leftrightharpoons N{O}^{},K_2$
$2NO\rightleftharpoons N_2+O_2;K_3$
$NO\rightleftharpoons \frac{1}{2}{N}_2+\frac{1}{2}{O}_2,K_4$
$K_1\times K_3=x;\sqrt{\left(K_1\right)}\times K_4=y;\sqrt{\left(K_3\right)}\times K_2=z$
What will be the value of $xyz$?

Answer 1

The given reactions are :-

(I) $N_2+O_2\rightleftharpoons 2NO$

$\Rightarrow K_1=\frac{{\left[NO\right]}^2}{\left[N_2\right]\left[O_2\right]}\to \left(I\right)$

(II) $1/2N_2+1/2O_2\rightleftharpoons NO$

$\Rightarrow K_2=\frac{\left[NO\right]}{{\left[N_2\right]}^{1/2}{\left[O_2\right]}^{1/2}}\to \left(II\right)$

(III) $2NO\rightleftharpoons N_2+O_2$

$\Rightarrow K_3=\frac{\left[N_2\right]\left[O_2\right]}{{\left[NO\right]}^2}\to \left(III\right)$

(IV) $NO\rightleftharpoons 1/2N_2+1/2O_2$

$K_4=\frac{{\left[N_2\right]}^{1/2}{\left[O_2\right]}^{1/2}}{\left[NO\right]}\to \left(IV\right)$

Now, By (I) & (III) $\Rightarrow K_1\times K_3=\left(I\right)=x$

By, (I) & (IV) $\Rightarrow \sqrt{K_1}\times K_4=y$

By, (III) & (II) $\Rightarrow \sqrt{K_3}\times K_2=1=z$

$\Rightarrow xyz=1$