Question

$K_1$ and $K_2$ are equilibrium constants for reaction (i) and (ii)
$N_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2NO\left(g\right)$ .....(i)
$NO\left(g\right)\rightleftharpoons \frac{1}{2}{N}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$ .....(ii)
Then :

• A. $K_1=(1/K_2{)}^2$
• B. $K_1=K_2^2$
• C. $K_1=1/K_2$
• D. $K_1=(K_2{)}^o$

Answer 1

The correct option is A $K_1=(1/K_2{)}^2$

$N_2+O_2\rightleftharpoons 2NO$ .....(i)
$NO\rightleftharpoons \frac{1}{2}{N}_2+\frac{1}{2}{O}_2$ .....(ii)
Equation (ii) is obtained by reversing equation (i) and dividing by 2.
$\therefore K_2=\frac{1}{(K_1{)}^{1/2}}$
$\Rightarrow (K_2{)}^2=\frac{1}{K_1}$
$\Rightarrow K_1=\frac{1}{(K_2{)}^2}={\left(\frac{1}{K_2}\right)}^2$