Question

$K_1\text{and}{K}_2$ are equilibrium constants for reactions (i) and (ii) respectively.
$N_{2\left(g\right)}+O_{2\left(g\right)}\rightleftharpoons 2N{O}_{\left(g\right)}$ ...(i)
$N{O}_{\left(g\right)}\rightleftharpoons \frac{1}{2}{N}_{2\left(g\right)}+\frac{1}{2}{O}_{2\left(g\right)}$ ...(ii)

• A. $K_1={\left(\frac{1}{K_2}\right)}^2$
• B. $K_1=K_2^2$
• C. $K_1=\frac{1}{K_2}$
• D. $K_1=(K_2{)}^1$

Answer 1

The correct option is A $K_1={\left(\frac{1}{K_2}\right)}^2$

Given,
$N_{2\left(g\right)}+O_{2\left(g\right)}\rightleftharpoons 2N{O}_{\left(g\right)}$ ...(i)
$N{O}_{\left(g\right)}\rightleftharpoons \frac{1}{2}{N}_{2\left(g\right)}+\frac{1}{2}{O}_{2\left(g\right)}$ ...(ii)
$\therefore K_1=\frac{[NO{]}^2}{\left[N_2\right]\left[O_2\right]}$
$K_2=\frac{\left[N_2{]}^{\frac{1}{2}}\right[O_2{]}^{\frac{1}{2}}}{\left[NO\right]}$
$\therefore K_2=\sqrt{\frac{1}{K_1}}\text{or}{K}_1={\left(\frac{1}{K_2}\right)}^2$