Question

$K_1$ and $K_2$ are equilibrium constant for reactions (i) and (ii)
$N_2(g)+O_2(g)\leftrightharpoons 2NO(g)$ ... ( i )
$NO(g)\leftrightharpoons \frac{1}{2}{N}_2(g)+\frac{1}{2}{O}_2(g)$ ... ( ii )
Then,

• A. $K_1={\left[\frac{1}{K_2}\right]}^2$
• B. $K_1=K_2^2$
• C. $K_1=\frac{1}{K_2}$
• D. $K_1=(K_2{)}^0$

Answer 1

Answer: (A)

Consider reaction (i),
$N_2(g)+O_2(g)\leftrightharpoons 2NO(g)\dots \text{ (i) }$
$K_1=\frac{[NO{]}^2}{\left[N_2\right]\left[O_2\right]}$
Now, consider reaction (ii),
$NO(g)\leftrightharpoons \frac{1}{2}{N}_2(g)+\frac{1}{2}{O}_2(g)$
$K_2=\frac{{\left[N_2\right]}^{1/2}{\left[O_2\right]}^{1/2}}{[NO]}$
$\frac{1}{K_2}=\frac{1}{\frac{{\left[N_2\right]}^{1/2}{\left[O_2\right]}^{1/2}}{[NO]}}=\frac{[NO]}{{\left[N_2\right]}^{1/2}{\left[O_2\right]}^{1/2}}$
${\left(\frac{1}{K_2}\right)}^2={\left\{\frac{[NO]}{{\left[N_2\right]}^{1/2}{\left[O_2\right]}^{1/2}}\right\}}^2=\frac{[NO{]}^2}{\left[N_2\right]\left[O_2\right]}=K_1$