Question

$K_1 $ and $ K_2$ are equilibrium constant for reactions (i) and (ii)
$ N_2 ( g ) + O_2 ( g ) \leftrightharpoons 2 NO ( 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),
$N_{2}(g)+O_{2}(g) \leftrightharpoons 2 N O(g) \ldots \text { (i) }$
$K_{1}=\frac{[N O]^{2}}{\left[N_{2}\right]\left[O_{2}\right]}$
Now, consider reaction (ii),
$NO ( g ) \leftrightharpoons \frac{1}{2} N _{2}(g)+\frac{1}{2} O_{2}(g)$
$K_{2}=\frac{\left[N_{2}\right]^{1 / 2}\left[O_{2}\right]^{1 / 2}}{[N O]}$
$\frac{1}{K_{2}}=\frac{1}{\frac{\left[N_{2}\right]^{1 / 2}\left[O_{2}\right]^{1 / 2}}{[N O]}}=\frac{[N O]}{\left[N_{2}\right]^{1 / 2}\left[O_{2}\right]^{1 / 2}}$
$\left(\frac{1}{K_{2}}\right)^{2}=\left\{\frac{[N O]}{\left[N_{2}\right]^{1 / 2}\left[O_{2}\right]^{1 / 2}}\right\}^{2}=\frac{[N O]^{2}}{\left[N_{2}\right]\left[O_{2}\right]}=K_{1}$