Question

Equilibrium constants $K_1$, and $K_2$ for the following equilibria I. $NO _{( g )}+\frac{1}{2} O _{2( g )} \rightleftharpoons NO _{2( g )}$ II. $2 NO _{2( g )} \rightleftharpoons 2 NO _{( g )}+ O _{2( g )}$ are related as :

• A. $K_1 = \sqrt{K_2}$
• B. $K_2 = \frac{1}{K_1}$
• C. $K_1 = 2 K_2$
• D. $K_2 = \frac{1}{K^2_1}$

Answer 1

Answer: (D)

$K_2 = \frac{1}{K^2_1}$

Answer 2

Given relations

$NO (g)+\frac{1}{2} O _{2}(g) \rightleftharpoons NO _{2}(g) ; K_{1} \,\,\,\,\,\,\dots(i)$
$2 NO _{2}(g) \rightleftharpoons 2 NO (g)+ O _{2}(g) ; K_{2} \ldots (ii)$

On comparing Eqs. (i) and (ii) and to make

E (i) = E (ii),

Reverse E (i) and multiply it with 2 , we get

$2 NO _{2}(g) \rightleftharpoons 2 NO (g)+ O _{2} ; \frac{1}{K_{1}^{2}}\,\,\,\,\,\,\dots(iii)$

Thus, E (ii) = E (iii) and, $K_{2}=\frac{1}{K_{1}^{2}}$