1. Tardigrade
  2. Question
  3. Chemistry
  4. At a given temperature, the equilibrium constants for the reactions, NO (g)+(1/2) O 2(g) leftharpoons NO 2(g) and 2 NO 2(g)leftharpoons 2 NO (g)+ O 2(g) are K1 and K2 respectively. If K1 is 4 × 10-3 then will be

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Q. At a given temperature, the equilibrium constants for the reactions,
$NO (g)+\frac{1}{2} O _{2}(g) \rightleftharpoons NO _{2}(g)$
and $2 NO _{2}(g)\rightleftharpoons 2 NO (g)+ O _{2}(g)$
are $K_{1}$ and $K_{2}$ respectively. If $K_{1}$ is $4 \times 10^{-3}$ then will be

Uttarkhand PMTUttarkhand PMT 2011

Solution:

$NO(g) + \frac{1}{2} O_2(g) \rightleftharpoons NO_2(g)$
$K_1 = 4 \times 10^{-3}$
$4 \times 10^{-3} = \frac{[NO_2]}{[NO][O_2]^{1/2}}$
On squaring, we get
$16 \times 10^{-6} = \frac{[NO_2]^2}{[NO]^2[O_2]}$ .... (i)
$2NO_2(g) \rightleftharpoons 2NO(g) + O_2(g) $
$K_2 = $?
$K_{2}=\frac{[N O]^{2}\left[O_{2}\right]}{\left[N O_{2}\right]^{2}}$....(ii)
From Eqs. (i) and (ii),
$K_{2}=\frac{1}{K_{1}}=\frac{1}{16 \times 10^{-6}}=6.25 \times 10^{4}$