Question bank/JEE Main/Chemistry

The equilibrium constant for the reaction, N2 (g) + O2
(g)[IMAGE] 2NO(g) at temperature T is 4 × 10-4. The
value NO(g)[IMAGE] N2 (g) +
[IMAGE] O2 (g) at the same temperature is:
N2 (g) + O2 (g)[IMAGE] 2NO(g), bl vfHkfØ fy, rki T ij
lkE;koLFkk fLFkjkad 4 × 10-4 gS fuEu vfHkfØ fy, leku rki ij Kc dk
eku gksxk NO(g)[IMAGE] $\frac{1}{2}$N2 (g) +
$\frac{1}{2}$ O2

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N2 (g) + O2 (g)[IMAGE] 2NO(g);
$K_{c} = \frac{\lbrack NO\rbrack^{2}}{\left\lbrack N_{2} \right\rbrack\left\lbrack O_{2} \right\rbrack} = 4 \times 10^{- 4}$
NO(g)[IMAGE] N2 (g) +
$K_{c} = \frac{\left\lbrack N_{2} \right\rbrack^{1/2}\left\lbrack O_{2} \right\rbrack^{1/2}}{\lbrack NO\rbrack} = \frac{1}{\sqrt{K_{0}}} = \frac{1}{\sqrt{4 \times 10^{- 4}}} = \frac{1}{2 \times 10^{- 2}} = \frac{100}{2} = 50$

Source: Grass-G / CH-26 Chemistry Paper 1 15 Jan.docx