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Q. $2 NOCl ( g ) \rightleftharpoons 2 NO ( g )+ Cl _{2}( g )$
In an experiment, $2.0$ moles of $NOCl$ was placed in a one-litre flask and the concentration of $NO$ after equilibrium established, was found to be $0.4\,mol / L$. The equilibrium constant at $30^{\circ} C$ is _______ $\times 10^{-4}$.

JEE MainJEE Main 2022Equilibrium

Solution:

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$\because x =0.4 \,M$
$\therefore[ NOCl ]_{ eq }=1.6 \,M$
${[ NO ]_{ eq }=0.4 \,M }$
${\left[ Cl _{2}\right]_{ eq }=0.2\, M }$
$\Rightarrow K _{ c }=\frac{[ NO ]^{2}\left[ Cl _{2}\right]}{[ NOCl ]^{2}}=\frac{[0.4]^{2}[0.2]}{[1.6]^{2}}$
$K _{ c }=\frac{32}{2.56} \times 10^{-3}$
$K _{ c }=12.5 \times 10^{-3}$
$K _{ c }=125 \times 10^{-4}$
Integer answer is $125$