1 mole of $N O_{2}$ and 2 moles of CO are introduced in a 1 litre closed vessel to get the following equilibrium: $N O_{2 (g)} + C O_{(g)} ⇌ N O_{(g)} + C O_{2 (g)}$ If at equilibrium, 25% of the initial amount of $C O_{(g)}$ is reacted, then $K_{c}$ for the above reaction is $1 / x$ The value of $x$ is________

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Detailed Solution

$N O_{2} + C O ⇌ N O + C O_{2}$

$\begin{matrix} \begin{array}{l} I n i t i a l \\ m o l e s \end{array} & 1 & 2 & 0 & 0 \\ E q . m o l e s & 1 - x & 2 - x & x & x \\ E q . m o l e s & 0.5 & 1.5 & 0.5 & 0.5 \end{matrix}$

Given that $x$ is 25% of 2 moles = 0.5

$K_{C} = \frac{[N O] [C O_{2}]}{[N O_{2}] [C O]} = \frac{0.5 x 0.5}{0.5 x 1.5} = \frac{5}{15} = \frac{1}{3}$

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