The decomposition of $N_{2}O_{4}$ to $NO_{2}$ is carried out at 280 K in chloroform. When equilibrium has been established, 0.2 mole of $N_{2}O_{4}$ and $2 imes 10^{- 3}$ mole of$NO_{2}$ are present in 2 L solution. The equilibrium constant for reaction $N_{2}O_{4} ightleftharpoons 2{NO}_{2}$ is

Question

The decomposition of $N_{2}O_{4}$ to $NO_{2}$ is carried out at 280 K in chloroform. When equilibrium has been established, 0.2 mole of $N_{2}O_{4}$ and $2 	imes 10^{- 3}$ mole of$NO_{2}$ are present in 2 L solution. The equilibrium constant for reaction $N_{2}O_{4} ightleftharpoons 2{NO}_{2}$ is

Dr. Abubacker · Accepted Answer

(d)
$K_{c} = \frac{\left\lbrack NO_{2} ightbrack^{2}}{\lbrack N_{2}O_{4}brack} = \frac{\left\lbrack 2 	imes \frac{10^{- 3}}{2} ightbrack^{2}}{\left\lbrack \frac{0.2}{2} ightbrack}$
$= \frac{10^{- 6}}{10^{- 1}} = 10^{- 5}$

Explanation

Reviewed by Dr. Abubacker

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