For a reaction equilibrium, $N_{2}O_{4}(g) ightleftharpoons 2NO_{2}(g)$, the concentrations of $N_{2}O_{4}\ and\ NO_{2}$ at equilibrium are $4.8 imes 10^{- 2}\ and\ 1.2 imes 10^{- 2}$ mol/L respectively. The value of $K_{c}$ for the reaction is

Question

For a reaction equilibrium, $N_{2}O_{4}(g) ightleftharpoons 2NO_{2}(g)$, the concentrations of $N_{2}O_{4}\ and\ NO_{2}$ at equilibrium are $4.8 	imes 10^{- 2}\ and\ 1.2 	imes 10^{- 2}$ mol/L respectively. The value of $K_{c}$ for the reaction is

Dr. Abubacker · Accepted Answer

(a)
According to law of mass action
$K_{c} = \frac{\left\lbrack NO_{2} ightbrack^{2}}{\lbrack N_{2}O_{4}brack} = \frac{\left\lbrack 1.2 	imes 10^{- 2} ightbrack^{2}}{4.8 	imes 10^{- 2}} = 0.3 	imes 10^{- 2} = 3 	imes 10^{- 3}$ mol/L

Q. For a reaction equilibrium, $N_{2}O_{4}(g) \rightleftharpoons 2NO_{2}(g)$, the concentrations of $N_{2}O_{4}\ and\ NO_{2}$ at equilibrium are $4.8 \times 10^{- 2}\ and\ 1.2 \times 10^{- 2}$ mol/L respectively. The value of $K_{c}$ for the reaction is

Explanation

Reviewed by Dr. Abubacker

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