Question

Consider the following dissociation reaction
$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$
What is the $K_{eq}$ for the reaction at equilibrium, if the concentration of $N_2{O}_4$ is $4\times {10}^{-2}$ mole$/$liter and that of $N{O}_2$ is $2\times {10}^{-2}$ mole$/$liter?

• A. $1\times {10}^{-2}$
• B. $2\times {10}^{-2}$
• C. $4\times {10}^{-2}$
• D. $4\times {10}^{-4}$
• E. $8\times {10}^{-2}$

Answer 1

The correct option is A $1\times {10}^{-2}$

The Keq expression for the following reaction; $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$ is
$K_{eq}=\frac{[N{O}_2{]}^2}{\left[N_2{O}_4\right]}$.
Note: The equilibrium constant is the ratio of product concentrations to the reactant concentrations. The concentration terms are raised to appropriate stoichiometric coefficients.
$K_{eq}=\frac{[N{O}_2{]}^2}{\left[N_2{O}_4\right]}=\frac{(2\times {10}^{-2}{)}^2}{4\times {10}^{-2}}=1\times {10}^{-2}$