The $K_{e q}$ expression for the following reaction; $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ is :

K_{e q} = \frac{[N_{2} O_{4}]}{[N O_{2}]}

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K_{e q} = \frac{[N_{2} O_{4}]}{[N O_{2}]^{2}}

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K_{e q} = \frac{[N O_{2}]}{[N_{2} O_{4}]}

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K_{e q} = \frac{[N O_{2}]^{2}}{[N_{2} O_{4}]}

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K_{e q} = \frac{[N_{2} O_{4}]^{2}}{[N O_{2}]}

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Solution

The correct option is E $K_{e q} = \frac{[N O_{2}]^{2}}{[N_{2} O_{4}]}$
The Keq expression for the following reaction; $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ is
$K_{e q} = \frac{[N O_{2}]^{2}}{[N_{2} O_{4}]}$ .
Note: The equilibrium constant is the ratio of product concentrations to the reactant concentrations. The concentration terms are raised to appropriate stoichiometric coefficients.

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Similar questions

Q. Consider the following dissociation reaction

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

What is the

K_{e q}

for the reaction at equilibrium, if the concentration of

N_{2} O_{4}

4 \times 10^{- 2}

mole

/

liter and that of

N O_{2}

2 \times 10^{- 2}

mole

/

liter?

Which of the following expressions is the correct equilibrium-constant expression for the equilibrium between dinitrogen tetroxide and nitrogen dioxide?

N₂O₄ (g) ⇌ 2NO₂ (g)

The reaction N $_{2}$ O $_{4}$ (g) $⇌$ 2NO $_{2}$ (g) is
carried out at 298 K and 20 bar. Five moles of each of N $_{2}$ O $_{4}$ and
NO $_{2}$ , are taken initially. Given: $Δ$ $_{f}$ G $^{\circ}$
(N $_{2}$ O $_{4}$ ) = 100 kJ mol $^{- 1}$ and $Δ$ $_{f}$ G $^{\circ}$
(NO $_{2}$ ) = 50 kJ mol $^{- 1}$ . The reaction proceeds at an initial pressure
of 20 bar. Determine the direction in which the reaction proceeds to achieve
equilibrium. Also determine the amounts of N $_{2}$ O $_{4}$ and NO $_{2}$ when the reaction attains equilibrium.