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Byju's Answer

Standard XII

Chemistry

Relation between Kp and Kc

N2O4g⇌ 2NO2gI...

Question

$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$
In this reaction the dissociation of $N_{2} O_{4}$ is $x$ then the how much mole will of $N_{2} O_{4}$ and $N O_{2}$ at equilibrium.

(1 - x)^{2}

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Solution

The correct option is A $(1 - x)^{2}$
$N_{2} O_{4} (g) ⇌ {2 N O}_{2} (g)$
initial $1$ $0$
At equilibrium $1 - x$ $x$
In this reaction the dissociation of $N_{2} O_{4}$ is $x$ . then ${(1 - x)}^{2}$ mole of $N_{2} O_{4}$ and ${N O}_{2}$ at equilibrium.

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Similar questions

Q. If, in the reaction

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g),

one mole of

N_{2} O_{4}

taken initially and

x

is that part of

N_{2} O_{4}

which dissociates, then the number of moles at equilibrium will be:

Q. If, in the reaction

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g),

one mole of

N_{2} O_{4}

taken initially and

x

is that part of

N_{2} O_{4}

which dissociates, then the number of moles at equilibrium will be:

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?

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.

Q. One mole of

N_{2} O_{4}

in a

1 L

flask decomposes to attain the equilibrium

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

At the equilibrium the mole fraction of

N O_{2}

1 / 2

. Hence

K_{C}

will be :

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N2O4(g)⇌2NO2(g)In this reaction the dissociation of N2O4 is x then the how much mole will of N2O4 and NO2 at equilibrium.

The correct option is A (1−x)2 N2O4(g)⇌2NO2(g)initial 1 0 At equilibrium 1−x xIn this reaction the dissociation of N2O4 is x. then (1−x)2 mole of N2O4 and NO2 at equilibrium.

$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$
In this reaction the dissociation of $N_{2} O_{4}$ is $x$ then the how much mole will of $N_{2} O_{4}$ and $N O_{2}$ at equilibrium.

The correct option is A $(1 - x)^{2}$
$N_{2} O_{4} (g) ⇌ {2 N O}_{2} (g)$
initial $1$ $0$
At equilibrium $1 - x$ $x$
In this reaction the dissociation of $N_{2} O_{4}$ is $x$ . then ${(1 - x)}^{2}$ mole of $N_{2} O_{4}$ and ${N O}_{2}$ at equilibrium.