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Standard XII

Chemistry

Reaction Quotient

8 moles of a ...

Question

8 moles of a gas $A B_{3}$ attain equilibrium in a closed container of volume $1 {d m}^{3}$ as,

$2 A B_{3} ⇌ A_{2} (g) + 3 B_{2} (g)$ .

If at equilibrium, 2 moles of $A_{2}$ is present then, the equilibrium constant is :

72 {m o l}^{2} L^{- 2}

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36 {m o l}^{2} L^{- 2}

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3 {m o l}^{2} L^{- 2}

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27 {m o l}^{2} L^{- 2}

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Solution

The correct option is D $27 {m o l}^{2} L^{- 2}$
$2 A B_{3} (g) ⇌ A_{2} (g) + 3 B_{2} (g)$

$t = 0$ $8$ $0$ $0$

At eq. $8 - 2 α$ $α$ $3 α$

Thus, $K_{c} = \frac{[A_{2}] {[B_{2}]}^{3}}{{[A B_{3}]}^{2}}$ ;

Given, $[A_{2}]$ = 2 moles

$\Rightarrow$ $α = 2$

$∴ [A B_{3}] = \frac{4}{1}; [A_{2}] = \frac{2}{1}; [B_{2}] = \frac{6}{1}$

Thus, $K_{c} = \frac{2 \times 6^{3}}{4^{2}} = 27 {m o l}^{2} L^{- 2}$

Hence, option D is correct.

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Similar questions

8

moles of a gas

A B_{3}

are introduced into a

1.0 d m^{3}

vessel.
It dissociates as

2 A B_{3} ⇌ A_{2} (g) + 3 B_{2} (g)

At equilibrium,

2

moles of

A_{2}

are found to be present. The equilibrium constant of the reaction is

8

mole of a gas

{A B}_{3}

are introduced into a

1.0 {d m}^{3}

vessel. It dissociates as:

2 A B_{3} (g) ⇌ A_{2} (g) + 3 B_{2} (g)

at equilibrium,

2

mole of

A_{2}

are found to be present. The equilibrium constant of the reaction is:

Q. A vessel of one litre capacity containing

1

mole of

{S O}_{3}

is heated till a state of equilibrium is attained.

2 {S O}_{3 (g)} ⇌ 2 {S O}_{2 (g)} + O_{2 (g)}

At equilibrium,

0.6

moles of

{S O}_{2}

had formed. The value of equilibrium constant is:

1

mole of

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and

3

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are placed in a closed container at a pressure of

4

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N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

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is:

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mole

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and

3

moles

H_{2}

are placed in a closed container at a pressure of

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atm at the same temperature when the equilibrium is attained

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8 moles of a gas AB3 attain equilibrium in a closed container of volume 1dm3 as, 2AB3⇌A2(g)+3B2(g). If at equilibrium, 2 moles of A2 is present then, the equilibrium constant is :

The correct option is D 27mol2L−2 2AB3(g)⇌A2(g)+3B2(g)t=0 8 0 0At eq. 8−2α α 3αThus, Kc=[A2][B2]3[AB3]2; Given, [A2] = 2 moles⇒ α=2 ∴[AB3]=41;[A2]=21;[B2]=61Thus, Kc=2×6342=27mol2L−2Hence, option D is correct.

8 moles of a gas $A B_{3}$ attain equilibrium in a closed container of volume $1 {d m}^{3}$ as,

$2 A B_{3} ⇌ A_{2} (g) + 3 B_{2} (g)$ .

If at equilibrium, 2 moles of $A_{2}$ is present then, the equilibrium constant is :