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Standard XII

Chemistry

Characteristics of Equilibrium Constant

Two moles of ...

Question

Two moles of $N H_{3}$ when put into a previously evacuated vessel (one litre), partially dissociate into $N_{2}$ and $H_{2}$ . If at equilibrium one mole of $N H_{3}$ is present , the equilibrium constant is :

\frac{3}{4} m o l^{2} l i t r e^{- 2}

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\frac{27}{64} m o l^{2} l i t r e^{- 2}

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\frac{27}{32} m o l^{2} l i t r e^{- 2}

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\frac{27}{16} m o l^{2} l i t r e^{- 2}

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Solution

The correct option is D $\frac{27}{16} m o l^{2} l i t r e^{- 2}$
$2 N H_{3} = N_{2} + 3 H_{2}$

Initial mole $2$ $0$ $0$

Final Mole $1$ $\frac{1}{2}$ $\frac{3}{2}$

Equilibrium constant

$= \frac{3}{2} \times \frac{3}{2} \times \frac{3}{2} \times \frac{1}{2}$

$= \frac{27}{16} m o l^{2} l t r^{- 2}$

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Similar questions

Q. Two moles of

{N H}_{3}

gas are introduced into a previously evacuated one litre vessel in which it partially dissociates at high temperature as

{2 N H}_{3} (g) ⇋ . N_{2} (g) + {3 H}_{2} (g)

. At equilibrium, one mole of

{N H}_{3} (g)

remain. The value of

K_{C}

is:

Q. At a certain temperature (T), the equilibrium constant

(K_{c})

is 1 for the reaction

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

If 2 moles of

N_{2},

4 moles of

H_{2},

6 moles of

N H_{3}

and 3 moles of inert gas are introduced into a two litre rigid vessel at constant temperature T. It has been found that equilibrium concentration of

H_{2}

and

N H_{3}

are equal then what is the equilibrium concentration of

N_{2}

(in M)?

2

moles of

N_{2}

are mixed with

6

moles of

H_{2}

in closed vessel of one litre capacity. If

50

N_{2}

is converted into

{N H}_{3}

at equilibrium, the value of

K_{c}

for the reaction,

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

Q. 1 mole of

N_{2}

and 3 moles of

H_{2}

are placed in 1L vessel. Find the concentration of

{N H}_{3}

at equilibrium if the equilibrium constant

(K_{c})

at 400K is

\frac{4}{27}

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

Q. In a three litre vessel, 2 moles of

N_{2}

and 3 moles of length

H_{2}

are taken. If at equilibrium x moles of

N_{2}

were obtained then the concentration of

N H_{3}

at equilibrium will be:

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Two moles of NH3 when put into a previously evacuated vessel (one litre), partially dissociate into N2 and H2. If at equilibrium one mole of NH3 is present , the equilibrium constant is :

The correct option is D 2716mol2litre−2 2NH3=N2+3H2Initial mole 2 0 0Final Mole 1 12 32Equilibrium constant =32×32×32×12=2716mol2ltr−2

Two moles of $N H_{3}$ when put into a previously evacuated vessel (one litre), partially dissociate into $N_{2}$ and $H_{2}$ . If at equilibrium one mole of $N H_{3}$ is present , the equilibrium constant is :