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Standard XII

Chemistry

Effect on Kc

In the revers...

Question

In the reversible reaction $2 N H_{3} (g) \Leftrightarrow N_{2} (g) + 3 H_{2} (g)$ when 1 mol of $N H_{3}$ is taken initially in a litre flask and 0.2 moles of $N_{2}$ are formed at equilibrium. Its equilibrium constant $K_{c}$ is:

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12 \times 10^{- 2}

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0.3

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0.27

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Solution

The correct option is B $12 \times 10^{- 2}$
The equilibrium reaction is $2 N H_{3} (g) \Leftrightarrow N_{2} (g) + 3 H_{2} (g)$ .
The concentrations of various species are given in the following table.

$N H_{3}$
$N_{2}$
$H_{2}$
Initial
1
0
0
change
-2x
x
3x
equilibrium
1-2x
x
3x
But $[N_{2}] = 0.2 = x$ .
Hence, the equilibrium concentrations of ammonia, nitrogen and hydrogen are 0.6M, 0.2M and 0.6M respectively.
The expression for the equilibrium constant will be $K_{c} = \frac{[N_{2}] [H_{2}]^{3}}{[N H_{3}^{2}]} = \frac{0.2 \times ({0.6}^{3})}{(0.6)^{2}} = 12 \times 10^{- 2}$ .

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Similar questions

Q. Assertion :For reaction

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

Unit of

K_{C} = L^{2} {m o l}^{- 2}

Reason: For the reaction

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

, Equilibrium constant

K_{C} = \frac{{[{N H}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}}

Q. Two moles of

N H_{3}

are introduced into one litre flask in which it dissociates at high temperature as follows:

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

. Determine

K_{C}

, if at equilibrium

1 m o l e

N H_{3}

remains.

Q. Consider the following reaction equilibrium

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

. Initially,

1 m o l

N_{2}

and

3 m o l

H_{2}

are taken in a

2

litre flask. At equilibrium state if the number of mol of

N_{2}

0.6

what is the total number mol of

N_{2}

0.6

what is the total number mol of all gases present in the flask.

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. 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)?

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	$N H_{3}$	$N_{2}$	$H_{2}$
Initial	1	0	0
change	-2x	x	3x
equilibrium	1-2x	x	3x

In the reversible reaction 2NH3(g)⇔N2(g)+3H2(g) when 1 mol of NH3 is taken initially in a litre flask and 0.2 moles of N2 are formed at equilibrium. Its equilibrium constant Kc is:

In the reversible reaction $2 N H_{3} (g) \Leftrightarrow N_{2} (g) + 3 H_{2} (g)$ when 1 mol of $N H_{3}$ is taken initially in a litre flask and 0.2 moles of $N_{2}$ are formed at equilibrium. Its equilibrium constant $K_{c}$ is: