426

You visited us 426 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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:

No worries! We‘ve got your back. Try BYJU‘S free classes today!

12 \times 10^{- 2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.27

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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 $K_{c} = \frac{[N_{2}] [H_{2}]^{3}}{[N H_{3}^{2}]} = \frac{0.2 \times ({0.6}^{3})}{(0.6)^{2}} = 12 \times 10^{- 2}$ .

Suggest Corrections

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)?

Join BYJU'S Learning Program

	$N H_{3}$	$N_{2}$	$H_{2}$
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: