Question

2 mols of NH3 were put into 10L vessel, partially dissociated into N2 and H2.If at equlibrium 1mol of NH3 is present then equlibrum constant is​

Answer 1

Answer: k_ck

c

for this reaction is 1.6875.

Explanation: We are given,

Volume of the vessel = 1L

For a reaction,

2NH_3\rightleftharpoons N_2+3H_22NH

3

⇌N

2

+3H

2

at t = 0 2 0 0

at t=t_{eq}t=t

eq

2-2x x 3x

at t = t 0 x 3x

where, 2x = change in moles of NH_3NH

3

It is given that at equilibrium, amount of NH_3NH

3

left is 1 mole,

\begin{gathered}2-2x=1\\2x=1\\x=0.5\end{gathered}

2−2x=1

2x=1

x=0.5

Concentration can be calculated as

C=\frac{m}{V}C=

V

m

So, the concentraction of NH_3NH

3

, H_2H

2

and N_2N

2

at equilibrium will be,

[NH_3]=\frac{1}{1}[NH

3

]=

1

1

= 1.0 mol/L

[N_2]=\frac{0.5}{1}[N

2

]=

1

0.5

= 0.5 mol/L

[H_2]=\frac{3\times 0.5}{1}[H

2

]=

1

3×0.5

= 1.5 mol/L

For a given reaction, k_ck

c

can be calculated as:

\begin{gathered}k_c=\frac{[N_2]^1[H_2]^3}{[NH_3]^1}\\k_c=\frac{(0.5)^1\text{mol/L }\times (1.5)^3\text{mol/L }}{(1)^1mol/L}\end{gathered}

k

c

=

[NH

3

]

1

[N

2

]

1

[H

2

]

3

k

c

=

(1)

1

mol/L

(0.5)

1

mol/L ×(1.5)

3

mol/L

k_ck

c

= 1.6875

I hope it is use full to you