Ncert Exercises & Solutions

Some reactions are written below in Column I and their equilibrium constants in terms of K_c are written in Column II.
Match the following reactions with the corresponding equilibrium constant.

Column I (Reaction)	Column II (Equilibrium constant)
A. $2 N_{2} (g) + 6 H_{2} (g) ⇌ 4 {NH}_{3} (g)$	1. $2 K_{c}$
B. $2 {NH}_{3} (g) ⇌ N_{2} (g) + 3 H_{2} (g)$	2. $K_{c}^{1 / 2}$
C. $\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) ⇌ {NH}_{3} (g)$	3. $\frac{1}{K_{c}}$
	4. $K_{c}^{2}$

Codes:

	A	B	C
1.	4	3	2
2.	1	2	3
3.	1	4	3
4.	4	1	3

Hint: Relations between Equilibrium Constants for a General Reaction and its Multiples.

For the reaction,

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

Equilibrium constant

K_{c} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}}

A. The given reaction

2 N_{2} (g) + 6 H_{2} (g) ⇌ 4 {NH}_{3} (g)

is twice the above reaction. Hence,

K = K_{c}^{2}

B. The reaction $2NH_3(g) \rightleftharpoons\ N_2(g) + \ 3H_2(g)$

is the reverse of the above reaction. Hence,

K = \frac{1}{K_{c}}

C. The reaction

\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) ⇌ {NH}_{3} (g)

is half of the above reaction. Hence,

K = \sqrt{K_{c}} = K_{c}^{1 / 2}

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