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.	$2 N_{2} (g) + 6 H_{2} (g) ⇌ 4 {NH}_{3} (g)$	(i)	$2 K_{c}$
B.	$2 {NH}_{3} (g) ⇌ N_{2} (g) + 3 H_{2} (g)$	(ii)	$K_{c}^{1 / 2}$
C.	$\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) ⇌ {NH}_{3} (g)$	(iii)	$\frac{1}{K_{c}}$
		(iv)	$K_{c}^{2}$

(Given that reference, Kc is the equilibrium constant for the ammonia synthesis.)

Codes:

	A	B	C
1.	(iv)	(iii)	(ii)
2.	(i)	(ii)	(iii)
3.	(i)	(iv)	(iii)
4.	(iv)	(i)	(iii)

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

Explanation:

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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