Match the following equilibria with the corresponding condition (i) Liquid ⇌ Vapour (a) Saturated solution (ii) Solid ⇌ Liquid (b) Boiling point (iii) Solid ⇌ Vapour (c) Sublimation point (iv) Solute (s) ⇌ Solute (solution) (d) Melting point (e) Unsaturated solution

(i)Liquid ⇌ Vapour (b)Boiling point (i) Liquid ⇌Vapour equilibrium exists at the boiling point. At boiling point, the rate of boiling is equal to the rate of condensation. (ii)Solid ⇌ Liquid (d)Melting point (ii) Solid ⇌Liquid equilibrium exists at the melting point. At the melting point, the rate of melting is equal to the rate of freezing. (iii)Solid ⇌ Vapour (c)Sublimation point (iii) Solid ⇌ Vapour equilibrium exists at the sublimation point. Ex:- I 2 s ⇌ I 2 vapour (iv)Solute(s) ⇌ Solute(solution) (a)Saturated solution (iv) Solute(s) ⇌Solute (solution) equilibrium exists in saturated solution. In a saturated solution, no more solute can be dissolved. (i) Liquid ⇌ Vapour (b) Boiling point (ii) Solid ⇌ Liquid (d) Melting point (iii) Solid ⇌ Vapour (c) Sublimation point (iv) Solute (s) ⇌ Solute (solution) (a) Saturated solution

Match standard free energy of the reaction with the corresponding equilibrium constant Column I Column II (i) ∆ G ⊝ > 0 (a) K > 1 (ii) ∆ G ⊝ < 0 (b) K = 1 (iii) ∆ G ⊝ = 0 (c) K = 0 (d) K < 1

As we know that ∆ G ⊝ = - RT ln K (i) If ∆ G ⊝ > 0 , i.e., ∆ G ⊝ is positive, then ln K is negative i.e., K < 1 (ii) If ∆ G ⊝ < 0 , i.e., ∆ G ⊝ is negative, then ln K is positive i.e., K > 1 As we know that ∆ G ⊝ = - RT ln K (iii) If ∆ G ⊝ = 0 , i.e., then ln K is zero i.e., K = 1 (i)-d, (ii)-a, (iii)-b Column I Column II (i) ∆ G ⊝ > 0 (d) K < 1 (ii) ∆ G ⊝ < 0 (a) K > 1 (iii) ∆ G ⊝ = 0 (b) K = 1

Match the following species with the corresponding conjugate acid Species Conjugate acid (i) NH 3 (p) CO 3 2 - (ii) HCO 3 - (q) NH 4 + (iii) H 2 O (r) H 3 O + (iv) HSO 4 - (s) H 2 SO 4 (t) H 2 CO 3

Conjugate acid is a chemical compound formed by the reception of a proton by a base. ( i ) NH 3 + H + → NH 4 + ( ii ) HCO 3 - + H + → H 2 CO 3 Conjugate acid is a chemical compound formed by the reception of a proton by a base. ( iii ) H 2 O + H + → H 3 O + ( iv ) HSO 4 - + H + → H 2 SO 4 Species Conjugate acid (i) NH 3 (q) NH 4 + (ii) HCO 3 - (t) H 2 CO 3 (iii) H 2 O (r) H 3 O + (iv) HSO 4 - (s) H 2 SO 4

Match the following graphical variation with their description A B (i) (a) Variation in product concentration with time (ii) (b) Reaction at equilibrium (iii) (c) Variation in reactant concentration with time

(i) This graph shows that, the reactant concentration decreases with time (ii) This graph shows that, the product concentration increases with time (iii) This graph shows that, at equilibrium the concentration of reactants and products are constant. A B (i) (c) Variation in reactant concentration with time (ii) (a) Variation in product concentration with time (iii) (b) Reaction at equilibrium

Match Column (I) with Column (II). Column I Column II (i) Equilibrium (a) ∆ G > 0 , K < 1 (ii) Spontaneous reaction (b) ∆ G = 0 (iii) Non-spontaneous reaction (c) ∆ G ⊝ = 0 (d) ∆ G < 0 , K > 1

∆ G is 0 , the reaction has achieved equilibrium; at this point, there is no longer any free energy left to drive the reaction. If ΔG ° = 0 , then K = 1 , and neither reactants nor products are favoured at equilibrium. ∆ G is negative, then the reaction is spontaneous and proceeds in the forward direction. If ΔG ° < 0 , then K > 1 , and products are favoured over reactants at equilibrium. ∆ G is positive, then the reaction is considered non-spontaneous. Instead, as reverse reaction would have a negative ∆ G , the products of the forward reaction shall be converted to the reactants. If ΔG ° > 0 , then K < 1 , and reactants are favoured over products at equilibrium. Column I Column II (i) Equilibrium (b) ∆ G = 0 (c) ∆ G ⊝ = 0 (ii) Spontaneous reaction (d) ∆ G < 0 , K > 1 (iii) Non-spontaneous reaction (a) ∆ G > 0 , K < 1

NCERT Solutions for Chapter: Equilibrium, Exercise 1: Matching Type

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NCERT Chemistry Solutions for Exercise - NCERT Solutions for Chapter: Equilibrium, Exercise 1: Matching Type

Attempt the practice questions on Chapter 7: Equilibrium, Exercise 1: Matching Type with hints and solutions to strengthen your understanding. NCERT Exemplar Chemistry - Class 11 solutions are prepared by Experienced Embibe Experts.

Questions from NCERT Solutions for Chapter: Equilibrium, Exercise 1: Matching Type with Hints & Solutions

MEDIUM

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 7 - Equilibrium>Matching Type>Q 38

Match the following equilibria with the corresponding condition

(i)	Liquid $⇌$ Vapour	(a)	Saturated solution
(ii)	Solid $⇌$ Liquid	(b)	Boiling point
(iii)	Solid $⇌$ Vapour	(c)	Sublimation point
(iv)	Solute (s) $⇌$ Solute (solution)	(d)	Melting point
		(e)	Unsaturated solution

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 7 - Equilibrium>Matching Type>Q 39

For the reaction : $N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$

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

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)
(i)	$2 N_{2} (g) + 6 H_{2} (g) ⇌ 4 {NH}_{3} (g)$	(a)	$2 K_{C}$
(ii)	$2 {NH}_{3} (g) ⇌ N_{2} (g) + 3 H_{2} (g)$	(b)	${K_{c}}^{\frac{1}{2}}$
(iii)	$\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) ⇌ {NH}_{3} (g)$	(c)	$\frac{1}{K_{c}}$
		(d)	$K_{c}^{2}$

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 7 - Equilibrium>Matching Type>Q 40

Match standard free energy of the reaction with the corresponding equilibrium constant

	Column I		Column II
(i)	$∆ G^{⊝} > 0$	(a)	$K > 1$
(ii)	$∆ G^{⊝} < 0$	(b)	$K = 1$
(iii)	$∆ G^{⊝} = 0$	(c)	$K = 0$
		(d)	$K < 1$

EASY

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 7 - Equilibrium>Matching Type>Q 41

Match the following species with the corresponding conjugate acid

Species		Conjugate acid
(i)	${NH}_{3}$	(p)	${CO}_{3}^{2 -}$
(ii)	${HCO}_{3}^{-}$	(q)	${NH}_{4}^{+}$
(iii)	$H_{2} O$	(r)	$H_{3} O^{+}$
(iv)	${HSO}_{4}^{-}$	(s)	$H_{2} {SO}_{4}$
		(t)	$H_{2} {CO}_{3}$

MEDIUM

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 7 - Equilibrium>Matching Type>Q 42

Match the following graphical variation with their description

A	B
(i)	(a)	Variation in product concentration with time
(ii)	(b)	Reaction at equilibrium
(iii)	(c)	Variation in reactant concentration with time

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 7 - Equilibrium>Matching Type>Q 43

Match Column (I) with Column (II).

Column I		Column II
(i)	Equilibrium	(a)	$∆ G > 0, K < 1$
(ii)	Spontaneous reaction	(b)	$∆ G = 0$
(iii)	Non-spontaneous reaction	(c)	$∆ G^{⊝} = 0$
		(d)	$∆ G < 0, K > 1$

NCERT Solutions for Chapter: Equilibrium, Exercise 1: Matching Type

NCERT Chemistry Solutions for Exercise - NCERT Solutions for Chapter: Equilibrium, Exercise 1: Matching Type

Questions from NCERT Solutions for Chapter: Equilibrium, Exercise 1: Matching Type with Hints & Solutions

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