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NCERT Solutions for Class 11 Chemistry Chapter 7 : Equilibrium

Q: Q1. What is a Buffer Solution defined as in Chemistry Chapter 7 Equilibrium Class 11 NCERT Solutions?

Answer: Chemistry Chapter 7 Class 11 Equilibrium NCERT Solutions defines Buffer solutions as solutions that withstand changes in pH caused by dilution or the addition of small amounts of acid or alkali. It is a water-based solvent-based solution composed of a weak acid and the conjugate base of the weak acid, or a weak base and the conjugate acid of the weak base.

Q: Q2. What are the key subtopics in Chemistry Chapter 7 Equilibrium Class 11 that could be tested?

Answer: Chapter 7 Class 11 Chemistry Equilibrium NCERT Solutions includes Equilibrium in physical processes, Equilibrium in Chemical Processes, the laws of chemical equilibrium, equilibrium constant, Ionic equilibrium in solution, Ionization of acids and bases, and Buffer Solutions. Students can now study and stay up to date on the latest CBSE syllabus by using the NCERT Solutions, which are available in PDF format.

Q: Q3. Is it necessary to practice all of the questions in Class 11 Equilibrium NCERT Solutions?

Answer: Chapter 7 Chemistry Class 11 Equilibrium NCERT Solutions are well-designed to help students understand that Equilibrium is the condition in which both reactants and products are present in concentrations that have no further tendency to change with time, resulting in no apparent change in the system's properties. It thoroughly covers all essential definitions, laws and numerical problems allowing students to grasp the concepts easily. All the questions, examples, and practice problems have been meticulously designed to help students gain a rapid and effective comprehension of the various topics.

Equilibrium Class 11 Chemistry Chapter 7 deals with the study of chemical equilibria. Chemistry Class 11 Chapter 7 Equilibrium explores key concepts such as Equilibrium in Physical Processes, Equilibrium in Chemical Processes, Homogeneous Equilibria, Heterogeneous Equilibria, Applications of Equilibrium Constants, Relationship between Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G, Acids, Bases and Salts, Ionization of Acids and Bases, Buffer Solutions, etc. Class 11 Chemistry Chapter 7 Equilibrium will help students to gain a thorough understanding of this concept and its practical applications.

Equilibrium Class 11 Chemistry Chapter 7 covers the aforementioned topics and the expert faculty of Toppr has created solutions related to each and every topic to aid students with their first-term exam preparations. Students will quickly learn about these concepts if they practice Chapter 7 Class 11 Chemistry Equilibrium NCERT Solutions. All of these solutions have been created with the new CBSE pattern in mind, ensuring that students have a thorough understanding of their exams.

Chemistry Chapter 7 Equilibrium Class 11 Questions and Answers can help you get good grades in tests and properly prepare for all of the important concepts. Several examples provided in the solutions will assist students with a better understanding of the different equilibrium concepts. Chapter 7 Class 11 Chemistry Equilibrium NCERT Solutions is available below in pdf format, and a few solutions are also included in the exercises. These solutions explain the topics covered with examples so that students can easily relate to the notion being discussed.

Table of Content

Introduction

Access NCERT Solutions for Class 11 Chemistry Chapter 7 : Equilibrium

Exercise Page Number 224

Question 2

What is

K c

for the following equilibrium when the equilibrium concentration of each substance is

[S O_{2}] = 0.60 M, [O_{2}] = 0.82 M, [S O_{3}] = 1.90 M 2 S O_{2} + O_{2} ⇌ 2 S O_{3}

Medium

Solution

Verified by Toppr

2 S O_{2} + O_{2} ⇌ 2 S O_{3}

K_{C} = \frac{[ S O _{3} ] ^{2}}{[ S O _{2} ] ^{2} [ O _{2} ]} = \frac{( 1 . 9 0 ) ^{2}}{( 0 . 6 0 ) ^{2} ( 0 . 8 2 )} = 12.23

Question 3

At a certain temperature and total pressure of

10^{5} P a

, iodine vapour contains

40 %

by volume of I atoms.

I_{2} (g) ⇌ 2 I (g)

Calculate

K_{p}

for the equilibrium.

Hard

Solution

Verified by Toppr

The partial pressure of iodine atoms is

P = \frac{1 0 ^{5} \times 4 0}{1 0 0} = 0.4 \times 1 0^{6} P a

The parital ressure of

I_{2}

molecules is

P^{'} = \frac{1 0 ^{5} \times 6 0}{1 0 0} = 0.6 \times 1 0^{5} P a

The value of the equilibium constant is

K_{P} = \frac{P ^{2}}{P ^{'}} = \frac{( 0 . 4 \times 1 0 ^{5} ) ^{2}}{0 . 6 \times 1 0 ^{5}} = 2.67 \times 1 0^{4}

Question 4

Write the expression for the equilibrium constant,

K_{c}

for each of the following reactions.
(i)

2 N O C l (g) ⇌ 2 N O (g) + C l_{2} (g)

(ii)

2 C u (N O_{3})_{2} (s) ⇌ 2 C u O (s) + 4 N O_{2} (g) + O_{2} (g)

(iii)

C H_{3} C O O C_{2} H_{5} (a q) + H_{2} O (1) ⇌ C H_{3} C O O H (a q) + C_{2} H_{5} O H (a q)

(iv)

F e^{3 +} (a q) + 3 O H^{-} (a q) ⇌ F e (O H)_{3} (s)

(v)

I_{2} (s) + 5 F_{2} ⇌ 2 I F_{5}

Hard

Solution

Verified by Toppr

(a) The expression for the equilibrium constant is

K_{c} = \frac{[ N O ] ^{2} [ C l _{2} ]}{[ N O C l ] ^{2}}

.
(b) The expression for the equilibrium constant is

K_{c} = [N O_{2}]^{4} [O_{2}]

.
(c) The expression for the equilibrium constant is

K_{c} = \frac{[ C H _{3} C O O H ] [ C _{2} H _{5} O H ]}{[ C H _{3} C O O C _{2} H _{5} ]}

.
(d) The expression for the equilibrium constant is

K_{c} = \frac{1}{[ F e ^{3 +} ] [ O H ^{-} ] ^{3}}

.
(e) The expression for the equilibrium constant is

K_{c} = \frac{[ I F _{5} ] ^{2}}{[ F _{2} ] ^{5}}

Question 5

Find out the value of

K_{C}

for each of the following equilibrium respectively from the value of

K_{P}

.
(i)

2 N O C l (g) ⇌ 2 N O (g) + C l_{2} (g); K_{P} = 1.8 \times 1 0^{- 2}

500

K
(ii)

C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g); K_{P} = 167

1073

4.4 \times 1 0^{- 4}

1.90

8.8 \times 1 0^{- 4}

3.8

4.4 \times 1 0^{4}

1.90

8.8 \times 1 9^{4}

3.8

Hard

Solution

Verified by Toppr

The relationship between

K_{p}

and

K_{c}

K_{p} = K_{c} (R T)^{Δ n}

For the reaction

2 N O C l (g) ⇌ 2 N O (g) + C l_{2} (g)

, the value of

Δ n

1 + 2 - 2 = 1

Substituting this value in the above expression, we get

K_{p} = K_{c} (R T)^{Δ n} = K_{c} (R T)^{1}

But

K_{P} = 1.8 \times 1 0^{- 2}, R = 0.082 L atm /mol. K, T = 500 K

Substituting values in the above expression, we get

1.8 \times 1 0^{- 2} = K_{c} (0.082 \times 500)^{1}

K_{c} = 4.4 \times 1 0^{- 4}

For the reaction

C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)

, the value of

Δ n

= 1

Substituting this value in the above expression, we get

K_{p} = K_{c} (R T)^{Δ n} = K_{c} (R T)^{1} .

But

K_{p} = 167, R = 0.082 L atm /mol. K, T = 1073 K

Substituting values in the above expression, we get

167 = K_{c} (0.082 \times 1073)^{1} .

K_{c} = 1.90

Question 6

For the following equilibrium,

K_{C} = 6.3 \times 1 0^{14}

1000 K

N O (g) + O_{3} (g) ⇌ N O_{2} (g) + O_{2} (g)

. Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is

K_{C}

, for the reverse reaction?

15.9 \times 1 0^{- 15}

1.59 \times 1 0^{- 15}

5 \times 1 0^{- 15}

9 \times 1 0^{- 15}

Hard

Solution

Verified by Toppr

The value of the equilibrium constant for the reverse reaction is obtained from the reciprocal of the value of the equilibrium constant of the forward reaction.

Hence,

K_{C (r e v e r s e)} = \frac{1}{K _{C}}

Substitute values in the above expression.

K_{C (r e v e r s e)} = \frac{1}{6 . 3 \times 1 0 ^{14}} = 1.59 \times 1 0^{- 15}

Question 7

Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression.

Medium

Solution

Verified by Toppr

Pure solids and liquids are not included in the equilibrium constant expression. This is because they do not affect the reactant amount at equilibrium in the reaction, so they are disregarded and kept at 1.

Question 8

Reaction between Nitrogen and Oxygen takes place as follows:-

2 N_{2} (g) + O_{2} (g) ⇌ 2 N_{2} O (g)

If a mixture of

0.482 m o l

N_{2}

and

0.933 m o l

O_{2}

is placed in a reaction vessel of volume

10 L

and allowed to form

N_{2} O

at a temperature for which

K e = 2.0 \times 1 0^{- 37}

, determine the composition of the equilibrium mixture.

Hard

Solution

Verified by Toppr

2 N_{2} + O_{2} ⇌ 2 N_{2} O (g)

initial concentration of

N_{2} [N_{2}] = \frac{0 . 4 8 2}{1 0}

= 0.0482

[O_{2}] = \frac{0 . 9 3 3}{2} = 0.0933

2 N_{2} + O_{2} ⇌ 2 N_{2} O

0.0482 - 2 x 0.0933 - x

2 x

2 \times 1 0^{- 37} = \frac{[ 2 x ] ^{2}}{( 0 . 0 4 8 2 - 2 x ) ^{2} ( 0 . 0 9 3 3 - x )}

2 \times 1 0^{- 37} = \frac{4 x ^{2}}{( 0 . 0 4 8 2 ) ^{2} ( 0 . 0 9 3 3 )}

x = 0.0329 \times 1 0^{- 19}

= 3.29 \times 1 0^{- 21}

[N_{2} O] = 2 \times 3.29 \times 1 0^{- 21} = 6.58 \times 1 0^{- 21}

[N_{2}] ≃ 0.0482

[O_{2}] ≃ 0.0933

Question 9

Nitric oxide reacts with bromine and gives nitrosyl-bromide as per reaction given below:

2 N O_{(g)} + B r_{2 (g)} ⇌ 2 N O B r_{(g)}

When

0.087

mole of

N O

and

0.0437

mole of

B r_{2}

are mixed is a closed container at constant temperature,

0.0518

mole of

N O B r

is obtained at equilibrium. Calculate the equilibrium amount of nitric oxide and bromine.

Hard

Solution

Verified by Toppr

2 N O_{(g)} + B r_{2 (g)} ⇌ 2 N O B r_{(g)}

Initial moles of

N O = 0.087

Initial moles of

B r_{2} = 0.0437

Final moles of

N O = 2 a

Final moles of

B r_{2} = a

∴

number of moles of

N O

at equilibrium

= (0.087 - 2 a)

Number of moles of

B r_{2}

at equilibrium $$= (0.0437 - a)

Given:

2 a = 0.0518

∴ a = \frac{0 . 0 5 1 8}{2} = 0.0259

At equilibrium,

Number of moles (amount) of

N O = 0.0870 - 0.0518 = 0.0352

mole
Number of moles (amount) of

B r_{2} = 0.0437 - 0.0259 = 0.0178

mole

Question 10

At 450K,

K_{p} = 2.0 \times 10^{10}

/bar for the given reaction at equilibrium.

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

What is

K_{c}

at this temperature?

Hard

Solution

Verified by Toppr

Δ n = n_{p r o d u c t s} - n_{r e a c t a n t s} = 2 - (2 + 1) = - 1

we know that

K_{p} = K_{c} (R T)^{Δ n}

∴ K_{c} = K_{p} (R T)^{- Δ n} = (2.0 \times 1 0^{10}) \times (0.0831 \times 450)^{1} = 7.48 \times 1 0^{11} M / m o l

Question 11

A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What of

K_{P}

for the given equilibrium?

2 H I (g) ⇌ H_{2} (g) + I_{2} (g)

Hard

Solution

Verified by Toppr

The equilibrium reaction is shown below.

2 H I (g) ⇌ H_{2} (g) + I_{2} (g)

	$H I$	$H_{2}$	$I_{2}$
Initial	0.2	0	0
Change	-2x	x	x
Equilibrium	0.2-2x	x	x

But the equilibrium pressure of

H I

0.04

atm.

0.2 - 2 x = 0.04

Hence,

2 x = 0.2 - 0.04 = 0.16

x = 0.08

.
The expression for the equilibrium constant is

K_{P} = \frac{P _{H_{2}} P _{I_{2}}}{P _{H I}^{2}} = \frac{0 . 0 8 \times 0 . 0 8}{0 . 0 4 ^{2}} = 4

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NCERT Solutions for Class 11 Chemistry Chapter 7 : Equilibrium

NCERT Solutions for Class 11 Chemistry Chapter 7 Equilibrium – Brief Overview

Equilibrium

Equilibrium is a condition of stability in which the system's variables such as pressure, temperature, and concentration do not change position throughout time. It remains in a stable state in which no changes take place and opposing forces cancel each other out. Physical processes and chemical reactions can both achieve equilibrium.

Law of Chemical Equilibrium

At a given temperature, the product of concentrations of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentrations of the reactants raised to their individual stoichiometric coefficients has a constant value. This is known as the Equilibrium Law or Law of Chemical Equilibrium.

Equilibrium Constant

The equilibrium constant for a general reaction, aA + bB ⇌ cC + dD is expressed as,

K_c = [C]^c [D]^d / [A]^a [B]^b

where [A], [B], [C] and [D] are the equilibrium concentrations of the reactants and products.

Acids, Bases and Salts

According to Arrhenius theory, acids are substances that dissociates in water to give hydrogen ions H⁺ (aq) and bases are substances that produce hydroxyl ions OH^- (aq).

When acids and bases are mixed in the proper proportions, they react to form salts.

pH Scale

The pH (potential of hydrogen) scale in chemistry is used to define the acidity or basicity of an aqueous solution. The pH of water is a measure of the relative number of free hydrogen and hydroxyl ions.

Acidic solution has pH < 7, Basic solution has pH > 7, Neutral solution has pH = 7

Buffer Solutions

The solutions which resist change in pH on dilution or with the addition of small amounts of acid or alkali are called Buffer Solutions. It is a water-based solvent-based solution composed of a weak acid and the conjugate base of the weak acid, or a weak base and the conjugate acid of the weak base.

Related Chapters

Chapter 1 : Some Basic Concepts of Chemistry

Chapter 2 : Structure of Atom

Chapter 3 : Classification of Elements and Periodicity in Properties

Chapter 4 : Chemical Bonding and Molecular Structure

Chapter 5 : States of Matter: Gases and Liquids

Chapter 6 : Thermodynamics

Frequently Asked Questions on NCERT Class 11 Chemistry Chapter 7 : Equilibrium

Q1. What is a Buffer Solution defined as in Chemistry Chapter 7 Equilibrium Class 11 NCERT Solutions?

Answer: Chemistry Chapter 7 Class 11 Equilibrium NCERT Solutions defines Buffer solutions as solutions that withstand changes in pH caused by dilution or the addition of small amounts of acid or alkali. It is a water-based solvent-based solution composed of a weak acid and the conjugate base of the weak acid, or a weak base and the conjugate acid of the weak base.

Q2. What are the key subtopics in Chemistry Chapter 7 Equilibrium Class 11 that could be tested?

Answer: Chapter 7 Class 11 Chemistry Equilibrium NCERT Solutions includes Equilibrium in physical processes, Equilibrium in Chemical Processes, the laws of chemical equilibrium, equilibrium constant, Ionic equilibrium in solution, Ionization of acids and bases, and Buffer Solutions. Students can now study and stay up to date on the latest CBSE syllabus by using the NCERT Solutions, which are available in PDF format.

Q3. Is it necessary to practice all of the questions in Class 11 Equilibrium NCERT Solutions?

Answer: Chapter 7 Chemistry Class 11 Equilibrium NCERT Solutions are well-designed to help students understand that Equilibrium is the condition in which both reactants and products are present in concentrations that have no further tendency to change with time, resulting in no apparent change in the system's properties. It thoroughly covers all essential definitions, laws and numerical problems allowing students to grasp the concepts easily. All the questions, examples, and practice problems have been meticulously designed to help students gain a rapid and effective comprehension of the various topics.

Related Chapters

Chapter 1 : Some Basic Concepts of Chemistry

Chapter 2 : Structure of Atom

Chapter 3 : Classification of Elements and Periodicity in Properties

Chapter 4 : Chemical Bonding and Molecular Structure

Chapter 5 : States of Matter: Gases and Liquids

Chapter 6 : Thermodynamics