NCERT Solutions for Class 11 Chemistry: Chapter 7 (Equilibrium)

NCERT Solutions for Class 11 Chemistry: Chapter 7 (Equilibrium) is provided on this page for the perusal of CBSE Class 11 Chemistry students. Detailed, step-by-step solutions for each and every intext and exercise question listed in Chapter 7 of the NCERT Class 11 Chemistry textbook can be accessed here. Furthermore, the NCERT Solutions provided on this page can also be downloaded as a PDF for free by clicking the download button provided below.

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NCERT Solutions for Chemistry – Class 11, Chapter 7: Equilibrium

“Equilibrium” is the seventh chapter in the NCERT Class 11 Chemistry textbook. Several important concepts such as equilibrium constants, buffer solutions and the common-ion effect are explained in this chapter. Along with structured questions, the NCERT Solutions for Class 11 Chemistry (Chapter 7) provided on this page come with detailed explanations to help students learn and understand the key concepts related to the chemical equilibrium, in a seamless manner. The important subtopics associated with Chapter 7  – “Equilibrium” are listed below.

Subtopics included in NCERT Class 11 Chemistry Chapter 7 – “Equilibrium”

1. Solid-liquid Equilibrium
   • Liquid-vapour Equilibrium
   • Solid – Vapour Equilibrium
   • Equilibrium Involving Dissolution Of Solid Or Gases In Liquids
   • General Characteristics Of Equilibria Involving Physical Processes
2. Equilibrium In Chemical Processes – Dynamic Equilibrium
3. Law Of Chemical Equilibrium And Equilibrium Constant
4. Homogeneous Equilibria
   • Equilibrium Constant In Gaseous Systems
5. Heterogeneous Equilibria
6. Applications Of Equilibrium Constants
   • Predicting The Extent Of A Reaction
   • Predicting The Direction Of The Reaction
   • Calculating Equilibrium Concentrations
7. Relationship Between Equilibrium Constant K, Reaction Quotient Q And Gibbs Energy G
8. Factors Affecting Equilibria
   • Effect Of Concentration Change
   • Effect Of Pressure Change
   • Effect Of Inert Gas Addition
   • Effect Of Temperature Change
   • Effect Of A Catalyst
9. Ionic Equilibrium In Solution
10. Acids, Bases And Salts
    • Arrhenius Concept Of Acids And Bases
    • The Bronsted-lowry Acids And Bases
    • Lewis Acids And Bases
11. Ionization Of Acids And Bases
    • The Ionization Constant Of Water And It’s Ionic Product
    • The Ph Scale
    • Ionization Constants Of Weak Acids
    • Ionization Of Weak Bases
    • The Relation Between Ka And Kb
    • Di- And Polybasic Acids And Di- And Polyacidic Bases
    • Factors Affecting Acid Strength
    • Common Ion Effect In The Ionization Of Acids And Bases
    • Hydrolysis Of Salts And The Ph Of Their Solutions
12. Buffer Solutions
13. Solubility Equilibria Of Sparingly Soluble Salts
    • Solubility Product Constant
    • Common Ion Effect On Solubility Of Ionic Salts

NCERT Solutions for Class 11 Chemistry Chapter 7

Q.1. A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.

a) What is the initial effect of the change on vapour pressure?

b) How do rates of evaporation and condensation change initially? 

c) What happens when equilibrium is restored finally and what will be the final vapour pressure?

 Ans.

(a) On increasing the volume of the container, the vapour pressure will initially decrease because the same amount of vapours are now distributed over a large space.

(b) On increasing the volume of the container, the rates of evaporation will increase initially because now more space is available. Since the amount of the vapours per unit volume decrease on increasing the volume, therefore, the rate of condensation will decrease initially.

(c) Finally, equilibrium will be restored when the rates of the forward and backward processes become equal. However, the vapour pressure will remain unchanged because it depends upon the temperature and not upon the volume of the container.

 

Q.2.What is Kc for the following equilibrium when the equilibrium concentration of each substance is: [SO₂]= 0.60M, [O₂] = 0.82M and [SO₃] = 1.90M ?

$2SO_{2}(g)+O_{2}(g) ⇌ 2SO_{3}(g)$

Ans.

As per the question,

$2SO_{2}(g)+O_{2}(g) ⇌ 2SO_{3}(g)$ (Given)

 

$K_{c}=frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]} =frac{(1.9)^{2}M^{2}}{(0.6)^{2}(0.82)M^{3}} =12.229M^{-1}$(approximately)

Hence, K for the equilibrium is 12.229 M^–1.

 

Q.3. At a certain temperature and total pressure of 10⁵ Pa, iodine vapour contains 40% by volume of I atoms

I₂(g) ⇌ 2I (g) Calculate K_p for the equilibrium

Ans.

Partial pressure of Iodine atoms (I)

$p_{I}=frac{40}{100} imes p_{total} =frac{40}{100} imes 10^{5} =4 imes 10^{4}Pa$

Partial pressure of I₂ molecules,

$p_{I}=frac{60}{100} imes p_{total} =frac{60}{100} imes 10^{5} =6 imes 10^{4}Pa$

Now, for the given reaction,

$K_{p}=frac{(p_{I})^{2}}{p_{I_{2}}} =frac{(4 imes 10^{4})^{2}Pa^{2}}{6 imes 10^{4}Pa} =2.67 imes 10^{4}Pa$

 

Q.4.Write the expression for the equilibrium constant, K_c for each of the following reactions:

(i)$2NOCl(g)leftrightarrow 2NO(g)+Cl_{2}(g)$

(ii)$2Cu(NO_{3})_{2}(s)leftrightarrow 2CuO(s)+4NO_{2}(g)+O_{2}(g)$

(iii)$CH_{3}COOC_{2}H_{5}(aq)+H_{2}O(1)leftrightarrow CH_{3}COOH(aq)+C_{2}H_{5}OH(aq)$

(iv)$Fe^{3+}(aq)+3OH^{-}(aq)leftrightarrow Fe(OH)_{3}(s)$

(v)$I_{2}(s)+5F_{2}leftrightarrow 2IF_{5}$

Ans.

(i) $K_{C}=frac{[NO_{g}]^{2}[Cl_{2_{(g)}}]}{[NOCl_{(g)}]^{2}}$

(ii)$K_{C}=frac{[CuO_{(s)}]^{2}[NO_{2_{(g)}}]^{4}[O_{2_{(g)}}]}{[Cu(NO_{3})_{2(g)}]^{2}} =[NO_{2(g)}]^{4}[O_{2(g)}]$

(iii)$K_{C}=frac{[CH_{3}COOH_{(aq)}][C_{2}H_{5}OH_{(aq)}]}{[CH_{3}COOC_{2}H_{5(aq)}][H_{2}O_{(l)}]} =frac{[CH_{3}COOH_{(aq)}][C_{2}H_{5}OH_{(aq)}]}{[CH_{3}COOC_{2}H_{5(aq)}]}$

(iv)$K_{C}=frac{[Fe(OH)_{3(s)}]}{[Fe^{3+}_{(aq)}][OH^{-}_{(aq)}]^{3}} =frac{1}{[Fe^{3+}_{(aq)}][OH^{-}_{(aq)}]^{3}}$

(v)$K_{C}=frac{[IF_{5}]^{2}}{[I_{2(s)}][F_{2}]^{5}} =frac{[IF_{5}]^{2}}{[F_{2}]^{5}}$

 

Q.5.Find out the value of Kc for each of the following equilibria from the value of Kp:

(i)$2NOCl(g) ⇌ 2NO(g)+Cl_{2}(g);: : : K_{p}=1.8 imes 10^{-2}: : at: 500K$

(ii)$CaCO_{3}(s) ⇌ CaO(s)+CO_{2}(g);: : : K_{p}=167: : at: 1073K$

Ans.

The relation between K_p and K_c is given as:

$K_{p}=K_{c}(RT)^{Delta n}$

(i) Given,

R = 0.0831 barLmol^–1K^–1

$Delta n = 3-2 =1$

T = 500 K

K_p=$1.8 imes 10^{-2}$

Now,

K_p = K_c (RT) ^∆n

$Rightarrow 1.8 imes 10^{-2}=K_{c}(0.0831 imes 500)^{1} Rightarrow K_{c}=frac{1.8 imes 10^{-2}}{0.0831 imes 500} =4.33 imes 10^{-4}(approximately)$

(ii) Here,

∆n =2 – 1 = 1

R = 0.0831 barLmol^–1K^–1

T = 1073 K

K_p= 167

Now,

K_p = K_c (RT) ^∆n

$Rightarrow 167=K_{c}{(0.0831 imes 1073)^{Delta n}} Rightarrow K_{c}=frac{167}{0.0831 imes 1073} =1.87(approximately)$

 

Q.6. For the following equilibrium, $K_{c}=6.3 imes 10^{14}: : : at: 1000K NO(g)+O_{3(g)} ⇌ NO_{2(g)}+O_{2(g)}$

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K_c, for the reverse reaction?

Ans.

For the reverse reaction, $K_{c}=frac{1}{K_{c }} =frac{1}{6.3 imes 10^{14}} =1.59 imes 10^{-15}$

 

Q.7. Explain why solids and pure liquids can be ignored while writing the equilibrium constant expression.

Ans.

This is because the molar concentration of a pure solid or liquid is independent of the amount present.

Mole concentration= $frac{Number: of: moles}{Volume} frac{Mass/molecular:mass}{Volume} =frac{Mass}{Volume, imes, Molecular :mass} =frac{Density}{Molecular: mass}$

Though the density of the solid and pure liquid is fixed and molar mass is also fixed.

$∴$Molar concentration are constant.

 

Q.8.

The reaction between N₂ and O₂ takes place as follows:

$2N_{2}(g)+O_{2} ⇌ 2N_{2}O(g)$

If a solution of 0.933 mol of oxygen and 0.482 mol of nitrogen is placed in a 10 L reaction vessel and allowed to form N₂O at a temperature for which K_c =$2.0 imes 10^{-37}$, determine the composition of the equilibrium solution.

Ans.

Let the concentration of N₂O at equilibrium be x.

The given reaction is:

2N₂(g)             +               O₂(g)               ⇌                2N₂O(g)

Initial conc.     0.482 mol                     0.933 mol                                  0

At equilibrium(0.482-x)mol            (1.933-x)mol                              x mol

$[N_{2}]=frac{0.482-x}{10}cdot [O_{2}]=frac{0.933-frac{x}{2}}{10},[N_{2}O]=frac{x}{10}$

The value of equilibrium constant  is extremely small. This means that only small amounts . Then,

$[N_{2}]=frac{0.482}{10}=0.0482molL^{-1} : :and:: [O_{2}]=frac{0.933}{10}=0.0933molL^{-1}$

Now,

$K_{c}=frac{[N_{2}O_{(g)}]^{2}}{[N_{2(g)}][O_{2(g)}]} Rightarrow 2.0 imes 10^{-37}=frac{(frac{x}{10})^{2}}{(0.0482)^{2}(0.0933)} Rightarrow frac{x^{2}}{100}=2.0 imes 10^{-37} imes (0.0482)^{2} imes (0.0933) Rightarrow x^{2}=43.35 imes 10^{-40} Rightarrow x=6.6 imes 10^{-20}$ $[N_{2}O]=frac{x}{10}=frac{6.6 imes 10^{-20}}{10} =6.6 imes 10^{-21}$

 

Q.9.

Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below:

2NO_(g)+Br_2(g) ⇌ 2NOBr_(g)

When 0.087 mol of NO and 0.0437 mol of Br₂ are mixed in a closed container at a constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate the equilibrium amount of NO and Br₂.

Ans.

The given reaction is:

2NO_(g)+Br_2(g)             ⇌                 2NOBr_(g)

2mol       1mol                                      2mol

Now, 2 mol of NOBr is formed from 2 mol of NO. Therefore, 0.0518 mol of NOBr is formed from 0.0518 mol of NO.

Again, 2 mol of NOBr is formed from 1 mol of Br.

Therefore, 0.0518 mol of NOBr is formed from $frac{0.0518}{2}$ mol or Br, or 0.0259 mol of NO.

The amount of NO and Br present initially is as follows:

[NO] = 0.087 mol [Br2] = 0.0437 mol

Therefore, the amount of NO present at equilibrium is:

[NO] = 0.087 – 0.0518 = 0.0352 mol

And, the amount of Br present at equilibrium is:

[Br₂] = 0.0437 – 0.0259 = 0.0178 mol

 

Q.10. At 450 K, K_p= $2.0 imes 10^{10}$/bar for the given reaction at equilibrium.

2SO₂(g)+O₂(g) ⇌ 2SO₃(g) What is K_c at this temperature?

Ans.

For the given reaction,

∆n = 2 – 3 = – 1

T = 450 K

R = 0.0831 bar L bar K^–1 mol^–1

K_p=$2.0 imes 10^{10}bar^{-1}$

We know that,

$K_{p}=K_{c}(RT)Delta n Rightarrow 2.0 imes 10^{10}bar^{-1}=K_{c}(0.0831L, bar, K^{-1}mol^{-1} imes 450K)^{-1} Rightarrow K_{c}=frac{2.0 imes 10^{10}bar^{-1}}{(0.0831L, bar, K^{-1}mol^{-1} imes 450K)^{-1}} =(2.0 imes 10^{10}bar^{-1})(0.0831L, bar, K^{-1}mol^{-1} imes 450K) =74.79 imes 10^{10}L, mol^{-1} =7.48 imes 10^{11}L, mol^{-1} =7.48 imes 10^{11}, M^{-1}$

 

Q.11. A sample of HI_(g) is placed in a flask at a pressure of 0.2 atm. At equilibrium, the partial pressure of HI_(g) is 0.04 atm. What is K_p for the given equilibrium?

2HI(g) ⇌ H₂(g)+I₂(g)

Ans.

The initial concentration of HI is 0.2 atm. At equilibrium, it has a partial pressure of 0.04 atm.

Therefore, a decrease in the pressure of HI is 0.2 – 0.04 = 0.16. The given reaction is:

2HI(g)               ⇌                         H₂(g)       +      I₂(g)

Initial conc.             0.2 atm                                           0                         0

At equilibrium        0.4 atm                                         0.16/2               0.16/2

=0.08atm         =0.08atm

Therefore,

$K_{p=}frac{p_{H_{2}} imes p_{I_{2}}}{p^{2}_{HI}} =frac{0.08 imes 0.08}{(0.04)^{2}} =frac{0.0064}{0.0016} =4.0$

Hence, the value of K_p for the given equilibrium is 4.0.

 

Q.12. A mixture of 1.57 mol of N₂, 1.92 mol of H₂ and 8.13 mol of NH₃ is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, K_c for the reaction

N₂(g)+3H₂(g) ⇌ 2NH₃(g) is $1.7 imes 10^{2}$

 Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Ans.

The given reaction is:

N₂(g)+3H₂(g) ⇌ 2NH₃(g)

The given concentration of various species is

[N₂]= $frac{1.57}{20}mol, L^{-1}$ [H₂]= $frac{1.92}{20}mol, L^{-1}$ [NH₃]= $frac{8.31}{20}mol, L^{-1}$

Now, reaction quotient Q_c is:

$Q=frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}} =frac{(frac{(8.13)}{20})^{2}}{(frac{1.57}{20})(frac{1.92}{20})^{3}} =2.4 imes 10^{3}$

Since, Q_c ≠ K_c, the reaction mixture is not at equilibrium.

Again, Q_c > K_c. Hence, the reaction will proceed in the reverse direction.

 

Q.13. The equilibrium constant expression for a gas reaction is,

$K_{c}=frac{[NH_{3}]^{4}[O_{2}]^{5}}{[NO]^{4}[H_{2}O]^{6}}$

Write the balanced chemical equation corresponding to this expression.

Ans.

The balanced chemical equation corresponding to the given expression can be written as:

4NO_(g)+6H₂o_(g) ⇌ 4NH_3(g)+5O_2(g)

 

Q.14. One mole of H₂O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium, 60% of water (by mass) reacts with CO according to the equation, 

H₂O(g)  +  CO(g) ⇌ H₂(g)   +  CO₂(g)

Calculate the equilibrium constant for the reaction.

Ans.

The given reaction is:

H₂O_(g)  +  CO_(g) ⇌ H_2(g)   +  CO_2(g)

Therefore, the equilibrium constant for the reaction,

K_c = ([H₂][CO₂])/([H₂O][CO]) = (0.4*0.4)/(0.6*0.6) = 0.444

 

Q.15. At 700 K, equilibrium constant for the reaction

H_2(g)+I_2(g) ⇌ 2HI_(g)

is 54.8. If 0.5 molL^–1 of HI_(g) is present at equilibrium at 700 K, what are the concentration of H_2(g) and I_2(g) assuming that we initially started with HI_(g) and allowed it to reach equilibrium at 700 K?

Ans.

It is given that equilibrium constant K_c for the reaction

H_2(g)+I_2(g)          ⇌                    2HI_(g)   is  54.8.

Therefore, at equilibrium, the equilibrium constant K^’_c for the reaction

2HI_(g)          ⇌                   H_2(g)+I_2(g)

[HI]=0.5 molL^-1 will be 1/54.8.

Let the concentrations of hydrogen and iodine at equilibrium be x molL^–1

[H₂]=[I₂]=x mol L^-1

Therefore, $frac{[H_{2}][I_{2}]}{[HI]^{2}}=K^{‘}_{c} Rightarrow frac{x imes x}{(0.5)^{2}}=frac{1}{54.8} Rightarrow x^{2}=frac{0.25}{54.8} Rightarrow x=0.06754 x=0.068molL^{-1}(approximately)$

Hence, at equilibrium, [H₂]=[I₂]=0.068 mol L^-1.

 

Q.16. What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?

2ICl_(g) ⇌ I_2(g)+Cl_2(g) ;  K_c =0.14

Ans.

The given reaction is:

2ICl_(g)                  ⇌                         I_2(g)      +   Cl_2(g)

Initial conc.                 0.78 M                                           0             0

At equilibrium      (0.78-2x) M                                       x M           x M

Now, we can write, $frac{[I_{2}][Cl_{2}]}{[IC]^{2}}=K_{c} Rightarrow frac{x imes x}{(0.78-2x)^{2}}=0.14 Rightarrow frac{x^{2}}{(0.78-2x)^{2}}=0.14 Rightarrow frac{x}{0.78-2x}=0.374 Rightarrow x=0.292-0.748x Rightarrow 1.748x=0.292 Rightarrow x=0.167$