NCERT Solutions for Class 11 Chemistry Chapter 7 Equilibrium

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. Ionisation of Acids and Bases
    • The Ionisation Constant of Water and Its Ionic Product
    • The pH Scale
    • Ionisation Constants of Weak Acids
    • Ionisation 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 Ionisation 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 is 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 vapours per unit volume decrease on increasing the volume, 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?

Q.9. Nitric oxide reacts with Br2 and gives nitrosyl bromide as per the reaction given below:

2NO(g)+Br2(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)+Br2(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

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 Kp 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,

Q.14. One mole of H2O and one mole of CO are taken in a 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) ⇌ H2(g)   +  CO2(g)

Compound	H20	CO	H2	CO2
Initial Conc.	0.1M	0.1M	0	0
Equilibrium Conc.	0.06M	0.06M	0.04M	0.04M

Therefore, the equilibrium constant for the reaction,

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

 

Q.15. At 700 K, the equilibrium constant for the reaction H2(g)+I2(g) ⇌ 2HI(g) is 54.8. If 0.5 molL–1 of HI(g) is present at equilibrium at 700 K, what is the concentration of H2(g) and I2(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 Kc for the reaction

H2(g)+I2(g)          ⇌                    2HI(g)   is  54.8.

Therefore, at equilibrium, the equilibrium constant K’c for the reaction

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

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) ⇌ I2(g)+Cl2(g) ;  Kc =0.14

Ans.

The given reaction is:

2ICl(g)                  ⇌                         I2(g)      +   Cl2(g)

Initial conc.                 0.78 M                                           0             0

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

Now, we can write,

Q.17. Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C2H6 when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?

C2H6(g) ⇌ C2H4(g) +H2(g)

Ans.

Let p be the pressure exerted by ethene and hydrogen gas (each) at equilibrium. Now, according to the reaction,

C2H6(g)                   ⇌               C2H4(g)    +     H2(g)

Initial conc.                 4.0 M                                           0                      0

At equilibrium          (4.0-p)                                            p                     p

We can write,

Q.18. Ethyl acetate is formed by the reaction between ethanol and acetic acid, and the equilibrium is represented as:

CH3COOH(l) + C2H5OH(l) ⇌ CH3COOC2H5(l) + H2O(l)

(i) Write the concentration ratio (reaction quotient) Qc for this reaction (note: water is not in excess and is not a solvent in this reaction).

(ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.    

(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after some time. Has equilibrium been reached?

Ans.

(i)Reaction quotient,

Q.20. One of the reactions that take place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO2.

 FeO (s) + CO (g) ⇌ Fe (s) + CO2 (g); Kp= 0.265 at 1050 K.

What is the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are: pCO = 1.4 atm and pCO2= 0.80 atm?

Ans.

For the given reaction,

FeO(g) + CO(g) ⇌ Fe(s)+CO2(g)

Initially,  pCO = 1.4 atm and pCO2= 0.80 atm

Qp=

Q.21. The equilibrium constant, Kc, for the reaction

N2(g)+3H2(g) ⇌ 2NH3  at 500 K is 0.061

At a specific time, from the analysis, we can conclude that the composition of the reaction mixture is 3.0 mol L –1 N2, 2.0 mol L–1 H2 and 0.5 mol L–1 NH3. Find out whether the reaction is at equilibrium or not. Find in which direction the reaction proceeds to reach equilibrium.

Ans.

N2(g)        +        3H2(g)                  ⇌                2NH3

At a particular time:  3.0 mol L-1          2.0mol L-1                              0.5 mol L-1

So,

Qc=

Q.25. Does the number of moles of reaction products increase, decrease or remain the same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?

(a) PCl5(g) ⇌ PCl3 +Cl2 (g)

(b) CaO(s) + CO2 (g) ⇌ CaCO3(s)

(c) 3Fe (s) + 4H2O (g) ⇌ Fe3O4 (s) +4H2 (g)

Ans.

(a) The number of moles of reaction products will increase. According to Le Chatelier’s principle, if pressure is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more. In the given reaction, the number of moles of gaseous products is more than that of gaseous reactants. Thus, the reaction will proceed in the forward direction. As a result, the number of moles of reaction products will increase.

(b) The number of moles of reaction products will decrease.

(c) The number of moles of reaction products remains the same.

 

Q.26. Which of the following reactions will get affected by increasing the pressure? Also, mention whether the change will cause the reaction to go in a forward or backward direction.

(I) COCl2 (g) ⇌ CO (g) +Cl2 (g)

(II) CH4 (g) +2S2 (g) ⇌ CS2 (g) + 2H2S (g)

(III) CO2 (g) +C (s) ⇌ 2CO (g)

(IV) 2H2 (g) +CO  (g) ⇌ CH3OH(g)

(V) CaCO3 (s) ⇌ CaO (s) + CO2 (g)

(VI) 4NH3 (g) + 5O2 (g) ⇌ 4NO (g) + 6H2O (g)

Ans.

When pressure is increased:

The reactions given in (i), (iii), (iv), (v), and (vi) will get affected.

Since the number of moles of gaseous reactants is more than that of gaseous products, the reaction given in (iv) will proceed in the forward direction

Since the number of moles of gaseous reactants is less than that of gaseous products, the reactions given in (i), (iii), (v), and (vi) will shift in the backward direction.

 

Q.27. The equilibrium constant for the following reaction is

Q.29. Describe the effect of:

a) Addition of H2

b) Addition of CH3OH

c) Removal of CO

d) Removal of CH3OH on the equilibrium of the reaction:

2H2 (g)+CO (g) ⇌ CH3OH (g)

Ans.

(a) According to Le Chatelier’s principle, on the addition of H2, the equilibrium of the given reaction will shift in the forward direction.

(b) On addition of CH3OH, the equilibrium will shift in the backward direction.

(c) On removing CO, the equilibrium will shift in the backward direction.

(d) On removing CH3OH, the equilibrium will shift in the forward direction.

 

Q.30. At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl5 is

Q.45. The first ionisation constant of H2S is 9.1 × 10–8. Calculate the concentration of HS– ion in its 0.1 M solution. How will this concentration be affected if the solution is 0.1 M in HCl also? If the second dissociation constant of H2S is 1.2 × 10–13, calculate the concentration of S2– under both conditions.

Ans.

To calculate [HS–]

To find [HS–]:

Case 1 – HCl is absent.

Now

Ka = ([H+][HS–])/[H2S] = 9.1×10-8 (given)

∴  x2/(0.1-x) = 9.1×10-8

But 0.1-x is approximately equal to 0.1. Substituting this value in the equation,

x2/0.1 = 9.1×10-8

x2 = 9.1×10-9

x = 9.54× 10-5 M

∴ The concentration of HS– is 9.54×10-5 M.

Case 2 – HCl is present

Now,

Ka = ([H+][HS–])/[H2S] = (y× (0.1+y))/(0.1-y) = 9.1×10-8 (given)

But (0.1 + y) and (0.1 – y) can be approximated to 0.1.

9.1× 10-8 = (0.1*y)/0.1

∴ y = [HS–] = 9.1× 10-8 M

To calculate [S2-]:

Case 1 – HCl is absent.

The dissociation of HS– is given by the equation:

HS– ⇌ H+ + S2-[HS–] = 9.1×10-5 M

[H+] = 9.54×10-5 M

Ka = ([H+][S2-])/[HS–] = 1.2×10-13 (given)

Ka = (9.1× 10-5 × [S2-])/9.1×10-5

∴ [S2-] = 1.2×10-13 M

Case 2 – HCl is present

[HS–] = 9.1×10-8 M

[H+] = 0.1 M

Ka = 1.2×10-13 M

= (0.1×[S2-])/ 9.1×10-8

Therefore, [S2-] = 1.092×10-19 M.

 

Q.46. The ionisation constant of acetic acid is 1.74 × 10–5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.

Ans.

CH3COOH ⇒ CH3COO– + H+