Question

NCERT Solutions Class 11 Chemistry Chapter 7 Equilibrium – Our NCERT Solutions is an amazing way to prepare for board exams and competitive exams.

In NCERT Solution Class 11 we have discussed all the important concepts in-depth for a complete understanding of the chapter. Our experts have discussed all the topics thoroughly. The topics discussed here are

• Equilibrium – the state of balance between two opposing forces or the situation which cancels out each other is called equilibrium.
• Equilibrium in Physical Processes – in a physical process the equilibrium represents the state in which the properties like temperature, pressure, and concentration exhibit no change during any duration of time.  
• Equilibrium in Chemical Processes (Dynamic Equilibrium) – In chemical equilibrium, the product and the reactants are in equilibrium. The product formed again reacts back to the reactants to form a new product hence remaining in constant equilibrium. There is no net change in the number of reactants and products.
• Homogeneous Equilibrium – such kind of equilibrium in which all species are present in the same phase. Example – gas-phase or solution reactions.
• Heterogeneous Equilibrium – Such equilibrium in which species exist in more than one phase is called heterogeneous equilibrium.
• Applications of Equilibrium Constants – One of the applications of the equilibrium constant is to predict the extent of a reaction, i.e. the degree of the disappearance of the reactants.
• Factors Affecting Equilibrium – there is several ways equilibrium is disturbed such as temperature, pressure, and concentration of the system. If any one of the above factors changes the equilibrium is disturbed.
•  Ionic Equilibrium in Solution – the equilibrium exhibited in which the unionized and the ionized molecules are in constant equilibrium with each other. This kind of ionic equilibrium is readily seen in weak electrolytes such as H2O and NaCl solution.
• Acids, Bases, and Salts –
  •  Acid – any substance or solution which liberates hydrogen when it reacts with metal, it tastes sour, changes blue litmus to red, reacts with bases to form a salt.
  • Bases – any substance or solution which feels slippery to touch, tastes bitter, changes the color of red litmus to blue, and reacts with an acid to form basic salt.
  • Salts – salt is a compound formed reaction of acid and bases. It is an ionic assembly of positively charged cations and negatively charges anions.

Our NCERT Solutions Class 11 Chemistry is provided in a systematic form such as everything is given in pointwise method with the use of chemical reaction, tips, and tricks.

Answer 1

NCERT Solutions Class 11 Chemistry Chapter 7 Equilibrium

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?

Answer:

(a) If the volume of the container is suddenly increased, then the vapour pressure would decrease initially. This is because the amount of vapour remains the same, but the volume increases suddenly. As a result, the same amount of vapour is distributed in a larger volume. 

(b) Since the temperature is constant, the rate of evaporation also remains constant. When the volume of the container is increased, the density of the vapour phase decreases. As a result, the rate of collisions of the vapour particles also decreases. Hence, the rate of condensation decreases initially. 

(c) When equilibrium is restored finally, the rate of evaporation becomes equal to the rate of condensation. In this case, only the volume changes while the temperature remains constant. The vapour pressure depends on temperature and not on volume. Hence, the final vapour pressure will be equal to the original vapour pressure of the system.

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

2SO₂ (g) + O₂ (g) \(\longleftrightarrow\) 2SO₃(g)

Answer:

The equilibrium constant (K_c) for the give reaction is:

Hence, K for the equilibrium is 12.239 M^–1.

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

I2 (g) \(\longleftrightarrow\) 2I (g)

Calculate K_p for the equilibrium.

Answer:

Partial pressure of I atoms,

Partial pressure of I₂ molecules,

Now, for the given reaction,

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

Answer:

5. Find out the value of K_c for each of the following equilibria from the value of K_p:

(i) 2NOCl(g) \(\longleftrightarrow \) 2NO(g) + Cl₂(g) : K_p = 1.8 x 10^-2 at 500 K

(ii) CaCO₃(s) \(\longleftrightarrow \) CaO(s) + CO₂(g): K_p = 167 at 1073 K

Answer:

The relation between K_p and K_c is given as: 

K_p = K_c (RT)^∆n

(a) Here, 

∆n = 3 – 2 = 1 

R = 0.0831 barLmol^–1K^–1

T = 500 K 

K_p = 1.8 × 10^–2 

Now,

K_p = K_c (RT)^∆n

(b) Here, 

∆n = 2 – 1 = 1 

R = 0.0831 barLmol^–1K^–1 

T = 1073 K 

K_p= 167 

Now,

K_p = K_c (RT)^∆n

6. For the following equilibrium,

K_c = 6.3 x 10¹⁴ at 1000 K

NO(g) + O₃(g) \(\longleftrightarrow\) NO₂(g) + O₂(g)

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

Answer:

It is given that K_c for the forward reaction is 6.3 x 10¹⁴

Then, K_c for the reverse reaction will be,

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

Answer:

For a pure substance (both solids and liquids),

Now, the molecular mass and density (at a particular temperature) of a pure substance is always fixed and is accounted for in the equilibrium constant. Therefore, the values of pure substances are not mentioned in the equilibrium constant expression.

Answer 2

8. Reaction between N₂ and O₂ takes place as follows:

2N₂(g) + O₂(g) \(\longleftrightarrow \) 2N₂O(g)

If a mixture of 0.482 mol of N₂ and 0.933 mol of O₂ is placed in a 10 L reaction vessel and allowed to form N₂O at a temperature for which Kc = 2.0 × 10^–37, determine the composition of equilibrium mixture.

Answer:

Let the concentration of N₂O at equilibrium be x. 

The given reaction is:

Therefore, at equilibrium, in the 10 L vessel:

The value of equilibrium constant i.e. K_c = 2.0 × 10^–37 is very small. Therefore, the amount of N₂ and O₂ reacted is also very small. Thus, x can be neglected from the expressions of molar concentrations of N₂ and O₂. Then,

Now,

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

2NO(g) + Br₂(g) \(\longleftrightarrow \) 2NOBr(g)

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

Answer:

The given reaction is:

Now, 2 mol of NOBr are formed from 2 mol of NO. 

Therefore, 0.0518 mol of NOBr are formed from 0.0518 mol of NO. 

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

Therefore, 0.0518 mol of NOBr are formed from \(\frac{0.0518}{2}\) mol of 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

10. At 450 K, K_p= 2.0 × 1010/bar for the given reaction at equilibrium.

2SO_2(g) + O_2(g) \(\longleftrightarrow \) 2SO_3(g)

What is K_c at this temperature?

Answer:

For the given reaction, 

∆n = 2 – 3 = – 1 

T = 450 K

R = 0.0831 bar L bar K^–1 mol^–1 

Kp = 2.0 × 10¹⁰ bar^–1 

We know that,

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 is Kp for the given equilibrium?

Answer:

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:

Therefore,

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) \(\longleftrightarrow \) 2NH₃(g) is 1.7 x 10²

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

Answer:

The given reaction is:

N₂(g) + 3H₂(g) \(\longleftrightarrow \) 2NH₃(g)

The given concentration of various species is 

Now, reaction quotient Q_c is:

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.

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

Write the balanced chemical equation corresponding to this expression.

Answer:

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

4NO_(g) + 6H₂O_(g) \(\longleftrightarrow \) 4NH_3(g) + 5O_2(g)

14. One mole of H₂O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation,

H₂O(g) + CO(g) \(\longleftrightarrow \) H₂(g) + CO₂(g)

Calculate the equilibrium constant for the reaction.

Answer:

The given reaction is:

Therefore, the equilibrium constant for the reaction,

Answer 3

15. At 700 K, equilibrium constant for the reaction

H_2(g) + I_2(g) \(\longleftrightarrow \) 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?

Answer:

It is given that equilibrium constant K_c for the reaction

H_2(g) + I_2(g) \(\longleftrightarrow \) 2HI_(g) is 54.8

Therefore, at equilibrium, the equilibrium constant \(K^+_1\) for the reaction

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

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

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

2ICI(g) \(\longleftrightarrow \) I_2(g) + Cl_2(g) ; K_c = 0.14

Answer:

The given reaction is:

Hence, at equilibrium,

[H₂] = [I₂] = 0.167 M

[HI] = (0.78 - 2 x 0.167) M

= 0.446 M

17. K_p = 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) \(\longleftrightarrow \) C₂H₄(g) + H₂(g)

Answer:

Let p be the pressure exerted by ethene and hydrogen gas (each) at equilibrium. 

Now, according to the reaction,

We can write,

Hence, at equilibrium,

[C₂H₆] - 4 - p = 4 - 0.38

= 3.62 atm

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

CH₃COOH (I) + C₂ H₅OH (l) \(\longleftrightarrow \) CH₃COOC₂H₅ (I) + 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 sometime. Has equilibrium been reached ?

Answer:

(i) Reaction quotient,

(ii) Let the volume of the reaction mixture be V. Also, here we will consider that water is a solvent and is present in excess. 

The given reaction is:

Therefore, equilibrium constant for the given reaction is:

(iii) Let the volume of the reaction mixture be V.

Therefore, the reaction quotient is,

Since Q_c < K_c, equilibrium has not been reached.

19. A sample of pure PCl₅ was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl₅ was found to be 0.5 × 10^–1 mol L^–1. If value of Kc is 8.3 × 10^–3, what are the concentrations of PCl₃ and Cl₂ at equilibrium?

PCl₅(g) \(\longleftrightarrow \) PCl₃(g) + Cl₂(g)

Answer:

Let the concentrations of both PCl₃ and Cl₂ at equilibrium be x molL^–1. The given reaction is:

It is given that the value of equilibrium constant , K_c is 8.3 x 10^-3.

Now we can write the expression for equilibrium as:

Therefore, at equilibrium,

[PCL₃] = [Cl₂] = 0.02 mol L^-1.

20. One of the reactions that takes 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 are the equilibrium partial pressures of CO and CO₂ at 1050 K if the initial partial pressures are: P_co = 1.4 atm and \(P_{co_2}\)= 0.80 atm?

Answer:

For the given reaction,

Since Q_p > K_p, the reaction will proceed in the backward direction. 

Therefore, we can say that the pressure of CO will increase while the pressure of CO₂ will decrease.

Now, let the increase in pressure of CO = decrease in pressure of CO₂ be p. 

Then, we can write,

Therefore, equilibrium partial of CO₂,\(P_{co_2}\)= 0.80 - 0339 = 0.461 atm.

And, equilibrium partial pressure of CO, Pco = 1.4 + 0.339 = 1.739 atm.

21. Equilibrium constant, K_c for the reaction 

N₂(g) + 3H₂(g) \(\longleftrightarrow \) 2NH₃(g) at 500K is 0.061.

At a particular time, the analysis shows that 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. Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium ?

Answer:

The given reaction is:

Now, we know that,

It is given that K_c = 0.061.

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

Since Q_c < K_c  , the reaction will proceed in the forward direction to reach equilibrium.

Answer 4

22. Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:

2BrCl (g) \(\longleftrightarrow \) Br₂ (g) + Cl₂ (g)

for which K_c= 32 at 500 K. 

If initially pure BrCl is present at a concentration of 3.3 × 10^-3 molL^-1 , what is its molar concentration in the mixture at equilibrium?

Answer:

Let the amount of bromine and chlorine formed at equilibrium be x. The given reaction is:

Now, we can write,

Therefore, at equilibrium,

23. At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO₂ in equilibrium with solid carbon has 90.55% CO by mass

C(s) + CO₂ (g) \(\longleftrightarrow \) 2CO (g)

Calculate K_c for this reaction at the above temperature.

Answer:

Let the total mass of the gaseous mixture be 100 g. 

Mass of CO = 90.55 g 

And, mass of CO₂ = (100 – 90.55) = 9.45 g

Now, number of moles of CO, n_co = \(\frac{90.55}{28}=3.234\, mol\)

Number of moles of CO₂, \(n_{co_2}\) = \(\frac{9.45}{44}= 0.215 \, mol\)

Partial pressure of CO,

Partial pressure of CO₂,

For the given reaction,

∆n = 2 – 1 = 1

We know that,

24. Calculate

a) ∆G° and 

b) the equilibrium constant for the formation of NO₂ from NO and O₂ at 298 K

NO (g) + 1/2 O₂ (g) \(\longleftrightarrow \) NO₂ (g)

Where:

∆_fG° (NO₂) = 52.0 kJ/mol 

∆_fG° (NO) = 87.0 kJ/mol 

∆_fG° (O₂) = 0 kJ/mol

Answer:

(a) For the given reaction, 

∆G° = ∆G°( Products) – ∆G°( Reactants) 

∆G° = 52.0 – {87.0 + 0} 

= – 35.0 kJ mol^–1

(b) We know that, 

∆G° = RT log K_c 

∆G° = 2.303 RT log K_c

Hence, the equilibrium constant for the given reaction K_c is 1.36 × 10⁶

24. Calculate 

a) ∆G° and 

b) the equilibrium constant for the formation of NO₂ from NO and O₂ at 298 K

NO(g) + 1/2 O₂(g) \(\longleftrightarrow \) NO₂(g)

Where:

∆fG° (NO₂) = 52.0 kJ/mol

∆fG° (NO) = 87.0 kJ/mol 

∆fG° (O₂) = 0 kJ/mol

Answer:

(a) For the given reaction,

∆G° = ∆G°( Products) – ∆G°( Reactants) 

∆G° = 52.0 – {87.0 + 0} 

= – 35.0 kJ mol^–1

(b) We know that,

∆G° = RT log K_c

∆G° = 2.303 RT log K_c

Hence, the equilibrium constant for the given reaction K_c is 1.36 × 10⁶

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

(a) PCl₅(g) \(\longleftrightarrow \) PCl₃(g) + Cl₂(g)

(b) CaO(s) + CO₂(g) \(\longleftrightarrow \) CaCO₃(s)

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

Answer:

(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.

26. Which of the following reactions will get affected by increasing the pressure? 

Also, mention whether change will cause the reaction to go into forward or backward direction.

Answer:

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

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

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

27. The equilibrium constant for the following reaction is 1.6 ×105 at 1024 K. 

H₂(g) + Br₂(g) \(\longleftrightarrow \) 2HBr(g)

Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024 K.

Answer:

Given, 

K_p for the reaction i.e., H_2(g) + Br_2(g) \(\longleftrightarrow \) 2HBr_(g)  is 1.6 x 10⁵

Therefore, for the reaction  2HBr_(g) \(\longleftrightarrow \) H_2(g) + Br_2(g) the equilibrium constant will be,

\(K^ {'}_p = \frac 1 {K_p}\)

\(= \frac 1 {1.6 \times10^{-6}}\)

Now, let p be the pressure of both H₂ and Br₂ at equilibrium.

Now, we can write,

Therefore, at equilibrium,

Answer 5

28. Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:

CH₄₍g) + H₂O(g) \(\longleftrightarrow \) CO(g) + 3H₂(g)

(a)Write as expression for K_p for the above reaction. 

(b)How will the values of K_p and composition of equilibrium mixture be affected by 

(i) Increasing the pressure 

(ii) Increasing the temperature 

(iii) Using a catalyst?

Answer:

(a) For the given reaction,

(b) 

(i) According to Le Chatelier’s principle, the equilibrium will shift in the backward direction. 

(ii) According to Le Chatelier’s principle, as the reaction is endothermic, the equilibrium will shift in the forward direction. 

(iii) The equilibrium of the reaction is not affected by the presence of a catalyst. A catalyst only increases the rate of a reaction. Thus, equilibrium will be attained quickly.

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) \(\longleftrightarrow \) CH₃OH(g)

Answer:

(a) According to Le Chatelier’s principle, on 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.

30. At 473 K, equilibrium constant K_c for decomposition of phosphorus pentachloride, PCl₅ is 8.3 ×10^-3 . If decomposition is depicted as,

PCl₅(g) \(\longleftrightarrow \) PCl₃(g) + Cl₂(g) ; ∆rH° = 124.0 kJmol^–1

a) Write an expression for K_c for the reaction. 

b) What is the value of K_c for the reverse reaction at the same temperature? 

c) What would be the effect on Kc if 

(i) more PCl₅ is added 

(ii) pressure is increased? 

(iii) The temperature is increased?

Answer:

(a)

(b) Value of Kc for the reverse reaction at the same temperature is:

(c) 

(i) K_c would remain the same because in this case, the temperature remains the same. 

(ii) K_c is constant at constant temperature. Thus, in this case, K_c would not change.

(iii) In an endothermic reaction, the value of K_c increases with an increase in temperature. Since the given reaction in an endothermic reaction, the value of K_c will increase if the temperature is increased.

28. Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:

CH₄₍g) + H₂O(g) \(\longleftrightarrow \) CO(g) + 3H₂(g)

(a)Write as expression for K_p for the above reaction. 

(b)How will the values of K_p and composition of equilibrium mixture be affected by 

(i) Increasing the pressure 

(ii) Increasing the temperature 

(iii) Using a catalyst?

Answer:

(a) For the given reaction,

(b) 

(i) According to Le Chatelier’s principle, the equilibrium will shift in the backward direction. 

(ii) According to Le Chatelier’s principle, as the reaction is endothermic, the equilibrium will shift in the forward direction. 

(iii) The equilibrium of the reaction is not affected by the presence of a catalyst. A catalyst only increases the rate of a reaction. Thus, equilibrium will be attained quickly.

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) \(\longleftrightarrow \) CH₃OH(g)

Answer:

(a) According to Le Chatelier’s principle, on 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.

30. At 473 K, equilibrium constant K_c for decomposition of phosphorus pentachloride, PCl₅ is 8.3 ×10^-3 . If decomposition is depicted as,

PCl₅(g) \(\longleftrightarrow \) PCl₃(g) + Cl₂(g) ; ∆rH° = 124.0 kJmol^–1

a) Write an expression for K_c for the reaction. 

b) What is the value of K_c for the reverse reaction at the same temperature? 

c) What would be the effect on Kc if 

(i) more PCl₅ is added (ii) pressure is increased? 

(iii) The temperature is increased?

Answer:

(a)

(b) Value of Kc for the reverse reaction at the same temperature is:

(c) 

(i) K_c would remain the same because in this case, the temperature remains the same. 

(ii) K_c is constant at constant temperature. Thus, in this case, K_c would not change.

(iii) In an endothermic reaction, the value of K_c increases with an increase in temperature. Since the given reaction in an endothermic reaction, the value of K_c will increase if the temperature is increased.

Answer 6

31. Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H₂. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,

CO(g) + H₂O(g) \(\longleftrightarrow \) CO₂(g) + H₂(g)

If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam such that \(P_{co}=P_{H_2O}= \) 4.0 bar, what will be the partial pressure of H₂ at equilibrium? 

K_p= 10.1 at 400°C

Answer:

Let the partial pressure of both carbon dioxide and hydrogen gas be p. The given reaction is:

It is given that K_p = 10.1

Now,

Hence, at equilibrium, the partial pressure of H₂ will be 3.04 bar.

32. Predict which of the following reaction will have appreciable concentration of reactants and products:

a) Cl₂ (g) \(\longleftrightarrow \) 2 Cl (g) ; K_c = 5 x 10^-39

b) Cl₂ (g) + 2 NO( g) \(\longleftrightarrow \) 2NOCl (g) ; Kc = 3.7 x 10⁸

c) Cl₂ + 2NO₂ (g) \(\longleftrightarrow \)  2NO₂Cl (g) ; K_c = 1.8

Answer:

If the value of K_c lies between 10^–3 and 10³ , a reaction has appreciable concentration of reactants and products. Thus, the reaction given in (c) will have appreciable concentration of reactants and products.

33. The value of K_c for the reaction 3O₂ (g) 2O₃ (g) is 2.0 × 10^–50 at 25°C. If the equilibrium concentration of O₂ in air at 25°C is 1.6 × 10^–2 , what is the concentration of O₃?

Answer:

The given reaction is:

Then, we have,

Hence, the concentration of O₃ is 2.86 x 10^-28 M.

34. The reaction, CO(g) + 3H₂(g) CH₄(g) + H₂O(g) is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H₂ and 0.02 mol of H₂O and an unknown amount of CH₄ in the flask. Determine the concentration of CH₄ in the mixture. The equilibrium constant, K_c for the reaction at the given temperature is 3.90.

Answer:

Let the concentration of methane at equilibrium be x.

It is given that K_c = 3.90.

Therefore,

Hence, the concentration of CH₄ at equilibrium is 5.85 × 10^–2 M.

35. What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species:

HNO₂, CN^-, HCIO₄, F^-, OH^-, \(CO^{2-}_3\) and S^-

Answer:

A conjugate acid-base pair is a pair that differs only by one proton. 

The conjugate acid-base for the given species is mentioned in the table below.

Species	Conjugate acid-base
HNO2	\(NO^-_2\) (base)
CN-	HCN (acid)
HClO4	\(ClO^-_4\) (base)
F–	HF (acid)
OH–	H2O (acid) /O2– (base)
\(CO^{2-}_3\)	\(HCO^-_3\) (acid)
S2-	HS– (acid)

36. Which of the followings are Lewis acids? H₂O, BF³, H⁺, and \(NH^+_4\)

Answer:

Lewis acids are those acids which can accept a pair of electrons. For example, BF³, H⁺, and \(NH^+_4\) are Lewis acids.

37. What will be the conjugate bases for the Brönsted acids: HF, H₂SO₄ and HCO₃?

Answer:

The table below lists the conjugate bases for the given Bronsted acids. 

Bronsted acid	Conjugate base
HF	F–
H2SO4	\(HSO^-_4\)
\(HCO^-_3\)	\(CO^{2-}_3\)

38. Write the conjugate acids for the following Brönsted bases: \(NH^-_2\) NH₃ and HCOO^–.

Answer:

The table below lists the conjugate acids for the given Bronsted bases.

Bronsted base	Conjugate acid
\(NH^-_2\)	NH3
NH3	\(NH^+_4\)
HCOO–	HCOO

39. The species: H₂O, \(HCO^-_3\), \(HSO^-_4\) and NH₃ can act both as Brönsted acids and bases. For each case give the corresponding conjugate acid and base.

Answer:

The table below lists the conjugate acids and conjugate bases for the given species.

Species	Conjugate acid	Conjugate base
H2O	H3O+	OH–
\(HCO^-_3\)	H2CO3	\(CO^{2-}_3\)
\(HSO^-_4\)	H2SO4	\(SO^{2-}_4\)
NH3	\(NH^+_4\)	\(NH^+_2\)

Answer 7

40. Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base:

(a) OH^– 

(b) F^– 

(c) H⁺ 

(d) BCl₃

Answer:

(a) OH^– is a Lewis base since it can donate its lone pair of electrons. 

(b) F^– is a Lewis base since it can donate a pair of electrons. 

(c) H⁺ is a Lewis acid since it can accept a pair of electrons. 

(d) BCl₃ is a Lewis acid since it can accept a pair of electrons.

 41. The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10^–3 M. what is its pH?

Answer:

Given,

[H⁺] = 3.8 x 10^-3 M

\(\therefore\) pH value of soft drink

42. The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.

Answer:

Given, 

pH = 3.76 

It is known that,

Hence, the concentration of hydrogen ion in the given sample of vinegar is 1.74 × 10^–4 M.

43. The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10^–4 , 1.8 × 10^–4 and 4.8 × 10^–9 respectively. Calculate the ionization constants of the corresponding conjugate base.

Answer:

It is known that,

\(K_p = \frac{K_w}{K_a}\)

Given, 

K_a of HF = 6.8 × 10^–4 

Hence, K_b of its conjugate base F^–

Given, 

K_a of HCOOH = 1.8 × 10^–4 

Hence, K_b of its conjugate base HCOO^–

Given, 

K_a of HCN = 4.8 × 10^–9 

Hence, K_b of its conjugate base CN

44. The ionization constant of phenol is 1.0 × 10^–10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01M in sodium phenolate?

Answer:

Ionization of phenol:

As the value of the ionization constant is very less, x wil be very small. 

Thus we can ignore x in the denominator.

Now, let be the degree of ionization of phenol in the presence of 0.01 M C₆H₅ONa.

Also,

45. The first ionization 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 S^2– under both conditions.

Answer:

(i) To calculate the concentration of HS^– ion:

Case I (in the absence of HCl): 

Let the concentration of HS^– be x M.

Case II (in the presence of HCl):

In the presence of 0.1 M of HCl, let [HS] be y M

(ii) To calculate the concentration of [S^2-]

Case I (in the absence of 0.1 M HCl):

Case II (in the presence of 0.1 M HCl):

Again, let the concentration of HS^– be X' M.

[HS^-] = 9.1 x 10^-8 (From first ionization, case II)

[H⁺] = 0.1 M (From HCl, case II)

 46. The ionization 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.

Answer:

Method 1

1) CH₃COOH \(\longleftrightarrow\) CH₃COO^- + H⁺  ; K_a = 1.74 x 10^-5

2) H₂O + H₂O \(\longleftrightarrow\) H₃O+ + OH^-   ;  K_w = 1.0 x 10^-14

Since K_a >> K_w,:

Method 2

Degree of dissociation,

c = 0.05 M 

K_a = 1.74 × 10^–5

Thus, concentration of CH₃COO– = c.a

Hence, the concentration of acetate ion in the solution is 0.00093 M and its Ph is 3.03.

47. It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pK_a.

Answer:

Let the organic acid be HA.

⇒ HA \(\longleftrightarrow\) H⁺ + A^-

Concentration of HA = 0.01 M pH 

= 4.15

Now,

Then,

Answer 8

48. Assuming complete dissociation, calculate the pH of the following solutions: 

(i) 0.003 M HCl 

(ii) 0.005 M NaOH 

(iii) 0.002 M HBr 

(iv) 0.002 M KOH

Answer:

(i) 0.003MHCl:

H₂O + HCl \(\longleftrightarrow\) H₃O⁺ +Cl^-

Since HCl is completely ionized,

Now,

Hence, the pH of the solution is 2.52.

(ii) 0.005MNaOH:

Hence, the pH of the solution is 11.70.

(iii) 0.002 HBr:

Hence, the pH of the solution is 2.69.

(iv) 0.002 M KOH:

Hence, the pH of the solution is 11.31.

49. Calculate the pH of the following solutions: 

a) 2 g of TlOH dissolved in water to give 2 litre of solution. 

b) 0.3 g of Ca(OH)₂ dissolved in water to give 500 mL of solution. 

c) 0.3 g of NaOH dissolved in water to give 200 mL of solution. 

d) 1mL of 13.6 M HCl is diluted with water to give 1 litre of solution.

Answer:

(a) For 2g of TlOH dissolved in water to give 2 L of solution:

(b) For 0.3 g of Ca(OH)₂ dissolved in water to give 500 mL of solution:

(c) For 0.3 g of NaOH dissolved in water to give 200 mL of solution:

(d) For 1mL of 13.6 M HCl diluted with water to give 1 L of solution:

13.6 × 1 mL = M₂ × 1000 mL 

(Before dilution) (After dilution)

13.6 × 10^–3 = M₂ × 1L M₂ 

= 1.36 × 10^–2 [H⁺] = 1.36 × 10^–2 pH = – log (1.36 × 10^–2) 

= (– 0.1335 + 2) 

= 1.866 1.87

50. The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pK_a of bromoacetic acid.

Answer:

Degree of ionization, α = 0.132 

Concentration, c = 0.1 M 

Thus, the concentration of H₃O⁺ = c.a

= 0.1 × 0.132 

= 0.0132

Now,

51. The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pK_a of bromoacetic acid.

Answer:

Degree of ionization, α = 0.132 

Concentration, c = 0.1 M

Thus, the concentration of H₃O⁺ = c.a

= 0.1 × 0.132 

= 0.0132

Now,

52. What is the pH of 0.001 M aniline solution? The ionization constant of aniline can be taken from Table 7.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

Answer:

K_b = 4.27 × 10^–10 

c = 0.001M pH

=? 

α =?

Now, 

Thus, the ionization constant of the conjugate acid of aniline is 2.34 × 10^–5.

53. Calculate the degree of ionization of 0.05M acetic acid if its pK_a value is 4.74. How is the degree of dissociation affected when its solution also contains (a) 0.01 M (b) 0.1 M in HCl ?

Answer:

When HCl is added to the solution, the concentration of H⁺ ions will increase. Therefore, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease.

Case I: When 0.01 M HCl is taken. 

Let x be the amount of acetic acid dissociated after the addition of HCl.

As the dissociation of a very small amount of acetic acid will take place, the values i.e., 0.05 – x and 0.01 + x can be taken as 0.05 and 0.01 respectively.

Case II: When 0.1 M HCl is taken. 

Let the amount of acetic acid dissociated in this case be X. As we have done in the first case, the concentrations of various species involved in the reaction are:

54. The ionization constant of dimethylamine is 5.4 × 10^–4. Calculate its degree of ionization in its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.1 M in NaOH?

Answer:

K_b = 5.4 x 10^-4

c = 0.02 M

Then, a = \(\sqrt{\frac{K_b}{c}}\)

= \(\sqrt{\frac{5.4 \times10^{-4}}{0.02 \, M}}\) 

= 0.1643

Now, if 0.1 M of NaOH is added to the solution, then NaOH (being a strong base) undergoes complete ionization.

And,

It means that in the presence of 0.1 M NaOH, 0.54% of dimethylamine will get dissociated.

Answer 9

55. Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below: 

(a) Human muscle-fluid, 6.83 

(b) Human stomach fluid, 1.2 

(c) Human blood, 7.38 

(d) Human saliva, 6.4.

Answer:

(a) Human muscle fluid 6.83: 

pH = 6.83 pH = – log [H⁺] 

\(\therefore\) 6.83 = – log [H⁺] 

[H⁺] =1.48 × 10^–7 M

(b) Human stomach fluid, 1.2: 

pH =1.2 

1.2 = – log [H⁺] 

\(\therefore\) [H⁺] = 0.063

(c) Human blood, 7.38: 

pH = 7.38 = – log [H⁺] 

\(\therefore\) [H⁺] = 4.17 × 10^–8 M

(d) Human saliva, 6.4: 

pH = 6.4 

6.4 = – log [H⁺] 

[H⁺] = 3.98 × 10^–7

56. The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.

Answer:

The hydrogen ion concentration in the given substances can be calculated by using the given relation: pH = –log [H⁺]

(i) pH of milk = 6.8 

Since, pH = –log [H⁺] 

6.8 = –log [H⁺] log 

[H⁺] = –6.8 

[H⁺] = anitlog(–6.8) 

= 1.5×19^–7 M

(ii) pH of black coffee = 5.0 

Since, pH = –log [H⁺] 

5.0 = –log [H⁺] log 

[H⁺] = –5.0 

[H⁺] = anitlog(–5.0) 

= 10–5 M

(iii) pH of tomato juice = 4.2 

Since, pH = –log [H⁺] 

4.2 = –log [H⁺] log 

[H⁺] = –4.2 

[H⁺] = anitlog(–4.2) 

= 6.31×10^–5 M

(iv) pH of lemon juice = 2.2 

Since, pH = –log [H⁺] 

2.2 = –log [H⁺] log 

[H⁺] = –2.2

[H⁺] = anitlog(–2.2) 

= 6.31×10^–3 M

(v) pH of egg white = 7.8 

Since, pH = –log [H⁺] 

7.8 = –log [H⁺] log

[H⁺] = –7.8 

[H⁺] = anitlog(–7.8) 

=1.58×10^–8 M

57. If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?

Answer:

58. The solubility of Sr(OH)₂ at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.

Answer:

Solubility of Sr(OH)₂ = 19.23 g/L 

Then, concentration of Sr(OH)₂

59. The ionization constant of propanoic acid is 1.32 × 10^–5 . Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?

Answer:

Let the degree of ionization of propanoic acid be a. 

Then, representing propionic acid as HA, we have:

In the presence of 0.1M of HCl, let α´ be the degree of ionization.

60. The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

Answer:

c = 0.1 M 

pH = 2.34

61. The ionization constant of nitrous acid is 4.5 × 10^–4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

Answer:

NaNO₂ is the salt of a strong base (NaOH) and a weak acid (HNO₂).

Now, If x moles of the salt undergo hydrolysis, then the concentration of various species present in the solution will be:

Therefore, degree of hydrolysis

= \(\frac{x}{0.04}\)

= \(\frac{.093\times10^{-5}}{.04}\)

= 2.325 × 10^–5

62. A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine

Answer:

pH = 3.44

We know that, pH = – log [H⁺]

63. Predict if the solutions of the following salts are neutral, acidic or basic: 

NaCl, KBr, NaCN, NH₄NO₃, NaNO₂ and KF

Answer:

(i) NaCl:

Therefore, it is a neutral solution. 

(ii) KBr:

Therefore, it is a neutral solution. 

(iii) NaCN:

Therefore, it is a basic solution. 

(iv) NH₄NO₃ :

Therefore, it is an acidic solution. 

(v) NaNO₂ :

Therefore, it is a basic solution. 

(vi) KF :

Therefore, it is a basic solution.

Answer 10

64. The ionization constant of chloroacetic acid is 1.35 × 10^–3. What will be the pH of 0.1M acid and its 0.1M sodium salt solution?

Answer:

It is given that K_a for ClCH₂COOH is 1.35 × 10^–3.

ClCH₂COONa is the salt of a weak acid i.e., ClCH₂COOH and a strong base i.e., NaOH.

(where x is the concentration of OH^- and ClCH₂COOH)

65. Ionic product of water at 310 K is 2.7 × 10^–14. What is the pH of neutral water at this temperature?

Answer:

Ionic product,

Hence, the pH of neutral water is 6.78.

66. Calculate the pH of the resultant mixtures: 

a) 10 mL of 0.2M Ca(OH)₂ + 25 mL of 0.1M HCl 

b) 10 mL of 0.01M H₂SO₄ + 10 mL of 0.01M Ca(OH)₂ 

c) 10 mL of 0.1M H₂SO₄ + 10 mL of 0.1M KOH

Answer:

(a) Moles of H₃O⁺ = \(\frac{25\times0.1}{1000}= .0025\,mol\)

Moles of OH^- = \(\frac{10\times0.2\times2}{1000}=.0040\,mol\)

Thus, excess of OH^- = .0015 mol

(b) Moles of H₃O⁺ = \(\frac{2\times10\times0.01}{1000}= .0002\,mol\)

Moles of OH^-  = \(\frac{2\times10\times.01}{1000}= .0002\,mol\)

Since there is neither an excess of H₃O^- or OH^-

(c) Moles of H₃O⁺  = \(\frac{2\times10\times0.01}{1000}= .0002\,mol\)

Moles of OH^- = \(\frac{10\times0.1}{1000}= .0001\,mol\)

Excess of H₃O⁺ = .001 mol 

The solution is neutral. Hence, pH = 7.

67. Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 7.9. Determine also the molarities of individual ions.

Answer:

(1) Silver chromate:

Let the solubility of Ag₂ CrO4 be s.

Molarity of Ag⁺ = 2s = 2 × 0.65 × 10^–4 = 1.30 × 10^–4 M

Molarity of CrO₄^2- = s = 0.65 × 10^–4 M

(2) Barium chromate:

Let the solubility of BaCrO₄ be s.

So, 

Molarity of Ba²⁺ = Molarity of CrO₄^2- = s = 1.09 x 10^-5 M

(3) Ferric hydroxide:

Let s be the solubility of Fe(OH)₃.

Thus [Fe³⁺] = s and [OH^-] = 3s 

Molarity of Fe³⁺ = s = 1.39 x 10^-10

Molarity of OH^- = 3s = 4.17 x 10^-10

(4) Lead chloride:

Let K_SP be the solubility of PbCl₂.

Molarity of PB²⁺ = s = 1.58 x 10^-2 M

Molarity of chloride = 2s = 3.16 x 10^-2 M

(5) Mercurous iodide:

Let s be the solubility of Hg₂I_{2 .}

Molarity of \(Hg^{2+}_2\) = s = 2.2 x 10^-10 M

Molarity of I- = 2s = 4.48 x 10^-10 M

68. The solubility product constant of Ag₂CrO₄ and AgBr are 1.1 × 10^–12 and 5.0 × 10^–13 respectively. Calculate the ratio of the molarities of their saturated solutions.

Answer:

Let s be the solubility of Ag₂CrO₄.

Let s´ be the solubility of AgBr.

Therefore, the ratio of the molarities of their saturated solution is

69. Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate K_sp = 7.4 × 10^–8).

Answer:

When equal volumes of sodium iodate and cupric chlorate solutions are mixed together, then the molar concentrations of both solutions are reduced to half i.e., 0.001 M. 

Then,

Now, the solubility equilibrium for copper iodate can be written as:

Ionic product of copper iodate:

Since the ionic product (1 × 10^–9) is less than K_sp (7.4 × 10^–8), precipitation will not occur.

70. The ionization constant of benzoic acid is 6.46 × 10^–5 and K_sp for silver benzoate is 2.5 × 10^–13. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?

Answer:

Since pH = 3.19,

Let the solubility of C₆H₅COOAg be x mol/L. 

Then,

Thus, the solubility of silver benzoate in a pH 3.19 solution is 1.66 × 10^–6 mol/L. 

Now, let the solubility of C₆H₅COOAg be x’ mol/L.

Hence, C₆H₅COOAg is approximately 3.317 times more soluble in a low pH solution.