Ionic Equilibrium, NCERT Solutions, Class 11, Chemistry | Additional Study Material for Class 11 PDF Download

Question 7.1: 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.

Question 7.2: 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 ?

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

Hence, K_cfor the equilibrium is . 12.239 mol^-1

Question 7.3: At a certain temperature and total pressure of 10⁵ Pa, iodine vapour contains 40% by volume of I atoms  . Calculate K_p for the equilibrium.

ANSWER: 

Partial pressure of I atoms,

ANSWER :

Question 7.5: Find out the value of K_c for each of the following equilibria from the value of K_p:

ANSWER :  The relation between K_p and K_c is given as:

K_p = K_c (RT)^Δn

(a) Here, Δn = 3 – 2 = 1

R = 0.0831 barLmol^–1K^–1

T = 500 K

K_p = 1.8 × 10^–2

Now,

K_p = K_c (RT) ^Δn

(b) Here, Δn = 2 – 1 = 1

R = 0.0831 barLmol^–1K^–1

T = 1073 K

K_p= 167

Now,

K_p = K_c (RT) ^Δn

Question 7.6: For the following equilibrium, 

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 is6.3 X10¹⁴.

Then, K_C for the reverse reaction will be,

Question 7.7: 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.

Question 7.8: Reaction between N₂ and O₂ takes place as follows:

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 K_c = 2.0 × 10^–37, determine the composition of equilibrium mixture.

ANSWER :  Let the concentration of N₂O at equilibrium be x.

The given reaction is

Question 7.9: Nitric oxide reacts with Br₂ and gives nitrosyl bromide as per reaction given below:

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 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

Question 7.10: At 450 K, K_p= 2.0 × 10¹⁰/bar for the given reaction at equilibrium.

What is K_c at this temperature?

ANSWER : 

For the given reaction,

Δn = 2 – 3 = – 1

T = 450 K

R = 0.0831 bar L bar K^–1 mol^–1

 = 2.0 × 10¹⁰ bar ^–1

We know that,

Question 7.11: 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 K_p 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

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

Question 7.12: 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 

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

ANSWER : 

The given reaction is:  

The given concentration of various species is

Write the balanced chemical equation corresponding to this expression.

ANSWER : 

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

Question 7.14: 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,

Therefore, the equilibrium constant for the reaction,

Question 7.15: At 700 K, equilibrium constant for the reaction

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 Kc for the reaction

Question 7.16: What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?

Hence, at equilibrium

Question 7.17: 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?

Hence, at equilibrium,

Question 7.18: Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:

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?

Question 7.19: 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 K_cis 8.3 × 10^–3, what are the concentrations of PCl₃ and Cl₂ at equilibrium?

Question 7.20: 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.

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 

And, equilibrium partial pressure of 

Question 7.21: Equilibrium constant, K_c for the reaction

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:

Question 7.22: Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium

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:

Question 7.23: 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

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

For the given reaction,

Δn = 2 – 1 = 1

We know that,

Question 7.24: Calculate a) ΔG°and b) the equilibrium constant for the formation of NO₂ from NO and O₂ at 298 K

where Δ_fG° (NO₂) = 52.0 kJ/mol

Δ_fG° (NO) = 87.0 kJ/mol

Δ_fG° (O₂) = 0 kJ/mol

ANSWER : 

(a) For the given reaction,

ΔG° = ΔG°( Products) – ΔG°( Reactants)

ΔG° = 52.0 – {87.0 0}

= – 35.0 kJ mol^–1

(b) We know that,

ΔG° = RT log K_c

ΔG° = 2.303 RT log K_c

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

Question 7.25: 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?

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.

Question 7.26: 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.

Question 7.27: The equilibrium constant for the following reaction is 1.6 ×10⁵ at 1024 K.

Question 7.28: Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:

(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.

Question 7.29: 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 :  

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.

Question 7.30: At 473 K, equilibrium constant K_c for decomposition of phosphorus pentachloride, PCl₅ is 8.3 ×10^-3. If decomposition is depicted as,

(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 K_c if (i) more PCl₅ is added (ii) pressure is increased? (iii) The temperature is increased?

ANSWER : 

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.

Question 7.32: Predict which of the following reaction will have appreciable concentration of reactants and products:

Question 7.34: The reaction  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.

Question 7.35: What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species:

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.

Question 7.36: Which of the followings are Lewis acids? H₂O, BF₃, H , and 

ANSWER :  Lewis acids are those acids which can accept a pair of electrons. For example, BF₃, H , and  are Lewis acids.

Question 7.37: 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.

Question 7.38: Write the conjugate acids for the following Brönsted bases: NH₂^–, NH₃ and HCOO^–.

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

Question 7.39: The species: H₂O, , 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

Question 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.

Question 7.41: The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10^–3 M. what is its pH?

ANSWER :  Given,

Question 7.42: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.

Question 7.43: 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,

Given,

K_a of HF = 6.8 × 10^–4

Hence, K_b of its conjugate base F^–

Question 7.44: The ionization constant of phenol is 1.0×10^–10. What is the concentration of phenolate ion in 0.05M solution of phenol? What will be its degree of ionization if the solution is also 0.01M in sodium phenolate?

ANSWER:  Ionization of phenol:

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

Question 7.45: The first ionization constant of H_2­­S 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.

Question 7.46: 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

pH = 3.03

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

Question 7.47: 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.

Question 7.49: 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

Question 7.50: 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.α

= 0.1 × 0.132

= 0.0132