Equilibrium Question Answers: NCERT Class 11 Chemistry

Welcome to the Chapter 7 - Equilibrium, Class 11 Chemistry NCERT Solutions page. Here, we provide detailed question answers for Chapter 7 - Equilibrium. The page is designed to help students gain a thorough understanding of the concepts related to natural resources, their classification, and sustainable development.

Our solutions explain each answer in a simple and comprehensive way, making it easier for students to grasp key topics Equilibrium and excel in their exams. By going through these Equilibrium question answers, you can strengthen your foundation and improve your performance in Class 11 Chemistry. Whether you’re revising or preparing for tests, this chapter-wise guide will serve as an invaluable resource.

Exercise 1

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?

Vapour pressure is the pressure exerted by the vapour in equilibrium with the liquid at a fixed temperature.

(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.

Q 10:

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

2SO₂(g) + O₂(g) ↔ 2SO₃ (g)

What is K_c at this temperature ?

For the given reaction,

Δn = 2 - 3 =-1

T = 450 K

R = 0.0831 bar L bar K^-1 mol^-1

K_p = 2.0 × 10¹⁰ bar^-1

We know that,

K_p = K_c (RT)Δn

Q 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 ?

2HI (g) ↔ H₂ (g) + I₂ (g)

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)

Intial concentration 0.2 atm 0 0

At equilibrium 0.04 atm 0.16/2 2.15/2

Therefore,

Therefore, 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 × 10².

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

The given reaction is:

N₂(g) + 3H₂ (g) ↔ 2NH₃ (g)

The given concetration of various species is:

N₂ = 1.57/20 molL^-1

H₂ = 1.92/20 molL^-1

NH₃= 8.13/20 molL^-1

Now, reaction quotient Qc is:

Qc = (8.13/20)² / (1.57/20) (1.92/20)³

= 2.4 x 10^-3

Since,Qc ≠ Kc the reaction mixture is not at equilibrium.

Again, Qc > Kc. Hence, the reaction will proceed in the reverse direction.

Q 13:

The equilibrium constant expression for a gas reaction is,

Write the balanced chemical equation corresponding to this expression.

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 40% 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.

The given reaction is:

H₂O (g) + CO (g) ↔ H₂ (g) + CO₂ (g)

Initial Concentration: 1/10 M 1/10 M 0 0

At equilibrium 1-0.4/10 M 1-0.4/10 M 0.4/10M 0.4/10M

= 0.06M 0.06M 0.04M 0.04M

Therefore, the equilibrium constant for the reaction,

Q 15:

At 700 K, equilibrium constant for the reaction:

H2 (g) + I2 (g) ↔ 2HI (g)

is 54.8. If 0.5 mol L^–1of HI(g) is present at equilibrium at 700 K, what are the concentration of H₂(g) and I₂(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?

It is given that equilibrium constant K_cfor the reaction

H₂ (g) + I₂ (g) ↔ 2HI (g) is 54.8.

Therefore, at equilibrium, the equilibrium constant K'_cfor the reaction

2HI (g) ↔ H₂ (g) + I₂ (g) will be 1/54.8

HI = 0.5 molL^-1

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

Hence, at equilibrium,

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

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, Kc = I₂ x Cl₂/ (ICl)²

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)

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. 4.0atm 0 0
At equilibrium 4.0-p p p

We can write,

Hence, at equilibrium,

C2H6 - 4 - p = 4 -.038

= 3.62 atm

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), Q_c, 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?

(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:

Q 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) ↔ PCl₃ (g) + Cl₂(g)

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

PCl₅ (g) ↔ PCl₃ (g) + Cl₂(g)

at equilibrium 0.5x10^-1molL^-1 xmolL^-1 xmolL^-1

it is given that the value of equilibrium constant, Kc is 8.3x10^–3

Now we can write the expression for equilibrium as:

Q 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 ?

2SO_2(g) + O_2(g) ↔ 2SO_3(g)

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

K_c= [SO₃]2 / [SO₂]2 [O₂]

= (1.90)² M² / (0.60)² (0.8321) M³

= 12.239 M^-1Approx.

Hence, K_cfor the equilibrium is 12.239 M^-1

Q 20:

One of the reaction 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 atm at 1050K

What are the equilibrium partial pressures of CO and CO₂at 1050 K if the initial partial pressures are: p_CO= 1.4 atm and P_co2 = 0.80 atm

Q 21:

Equilibrium constant, Kc for the reaction

N2 (g) + 3H2 (g) ↔ 2NH₃ (g) at 500 K 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?

The given reaction is:

N₂ (g) + 3H₂ (g) ↔ 2NH₃ (g)

at a particular time: 3.0molL^-1 2.0 molL^-1 0.5molL^-1

Now, we know that,

Qc = [NH₃]² / [N₂][H₂]³

= (0.5)² / (3.0)(2.0)³

= 0.0104

It is given that K_c = 0.061

Since,Qc ≠ Kc, the reaction mixture is not at equilibrium.

Again, Qc < Kc, the reaction will proceed in the forward direction to reach equilibrium.

Q 22:

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

2BrCl (g) ↔ 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 mol L^–1, what is its molar concentration in the mixture at equilibrium?

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

2BrCl (g) ↔ Br₂ (g) + Cl₂ (g)

Initial Conc. 3.3x10^-3 0 0

at equilibrium 3.3x10^-3 -2x x x

Now, we can write,

Kc = [Br₂][Cl₂] / [BrCl]²

⇒ (x) x (x) / (3.3x10^-3 -2x)²= 32

⇒ x / (3.3x10^-3 -2x) = 5.66

⇒ x = 18.678x10^-3 - 11.32x

⇒ x + 11.32x = 18.678x10^-3

⇒ 12.32x = 18.678x10^-3

⇒ x = 1.5 x 10^-3

Therefore, at equilibrium,

[BrCl] = 3.3x10^-3- (2 x 1.5 x 10^-3)

= 3.3x10^-3 - 3.0x10^-3

= 0.3 x 10^-3

= 3.0 x 10^-4 mol L^-1

Q 23:

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

C (s) + CO₂ (g) ↔ 2CO (g)

Calculate K_c for this reaction at the above temperature.

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 = 90.55/28 = 3.234 mol

Number of moles of CO₂, n_co2 = 9.45/44 = 0.215 mol

Partial pressure of CO,

For the given reaction,

Δn = 2 - 1 = 1

we know that,

Kp = Kc (RT)^Δn

⇒ 14.19 = Kc (0.082 x 1127)¹

⇒ Kc = 14.19 / 0.082 x 1127

= 0.154 (arrpox.)

Q 24:

Calculate a) ΔG0 and b) the equilibrium constant for the formation of NO₂ from NO and O₂ at 298K

NO (g) + ½ O₂ (g) ↔ NO₂ (g)

where Δ_fG⁰ (NO₂) = 52.0 kJ/mol

Δ_fG⁰ (NO) = 87.0 kJ/mol

Δ_fG⁰ (O₂) = 0 kJ/mol

(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

Q 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) ↔ PCl₃ (g) + Cl₂ (g)

(b) CaO (s) + CO₂ (g) ↔ CaCO₃ (s)

(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.

Q 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.

(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) 4 NH₃ (g) + 5O₂ (g) ↔ 4NO (g) + 6H₂O(g)

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.

Q 27:

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

H₂(g) + Br₂(g) ↔ 2HBr(g)

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

Given,k_p for the reaction i.e., H₂(g) + Br₂(g) ↔ 2HBr(g) is 1.6 ×10⁵

Therefore, for the reaction

2HBr(g) ↔ H₂(g) + Br₂(g)

the equilibrium constant will be,

K'_p = 1/K_p

= 1/1.6 ×10⁵

= 6.25x10^-6

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

2HBr(g) ↔ H₂(g) + Br₂(g)

Initial Conc. 10 0 0

at equilibrium 10-2p p p

Now, we can write,

Q 28:

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

CH₄ (g) + H₂O (g) ↔ CO (g) + 3H₂ (g)

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

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

(i) increasing the pressure

(ii) increasing the temperature

(iii) using a catalyst ?

(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.

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)

(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.

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

Q 3:

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

I_2(g) ↔ 2l_(g)

Calculate K_pfor the equilibrium.

Partial pressure of I atoms,

p_I= p_total x 40/100

= 10⁵ x 40/100

= 4 x 10⁴ Pa

Partial pressure of I₂ molecules,

p_I_₂ = p_total x 60/100

= 10⁵ x 60/100

= 6 x 10⁴ Pa

Now, for the given reaction,

Kp = (pI)² / pI₂

= (4 x 10⁴ )² Pa² / 6 x 10⁴ Pa

= 2.67 x 10⁴ Pa

Q 30:

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

PCl₅ (g) ↔ PCl₃ (g) + Cl₂ (g) Δ_rH⁰ = 124.0 kJ mol^–1

(a) write an expression for Kc for the reaction.

(b) what is the value of Kc 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 ?

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

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

Q 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) ↔ 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_H2O = 4.0 bar, what will be the partial pressure of H₂ at equilibrium? K_p= 10.1 at 400°C.

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

CO (g) + H₂O (g) ↔ CO₂ (g) + H₂ (g)

Initial Conc. 4.0 bar 4.0 bar 0 0

At equilibrium 4.0-p 4.0-p p p

It is given that K_p = 10.1

Now,

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

Q 32:

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

a) Cl₂ (g) ↔ 2Cl (g) Kc = 5 ×10^–39

b) Cl₂(g) + 2NO (g) ↔ 2NOCl (g) Kc = 3.7 × 10⁸

c) Cl₂ (g) + 2NO₂ (g) ↔ 2NO₂Cl (g) Kc = 1.8

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.

Q 33:

The value of Kc 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?

The given reaction is:

3O₂ (g) ↔ 2O₃ (g)

Then, Kc = [O₃ (g)]² / [O₂ (g)]³

It is given that Kc = 2.0 ×10^–50and O₂ (g) = 1.6 ×10^–2

Then we have,

2.0 ×10^–50 = [O₃ (g)]²/ [1.6 ×10^–2]³

⇒ [O₃ (g)]² = [2.0 ×10^–50] x [1.6 ×10^–2]³

⇒ [O₃ (g)]²= 8.192 x 10^-56

⇒ O₃ (g) = 2.86x10^-28M

Hence, the concentration of O₂ (g) = 2.86x10^-28M

Q 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 H₂and 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.

Let the concentration of methane at equilibrium be x.

CO(g) + 3H₂(g) ↔ CH₄(g) + H₂O(g)

At equilibrium 0.3/1 M 0.1/1 M x 0.02/1 M

It is given that K_c= 3.90.

Therefore,

Kc = [CH₄(g)] [ H₂O(g)] / [ CO(g) ] [H₂(g)]³

⇒ [x] [0.02] / (0.3) (0.1)³ = 3.90

⇒ x = (3.90) (0.3) (0.1)³/ [0.02]

⇒ x = 0.00117 / 0.02

= 0.0585 M

= 5.85 x 10^-2

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

Q 35:

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

HNO₂, CN^– , HClO₄, F^–, OH^–, CO^2–₃ and S^-

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) / O^2- (base)
CO^2–₃	HCO^–₃(Acid)
S^-	HS^-(Acid)

Q 36:

Which of the followings are Lewis acids? H₂O, BF₃, H⁺ and NH⁺₄

Lewis acids are those acids which can accept a pair of electrons. For example, BF₃, H⁺, and NH⁺₄ are Lewis acids.

Q 37:

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

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

Bronsted acid	Conjugate base
HF	F^-
H₂SO₄	HSO^-₄
HCO₃	CO^2-₃

Q 38:

Write the conjugate acids for the following Brönsted bases: NH^–₂ , NH₃ and HCOO^–.

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

Bronsted base	Conjugate acid
NH^–₂	NH₃
NH₃	NH⁺₄
HCOO^–	HCOOH

Q 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.

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

Q 4:

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

(i) 2NOCl (g) ↔ 2NO (g) + Cl₂ (g)

(ii) 2Cu(NO₃)₂ (s) ↔ 2CuO (s) + 4NO₂ (g) + O₂ (g)

(iii) CH₃COOC₂H₅(aq) + H₂O(l) ↔ CH₃COOH (aq) + C₂H₅OH (aq)

(iv) Fe³⁺ (aq) + 3OH^– (aq) ↔ Fe(OH)₃ (s)

(v) I₂ (s) + 5F₂ ↔ 2IF₅

Q 40:

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

(a) OH^–

(b) F^–

(d) BCl₃

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

(b) F^{- i}s a Lewis base since it can donate a pair of electrons.

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

Q 41:

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

Given that:

[H⁺] = 3.8 × 10^–3

∴ pH value of soft drink

= -log [H⁺]

= -log [3.8 × 10^–3]

= -log [3.8] - log[10^–3]

= -log [3.8] + 3

= -0.58 + 3

= 2.42

Q 42:

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

Given that:

pH = 3.76

It is known that,

pH = -log [H⁺]

⇒ log [H⁺] = -pH

⇒ [H⁺] = antilog (-pH)

= antilog(-3.76)

= 1.74x10^-4 M

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

Q 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.

It is known that,

K_b = K_w / K_a

Given

K_a of HF = 6.8 × 10^-4

Hence, K_b of its conjugate base F- = K_w / K_a

= 10^-14 / 6.8 × 10^-4

= 1.5 x 10^-11

Given, K_a of HCOOH = 1.8 × 10^-4

Hence, K_b of its conjugate base HCOO^-= K_w / K_a

= 10^-14 / 1.8 × 10^-4

= 5.6x10^-11

Given, K_a of HCN = 4.8 × 10^-9

Hence, Kb of its conjugate base CN^-= K_w / K_a

= 10^-14 / 4.8 × 10^-9

= 2.8 x 10^-6

Q 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?

Ionization of phenol:

C₆H₅OH + H₂O ↔ C₆H₅O^- + H₃O⁺

Initial Conc. 0.05 0 0

At equilibrium 0.05-x x x

Ka = [C₆H₅O^- x H₃O⁺] / C₆H₅OH

Ka = x x / 0.05-x

Ka = x² / 0.05-x

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

we can ignore x in the denominator

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

Q 45:

The first ionization constant of H₂S is 9.1 × 10^–8. Calculate the concentration of HS^– ion in its 0.1M solution. How will this concentration be affected if the solution is 0.1M in HCl also ? If the second dissociation constant of H₂S is 1.2 × 10^–13, calculate the concentration of S^2– under both conditions.

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

Case I (in the absence of HCl):

Let the concentration of HS^- be x M.

H₂S ↔ H⁺ + HS^-

C_i 0.1 0 0

C_f 0.1-x x x

Then K_a1 = [H⁺] [ HS^-] / H₂S

9.1 × 10^–8= xx / 0.1-x

(9.1 × 10^–8) (0.1-x) = x²

Taking 0.1 - x M ; 0.1M, we have

(9.1 × 10^–8) (0.1) = x²

9.1 x 10^-9= x²

Case II (in the presence of HCl):

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

Then, H₂S ↔ H⁺ + HS^-

C_i 0.1 0 0

C_f 0.1-y y y

Also, HCI ↔ H⁺ + CI^-

0.1 0.1

Now, K_a1 = [H⁺] [ HS^-] / H₂S

K_a1 = [y] [0.1+y] / [0.1-y]

9.1 × 10^–8= y x 0.1 / 0.1 (∵ 0.1-y; 0.1M) (and 0.1+y; 0.1M)

9.1 × 10^–8= y

⇒ [ HS^-] = 9.1 × 10^–8

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

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

HS^- ↔ H⁺ + S^2-

HS^- = 9.54 x 10^-5 M (From first ionization, case I)

Let S^2-be X.

Also, [H⁺] = 9.54 x 10^-5 M (From first ionization, case I)

K_a2 = (9.54 x 10^-5) (X) / (9.54 x 10^-5)

1.2x10-13 = X = S^2-

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

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

Q 46:

The ionization constant of acetic acid is 1.74 x 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.

1) CH₃COOH ↔ CH₃COO^- + H⁺ K_a = 1.74x10^-5

2) H₂O + H₂O ↔ H₃O⁺ + OH^- K_w = 1.0x10^-14

Since K_a > K_w

CH₃COOH + H₂O ↔ CH₃COO^- + H₃O⁺

Ci = 0.05 0 0

0.05- 0.05α 0.05α 0.05α

k_a = 0.05α x 0.05α / 0.05- 0.05α

= ~~0.05~~α x 0.05α / ~~0.05~~ (1-α)

= 0.05α² / (1-α)

⇒ 1.74x10^-5 = 0.05α² / (1-α)

⇒ 1.74x10^-5 - 1.74x10^-5 α = 0.05α²

⇒ 0.05α² + 1.74x10^-5 α - 1.74x10^-5 =0

D = b²- 4ac

= (1.74x10^-5)² - 4 (0.05) (1.74x10^-5)

= 3.02x10^-25 + 0.348x10^-5

Method 2:

Degree of Dissociation,

CH₃COOH ↔ CH₃COO^- + H⁺

Thus, concentration of CH₃COO^- = c.α

= 0.05x1.86x10^-2

= 0.93x10^-2

=.00093M

Since [oAc-] = [H⁺]

[H⁺] = .00093 = 0.093x10-2

pH = -log[H⁺]

= -log (0.93x10^-2)

∴ pH = 3.03

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

Q 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.

Let the organic acid be HA.

⇒ HA ↔ H⁺ + A^-

Concentration of HA = 0.01 M

pH = 4.15

-log[H⁺] = 4.15

[H⁺] = 7.08x10^-5

Now, K_a = [H⁺] [A^-] / [HA]

[H⁺] = [A^-] = 7.08x10^-5

[HA] = 0.01

Then,

K_a = 7.08x10^-5x 7.08x10^-5/ 0.01

K_a = 5.01x10^-7

pK_a= -log K_a

= -log(5.01x10^-7)

pK_a = 6.3001

Q 48:

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

(a) 0.003 M HCl

(b) 0.005 M NaOH

(d) 0.002 M KOH

(i) 0.003MHCl:

H₂O + HCl ↔ H3O⁺ + Cl^-

Since HCl is completely ionized,

[H₃O⁺] = [ HCl]

⇒ [H₃O⁺] = = 0.003

Now

pH = -log [H₃O⁺] = -log (0.003)

= 2.52

Hence, the pH of the solution is 2.52.

(b) 0.005 M NaOH

NaOH_(aq) ↔ Na⁺_(aq) + HO^-_(aq)

[NaOH] = [ HO^-]

⇒ [ HO^-] = 0.05

pOH = -log[ HO^-] = -log (0.05)

= 2.30

∴ pH = 14 - 2.30 = 11.70

Hence, the pH of the solution is 11.70.

HBr + H₂O ↔ H₃O⁺ + Br^-

[HBr] = [H₃O⁺]

⇒ [H₃O⁺] = 0.002

∴ pH = -log [H₃O⁺] = -log (0.002)

= 2.69

Hence, the pH of the solution is 2.69.

(d) 0.002 M KOH

KOH_(aq) ↔ K⁺_(aq) + OH^-_(aq)

[OH^-] = [KOH]

⇒ [OH^-] = 0.002

Now pOH = -log[OH^-] = -log (0.002)

= 2.69

∴ pH = 14-2.69 = 11.31

Hence, the pH of the solution is 11.31.

Q 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.

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

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

[TIOH(aq)] = 2/2 g/L

= 2/2 x 1/221 M

= 1/221 M

TIOH_(aq) → TI⁺_(aq) + OH^-_(aq)

OH^-_(aq)= TIOH_(aq)= 1/221M

Kw = [H⁺] [OH^-]

10^-14= [H⁺] [1/221]

[H⁺] = 221x10^-14

⇒ pH = -log [H⁺] = -log ( 221x10^-14)

= 11.65

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

Ca(OH)₂ → Ca²⁺ + 2OH^-

[Ca(OH)₂] = 0.3x1000/500 = 0.6M

OH^-_(aq) = 2 x [Ca(OH)_2(aq)] = 2 x 0.6 = 1.2M

[H+] = Kw / OH^-_(aq)

= 10^-14/1.2 M

= 0.833 x 10^-14

pH = -log(0.833 x 10^-14)

= -log(8.33 x 10^-13)

= (-0.902 + 13)

= 12.098

NaOH → Na⁺_(aq) + OH^-_(aq)

[NaOH] = 0.3 x 1000/200 = 1.5M

[OH^-_(aq)] = 1.5M

Then [H+] = 10^-14/ 1.5

= 6.66 x 10^-13

pH = -log ( 6.66 x 10^-13)

= 12.18

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

13.6 x 1 mL = M₂x 1000 mL

(Before dilution) (after dilution)

13.6 x 10^-3 = M₂ x 1L

M₂ = 1.36 x 10^-2

[H⁺] = 1.36 × 10^-2

pH = - log (1.36 × 10^-2)

= (- 0.1335 + 2)

= 1.866 = 1.87

Q 5:

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

(i) 2NOCl (g) ↔ 2NO (g) + Cl₂ (g); Kp = 1.8 × 10^–2 at 500 K

(ii) CaCO₃ (s) ↔ CaO(s) + CO₂(g); Kp = 167 at 1073 K

(a) 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

Kp = 1.8 x 10^-2

Now, K_p = K_c (RT)^Δn

⇒ K_c = 1.8 x 10^-2/ (0.0831x500)

= 4.33 x 10^-4(approx.)

(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

⇒ K_c = 167 / (0.0831x1073)

= 1.87 (approx.)

Q 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.

Degree of ionization, α = 0.132

Concentration, c = 0.1 M

Thus, the concentration of H₃O⁺ = c.α

= 0.1 × 0.132

= 0.0132

pH = -log [H⁺]

= -log (0.0132)

= 1.879 : 1.88

Now,

K_a = Cα²

= 0.1 x (0.132)²

K_a = 0.0017

pK_α = 2.75

Q 51:

The pH of 0.005M codeine (C₁₈H₂₁NO₃) solution is 9.95. Calculate its ionization constant and pKb.

c = 0.005

pH = 9.95

pOH = 4.05

pH = -log (4.05)

4.05 = - log [OH^-]

[OH^-] = 8.91x 10^-5

cα = 8.91x 10^-5

α = 8.91x 10^-5/ 5 x 10^-3 = 1.782 x 10^-2

Thus , K_b = cα²

= 0.005 x (1.782)² x 10^-4

= 0.005 x 3.1755 x 10^-4

= 0.0158 x 10^-4

K_b = 1.58 x 10^-6

PK_b = -logK_b

= -log (1.58 x 10^-6)

= 5.80

Q 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.

K_b = 4.27 x 10^-10

c = 0.001M

pH = ?

α = ?

K_b = cα²

4.27 x 10^-10 = 0.001 x α²

4270 x 10^-10= α²

α = 65.34 x 10^-4

Then (anion) = cα = 0.001 x 65.34 x 10^-4

= 0.65 x 10^-5

pOH = -log ( 0.65 x 10^-5)

= 6.187

pH = 7.813

Now

K_a x K_b = Kw

∴ 4.27 x 10^-10 x Ka = Kw

K_a = 10^-14 / 4.27 x 10^-10

= 2.34 x 10^-5

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

Q 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?

c = 0.05 M

pK_a= 4.74

pK_a= -log (K_a)

K_a= 1.82 x 10^-5

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.

CH₃COOH ↔ H⁺ + CH₃COO^-

Initial Conc. 0.05M 0 0

After dissociation 0.05-x 0.01+x x

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.

K_a = [CH₃COO^-] [ H⁺] / [CH₃COOH]

∴ K_a = (0.01) (x) / (0.05)

x = 1.82 x 10^-5 x(multiply) 0.05 / 0.01

x = 1.82 x 10^-3 x(multiply) 0.05

Now

α = Amount of acid dissociated / amount of acid taken

= 1.82 x 10^-3x 0.05 / 0.05

= 1.82 x 10^-3

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:

[CH₃COOH] = 0.05 - X : 0.05M

[CH₃COO^-] = X

[ H⁺] = 0.1+X ; 0.1M

K_a = [CH₃COO^-] [ H⁺] / [CH₃COOH]

∴ K_a = (0.1) (X) / (0.05)

x = 1.82 x 10^-5 x(multiply) 0.05 / 0.1

x = 1.82 x 10^-4 x(multiply) 0.05

Now

α = Amount of acid dissociated / amount of acid taken

= 1.82 x 10^-4x 0.05 / 0.05

= 1.82 x 10^-4

Q 54:

The ionization constant of dimethylamine is 5.4 x 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?

Kb = 5.4x10-4

c = 0.02M

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

NaOH_(aq) ↔ Na⁺_(aq) + OH^- _(aq)

0.1M 0.1M

and

(CH₃)₂ NH + H₂O ↔ (CH₃)₂ NH⁺₂ + O^-H

0.02-x x x

Then (CH₃)₂ NH⁺₂= x

[OH^-] = x + 0.1 ; 0.1

⇒ K_b = [(CH₃)₂ NH⁺₂ ] [OH^-] / [CH₃)₂ NH]

5.4x10-4 = x x 0.1 / 0.02

x = 0.0054

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

Q 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

(d) Human saliva, 6.4.

a) Human muscle fluid 6.83:

pH = 6.83

pH = - log [H⁺]

∴6.83 = - log [H⁺]

[H⁺] =1.48 x 10^-7 M

(b) Human stomach fluid, 1.2:

pH =1.2

1.2 = - log [H⁺]

∴[H⁺] = 0.063

pH = 7.38 = - log [H⁺]

∴ [H⁺] = 4.17 x 10^-8 M

(d) Human saliva, 6.4:

pH = 6.4

6.4 = - log [H⁺]

[H⁺] = 3.98 x 10^-7

Q 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.

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)

= 1.5x10-7M

(ii) pH of black coffee = 5.0

Since, pH = -log [H⁺]

5.0 = -log [H⁺]

log [H⁺] = -5.0

[H⁺] = anitlog(-5.0)

= 10^-5M

(iii) pH of tomato juice = 4.2

Since, pH =-log [H⁺]

4.2 =-log [H⁺]

log [H⁺] = -4.2

[H⁺] = anitlog(-4.2)

= 6.31x10^-5M

(iv) pH of lemon juice = 2.2

Since, pH = -log [H⁺]

2.2 = -log [H⁺]

log [H⁺] = -2.2

[H⁺] = anitlog(-2.2)

=6.31x10^-3M

(v) pH of egg white = 7.8

Since, pH = -log [H⁺]

7.8 = -log[H⁺]

log[H⁺] = -7.8

[H⁺] = anitlog(-7.8)

= 1.58x10^-8M

Q 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?

[KOH_(aq)] = 0.561 / (1/5)g/L

= 2.805 g/L

= 2.805 x 1/56.11 M

= 0.05M

KOH_(aq) → K⁺_(aq) + OH^-_(aq)

[OH^-] = 0.05M = [K⁺]

[H^-] [H⁺] = K_w

[H⁺] K_w / [OH^-]

= 10^-14 / 0.05 = 2x10^-13 M

∴ pH = 12.70

Q 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.

Solubility of Sr(OH)₂ = 19.23 g/L

Then, concentration of Sr(OH)₂

= 19.23 / 121.63 M

= 0.1581 M

Sr(OH)_2(aq)→ Sr²⁺_(aq)+ 2 (OH^-)_(aq)

∴Sr²⁺= 0.1581M

[OH^-] = 2 x 0.1581M = 0.3126 M

Now

Kw = [OH^-] [H⁺]

10^-14 / 0.3126 = [H⁺]

⇒ [H⁺] = 3.2 x 10^-14

∴ pH = 13.495 = 13.50

Q 59:

The ionization constant of propanoic acid is 1.32 x 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?

Let the degree of ionization of propanoic acid be α.

Then, representing propionic acid as HA, we have:

HA + H₂O ↔ H₃O⁺ + A^-

(0.05-0.0α) ≈ 0.05 0.05α 0.05α

K_a = [H₃O⁺] [A^-] / [HA]

= (0.05α)(0.05α) / 0.05

= 0.05 α²

Then, [H₃O⁺] = 0.05α = 0.05 x 1.63 x 10^-2 = Kb . 15 x 10^-4 M

∴ pH = 3.09

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

Then, [H₃O⁺] = 0.01

[A^-] = 0.05α'

[HA] = 0.05

Ka = 0.01 x 0.05α' / 0.05

⇒ 1.32 x 10^-5 = 0.1 α'

α' = 1.32 x 10^-3

Q 6:

For the following equilibrium, K_c= 6.3 × 10¹⁴ at 1000 K

NO (g) + O₃ (g) ↔ 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?

It is given that K_cfor the forward reaction is 6.3 × 10¹⁴

Then, K_c for the reverse reaction will be,

K^’_c = 1 / K_c

= 1 / 6.3 × 10¹⁴

= 1.59 x 10^-15

Q 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.

c = 0.1 M

pH = 2.34

-log [H⁺] = pH

-log [H⁺] = 2.34

[H⁺] = 4.5 x 10^-3

also

[H⁺] = cα

4.5 x 10^-3= 0.1 α

α = 4.5 x 10^-3 / 0.1

α = 45 x 10^-3

Then,

K_a = cα²

= 0.1 (45 x 10^-3)²

= 202.5 x 10^-6

= 2.02 x 10^-4

Q 61:

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

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

NO^-₂ + H₂O ↔ HNO₂ + OH^-

K_h = [ HNO₂ ] [ OH^-] / [NO^-₂]

⇒ K_w / k_a = 10^-14 / 4.5 x 10^-4= 0.22 x 10^-10

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

[NO^-₂ ] = 0.04 - x ; 0.04

[ HNO₂ ] = x

[ OH^-] = x

K_h = x² / 0.04 = 0.22 x 10^-10

x² = 0.0088 x 10^-10

x = 0.093 x 10^-5

∴ [ OH^-] = 0.093 x 10^-5M

[H₃O⁺] = 10^-14 / 0.093 x 10^-5= 10.75 x 10^-9M

⇒ pH = -log(10.75 x 10^-9)

= 7.96

Therefore, degree of hydrolysis

= x / 0.04 = (0.093 x 10^-5 ) / 0.04 = 2.325 x 10^-5

Q 62:

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

Given, pH = 3.44

We know that

pH = -log [H⁺]

∴ [H⁺] = 3.63x10^-4

Then K_h = ( 3.63x10^-4)² / 0.02 (∵ Concentration is 0.02M)

⇒ K_h = 6.6 x 10^-6

Now , K_h = K_w / K_a

⇒ K_a= K_w / K_h

= 10^-14 / 6.6 x 10^-6

= 1.51 x 10^-9

Q 65:

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

Ionic product,

K_w = [H⁺] [ OH^-]

Let [H⁺] = x.

Since [H⁺] = [ OH^-] , K_w = x²

⇒ K_w at 310K is 2.7x 10^-14

∴ 2.7x 10^-14 = x²

⇒ x = 1.64 x 10^-7

⇒ [H⁺] = 1.64 x 10^-7

⇒ pH = -log [H⁺]

= -log (1.64 x 10^-7)

= 6.78

Hence, the pH of neutral water is 6.78.

Q 7:

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

: For a pure substance (both solids and liquids),

Pure Substance = Number of moles / Volume

= Mass/Molecular Mass / Volume

= Mass / Volume x Molecular Mass

= Density / Molecular Mass

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

Q 72:

What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, K_sp is 9.1 x 10^-6).

CaSO_4(s) ↔ Ca²⁺ _(aq) + SO^2-_4(aq)

K_sp = [ Ca²⁺] [ SO^2-]

Let the solubility of CaSO₄ be s.

Then, K_sp = s²

9.1 x 10^-6= s²

s = 3.02 x 10^-3 mol/L

Molecular mass of CaSO₄ = 136 g/mol

Solubility of CaSO₄ in gram/L = 3.02 × 10^-3 × 136 = 0.41 g/L

This means that we need 1L of water to dissolve 0.41g of CaSO₄

Therefore, to dissolve 1g of CaSO₄ we require = 1/0.41 L = 2.44 Lof water.

Q 8:

Reaction between N₂ and O_2– takes place as follows:

2N₂ (g) + O₂ (g) ↔ 2N₂O (g)

If a mixture of 0.482 mol 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.

Let the concentration of N₂O at equilibrium be x. The given reaction is:

2N₂ (g) + O₂ (g) ↔ 2N₂O (g)

Intial Concentration: 0.482 mol 0.933 mol 0 mol

At equilibrium (0.482-x) mol (0.933 -x) mol x mol

Therefore, at equilibrium, in the 10 L vessel:

N₂ = 0.482-x / 10

O₂ = 0.933-x/2 / 10

N₂O = x / 10

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,

N2 = 0.482/10 = 0.0482 molL-1 and O2 = 0.933/10 = 0.0933 molL-1

Now,

Q 9:

Nitric oxide reacts with Br₂ and gives nitrosyl bromide as per 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 constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br₂ .

The given reaction is:

2NO (g) + Br₂ (g) ↔ 2NOBr (g)

2 mol 1 mol 2 mol

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

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- Our solutions break down complex problems into simple steps, provide clear explanations, and include relevant examples to help students grasp the concepts easily.
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- Yes, practice regularly, understand the concepts before memorizing, solve additional problems, and refer to our step-by-step solutions for better understanding.

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Equilibrium Question Answers: NCERT Class 11 Chemistry

Exercise 1

Frequently Asked Questions about Equilibrium - Class 11 Chemistry

Exam Preparation Tips for Equilibrium

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Equilibrium Question Answers: NCERT Class 11 Chemistry

Exercise 1

Frequently Asked Questions about Equilibrium - Class 11 Chemistry

Exam Preparation Tips for Equilibrium

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Effective Tips to Avoid Nervous Breakdown during CBSE Board Exam

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