NCERT Solutions Class 11 Chemistry Chapter 6 - Equilibrium

NCERT Solutions for Class 11 Chemistry Chapter 6 (Equilibrium) help students understand reversible reactions and the conditions under which reactants and products remain in equilibrium. The chapter covers chemical and ionic equilibrium, equilibrium constants, Le Chatelier’s principle, and buffer solutions. Students can also learn how equilibrium concepts apply to processes such as ammonia production through the Haber process and the regulation of pH in human blood.

NCERT Solutions for Class 11 Chemistry Chapter 6 by ALLEN explain how to use equilibrium constants $(K_{(c)} an d K_{(p)}$ to solve numerical problems and predict the effects of changes in temperature, pressure, and concentration. Students can practise questions on chemical equilibrium, pH, buffer solutions, and solubility equilibrium. These solutions help students revise NCERT concepts, practise textbook questions, and prepare for school exams, JEE Main, JEE Advanced, and NEET.

1.0Class 11 Chemistry Chapter 6 Equilibrium: Key Concepts

This chapter explains how physical and chemical systems reach equilibrium and how changes in conditions affect it. The key concepts include:

Dynamic Equilibrium: Understanding that the forward and backward reactions occur at the same rate at equilibrium.
Law of Chemical Equilibrium: Learning how to calculate equilibrium constants $(K_{(c)} an d K_{(p)}$ and use the relation $(K_{p} = K_{c} (RT)^{Δ n})$ .
Homogeneous and Heterogeneous Equilibrium: Understanding equilibrium in systems with one or more phases.
Le Chatelier’s Principle: Learning how changes in concentration, pressure, and temperature affect the position of equilibrium.
Ionic Equilibrium: Understanding how weak acids and weak bases form ions in water.
pH Scale: Learning how to calculate the pH of a solution using its hydrogen ion concentration.
Ionisation Constants: Understanding acid ionisation constant $(K_{a})$ , base ionisation constant $(K_{b})$ , and the common ion effect.
Buffer Solutions: Learning how buffer solutions help prevent large changes in pH when small amounts of acid or base are added.
Solubility Product ((K_{sp})): Learning how to calculate the solubility of sparingly soluble salts and predict when a precipitate will form.

2.0Detailed NCERT Textbook Solutions : Class 11 Chemistry Chapter 6

NCERT EXERCISE

A liquid is in equilibrium with its vapour in a sealed container at a fixed tempoperature. The volume of the container is suddenly increased. (a) What is the initial effect of the change on vapour pressure? (b) How do rates of evaporation and condensation change initially? (c) What happens when equilibrium is restored finally and what will be the final vapour pressure?

Ans.

(a) 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.

What is Kc for the following equilibrium when the equilibrium concentration of each substance is:

[SO_{2}] = 0.60 M, [O_{2}] = 0.82 M and [SO_{3}] = 1.90 M ?

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

Ans. The equilibrium constant $(K_{c})$ for the give reaction is:

K_{C} = \frac{[ SO _{3} ] ^{2}}{[ SO _{2} ] ^{2} ( 0.821 ) M ^{3}} = \frac{( 1.90 ) ^{2} M ^{2}}{( 0.60 ) ^{2} ( 0.821 ) M ^{3}} = 12.239 M^{- 1} (approximately)

Hence, $K_{c}$ for the equilibrium is $12.239 M^{- 1}$ .

At a certain temperature and total pressure of $1 0^{5} Pa$ , iodine vapour contains 40% by volume of I atoms

I_{2} (g) ⇌ 2 I (g)

Calculate $K_{p}$ for the equilibrium. Ans. Partial pressure of I atoms,

p_{1} = \frac{40}{100} \times p_{total} = \frac{40}{100} \times 1 0^{5} = 4 \times 1 0^{4} Pa

Partial pressure of $I_{2}$ molecules,

p_{1_{2}} = \frac{60}{100} \times p_{total} = \frac{60}{100} \times 1 0^{5} = 6 \times 1 0^{4} Pa

Now, for the given reaction,

Write the expression for the equilibrium constant, $K_{c}$ for each of the following reactions:

$(i) 2NOCl (g) ⇌ 2NO (g) + C l_{2} (g) (ii) 2Cu (N O_{3})_{2} (s) ⇌ 2Cu (s) + 4N O_{2} (g) + O_{2} (g) (iii) C H_{3} COO C_{2} H_{5} (aq) + H_{2} O (l) ⇌ C H_{3} COOH (aq) + C_{2} H_{5} OH (aq) (iv) F e^{3 +} (aq) + 3O H^{-} (aq) ⇌ Fe (OH)_{3} (s) (v) I_{2} (s) + 5 F_{2} ⇌ 2I F_{5}$

Ans.

$(i) K_{c} = \frac{[ N O _{(g)} ] ^{2} [ C l _{2 (g)} ]}{[ NOC l _{(g)} ] ^{2}} (ii) K_{c} = \frac{[ Cu O _{(s)} ] ^{2} [ N O _{2 (g)} ] ^{4} [ O _{2 (g)} ]}{[ Cu ( N O _{3} ) _{2 (s)} ] ^{2}} = [N O_{2 (g)}]^{4} [O_{2 (g)}]$

$(iii) K_{c} = \frac{[ C H _{3} COO H _{(aq)} ] [ C _{2} H _{5} O H _{(aq)} ]}{[ C H _{3} COO C _{2} H _{5 (aq)} ] [ H _{2} O _{(l)} ]} = \frac{[ C H _{3} COO H _{(aq)} ] [ C _{2} H _{5} O H _{(aq)} ]}{[ C H _{3} COO C _{2} H _{5 (aq)} ]} (iv) K_{c} = \frac{[ Fe ( OH ) _{3 (s)} ]}{[ F e _{(aq)}^{3 +} ] [ O H _{(aq)}^{-} ] ^{3}} = \frac{1}{[ F e _{(aq)}^{3 +} ] [ O H _{(aq)}^{-} ] ^{3}} (v) K_{c} = \frac{[ I F _{5} ] ^{2}}{[ I _{2 (s)} ] [ F _{2} ] ^{5}} = \frac{[ I F _{5} ] ^{2}}{[ F _{2} ] ^{5}}$

Find out the value of $K_{c}$ for each of the following equilibria from the value of $K_{p}$ :

$(i) 2NOCl (g) ⇌ 2NO (g) + C l_{2} (g); K_{p} = 1.8 \times 1 0^{- 2} at 500 K (ii) CaC O_{3} (s) ⇌ CaO (s) + C l_{2} (g); K_{p} = 167 at 1073 K Ans. The relation between K_{p} and K_{c} is given as: K_{p} = K_{c} (RT)^{Δ n} (a) Here, Δ n = 3 - 2 = 1 R = 0.0831 bar L mo l^{- 1} K^{- 1} T = 500 K K_{p} = 1.8 \times 1 0^{- 2} Now, K_{p} = K_{c} (RT)^{Δ n} \Rightarrow 1.8 \times 1 0^{- 2} = K_{c} (0.0831 \times 500)^{1} \Rightarrow K_{c} = \frac{1.8 \times 1 0 ^{- 2}}{0.0831 \times 500} = 4.33 \times 1 0^{- 4} (approximately) (b) Here, Δ n = 2 - 1 = 1 R = 0.0831 bar L mo l^{- 1} K^{- 1} T = 1073 K K_{p} = 167 Now, K_{p} = K_{c} (RT)^{Δ n} \Rightarrow 167 = K_{c} (0.0831 \times 1073)^{Δ n} \Rightarrow K_{c} = \frac{167}{0.0831 \times 1073} = 1.87 (approximately)$

$6. For the following equilibrium, K_{c} = 6.3 \times 1 0^{14} at 1000 K NO (g) + O_{3} (g) ⇌ N O_{2} (g) + O_{2} (g) Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K_{c} for the reverse reaction? Ans. It is given that K_{c} for the forward reaction is 6.3 \times 1 0^{14} . Then, K_{c} for the reverse reaction will be, K_{c}^{'} = \frac{1}{K _{c}} = \frac{1}{6.3 \times 1 0 ^{14}} = 1.59 \times 1 0^{- 15}$

$7. Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression? Ans. For a pure substance (both solids and liquids), [Pure substance] = \frac{Number of moles}{Volume} = \frac{Mass / molecular mass}{Volume} = \frac{Mass}{Volume \times molecular mass} = \frac{Density}{Molecular mass} Now, the molecular mass and density (at a particular temperature) of a pure substance is always fixed and is accounted for in the equilibrium constant. Therefore, the values of pure substances are not mentioned in the equilibrium constant expression.$

Reaction between $N_{2}$ and $O_{2}$ takes place as follows:

$2 N_{2} (g) + O_{2} (g) ⇌ 2 N_{2} O (g)$

If a mixture of 0.482 mol of $N_{2}$ and 0.933 mol of $O_{2}$ is placed in a 10 L reaction vessel and allowed to form $N_{2} O$ at a temperature for which Kc = 2.0 × 10–37, determine the composition of equilibrium mixture.

Ans. Let the concentration of $N_{2} O$ at equilibrium be x.

The given reaction is:

	2N₂(g) + O₂(g) ⇌ 2N₂O(g)
Initial conc.	0.482 mol 0.933 mol 0
At equil.	(0.482–x) mol (0.933–x) mol x mol

Therefore, at equilibrium, in the 10 L vessel:

$R = 0.0831 barL mol^{- 1} K^{- 1} T = 500 K K_{p} = 1.8 \times 1 0^{- 2}$

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

The value of equilibrium constant i.e., $K_{c} = 2.0 \times 1 0^{-} 37$ is very small. Therefore, the amount of $N_{2}$ and $O_{2}$ reacted is also very small. Thus, x can be neglected from the expressions of molar concentrations of $N_{2}$ and $O_{2}$ Then,

$Now, K_{c} = \frac{[ N _{2} O _{(g)} ] ^{2}}{[ N _{2 (g)} ] ^{2} [ O _{2 (g)} ]} \Rightarrow 2.0 \times 1 0^{- 37} = \frac{( \frac{x}{10} ) ^{2}}{( 0.0482 ) ^{2} ( 0.0933 )} \Rightarrow \frac{x ^{2}}{100} = 2.0 \times 1 0^{- 37} \times (0.0482)^{2} \times (0.0933) \Rightarrow x^{2} = 43.35 \times 1 0^{- 40} \Rightarrow x = 6.6 \times 1 0^{- 20} \Rightarrow [N_{2} O] = \frac{x}{10} = \frac{6.6 \times 1 0 ^{- 20}}{10} = 6.6 \times 1 0^{- 21}$

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

$2NO (g) + B r_{2} (g) ⇌ 2NOBr (g)$

When 0.087 mol of NO and 0.0437 mol of $B r_{2}$ are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount NO and $B r_{2} .$

Ans. The given reaction is:

$2NO (g) 2 mol + B r_{2} (g) 1 mol ⇌ 2NOBr (g) 2 mol$

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 $[B r_{2}]$ = 0.0437 mol

Therefore, the amount of NO present at equilibrium is:

$[NO] = 0.087 mol [B r_{2}] = 0.0437 mol$

[NO] = 0.087 – 0.0518 = 0.0352 mol

And, the amount of Br present at equilibrium is:

$[B r_{2}]$ = 0.0437 – 0.0259 = 0.0178 mol

10. At $450 K, K_{p} = 2.0 \times 1 0^{10}$ /bar for the given reaction at equilibrium.

$2S O_{2} (g) + O_{2} (g) ⇌ 2S O_{3} (g)$

What is $K_{c}$ at this temperature?

Ans. For the given reaction, Δn = 2 – 3 = – 1

T = 450 K

$R = 0.0831 bar L bar K^{- 1} mo l^{- 1} K_{P} = 2.0 \times 1 0^{10} ba r^{- 1} We know that, K_{P} = K_{C} (RT) Δ n \Rightarrow 2.0 \times 1 0^{10} ba r^{- 1} = K_{C} (0.083 1 L bar K^{- 1} mo l^{- 1} \times 450 K)^{- 1} \Rightarrow K_{c} = \frac{2.0 \times 1 0 ^{10} ba r ^{- 1}}{( 0.0831 L bar K ^{- 1} mo l ^{- 1} \times 450 K ) ^{- 1}} = (2.0 \times 1 0^{10} ba r^{- 1}) (0.083 1 L bar K^{- 1} mo l^{- 1} \times 450 K) = 74.79 \times 1 0^{10} L mo l^{- 1} = 7.48 \times 1 0^{11} L mo l^{- 1} = 7.48 \times 1 0^{11} M^{- 1}$

A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of $H I_{(} g)$ is 0.04 atm. What is Kp for the given equilibrium?

$2HI (g) ⇌ H_{2} (g) + I_{2} (g)$

Ans. The initial concentration of HI is 0.2 atm. At equilibrium, it has a partial pressure of 0.04 atm. Therefore, a decrease in the pressure of HI is 0.2 – 0.04 = 0.16. The given reaction is:

$Initial conc. At equilib. 2HI (g) 0.2 atm 0.04 atm ⇌ H_{2} (g) 0 \frac{0.16}{2} = 0.08 atm + I_{2} (g) 0 \frac{0.15}{2} = 0.08 atm$

$Therefore, = \frac{0.80}{1.4} = \frac{0.08 \times 0.08}{( 0.04 ) ^{2}} = \frac{0.0064}{0.0016} = 4.0$

Hence, the value of $K_{p}$ for the given equilibrium is 4.0.

A mixture of 1.57 mol of $N_{2}$ 1.92 mol of $H_{2}$ and 8.13 mol of $N H_{3}$ is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, $K_{c}$ for the reaction

$N 2 (g) + 3 H 2 (g) ⇌ 2 NH_{3} (g) is 1.7 \times 1 0^{2}$

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

Ans. The given reaction is:

$N 2 (g) + 3 H 2 (g) ⇌ 2 NH_{3} (g)$

The given concentration of various species is

$[N 2] = \frac{1.57}{20} molL^{- 1} [H 2] = \frac{1.92}{20} molL^{- 1} [NH_{3}] = \frac{8.13}{20} molL^{- 1}$

Now, reaction quotient $Q_{c}$ is:

$Q C = \frac{[ NH 3 ] ^{2}}{[ N 2 ] [ H 2 ] ^{3}} = \frac{( \frac{( 8.13 )}{20} )}{( \frac{1.57}{20} ) ( \frac{1.92}{20} ) ^{3}} = 2.4 \times 1 0^{3}$

Since QC ≠ KC, the reaction mixture is not at equilibrium.

Again, QC ≠ KC . Hence, the reaction will proceed in the reverse direction.

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

$K_{C} = \frac{[ N H _{3} ] ^{4} [ O _{2} ] ^{5}}{[ NO ] ^{4} [ H _{2} O _{2} ] ^{6}}$

Write the balanced chemical equation corresponding to this expression.

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

$4 NO (g) + 6 H 2 O (g) ⇌ 4 NH 3 (g) + 5 O_{2 (g)}$

One mole of $H_{2 O}$ 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,

$H 2 O (g) + CO (g) ⇌ H 2 (g) + CO_{2 (g)}$

Calculate the equilibrium constant for the reaction.

Ans. The given reaction is:

$Initial conc. At equilib. H_{2} O_{(g)} + \frac{1}{10} M \frac{1 - 0.4}{10} M = 0.06 M C O_{(g)} \frac{1}{10} M \frac{1 - 0.4}{10} M = 0.06 M ⇌ H_{2 (g)} + 0 \frac{0.4}{10} M = 0.04 M C O_{2 (g)} 0 \frac{0.4}{10} M = 0.04 M$

$Therefore, the equilibrium constant for the reaction, K_{C} = \frac{[ H _{2} ] [ C O _{2} ]}{[ H _{2} O ] [ CO ]} = \frac{0.04 \times 0.04}{0.06 \times 0.06} = 0.444 (approximately)$

At 700 K , equilibrium constant for the reaction

$H_{2 (g)} + I_{2 (g)} ⇌ 2 HI (g)$ is 54.8. If 0.5 $\sim m o l \sim L^{- 1} o f H I g$

is present at equilibrium at 700 K, what are the concentration of $H_{2 (\sim g)}$ and $I_{2 (\approx g)}$ assuming that we initially started with $H I_{g}$ and allowed it to reach equilibrium at 700 K?

Ans. It is given that equilibrium constant $K_{c}$ for the reaction $H_{2 (g)} + I_{2 (g)} ⇌ 2 HI_{(g)}$ is 54.8. Therefore, at equilibrium, the equilibrium constant $K_{c}$ for the reaction

$2 HI_{(g)} ⇌ H_{2 (g)} + I_{2 (g)}$ will be $\frac{1}{54.8}$ .

Let the concentrations of hydrogen and iodine at equilibrium be $x m o l L^{- 1}$ .

$[H_{2}] = [I_{2}] = x mol L^{- 1} Therefore, \frac{[ H _{2} ] [ I _{2} ]}{[ HI ] ^{2}} = K_{C}^{'} \Rightarrow \frac{xxx}{( 0.5 ) ^{2}} = \frac{1}{54.8} \Rightarrow x^{2} = \frac{0.25}{54.8} \Rightarrow x = 0.06754 x = 0.068 mol L^{- 1} (approximately) Hence, at equilibrium, [H_{2}] = [I_{2}] = 0.068 mol L^{- 1}$

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

$2 ICl (g) ⇌ I 2 (g) + Cl 2 (g); KC = 0.14$

Ans. The given reaction is:

$2 ICl (g) ⇌ I 2 (g) + Cl 2 (g)$

$Initial conc. equilibrium 0.78 M (0.78 - 2 x) 0 M \times M 0 M$

$Now, we can write, \frac{[ I _{2} ] [ C l _{2} ]}{[ ICl ] ^{2}} = K_{C} \Rightarrow \frac{x \times x}{( 0.78 - 2 x ) ^{2}} = 0.14 \Rightarrow \frac{x ^{2}}{( 0.78 - 2 x ) ^{2}} = 0.14 \Rightarrow \frac{x}{0.78 - 2 x} = 0.374 \Rightarrow x = 0.292 - 0.748 x \Rightarrow 1.748 x = 0.292 \Rightarrow x = 0.167 Hence, at equilibrium, [ICl] = [I_{2}] = 0.167 M [ICl] = (0.78 - 2 \times 0.167) M = 0.446 M$

Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of $C_{2} H_{6}$ when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?

$C 2 H 6 (g) ⇌ C 2 H 4 (g) + H_{2 (g)}$

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

Now, according to the reaction,

	C₂H₆₍g₎ ⇌ C₂H₄₍g₎ + H₂₍g₎
Initial conc.	0.4 atm 0 0
Equi.	4.0 – p p p

We can write,

$\Rightarrow \frac{p C 2 H 2 \times p H 2}{p C 2 H 6} = K_{p} \Rightarrow \frac{p \times p}{4.0 - p} = 0.04 \Rightarrow p^{2} + 0.16 - 0.04 p \Rightarrow p^{2} + 0.04 p - 0.16 = 0$

Hence, at equilibrium

$[ICl] = [I_{2}] = 0.167 M [ICl] = (0.78 - 2 \times 0.167) M = 0.446 M$

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

CH₃COOH(l) + C₂H₅OH CH₃ ⇌ COOC₂H₅(l) + H₂O(l)
Initial conc.	1/V M 0.18/V M 0 0
Equi.	(1 – 0.171)/V (0.18 – 0.171)/V 0.171/V M 0.171/V M

$= \frac{0.829}{V} M = \frac{0.009}{V} M$

(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?

(i) Reaction quotient,

$Q_{C} = \frac{[ C H _{3} COO C _{2} H _{5} ] [ H _{2} O ]}{[ C H _{3} COOH ] [ C _{2} H _{5} OH ]}$

(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:

CH₃COOH(l) + C₂H₅OH CH₃ ⇌ COOC₂H₅(l) + H₂O(l)
Initial conc.	1/V M 0.18/V M 0 0
Equi.	(1 – 0.171)/V (0.18 – 0.171)/V 0.171/V M 0.171/V M

Therefore, equilibrium constant for the given reaction is:

$Q_{C} = \frac{[ C H _{3} COO C _{2} H _{5} ] [ H _{2} O ]}{[ C H _{3} COOH ] [ C _{2} H _{5} OH ]} = \frac{\frac{0.171}{V} \times \frac{0.171}{V}}{\frac{0.829}{V} \times \frac{0.009}{V}} = 3.919 = (approximately)$

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

CH₃COOH(l) + C₂H₅OH CH₃ ⇌ COOC₂H₅(l) + H₂O(l)
Initial conc.	1.0/V M 0.5/V M 0 0
Equi.	(10 – 0.214)/V (0.5 – 0.214)/V 0.214/V M 0.214/V M

Therefore, the reaction quotient is,

$Q_{C} = \frac{[ C H _{3} COO C _{2} H _{5} ] [ H _{2} O ]}{[ C H _{3} COOH ] [ C _{2} H _{5} OH ]} = \frac{\frac{0.214}{V} \times \frac{0.214}{V}}{\frac{0.786}{V} \times \frac{0.286}{V}} = 0.2037 = 0204 (approximately) Since Q_{C} < K_{C}, equilibrium has not been reached.$

A sample of pure $PC l_{5}$ was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of $PC l_{5}$ 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 $PC l_{3}$ and $C l_{2}$ mat equilibrium?

$PCl 5 (g) ⇌ PCl 3 (g) + Cl_{2 (g)}$

Ans. Let the concentrations of both $PC l_{3}$ and $C l_{2}$ at equilibrium be x $m o l L^{-} 1.$

The given reaction is:

PCl₅₍g₎ ⇌ PCl₃₍g₎ + Cl₂₍g₎
Equi.	0.5 × 10⁻¹ mol⁻¹ x mol L⁻¹ x mol L⁻¹

It is given that the value of equilibrium constant, $K_{C}$ is 8.3 × 10–3.

Now we can write the expression for equilibrium as:

$\frac{[ PC l _{3} ] [ C l _{2} ]}{[ PC l _{5} ]} = K_{C}$

$\Rightarrow \frac{x \times x}{0.5 \times 1 0 ^{- 1}} = 8.3 \times 1 0^{- 3} \Rightarrow x^{2} = 4.15 \times 1 0^{- 4} \Rightarrow x = 2.04 \times 1 0^{- 2} = 0.0204 = 0.02 (approximately) Therefore, at equilibrium, [PC l_{3}] [C l_{2}] = 0.02 mol L^{- 1} .$

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 $C O_{2} .$

$FeO (s) + CO (g) ⇌ Fe (s) + CO 2 (g); K_{p} = 0.265 at 1050 K$

What are the equilibrium partial pressures of CO and $C O_{2}$ at 1050 K if the initial partial pressures are:

$p CO = 1.4 atm and p CO_{2} = 0.80 atm ?$

Ans. For the given reaction,

FeO₍g₎ + CO₍g₎ ⇌ Fe₍s₎ + CO₂₍g₎
Initially 1.4 atm 0.80 atm

$Q p = \frac{p CO 2}{p _{CO}} = \frac{0.80}{1.4} = 0.571$

It is given that Kp = 0.265 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 $C O_{2}$ will decrease.

Now, let the increase in pressure of CO = decrease in pressure of $C O_{2}$ be p.

$Then, we can write, K_{p} = \frac{p _{C O_{2}}}{p _{CO}} \Rightarrow 0.265 = \frac{0.80 - p}{1.4 + p} \Rightarrow 0.371 + 0.265 p = 0.80 - p \Rightarrow 1.265 p = 0.429 \Rightarrow p = 0.339 atm Therefore, equilibrium partial of C O_{2}, p_{C O_{2}} = 0.80 - 0.339 = 0.461 atm And, equilibrium partial pressure of CO, p_{CO} = 1.4 + 0.339 = 1.739 atm$

Equilibrium constant, $K_{c}$ for the reaction $N_{2} (g) + H_{2} (g) ⇌ 2N H_{3} (g)$ at 500 K is 0.061. At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol $L^{- 1} N_{2}, 2.0 mol L^{- 1} H_{2} and 0.5 mol L^{- 1} N H_{3} .$ Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?

Ans. The given reaction is:

N₂₍g₎ + 3H₂₍g₎ ⇌ 2NH₃₍g₎
At a particular time	3.0 mol L⁻¹ 2.0 mol L⁻¹ 0.5 mol L⁻¹

Now, we know that,

$Q_{C} = \frac{[ NH 3 ] ^{2}}{[ N 2 ] [ NH _{2} ] ^{3}} = \frac{( 0.5 ) ^{2}}{( 3.0 ) ( 2.0 ) ^{3}} = 0.0104$

It is given that $K_{C}$ = 0.061

Since $Q_{c} \neq = K_{c},$ the reaction is not at equilibrium.

Since $Q_{c} \neq = K_{c},$ the reaction will proceed in the forward direction to reach equilibrium.

Bromine monochloride, $B_{r} Cl$ decomposes into bromine and chlorine and reaches the equilibrium:

$2 BrCl (g) ⇌ Br_{2} (g) + Cl_{2 (g)}$

for which $K_{c}$ = 32 at 500 K. If initially pure $B r Cl$ is present at a concentration of $3.3 \times 1 0^{3}$ $m o l L^{-} 1$ what is its molar concentration in the mixture at equilibrium?

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

2BrCl₍g₎ ⇌ Br₂₍g₎ + Cl₂₍g₎
Initial conc.	3.3×10⁻³ 0 0
Equi.	3.3×10⁻³ – 2x x x

Now, we can write,

$\frac{[ B r _{2} ] [ C l _{2} ]}{[ B r C l ^{2} ]} = K_{c}$

$\Rightarrow \frac{x \times x}{( 3.3 \times 1 0 ^{- 3} - 2 x ) ^{2}} = 5.66 \Rightarrow x = 18.678 \times 1 0^{- 3} - 11.32 x \Rightarrow 12.32 x = 18.678 \times 1 0^{- 3} \Rightarrow x = 1.5 \times 1 0^{- 3} Therefore, at equilibrium, [B r Cl] = 3.3 \times 1 0^{- 3} - (2 \times 1.5 \times 1 0^{- 3}) = 3.3 \times 1 0^{- 3} - 3.0 \times 1 0^{- 3} = 0.3 \times 1 0^{- 3} = 3.0 \times 1 0^{- 4} m o l L^{- 1}$

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) ⇌ 2CO(g)

Calculate Kc for this reaction at the above temperature.

Ans. 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,

p₍CO₎ = n₍CO₎/(n₍CO₎ + n₍CO₂₎) × pₜₒₜₐₗ

= 3.234/(3.234 + 0.215) × 1 = 0.935 atm

Partial pressure of CO₂,

p₍CO₂₎ = n₍CO₂₎/(n₍CO₎ + n₍CO₂₎) × pₜₒₜₐₗ

= 0.215/(3.234 + 0.215) × 1

= 0.062 atm

Therefore,

Kₚ = (p₍CO₎)² / p₍CO₂₎

= (0.938)² / 0.062

= 14.19

For the given reaction,

Δn = 2 − 1 = 1

We know that,

$K_{p} = K ꜀ (RT)^{Δ n}$

⇒ 14.19 = K꜀(0.082 × 1127)¹

⇒ K꜀ = 14.19 / (0.082 × 1127)

= 0.154 (approximately)

#Please refer attached PDF for the next set of questions and solutions.

3.0NCERT Class 11 Biology – Chapter-Wise Solutions & Explanations

Find Class 11 Chemistry NCERT Solutions with chapter-wise answers to textbook questions. Revise important topics, practise questions, and prepare for your school exams.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Some Basic Concepts of Chemistry
Chapter 2	Structure of Atom
Chapter 3	Classification of Elements and Periodicity in Properties
Chapter 4	Chemical Bonding and Molecular Structure
Chapter 5	Chemical Thermodynamics
Chapter 7	Redox Reactions
Chapter 8	Organic Chemistry – Some Basic Principles and Techniques
Chapter 9	Hydrocarbons

4.0Key Features of NCERT Solutions for Class 11 Chemistry Chapter 6

Equilibrium Calculations: Learn how to solve equilibrium concentration problems using quadratic equations and suitable approximations.
Reaction Quotient and Equilibrium Constant: Understand the difference between $(Q_{c}) an d (K_{c})$ and how to use them to predict the direction of a reaction.
Buffer Solutions and pH: Study buffer solutions and learn how to calculate their pH using the Henderson–Hasselbalch equation.
Solubility Equilibrium: Practise questions based on the solubility product constant $(K_{s p})$ , common ion effect, and related equilibrium calculations.