NCERT Solutions for Class 11 Chemistry Chapter 6: Equilibrium – Free PDF Download

Equilibrium is one of the most conceptually important chapters in the Class 11 Chemistry syllabus, explaining why reversible physical and chemical processes settle into a stable, dynamic balance rather than running to completion. It builds the foundation for the law of chemical equilibrium, equilibrium constants K_c and K_p, Le Chatelier’s principle, and the vast topic of ionic equilibrium — acids, bases, pH, ionization constants, and the common-ion effect. Below you will find complete, independently worked NCERT Solutions for every exercise question of Class 11 Chemistry Chapter 6: Equilibrium, with full step-by-step derivations rather than just final answers.

NCERT Solutions for Class 11 Chemistry Chapter 6: Equilibrium

Q6.1: Effect of Sudden Volume Increase on Liquid–Vapour Equilibrium

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 the rates of evaporation and condensation change initially? (c) What happens when equilibrium is restored finally, and what will be the final vapour pressure?

(a) The moment the volume is increased, the same number of vapour molecules now occupy a larger space, so the vapour pressure initially decreases. (b) The rate of evaporation depends only on the temperature and the exposed liquid surface, both of which are unchanged, so it initially stays the same. The rate of condensation, however, depends on how often vapour molecules strike the liquid surface; since the vapour is now more spread out (lower density), collisions with the surface become less frequent, so the rate of condensation initially decreases. (c) Because the rate of evaporation now exceeds the rate of condensation, more liquid evaporates until the two rates become equal again and a new dynamic equilibrium is established. Since vapour pressure is a function of temperature alone (not volume) for a pure liquid, the system returns to exactly the same vapour pressure as before the volume change once equilibrium is restored.

Final answer: vapour pressure initially drops, condensation rate initially drops while evaporation rate is unchanged, and the final equilibrium vapour pressure equals the original vapour pressure because it depends only on temperature.

Q6.2: Calculating K_c for the SO₂–O₂–SO₃ Equilibrium

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) ↔ 2SO₃(g)

By the law of mass action, K_c = [SO₃]² / ([SO₂]²[O₂]). Substituting the given values: K_c = (1.90)² / [(0.60)² × 0.82] = 3.61 / 0.2952.

K_c ≈ 12.2 M⁻¹.

Q6.3: K_p for Iodine Vapour Dissociation

At a certain temperature and total pressure of 10⁵ Pa, iodine vapour contains 40% by volume of I atoms. I₂(g) ↔ 2I(g). Calculate K_p for the equilibrium.

Since I atoms make up 40% of the mixture by volume (equivalent to mole/pressure fraction), the partial pressure of I is 40% of 10⁵ Pa = 4 × 10⁴ Pa, and the partial pressure of I₂ is the remaining 60%: 6 × 10⁴ Pa. Then K_p = (p_I)² / p_I2 = (4 × 10⁴)² / (6 × 10⁴) = 16 × 10⁸ / 6 × 10⁴.

K_p ≈ 2.67 × 10⁴ Pa.

Q6.4: Writing K_c Expressions for Five Reactions

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(g) ↔ 2IF₅(g)

Pure solids and pure liquids are omitted from equilibrium-constant expressions because their “concentration” (density ÷ molar mass) is a fixed constant already absorbed into K itself. Applying the law of mass action to only the gaseous/aqueous species that can genuinely vary in concentration: (i) K_c = [NO]²[Cl₂] / [NOCl]². (ii) Since Cu(NO₃)₂ and CuO are both solids, K_c = [NO₂]⁴[O₂]. (iii) Here water is a genuine reactant (not present in large excess as a solvent), so it is included: K_c = [CH₃COOH][C₂H₅OH] / ([CH₃COOC₂H₅][H₂O]). (iv) Fe(OH)₃ is a solid and is omitted, and OH⁻ carries a stoichiometric coefficient of 3, so K_c = 1 / ([Fe³⁺][OH⁻]³). (v) I₂ is a solid and is omitted: K_c = [IF₅]² / [F₂]⁵.

The five K_c expressions are as derived above, with all pure solids correctly excluded.

Q6.5: Converting K_p to K_c for Two Equilibria

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); K_p = 1.8 × 10⁻² at 500 K (ii) CaCO₃(s) ↔ CaO(s) + CO₂(g); K_p = 167 at 1073 K

Using K_p = K_c(RT)^Δn, so K_c = K_p / (RT)^Δn, with R = 0.0831 L bar K⁻¹ mol⁻¹. For (i), Δn = 3 − 2 = 1, so K_c = 1.8 × 10⁻² / (0.0831 × 500) = 1.8 × 10⁻² / 41.55. For (ii), Δn = 1 − 0 = 1, so K_c = 167 / (0.0831 × 1073) = 167 / 89.17.

(i) K_c ≈ 4.33 × 10⁻⁴ mol L⁻¹; (ii) K_c ≈ 1.87 mol L⁻¹.

Q6.6: K_c for the Reverse Reaction

For the equilibrium NO(g) + O₃(g) ↔ NO₂(g) + O₂(g), K_c = 6.3 × 10¹⁴ at 1000 K. Both the forward and reverse reactions are elementary bimolecular reactions. What is K_c for the reverse reaction?

The equilibrium constant of the reverse reaction is simply the reciprocal of the forward one: K_c′ = 1 / K_c = 1 / (6.3 × 10¹⁴).

K_c′ ≈ 1.59 × 10⁻¹⁵.

Q6.7: Why Pure Liquids and Solids Are Omitted from K

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

The “concentration” of a pure solid or pure liquid, defined as its density divided by its molar mass, is a fixed physical property at a given temperature — it does not change no matter how much of the solid or liquid is present, because adding more of it only increases the amount of substance, not its concentration per unit volume. Since this quantity is already a constant, it is mathematically absorbed into the equilibrium constant itself rather than written out separately.

Pure solids and liquids have fixed, unchanging molar concentrations, so they are merged into the value of K rather than appearing explicitly in its expression.

Q6.8: Equilibrium Composition for N₂–O₂–N₂O with a Tiny K_c

2N₂(g) + O₂(g) ↔ 2N₂O(g). If a mixture of 0.482 mol N₂ and 0.933 mol O₂ is placed in a 10 L vessel and allowed to reach equilibrium at a temperature where K_c = 2.0 × 10⁻³⁷, determine the composition of the equilibrium mixture.

Initial concentrations are [N₂] = 0.0482 M and [O₂] = 0.0933 M. Because K_c is astronomically small, only a minuscule amount of N₂O forms, so [N₂] and [O₂] remain essentially unchanged at equilibrium. Let [N₂O] = x. Then K_c = x² / ([N₂]²[O₂]), so x² = 2.0 × 10⁻³⁷ × (0.0482)² × 0.0933 = 2.0 × 10⁻³⁷ × 2.324 × 10⁻³ × 0.0933 ≈ 4.33 × 10⁻⁴¹. Taking the square root, x = [N₂O] ≈ 6.6 × 10⁻²¹ M.

At equilibrium, [N₂] ≈ 0.0482 M, [O₂] ≈ 0.0933 M, and [N₂O] ≈ 6.6 × 10⁻²¹ M — confirming the reaction barely proceeds forward.

Q6.9: Equilibrium Amounts of NO and Br₂ in the NOBr Reaction

2NO(g) + Br₂(g) ↔ 2NOBr(g). When 0.087 mol NO and 0.0437 mol Br₂ are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate the equilibrium amounts of NO and Br₂.

Since 2 mol NOBr forms from 2 mol NO, the moles of NO consumed equal the moles of NOBr formed: 0.0518 mol. Since 2 mol NOBr forms from 1 mol Br₂, the Br₂ consumed is 0.0518 / 2 = 0.0259 mol. Subtracting from the initial amounts: NO remaining = 0.087 − 0.0518 = 0.0352 mol; Br₂ remaining = 0.0437 − 0.0259 = 0.0178 mol.

At equilibrium, NO = 0.0352 mol and Br₂ = 0.0178 mol.

Q6.10: K_c from K_p for the SO₂ Oxidation Equilibrium

2SO₂(g) + O₂(g) ↔ 2SO₃(g). At 450 K, K_p = 2.0 × 10¹⁰ bar⁻¹. What is K_c at this temperature?

Here Δn = 2 − 3 = −1, so K_p = K_c/(RT), which rearranges to K_c = K_p × RT. Substituting: K_c = 2.0 × 10¹⁰ × 0.0831 × 450 = 2.0 × 10¹⁰ × 37.4.

K_c ≈ 7.48 × 10¹¹ mol⁻¹ L. (Note: because Δn is negative, K_c must be obtained by multiplying K_p by RT, not dividing — a common source of error.)

Q6.11: K_p for the Dissociation of HI

A sample of HI(g) is placed in a flask at a pressure of 0.2 atm. At equilibrium, the partial pressure of HI(g) is 0.04 atm. What is K_p for 2HI(g) ↔ H₂(g) + I₂(g)?

The drop in HI pressure is 0.2 − 0.04 = 0.16 atm. Since 2 mol HI produces 1 mol each of H₂ and I₂, the partial pressures of H₂ and I₂ at equilibrium are each half of that drop: 0.16 / 2 = 0.08 atm. Then K_p = (p_H2 × p_I2) / (p_HI)² = (0.08 × 0.08) / (0.04)² = 0.0064 / 0.0016.

K_p = 4.

Q6.12: Checking Whether an NH₃ Mixture Is at Equilibrium

A mixture of 1.57 mol N₂, 1.92 mol H₂, and 8.13 mol NH₃ is placed in a 20 L vessel at 500 K, where K_c for N₂(g) + 3H₂(g) ↔ 2NH₃(g) is 1.7 × 10². Is the mixture at equilibrium? If not, in which direction does the reaction proceed?

Concentrations: [N₂] = 1.57/20 = 0.0785 M, [H₂] = 1.92/20 = 0.096 M, [NH₃] = 8.13/20 = 0.4065 M. The reaction quotient is Q_c = [NH₃]² / ([N₂][H₂]³) = (0.4065)² / (0.0785 × (0.096)³) = 0.1652 / (0.0785 × 8.847 × 10⁻⁴) = 0.1652 / 6.945 × 10⁻⁵.

Q_c ≈ 2.4 × 10³, which is far greater than K_c = 1.7 × 10². Since Q_c > K_c, the mixture is not at equilibrium and the reaction will proceed in the reverse (backward) direction, converting NH₃ back into N₂ and H₂.

Q6.13: Deriving the Balanced Equation from a K_c Expression

The equilibrium constant expression for a gas reaction is K_c = [NH₃]⁴[O₂]⁵ / ([NO]⁴[H₂O]⁶). Write the balanced chemical equation corresponding to this expression.

Since NH₃ and O₂ appear in the numerator, they are the products, each with the stoichiometric coefficient shown as its exponent (4 and 5); NO and H₂O, in the denominator, are the reactants with coefficients 4 and 6. Checking atom balance: N: 4 (from NO) = 4 (in NH₃) ✓; H: 6 × 2 = 12 (from H₂O) = 4 × 3 = 12 (in NH₃) ✓; O: 4 + 6 = 10 (from NO + H₂O) = 5 × 2 = 10 (in O₂) ✓.

4NO(g) + 6H₂O(g) ↔ 4NH₃(g) + 5O₂(g).

Q6.14: K_c for the Water-Gas Shift Reaction

One mole of H₂O and one mole of CO are taken in a 10 L vessel and heated to 725 K. At equilibrium, 40% of the water (by mass) reacts with CO according to H₂O(g) + CO(g) ↔ H₂(g) + CO₂(g). Calculate the equilibrium constant for the reaction.

Initial concentrations: [H₂O] = [CO] = 1/10 = 0.1 M. Since 40% reacts, the amount converted is 0.4 × 0.1 = 0.04 M. At equilibrium: [H₂O] = [CO] = 0.1 − 0.04 = 0.06 M, and [H₂] = [CO₂] = 0.04 M. Then K_c = ([H₂][CO₂]) / ([H₂O][CO]) = (0.04 × 0.04) / (0.06 × 0.06) = 0.0016 / 0.0036.

K_c ≈ 0.44.

Q6.15: Finding [H₂] and [I₂] at Equilibrium from HI Decomposition

At 700 K, K_c for H₂(g) + I₂(g) ↔ 2HI(g) is 54.8. If 0.5 mol L⁻¹ of HI is present at equilibrium, starting purely from HI, what are the equilibrium concentrations of H₂ and I₂?

Since the system was built up starting only from HI, equal amounts of H₂ and I₂ must have formed; let each equal x. For the decomposition direction, K_c′ = 1/K_c = 1/54.8, and K_c′ = ([H₂][I₂]) / [HI]² = x² / (0.5)². So x² = 0.25/54.8 = 4.562 × 10⁻³, giving x = √(4.562 × 10⁻³).

[H₂] = [I₂] ≈ 0.0676 mol L⁻¹.

Q6.16: Equilibrium Concentrations for the ICl Decomposition

2ICl(g) ↔ I₂(g) + Cl₂(g); K_c = 0.14. What is the equilibrium concentration of each substance when the initial concentration of ICl was 0.78 M?

Let x = [I₂] = [Cl₂] formed at equilibrium; since 2 mol ICl is consumed per mole of I₂ (or Cl₂) formed, [ICl] at equilibrium = 0.78 − 2x. Then K_c = x² / (0.78 − 2x)² = 0.14, so √0.14 = 0.374 = x / (0.78 − 2x). Solving: x = 0.374(0.78 − 2x) = 0.2917 − 0.748x, giving 1.748x = 0.2917, so x ≈ 0.167.

[I₂] = [Cl₂] ≈ 0.167 M, and [ICl] = 0.78 − 2(0.167) ≈ 0.446 M.

Q6.17: Equilibrium Pressure of C₂H₆ from Ethane Cracking

K_p = 0.04 atm at 899 K for C₂H₆(g) ↔ C₂H₄(g) + H₂(g). What is the equilibrium pressure of C₂H₆ when it is placed in a flask at 4.0 atm and allowed to reach equilibrium?

Let p = the pressure of C₂H₄ (= pressure of H₂) formed at equilibrium. Then K_p = p² / (4 − p) = 0.04, giving p² + 0.04p − 0.16 = 0. Using the quadratic formula, p = [−0.04 ± √(0.0016 + 0.64)] / 2 = [−0.04 ± 0.801] / 2, taking the positive root, p ≈ 0.38 atm.

p(C₂H₆) at equilibrium = 4 − 0.38 = 3.62 atm.

Q6.18: Reaction Quotient and Equilibrium Constant for Ethyl Acetate Formation

Ethyl acetate is formed by the reaction of ethanol and acetic acid: CH₃COOH(l) + C₂H₅OH(l) ↔ CH₃COOC₂H₅(l) + H₂O(l), where water is not present in excess. (i) Write Q_c for this reaction. (ii) At 293 K, starting with 1.00 mol acetic acid and 0.18 mol ethanol, 0.171 mol ethyl acetate is present at equilibrium; calculate K_c. (iii) Starting instead with 0.5 mol ethanol and 1.0 mol acetic acid at 293 K, 0.214 mol ethyl acetate is found after some time. Has equilibrium been reached?

(i) Since water is treated as a genuine participant here, Q_c = ([CH₃COOC₂H₅][H₂O]) / ([CH₃COOH][C₂H₅OH]). (ii) Because the numbers of moles are the same on both sides of the reaction, the common volume V cancels out of the K_c expression. At equilibrium: acetic acid = 1 − 0.171 = 0.829 mol, ethanol = 0.18 − 0.171 = 0.009 mol, ester = water = 0.171 mol. So K_c = (0.171 × 0.171) / (0.829 × 0.009) = 0.02924 / 0.007461. (iii) At this snapshot: acetic acid = 1 − 0.214 = 0.786 mol, ethanol = 0.5 − 0.214 = 0.286 mol, ester = water = 0.214 mol. So Q_c = (0.214 × 0.214) / (0.786 × 0.286) = 0.04580 / 0.22480.

(ii) K_c ≈ 3.92. (iii) Q_c ≈ 0.204, which is much less than K_c = 3.92, so equilibrium has NOT yet been reached — the reaction will continue to move forward.

Q6.19: Equilibrium Concentrations of PCl₃ and Cl₂ from PCl₅ Decomposition

PCl₅(g) ↔ PCl₃(g) + Cl₂(g). A sample of pure PCl₅ was placed in an evacuated vessel at 473 K; at equilibrium, [PCl₅] = 0.05 mol L⁻¹ and K_c = 8.3 × 10⁻³. What are the equilibrium concentrations of PCl₃ and Cl₂?

Let x = [PCl₃] = [Cl₂] at equilibrium. Then K_c = x² / [PCl₅] = x² / 0.05 = 8.3 × 10⁻³, so x² = 4.15 × 10⁻⁴, giving x = √(4.15 × 10⁻⁴).

[PCl₃] = [Cl₂] ≈ 0.0204 mol L⁻¹.

Q6.20: Equilibrium Partial Pressures in Iron Ore Reduction

FeO(s) + CO(g) ↔ Fe(s) + CO₂(g); K_p = 0.265 at 1050 K. What are the equilibrium partial pressures of CO and CO₂ if the initial partial pressures are p_CO = 1.4 atm and p_CO2 = 0.80 atm?

First check the direction: Q_p = p_CO2/p_CO = 0.80/1.4 = 0.571, which is greater than K_p = 0.265, so the reaction shifts backward, converting some CO₂ back into CO. Let p = the decrease in p_CO2 (equal to the increase in p_CO). Then K_p = (0.80 − p) / (1.4 + p) = 0.265, so 0.80 − p = 0.371 + 0.265p, giving 0.429 = 1.265p, so p ≈ 0.339 atm.

At equilibrium, p_CO2 = 0.80 − 0.339 ≈ 0.461 atm and p_CO = 1.4 + 0.339 ≈ 1.739 atm.

Q6.21: Direction of Approach to Equilibrium for Ammonia Synthesis

K_c for N₂(g) + 3H₂(g) ↔ 2NH₃(g) at 500 K is 0.061. A reaction mixture contains 3.0 mol L⁻¹ N₂, 2.0 mol L⁻¹ H₂, and 0.5 mol L⁻¹ NH₃. Is the mixture at equilibrium? If not, in which direction does it proceed?

Q_c = [NH₃]² / ([N₂][H₂]³) = (0.5)² / (3.0 × (2.0)³) = 0.25 / 24.

Q_c ≈ 0.0104, which is less than K_c = 0.061. Since Q_c < K_c, the mixture is not at equilibrium, and the reaction proceeds in the forward direction to form more NH₃.

Q6.22: Equilibrium Concentration of BrCl from Decomposition

2BrCl(g) ↔ Br₂(g) + Cl₂(g); K_c = 32 at 500 K. If pure BrCl is initially present at 3.3 × 10⁻³ mol L⁻¹, what is its molar concentration at equilibrium?

Let x = [Br₂] = [Cl₂] formed. Then [BrCl] at equilibrium = 3.3 × 10⁻³ − 2x, and K_c = x² / (3.3 × 10⁻³ − 2x)² = 32, so √32 = 5.657 = x / (3.3 × 10⁻³ − 2x). Solving: x = 5.657(3.3 × 10⁻³ − 2x) = 0.01867 − 11.314x, so 12.314x = 0.01867, giving x ≈ 1.516 × 10⁻³ M.

[BrCl] at equilibrium = 3.3 × 10⁻³ − 2(1.516 × 10⁻³) ≈ 2.7 × 10⁻⁴ mol L⁻¹.

Q6.23: K_c for the Boudouard Equilibrium from Percentage Composition

At 1127 K and 1 atm, a gaseous mixture of CO and CO₂ in equilibrium with solid carbon contains 90.55% CO by mass: C(s) + CO₂(g) ↔ 2CO(g). Calculate K_c at this temperature.

Assume 100 g of gas mixture: CO = 90.55 g, CO₂ = 9.45 g. Moles: CO = 90.55/28 = 3.234 mol, CO₂ = 9.45/44 = 0.215 mol; total = 3.449 mol. Mole (= pressure) fractions give p_CO = (3.234/3.449) × 1 ≈ 0.938 atm and p_CO2 = (0.215/3.449) × 1 ≈ 0.062 atm. Then K_p = (p_CO)² / p_CO2 = (0.938)² / 0.062 ≈ 14.2. Converting with Δn = 2 − 1 = 1: K_c = K_p / (RT) = 14.2 / (0.0831 × 1127) = 14.2 / 93.65.

K_c ≈ 0.15 mol L⁻¹.

Q6.24: ΔG° and K for the Formation of NO₂ from NO

Calculate (a) Δ_rG° and (b) the equilibrium constant for NO(g) + ½O₂(g) ↔ NO₂(g) at 298 K, given Δ_fG°[NO₂] = 52.0 kJ/mol, Δ_fG°[NO] = 87.0 kJ/mol, Δ_fG°[O₂] = 0 kJ/mol.

(a) Δ_rG° = Δ_fG°(products) − Δ_fG°(reactants) = 52.0 − (87.0 + 0) = −35.0 kJ/mol. (b) Using ΔG° = −RT ln K: ln K = 35000 / (8.314 × 298) = 35000 / 2477.6 ≈ 14.13, so K = e^14.13.

(a) Δ_rG° = −35.0 kJ/mol. (b) K ≈ 1.37 × 10⁶, a large value consistent with the strongly negative ΔG°.

Q6.25: Effect of Decreasing Pressure on Three Equilibria

Does the number of moles of reaction products increase, decrease, or remain the same when each of the following equilibria is subjected to a decrease in pressure (by increasing volume)? (a) PCl₅(g) ↔ PCl₃(g) + Cl₂(g) (b) CaO(s) + CO₂(g) ↔ CaCO₃(s) (c) 3Fe(s) + 4H₂O(g) ↔ Fe₃O₄(s) + 4H₂(g)

By Le Chatelier’s principle, decreasing pressure shifts an equilibrium toward the side with the greater number of gas moles. (a) Reactant side has 1 gas mole, product side has 2, so the equilibrium shifts forward. (b) Reactant side has 1 gas mole (CO₂), product side has 0 (CaCO₃ is solid), so it shifts backward, away from the product. (c) Both sides have 4 gas moles (4H₂O vs 4H₂), so pressure has no effect.

(a) Products increase. (b) Products (CaCO₃) decrease. (c) Products remain the same.

Q6.26: Which Reactions Are Affected by Increasing Pressure

Which of the following reactions will be affected by increasing the pressure, and in which direction? (i) COCl₂(g) ↔ CO(g) + Cl₂(g) (ii) CH₄(g) + 2S₂(g) ↔ CS₂(g) + 2H₂S(g) (iii) CO₂(g) + C(s) ↔ 2CO(g) (iv) 2H₂(g) + CO(g) ↔ CH₃OH(g) (v) CaCO₃(s) ↔ CaO(s) + CO₂(g) (vi) 4NH₃(g) + 5O₂(g) ↔ 4NO(g) + 6H₂O(g)

Increasing pressure shifts equilibrium toward the side with fewer gas moles. Counting gas moles on each side: (i) 1 → 2, shifts backward. (ii) 3 → 3, equal, unaffected. (iii) 1 → 2, shifts backward. (iv) 3 → 1, shifts forward. (v) 0 → 1, shifts backward. (vi) 9 → 10, shifts backward.

Reactions (i), (iii), (v), and (vi) shift backward; reaction (iv) shifts forward; reaction (ii) is unaffected because gas moles are equal on both sides.

Q6.27: Equilibrium Pressures from Pure HBr Introduced into a Vessel

The equilibrium constant for H₂(g) + Br₂(g) ↔ 2HBr(g) is 1.6 × 10⁵ at 1024 K. Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024 K.

Working with the reverse (decomposition) direction, 2HBr(g) ↔ H₂(g) + Br₂(g), K_p′ = 1/(1.6 × 10⁵) = 6.25 × 10⁻⁶. Let p = pressure of H₂ = pressure of Br₂ formed. Then K_p′ = p² / (10 − 2p)², so √(6.25 × 10⁻⁶) = 2.5 × 10⁻³ = p / (10 − 2p). Solving: p = 2.5 × 10⁻³(10 − 2p) = 0.025 − 0.005p, so 1.005p = 0.025, giving p ≈ 0.0249 bar.

p(H₂) = p(Br₂) ≈ 0.025 bar, and p(HBr) = 10 − 2(0.025) ≈ 9.95 bar — because K_p strongly favours HBr formation, only a tiny fraction of it decomposes back to H₂ and Br₂.

Q6.28: K_p Expression and Le Chatelier Effects for Steam-Methane Reforming

Dihydrogen is obtained from natural gas by the endothermic reaction CH₄(g) + H₂O(g) ↔ CO(g) + 3H₂(g). (a) Write an expression for K_p. (b) How will K_p and the composition of the equilibrium mixture be affected by (i) increasing pressure, (ii) increasing temperature, and (iii) using a catalyst?

(a) K_p = (p_CO × p_H2³) / (p_CH4 × p_H2O). (b)(i) The gas moles go from 2 (reactants) to 4 (products), so increasing pressure shifts the composition backward toward CH₄ and H₂O; however, because K_p depends only on temperature, its numerical value is unchanged by pressure. (ii) Since the reaction is endothermic, raising the temperature shifts equilibrium forward and genuinely increases the value of K_p. (iii) A catalyst speeds up the attainment of equilibrium equally in both directions but changes neither K_p nor the equilibrium composition.

K_p = p_COp_H2³/(p_CH4p_H2O); pressure shifts composition backward but leaves K_p unchanged, higher temperature increases K_p and shifts forward, and a catalyst affects neither K_p nor composition.

Q6.29: Le Chatelier Effects on Methanol Synthesis

Describe the effect of (a) addition of H₂, (b) addition of CH₃OH, (c) removal of CO, and (d) removal of CH₃OH on the equilibrium 2H₂(g) + CO(g) ↔ CH₃OH(g).

Le Chatelier’s principle predicts that a system disturbed from equilibrium shifts to partially counteract the disturbance. (a) Adding H₂ (a reactant) shifts the equilibrium forward, consuming some of the added H₂. (b) Adding CH₃OH (the product) shifts the equilibrium backward. (c) Removing CO (a reactant) shifts the equilibrium backward, to partially replace the lost CO. (d) Removing CH₃OH (the product) shifts the equilibrium forward, to partially replace it.

(a) Forward shift. (b) Backward shift. (c) Backward shift. (d) Forward shift.

Q6.30: K_c, Reverse K_c, and Temperature Effects for PCl₅ Decomposition

At 473 K, K_c for PCl₅(g) ↔ PCl₃(g) + Cl₂(g), Δ_rH° = 124.0 kJ mol⁻¹, is 8.3 × 10⁻³. (a) Write the K_c expression. (b) What is K_c for the reverse reaction? (c) What is the effect on K_c of (i) adding more PCl₅, (ii) increasing pressure, (iii) increasing temperature?

(a) K_c = ([PCl₃][Cl₂]) / [PCl₅]. (b) The reverse equilibrium constant is the reciprocal: K_c′ = 1 / (8.3 × 10⁻³) ≈ 120.5. (c) Since K depends only on temperature: (i) adding more PCl₅ shifts the equilibrium position forward but leaves K_c unchanged. (ii) Increasing pressure shifts the composition backward (toward fewer gas moles) but again leaves K_c unchanged. (iii) Because the reaction is endothermic (ΔH° positive), raising the temperature genuinely increases K_c.

(a) K_c = [PCl₃][Cl₂]/[PCl₅]. (b) K_c′ ≈ 1.2 × 10². (c) K_c is unaffected by adding PCl₅ or by pressure, but increases with rising temperature.

Q6.31: Partial Pressure of H₂ in the Water-Gas Shift Reaction

CO(g) + H₂O(g) ↔ CO₂(g) + H₂(g). If a vessel at 400°C is charged with an equimolar mixture such that p_CO = p_H2O = 4.0 bar, what is the partial pressure of H₂ at equilibrium given K_p = 10.1?

Let p = the pressure of CO₂ (= pressure of H₂) formed at equilibrium. Then K_p = p² / (4 − p)² = 10.1, so √10.1 = 3.178 = p / (4 − p). Solving: p = 3.178(4 − p) = 12.712 − 3.178p, giving 4.178p = 12.712, so p ≈ 3.04 bar.

The equilibrium partial pressure of H₂ is approximately 3.04 bar.

Q6.32: Identifying the Reaction with Appreciable Reactant and Product Concentrations

Predict which of the following reactions will have appreciable concentrations of both reactants and products at equilibrium: (a) Cl₂(g) ↔ 2Cl(g); K_c = 5 × 10⁻³⁹ (b) Cl₂(g) + 2NO(g) ↔ 2NOCl(g); K_c = 3.7 × 10⁸ (c) Cl₂(g) + 2NO₂(g) ↔ 2NO₂Cl(g); K_c = 1.8

As a rule of thumb, an equilibrium contains appreciable amounts of both reactants and products only when K_c is roughly between 10⁻³ and 10³. Reaction (a) has a vanishingly small K_c, so it barely proceeds beyond reactants. Reaction (b) has an enormous K_c, so it goes essentially to completion. Only reaction (c), with K_c = 1.8, falls within the moderate range.

Only reaction (c), Cl₂ + 2NO₂ ↔ 2NO₂Cl, has appreciable concentrations of both reactants and products at equilibrium.

Q6.33: Equilibrium Ozone Concentration in Air

K_c for 3O₂(g) ↔ 2O₃(g) is 2.0 × 10⁻⁵⁰ at 25°C. If [O₂] in air is 1.6 × 10⁻² M, what is the equilibrium concentration of O₃?

K_c = [O₃]² / [O₂]³, so [O₃]² = K_c × [O₂]³ = 2.0 × 10⁻⁵⁰ × (1.6 × 10⁻²)³ = 2.0 × 10⁻⁵⁰ × 4.096 × 10⁻⁶ = 8.19 × 10⁻⁵⁶. Taking the square root gives [O₃].

[O₃] ≈ 2.86 × 10⁻²⁸ M, showing that even though ozone is vital in the upper atmosphere, its equilibrium concentration relative to O₂ is extraordinarily small.

Q6.34: Equilibrium Methane Concentration in the Sabatier-Type Reaction

CO(g) + 3H₂(g) ↔ CH₄(g) + H₂O(g) is at equilibrium at 1300 K in a 1 L flask containing 0.30 mol CO, 0.10 mol H₂, 0.02 mol H₂O, and an unknown amount of CH₄. K_c = 3.90. Determine [CH₄].

Let x = [CH₄]. Then K_c = ([CH₄][H₂O]) / ([CO][H₂]³), so 3.90 = (x × 0.02) / (0.30 × (0.10)³) = 0.02x / (0.30 × 0.001) = 0.02x / 3 × 10⁻⁴. Solving, x = 3.90 × 3 × 10⁻⁴ / 0.02.

[CH₄] ≈ 5.85 × 10⁻² mol L⁻¹.

Q6.35: Identifying Conjugate Acid–Base Pairs

What is meant by a conjugate acid–base pair? Find the conjugate acid or base for the following species: HNO₂, CN⁻, HClO₄, F⁻, OH⁻, CO₃²⁻, and S²⁻.

A conjugate acid–base pair consists of two species that differ from each other by exactly one proton (H⁺) — for example, HCl and Cl⁻, or H₃O⁺ and H₂O. Applying this: HNO₂ (an acid) loses a proton to give its conjugate base NO₂⁻. CN⁻ (a base) gains a proton to give its conjugate acid HCN. HClO₄ loses a proton to give ClO₄⁻. F⁻ gains a proton to give HF. OH⁻, acting as a base, gains a proton to give H₂O. CO₃²⁻ gains a proton to give HCO₃⁻. S²⁻ gains a proton to give HS⁻.

Conjugate base of HNO₂ is NO₂⁻; conjugate acid of CN⁻ is HCN; conjugate base of HClO₄ is ClO₄⁻; conjugate acid of F⁻ is HF; conjugate acid of OH⁻ is H₂O; conjugate acid of CO₃²⁻ is HCO₃⁻; conjugate acid of S²⁻ is HS⁻.

Q6.36: Identifying Lewis Acids

Which of the following are Lewis acids? H₂O, BF₃, H⁺, and NH₄⁺.

A Lewis acid is a species that can accept an electron pair. BF₃ has an incomplete octet on boron (only six electrons), giving it a vacant orbital that readily accepts an electron pair, making it a classic Lewis acid. H⁺ has a completely empty 1s orbital and readily accepts a lone pair, making it a Lewis acid as well. H₂O, by contrast, has lone pairs to donate and acts as a Lewis base. NH₄⁺, although positively charged, has nitrogen with a complete octet and no accessible vacant orbital, so it does not behave as a simple Lewis acid in the same direct sense.

BF₃ and H⁺ are Lewis acids (electron-pair acceptors); H₂O is a Lewis base.

Q6.37: Conjugate Bases of Three Brønsted Acids

What are the conjugate bases for the Brønsted acids HF, H₂SO₄, and HCO₃⁻?

Removing one proton from each acid gives its conjugate base: HF loses H⁺ to give F⁻; H₂SO₄ loses H⁺ to give HSO₄⁻; HCO₃⁻ loses H⁺ to give CO₃²⁻.

Conjugate bases: F⁻ (from HF), HSO₄⁻ (from H₂SO₄), and CO₃²⁻ (from HCO₃⁻).

Q6.38: Conjugate Acids of Three Brønsted Bases

Write the conjugate acids for the Brønsted bases NH₂⁻, NH₃, and HCOO⁻.

Adding one proton to each base gives its conjugate acid: NH₂⁻ gains H⁺ to give NH₃; NH₃ gains H⁺ to give NH₄⁺; HCOO⁻ gains H⁺ to give HCOOH.

Conjugate acids: NH₃ (from NH₂⁻), NH₄⁺ (from NH₃), and HCOOH (from HCOO⁻).

Q6.39: Conjugate Acids and Bases of Four Amphoteric Species

The species H₂O, HCO₃⁻, HSO₄⁻, and NH₃ can act as both Brønsted acids and bases. Give the corresponding conjugate acid and conjugate base for each.

Each species can either donate a proton (acting as an acid, forming its conjugate base) or accept one (acting as a base, forming its conjugate acid). For H₂O: conjugate acid H₃O⁺, conjugate base OH⁻. For HCO₃⁻: conjugate acid H₂CO₃, conjugate base CO₃²⁻. For HSO₄⁻: conjugate acid H₂SO₄, conjugate base SO₄²⁻. For NH₃: conjugate acid NH₄⁺, conjugate base NH₂⁻.

H₂O → H₃O⁺/OH⁻; HCO₃⁻ → H₂CO₃/CO₃²⁻; HSO₄⁻ → H₂SO₄/SO₄²⁻; NH₃ → NH₄⁺/NH₂⁻.

Q6.40: Classifying Four Species as Lewis Acids or Bases

Classify the following species into Lewis acids and Lewis bases, and show how each acts as such: (a) OH⁻ (b) F⁻ (c) H⁺ (d) BCl₃.

A Lewis base donates an electron pair; a Lewis acid accepts one. (a) OH⁻ has lone pairs on oxygen available to donate, so it is a Lewis base. (b) F⁻ similarly has lone pairs to donate, so it is a Lewis base. (c) H⁺ has an empty 1s orbital that accepts an electron pair, so it is a Lewis acid. (d) BCl₃ has boron with only six electrons around it (an incomplete octet), giving it a vacant orbital that accepts an electron pair, so it is a Lewis acid.

Lewis bases: OH⁻ and F⁻ (electron-pair donors). Lewis acids: H⁺ and BCl₃ (electron-pair acceptors).

Q6.41: pH from Hydrogen Ion Concentration in Soft Drink

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

pH = −log[H⁺] = −log(3.8 × 10⁻³) = 3 − log(3.8) = 3 − 0.5798.

pH ≈ 2.42, confirming the soft drink is fairly acidic.

Q6.42: Hydrogen Ion Concentration in Vinegar from pH

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

Since pH = −log[H⁺], [H⁺] = 10^−pH = 10^−3.76 = 10^0.24 × 10⁻⁴.

[H⁺] ≈ 1.74 × 10⁻⁴ mol L⁻¹.

Q6.43: K_b of the Conjugate Bases of Three Weak Acids

The ionization constants of HF, HCOOH, and HCN at 298 K are 6.8 × 10⁻⁴, 1.8 × 10⁻⁴, and 4.8 × 10⁻⁹ respectively. Calculate the ionization constants of their conjugate bases.

Using K_b = K_w/K_a with K_w = 1.0 × 10⁻¹⁴: K_b(F⁻) = 10⁻¹⁴/(6.8 × 10⁻⁴); K_b(HCOO⁻) = 10⁻¹⁴/(1.8 × 10⁻⁴); K_b(CN⁻) = 10⁻¹⁴/(4.8 × 10⁻⁹).

K_b(F⁻) ≈ 1.47 × 10⁻¹¹; K_b(HCOO⁻) ≈ 5.56 × 10⁻¹¹; K_b(CN⁻) ≈ 2.08 × 10⁻⁶ — the weakest acid (HCN) has, as expected, the strongest conjugate base.

Q6.44: Phenolate Ion Concentration and the Common-Ion Effect

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

For pure phenol, C₆H₅OH + H₂O ↔ C₆H₅O⁻ + H₃O⁺. Let x = [C₆H₅O⁻]. Since ionization is small, K_a = x²/0.05 = 1.0 × 10⁻¹⁰, so x² = 5.0 × 10⁻¹², giving x ≈ 2.24 × 10⁻⁶ M. When 0.01 M sodium phenolate (a strong electrolyte) is also present, it supplies an initial 0.01 M of C₆H₅O⁻, suppressing further ionization of phenol by the common-ion effect. Let y = additional ionization. Then K_a = (0.01 + y)(y)/0.05 ≈ (0.01)(y)/0.05 = 1.0 × 10⁻¹⁰, so y = 1.0 × 10⁻¹⁰ × 0.05/0.01 = 5.0 × 10⁻¹⁰. The degree of ionization is then y/0.05.

Without sodium phenolate, [C₆H₅O⁻] ≈ 2.24 × 10⁻⁶ M. With 0.01 M sodium phenolate present, the degree of ionization of phenol drops to about 1 × 10⁻⁸ (roughly 10,000 times smaller), a clear demonstration of the common-ion effect.

Q6.45: pH and Degree of Hydrolysis of Sodium Nitrite Solution

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

NaNO₂ is the salt of a weak acid (HNO₂) and a strong base (NaOH), so the nitrite ion hydrolyses: NO₂⁻ + H₂O ↔ HNO₂ + OH⁻. Its hydrolysis constant is K_b = K_w/K_a = (1.0 × 10⁻¹⁴)/(4.5 × 10⁻⁴) ≈ 2.22 × 10⁻¹¹. Using [OH⁻] = √(K_b × C) = √(2.22 × 10⁻¹¹ × 0.04) = √(8.89 × 10⁻¹³) ≈ 9.43 × 10⁻⁷ M. Then pOH = −log(9.43 × 10⁻⁷) ≈ 6.03, so pH = 14 − 6.03 ≈ 7.97. The degree of hydrolysis is h = √(K_b/C) = √(2.22 × 10⁻¹¹/0.04) = √(5.56 × 10⁻¹⁰).

pH ≈ 7.97 (mildly basic, as expected for the salt of a weak acid and a strong base), and the degree of hydrolysis h ≈ 2.36 × 10⁻⁵.

Q6.46: Degree of Dissociation and pH of Acetic Acid

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 and the concentration of acetate ion and pH of the solution.

For a weak acid at low degree of dissociation, K_a ≈ α²C, so α = √(K_a/C) = √(1.74 × 10⁻⁵/0.05) = √(3.48 × 10⁻⁴) ≈ 0.01865, i.e. about 1.87%. The acetate ion concentration is [CH₃COO⁻] = αC = 0.01865 × 0.05 ≈ 9.33 × 10⁻⁴ M, which equals [H⁺]. Then pH = −log(9.33 × 10⁻⁴) ≈ 3.03.

Degree of dissociation α ≈ 0.0187 (1.87%); [CH₃COO⁻] ≈ 9.33 × 10⁻⁴ M; pH ≈ 3.03.

Q6.47: Finding K_a and pK_a of an Organic Acid from Its pH

The pH of a 0.01 M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid, and its pK_a.

[H⁺] = 10^−4.15 = 10^0.85 × 10⁻⁵ ≈ 7.08 × 10⁻⁵ M. Since HA ↔ H⁺ + A⁻ dissociate 1:1, [A⁻] = [H⁺] ≈ 7.08 × 10⁻⁵ M. The undissociated acid remaining is [HA] ≈ 0.01 − 7.08 × 10⁻⁵ ≈ 9.93 × 10⁻³ M. Then K_a = [H⁺][A⁻]/[HA] = (7.08 × 10⁻⁵)²/(9.93 × 10⁻³) ≈ 5.05 × 10⁻⁷, and pK_a = −log(5.05 × 10⁻⁷) ≈ 6.30.

[A⁻] ≈ 7.08 × 10⁻⁵ M; K_a ≈ 5.0 × 10⁻⁷; pK_a ≈ 6.30.

Q6.48: pH of Four Strong Acid and Base Solutions

Assuming complete dissociation, calculate the pH of the following solutions: (a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH.

Since strong acids and bases dissociate completely, [H⁺] or [OH⁻] equals the stated molarity directly. (a) [H⁺] = 3 × 10⁻³ M, pH = 3 − log 3 = 3 − 0.477. (b) [OH⁻] = 5 × 10⁻³ M, pOH = 3 − log 5 = 3 − 0.699, pH = 14 − pOH. (c) [H⁺] = 2 × 10⁻³ M, pH = 3 − log 2 = 3 − 0.301. (d) [OH⁻] = 2 × 10⁻³ M, pOH = 3 − log 2 = 2.699, pH = 14 − pOH.

(a) pH ≈ 2.52. (b) pH ≈ 11.70. (c) pH ≈ 2.70. (d) pH ≈ 11.30.

Class 11 Chemistry Chapter 6 – Notes and Extra Questions

Once you have worked through every exercise question above, it is worth reinforcing your understanding of Equilibrium with focused revision notes and additional practice beyond the textbook. Key ideas to revisit include the distinction between K_c and K_p, how to build an ICE (Initial–Change–Equilibrium) table for any reversible reaction, the direction rules of Le Chatelier’s principle, and the full ionic-equilibrium toolkit — pH, pOH, K_a, K_b, K_w, the common-ion effect, and salt hydrolysis. Check the Class 11 Chemistry Revision Notes and Extra Questions section on this site for a chapter-wise summary sheet and additional numerical practice problems modelled on this exact exercise pattern, useful for both school exams and competitive exams such as JEE and NEET.

Frequently Asked Questions

What is the difference between K_c and K_p, and how are they related?

K_c is the equilibrium constant expressed in terms of molar concentrations, while K_p is expressed in terms of partial pressures of gaseous species. They are related by K_p = K_c(RT)^Δn, where Δn is the difference between the moles of gaseous products and gaseous reactants, R is the gas constant (0.0831 L bar K⁻¹ mol⁻¹), and T is the absolute temperature. When Δn = 0, K_p and K_c are numerically equal.

How does Le Chatelier’s principle predict the direction of a shift in equilibrium?

Le Chatelier’s principle states that if a system at equilibrium is subjected to a change in concentration, pressure, volume, or temperature, the equilibrium shifts in the direction that partially counteracts that change. For example, adding a reactant shifts equilibrium forward; increasing pressure shifts equilibrium toward the side with fewer gas moles; and raising the temperature shifts equilibrium in the endothermic direction. Note that only a genuine temperature change alters the numerical value of K itself — changes in concentration, pressure, or the addition of a catalyst change only the position of equilibrium, not K.

What does the pH scale actually measure, and why does pure water have a pH of 7?

The pH scale measures the concentration of hydrogen ions in a solution on a logarithmic basis: pH = −log₁₀[H⁺]. Pure water self-ionizes very slightly, 2H₂O ↔ H₃O⁺ + OH⁻, producing equal concentrations of H⁺ and OH⁻, each equal to 1.0 × 10⁻⁷ M at 25°C. Taking the negative logarithm of 10⁻⁷ gives a pH of exactly 7, which is why neutral water sits at the midpoint of the 0–14 pH scale at that temperature.

What is the common-ion effect, and why does it suppress the ionization of a weak acid?

The common-ion effect is the suppression of the ionization (or solubility) of a weak electrolyte when a second source of one of its own ions is added to the solution. For a weak acid like phenol, adding sodium phenolate (which fully dissociates to supply extra phenolate ions) increases [C₆H₅O⁻] in the solution. Because the equilibrium constant K_a = [C₆H₅O⁻][H⁺]/[C₆H₅OH] must stay fixed at a given temperature, this extra phenolate ion forces [H⁺] (and hence the degree of ionization of phenol) to decrease sharply, exactly as seen in Q6.44 above, where the degree of ionization dropped roughly 10,000-fold in the presence of the common ion.