CBSE Class 11 Chemistry Revision Notes Chapter 6: Equilibrium

Equilibrium is a dynamic state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant over time. Chemical equilibria are fundamental to biological processes, industrial chemistry, and environmental science, from oxygen transport in blood to the manufacture of ammonia. These class 11 equilibrium notes cover every topic in NCERT Chapter 6, Reprint 2026-27, and the class 11 equilibrium ncert notes are also available for download at ncert.nic.in.

Class 11 Chemistry Chapter 6 Equilibrium is one of the most important chapters in the CBSE syllabus. It carries significant weightage in board exams and forms the foundation for physical chemistry in Class 12. These chemistry class 11 equilibrium notes follow the NCERT 2026-27 chapter sequence and cover every topic: physical equilibrium, chemical equilibrium, Kc, Kp, Le Chatelier's principle, ionic equilibrium, acids and bases, pH, buffer solutions, and solubility product, with worked examples, comparison tables, and exam-ready explanations. 

Use these notes alongside Class 12 Chemistry Chapter 6 Revision Notes to build a complete two-year picture of chemical equilibrium.

Key Takeaways — Class 11 Equilibrium Short Notes

Overview of Class 11 Equilibrium Notes

Class 11 equilibrium chapter notes are built around two equally important halves. The first half covers physical and chemical equilibrium, the law of chemical equilibrium, equilibrium constants, applications of Kc and Kp, reaction quotient, and Le Chatelier's principle.

The second half covers ionic equilibrium: acids and bases, ionisation constants, pH calculations, buffer solutions, hydrolysis of salts, and solubility product. Students often focus more on the first half and underestimate ionic equilibrium. Board papers consistently draw from both halves, and entrance exams particularly target pH calculations, buffer problems, and Ksp numericals.

Introduction to Equilibrium: Class 11 Chemistry

When a reversible reaction takes place in a closed vessel, the concentrations of reactants decrease and those of products increase initially. The rate of the forward reaction slows as reactants are consumed. The rate of the reverse reaction increases as products accumulate.

Eventually, the two rates become equal. No further change in concentration occurs. This is the point of dynamic chemical equilibrium.

The word dynamic is important. At equilibrium, both reactions still occur simultaneously. The system has no net change in composition. This is why equilibrium is described as dynamic and stable.

Equilibrium applies to both physical processes and chemical processes. In every case, the condition is a closed system at constant temperature.

Equilibrium in Physical Processes

Physical equilibrium involves a balance between two phases of the same substance. The NCERT chapter covers three main types.

Solid-Liquid Equilibrium

Ice and water in a perfectly insulated flask at 273 K and 1 atm are at equilibrium. The rate at which molecules leave the ice surface equals the rate at which water molecules rejoin the ice. The temperature at which this happens for any pure substance at 1 atm is its normal melting point or normal freezing point.

Liquid-Vapour Equilibrium

In a closed container, water molecules escape the liquid surface (evaporation) and return from the vapour phase (condensation). Initially, evaporation dominates. As vapour builds up, the rate of condensation increases until it equals the rate of evaporation.

The pressure at this point is the equilibrium vapour pressure, constant at a given temperature and higher at higher temperatures. The temperature at which vapour pressure equals 1.013 bar is the normal boiling point.

Different liquids have different vapour pressures at the same temperature. Higher vapour pressure means more volatile and lower boiling point.

Solid-Vapour Equilibrium

Solids that sublime, like iodine, camphor, and NH₄Cl, establish equilibrium between the solid and vapour phases in a closed vessel. The intensity of colour for iodine increases until equilibrium, then stays constant.

General characteristics of physical equilibria:

Dissolution Equilibrium: Solids and Gases in Liquids

Solids Dissolving in Liquids

When sugar dissolves in water in a closed system, a point is reached where the rate of dissolution equals the rate of crystallisation. The solution is saturated. The concentration of solute in this saturated solution is constant at a given temperature.

This is confirmed by radioactive tracer experiments. Radioactive sugar mixed into a saturated non-radioactive sugar solution eventually distributes evenly between solid and solution phases, proving ongoing exchange even at equilibrium.

Gases Dissolving in Liquids

CO₂ gas over water in a sealed bottle is in equilibrium with dissolved CO₂:

CO₂(g) ⇌ CO₂(aq)

This follows Henry's Law: the mass of gas dissolved in a given solvent at constant temperature is proportional to the pressure of the gas above the solvent. When a soda bottle is opened, pressure drops, solubility decreases, and CO₂ escapes.

Dynamic Nature of Chemical Equilibrium

The Haber process for ammonia synthesis is the clearest proof that chemical equilibrium is dynamic. N₂ and H₂ react in a sealed vessel until concentrations become constant. If the same experiment is repeated with deuterium (D₂) instead of H₂, and both equilibrium mixtures are combined, mass spectrometer analysis shows scrambled isotope forms: NH₂D, NHD₂, ND₃. This is only possible if the forward and reverse reactions continue even after equilibrium is reached.

Equilibrium can be approached from either direction. Whether you start with N₂ and H₂, or with pure NH₃, the same equilibrium composition is reached at the same temperature and pressure.

Law of Chemical Equilibrium and Equilibrium Constant Kc

For the general reversible reaction:

aA + bB ⇌ cC + dD

The equilibrium constant expression is:

Kc = [C]^c [D]^d / [A]^a [B]^b

At a given temperature, this ratio is always constant, regardless of initial concentrations or which direction you start from. This is the Law of Chemical Equilibrium, established by Guldberg and Waage in 1864.

Important rules for Kc:

What Kc tells you about the reaction:

• Kc > 10³: Products dominate at equilibrium; reaction goes nearly to completion
• Kc < 10⁻³: Reactants dominate; reaction barely proceeds forward
• Kc between 10⁻³ and 10³: Appreciable amounts of both reactants and products present

Kc and Kp Class 11: Relationship Between the Two Constants

For reactions involving gases, the equilibrium constant can also be expressed in terms of partial pressures:

Kp = Kc(RT)^Δn

Where:

• R = 0.0831 bar L mol⁻¹ K⁻¹
• T = temperature in Kelvin
• Δn = moles of gaseous products minus moles of gaseous reactants

When Δn = 0: Kp = Kc (example: H₂ + I₂ ⇌ 2HI) When Δn > 0: Kp > Kc (example: PCl₅ ⇌ PCl₃ + Cl₂, Δn = 1) When Δn < 0: Kp < Kc (example: N₂ + 3H₂ ⇌ 2NH₃, Δn = −2)

Pressure must be expressed in bar for Kp calculations because the standard state for pressure is 1 bar.

Homogeneous and Heterogeneous Equilibria

For CaCO₃(s) ⇌ CaO(s) + CO₂(g):

K'c = [CO₂] and Kp = pCO₂

At a given temperature, there is always a fixed pressure of CO₂ above a mixture of CaCO₃ and CaO. Experimentally, at 1100 K, this pressure is 2.0 × 10⁵ Pa, giving Kp = 2.00.

Applications of Equilibrium Constant

The equilibrium constant has three main applications in class 11 equilibrium numericals and conceptual questions.

1. Predicting the extent of reaction: High Kc means reaction nearly complete. Low Kc means reaction barely proceeds. Intermediate Kc means both reactants and products present significantly.
2. Predicting the direction of reaction using Qc: The reaction quotient Qc uses the same expression as Kc but with concentrations at any point.

• Qc < Kc: reaction proceeds forward (more products form)
• Qc > Kc: reaction proceeds in reverse (more reactants form)
• Qc = Kc: system is at equilibrium; no net change

1. Calculating equilibrium concentrations: This is the most common numerical application, using initial concentrations and Kc to find equilibrium concentrations via ICE tables.

Class 11 Equilibrium Numericals: Solved Examples

Solved Example 1: Calculating Kc

For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at 500 K: [N₂] = 1.5 × 10⁻² M, [H₂] = 3.0 × 10⁻² M, [NH₃] = 1.2 × 10⁻² M

Kc = [NH₃]² / [N₂][H₂]³ Kc = (1.2 × 10⁻²)² / (1.5 × 10⁻²)(3.0 × 10⁻²)³ Kc = 1.44 × 10⁻⁴ / 4.05 × 10⁻⁷ = 3.55 × 10²

Solved Example 2: ICE Table Method

For CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g), Kc = 4.24 at 800 K. Initial concentrations of CO and H₂O are both 0.10 M.

Kc = x² / (0.10 − x)² = 4.24 Taking square root: x / (0.10 − x) = 2.06 x = 0.206 − 2.06x → 3.06x = 0.206 → x = 0.067 M [CO₂] = [H₂] = 0.067 M; [CO] = [H₂O] = 0.033 M

Solved Example 3: Predicting Direction Using Qc

For 2A ⇌ B + C, Kc = 2 × 10⁻³. At a given time, [A] = [B] = [C] = 3 × 10⁻⁴ M.

Qc = [B][C] / [A]² = (3 × 10⁻⁴)(3 × 10⁻⁴) / (3 × 10⁻⁴)² = 1

Since Qc (1) > Kc (2 × 10⁻³), reaction proceeds in the reverse direction.

Relation Between K, Q, and Gibbs Energy

ΔG° = −RT ln K

• ΔG < 0: spontaneous forward reaction; K > 1
• ΔG > 0: non-spontaneous forward reaction; K < 1
• ΔG = 0: system at equilibrium; no free energy available to drive the reaction

The relationship ΔG = ΔG° + RT ln Q connects any state of the system to its equilibrium constant. At equilibrium, ΔG = 0 and Q = K, which gives ΔG° = −RT ln K. This connects thermodynamics and equilibrium directly.

Le Chatelier Principle Class 11 Notes

Le Chatelier's principle: If a system at equilibrium is subjected to a change in concentration, pressure, volume, or temperature, it shifts in the direction that counteracts the effect of that change.

Effect of Concentration Change

Adding a reactant shifts equilibrium right. Removing a product shifts equilibrium right. In ammonia manufacture, ammonia is continuously removed by liquefaction, keeping Qc < Kc and the reaction moving forward. In lime kiln production of CaO from CaCO₃, CO₂ is continuously removed, driving the reaction to completion.

Effect of Pressure and Volume Change

Increasing pressure shifts equilibrium toward the side with fewer moles of gas. For N₂ + 3H₂ ⇌ 2NH₃, there are 4 moles of gas on the left and 2 on the right, so increased pressure favours ammonia formation. If Δn = 0, pressure change has no effect on equilibrium position.

Effect of Inert Gas Addition

At constant volume, adding an inert gas like argon does not change partial pressures or molar concentrations of reacting species. Equilibrium is undisturbed.

Effect of Temperature

Temperature is the only factor that changes the value of K itself.

For ammonia synthesis (ΔH = −92.38 kJ mol⁻¹), low temperature favours high yield but slows the reaction. A catalyst (iron) is used to reach equilibrium faster without changing K.

Effect of Catalyst

A catalyst lowers activation energy for both forward and reverse reactions equally. It does not change Kc, Kp, or the equilibrium composition. It only speeds up the time to reach equilibrium. In the Contact process for H₂SO₄, V₂O₅ is used as a catalyst for the slow oxidation of SO₂ to SO₃.

Ionic Equilibrium Class 11 Notes

When weak electrolytes dissolve in water, they partially ionise and establish equilibrium between ions and unionised molecules in solution. This is ionic equilibrium.

Acids, Bases and Salts: Class 11 Equilibrium Chemistry Notes

A conjugate acid-base pair differs by exactly one proton. When HCl donates a proton to H₂O, Cl⁻ is the conjugate base of HCl, and H₃O⁺ is the conjugate acid of H₂O.

Key rule: Strong acid gives very weak conjugate base. Weak acid gives stronger conjugate base.

Ka × Kb = Kw for conjugate pairs at 298 K

So if Ka of NH₄⁺ = 5.6 × 10⁻¹⁰, then Kb of NH₃ = 10⁻¹⁴ / 5.6 × 10⁻¹⁰ = 1.8 × 10⁻⁵

Ionisation of Water and Ionic Product Kw

Pure water self-ionises:

H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)

Kw = [H⁺][OH⁻] = 1 × 10⁻¹⁴ at 298 K

At 298 K: [H⁺] = [OH⁻] = 1 × 10⁻⁷ M

Kw is temperature-dependent. It increases with temperature, which means water becomes slightly more acidic at higher temperatures.

pH Scale and pOH: Class 11 Equilibrium Notes

pH = −log[H⁺] pOH = −log[OH⁻] pH + pOH = pKw = 14 at 298 K

The pH scale is logarithmic. A change of 1 unit means a 10× change in [H⁺]. A change of 2 units means a 100× change.

Ionisation Constants of Weak Acids and Bases

For weak acid HX:

HX(aq) + H₂O(l) ⇌ H₃O⁺(aq) + X⁻(aq)

Ka = [H⁺][X⁻] / [HX]

pKa = −log Ka (larger Ka = smaller pKa = stronger acid)

For weak base MOH:

Kb = [M⁺][OH⁻] / [MOH]

pKb = −log Kb

For conjugate pairs: Ka × Kb = Kw → pKa + pKb = 14

Solved Example: Weak Acid pH

Calculate the pH of 0.1 M acetic acid. Ka = 1.74 × 10⁻⁵

Ka = cα² (where α is degree of ionisation, c = 0.1 M) α = √(Ka/c) = √(1.74 × 10⁻⁵ / 0.1) = √(1.74 × 10⁻⁴) = 0.0132 [H⁺] = cα = 0.1 × 0.0132 = 1.32 × 10⁻³ M pH = −log(1.32 × 10⁻³) = 2.88

Common ion effect: Adding acetate ions (CH₃COO⁻) to acetic acid solution suppresses ionisation. The degree of ionisation decreases. This is a direct application of Le Chatelier's principle to ionic equilibrium.

Hydrolysis of Salts

Salts undergo hydrolysis in water when their constituent ions interact with water molecules to re-form the parent acid or base.

For CH₃COONa: Acetate ion hydrolyses to give acetic acid and OH⁻. Acetic acid (Ka = 1.8 × 10⁻⁵) stays mostly unionised. OH⁻ builds up giving a basic solution.

For NH₄Cl: NH₄⁺ hydrolyses to give NH₄OH and H⁺. NH₄OH (Kb = 1.77 × 10⁻⁵) stays largely unionised. H⁺ builds up giving an acidic solution.

Buffer Solution Class 11: Class 11 Equilibrium Important Topics

A buffer solution resists changes in pH when small amounts of acid or base are added, or when diluted.

Henderson-Hasselbalch equation:

pH = pKa + log [Salt] / [Acid]

When [Salt] = [Acid], pH = pKa. This is the most useful design rule for buffer preparation.

Solved Example: Buffer pH

Calculate the pH of a buffer containing 0.1 M CH₃COOH and 0.1 M CH₃COONa. pKa of acetic acid = 4.74.

pH = 4.74 + log (0.1/0.1) = 4.74 + log 1 = 4.74 + 0 = 4.74

Buffer pH is not affected by dilution because the ratio [Salt]/[Acid] stays constant.

When acid is added to an acidic buffer, the acetate ions react with H⁺ to form more acetic acid, absorbing the added acid. When base is added, acetic acid reacts with OH⁻, neutralising the added base. In both cases, pH changes minimally.

Solubility Product Class 11: Solubility Equilibria and Ksp

For a sparingly soluble salt dissolving in water:

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq)

Ksp = [Ba²⁺][SO₄²⁻]

Ksp at 298 K for BaSO₄ = 1.1 × 10⁻¹⁰

Molar solubility S: [Ba²⁺] = [SO₄²⁻] = S

So S² = 1.1 × 10⁻¹⁰ → S = 1.05 × 10⁻⁵ mol L⁻¹

For general salt MₓYy:

Ksp = (xS)^x(yS)^y = x^x · y^y · S^(x+y)

Common ion effect on solubility: Adding a common ion like SO₄²⁻ to a BaSO₄ solution decreases solubility. The ionic product exceeds Ksp and precipitation occurs. This principle is used in gravimetric analysis and water treatment.

Solved Example: Ksp

Calculate the solubility of A₂X₃ if Ksp = 1.1 × 10⁻²³.

A₂X₃ → 2A³⁺ + 3X²⁻ [A³⁺] = 2S; [X²⁻] = 3S Ksp = (2S)²(3S)³ = 4S² × 27S³ = 108S⁵ 108S⁵ = 1.1 × 10⁻²³ S⁵ = 1.02 × 10⁻²⁵ S = 1.0 × 10⁻⁵ mol L⁻¹

Class 11 Equilibrium Important Topics: Formula Quick Reference

• Equilibrium constant: Kc = [products]^n / [reactants]^n (stoichiometric coefficients as powers)
• Kp and Kc: Kp = Kc(RT)^Δn
• Reaction quotient: Qc, same form as Kc, concentrations need not be at equilibrium
• Ionic product of water: Kw = [H⁺][OH⁻] = 10⁻¹⁴ at 298 K
• pH: pH = −log[H⁺]; pH + pOH = 14
• Weak acid ionisation: Ka = cα²/(1−α) ≈ cα² when α is small
• Weak base ionisation: Kb = cα²/(1−α) ≈ cα² when α is small
• Conjugate pair: Ka × Kb = Kw → pKa + pKb = 14
• Buffer: pH = pKa + log [Salt]/[Acid]
• Salt hydrolysis: pH = 7 + ½(pKa − pKb) for weak acid-weak base salt
• Solubility product: Ksp = (xS)^x(yS)^y
• Gibbs energy: ΔG° = −RT ln K

Most Asked Class 11 Equilibrium Important Questions

1-Mark and Short Answer Type

1. State Le Chatelier's principle.
2. What is the effect of a catalyst on the equilibrium constant?
3. Define buffer solution.
4. Write the expression for Kp for N₂ + 3H₂ ⇌ 2NH₃.
5. What is the pH of a neutral solution at 298 K?
6. Define degree of ionisation.
7. What happens to Kc when the equation is reversed?
8. Define conjugate acid-base pair with one example.
9. Give one example of a heterogeneous equilibrium.
10. What is the common ion effect?

2-Mark and Application Type

1. For the reaction PCl₅ ⇌ PCl₃ + Cl₂, predict the effect of increasing pressure.
2. Calculate Kp if Kc = 0.5 at 300 K for a reaction where Δn = 2.
3. Why does Kc not change when a catalyst is added?
4. Distinguish between physical and chemical equilibrium with one example each.
5. How does temperature affect Kc for an endothermic reaction?

5-Mark and Numerical Type

1. 13.8 g of N₂O₄ is placed in a 1 L vessel at 400 K. At equilibrium, total pressure is 9.15 bar. Calculate Kc and Kp.
2. Calculate the pH of 0.02 M HF solution. Ka = 3.2 × 10⁻⁴.
3. Calculate the solubility of Ni(OH)₂ in 0.1 M NaOH. Ksp = 2.0 × 10⁻¹⁵.
4. For CO + H₂O ⇌ CO₂ + H₂, Kc = 4.24 at 800 K. Find equilibrium concentrations starting with 0.1 M each of CO and H₂O.
5. Calculate the pH of a buffer made with 0.2 M NH₄Cl and 0.1 M NH₃. pKb = 4.75.

Quick Revision: Class 11 Equilibrium Short Notes

• Equilibrium is dynamic: both forward and reverse reactions continue at equal rates
• Kc is constant at constant temperature; only temperature changes K
• Qc < Kc: forward; Qc > Kc: reverse; Qc = Kc: equilibrium
• Le Chatelier's principle predicts direction of shift
• Catalyst speeds up both reactions equally and does not change K or equilibrium composition
• Pressure affects only reactions with Δn ≠ 0
• Inert gas at constant volume has no effect on equilibrium
• Strong acid has very weak conjugate base; weak acid has stronger conjugate base
• Ka × Kb = Kw = 10⁻¹⁴ at 298 K for any conjugate pair
• pH + pOH = 14 always at 298 K
• Salts of strong acid + strong base are neutral; others hydrolyse
• Buffer pH = pKa when [Salt] = [Acid]
• Precipitation occurs when ionic product > Ksp
• Common ion effect reduces solubility and degree of ionisation
• ΔG° = −RT ln K: negative ΔG° means K > 1 means spontaneous forward reaction