Equilibrium covers the dynamic state where forward and backward reactions run at equal rates, introducing the equilibrium constants Kc and Kp, the reaction quotient Q, and Le Chatelier’s principle for how concentration, pressure and temperature shift a reaction. The second half is ionic equilibrium: acids and bases, pH and Kw, weak-acid ionisation, buffers and the solubility product Ksp. It is a high-yield NEET chapter because its numerical and conceptual ideas recur across physical and inorganic chemistry.
Table of Contents
- Key Concepts - Equilibrium, Kc and Kp, Q, Le Chatelier, acids and bases, pH, buffers, Ksp
- Weightage in Board & Entrance Exams
- Important Definitions
- Solved Examples
- Important Questions for Board Exams
- Quick Revision Points
Key Concepts
1. Equilibrium - Physical and Chemical
Equilibrium is the state of a reversible process at which the rate of the forward process equals the rate of the backward process, so the observable properties become constant with time.
It is dynamic, not static - both processes keep happening, just at equal rates. Equilibrium is possible only in a closed system at constant temperature.
Physical Equilibrium
- Solid ⇌ Liquid (melting/freezing at the melting point): rate of melting = rate of freezing.
- Liquid ⇌ Vapour (in a closed vessel): vapour pressure becomes constant.
- Solid/Gas ⇌ Solution (dissolution of sugar or CO₂ in a sealed bottle): concentration stops changing.
Chemical Equilibrium
In a reversible reaction such as H₂(g) + I₂(g) ⇌ 2HI(g), reactant and product concentrations become constant once the forward and backward rates are equal.
2. Law of Chemical Equilibrium and Equilibrium Constant
The law of mass action states that the rate of a reaction is proportional to the product of the molar concentrations of the reactants, each raised to the power of its stoichiometric coefficient.
For a general reaction aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentration is:
Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ
- Kc is constant at a given temperature.
- A large Kc (much greater than 1) means products are favoured; a small Kc (much less than 1) means reactants are favoured.
- Pure solids and pure liquids are not included (their activity = 1).
3. Equilibrium Constant in Gaseous Systems (Kp) and the Kc–Kp Relation
For gaseous reactions it is convenient to express the equilibrium constant in terms of partial pressures. For aA + bB ⇌ cC + dD:
Kp = (p_C)ᶜ(p_D)ᵈ / (p_A)ᵃ(p_B)ᵇ
Using the ideal gas equation, Kp and Kc are related by:
Kp = Kc (RT)^Δn
- Δn = (moles of gaseous products) − (moles of gaseous reactants)
- If Δn = 0, then Kp = Kc (e.g. H₂ + I₂ ⇌ 2HI).
- If Δn is positive, Kp is greater than Kc; if Δn is negative, Kp is less than Kc.
- R = 0.0821 L·atm·K⁻¹·mol⁻¹ when pressures are in atm.
4. Characteristics of the Equilibrium Constant
- For the reverse reaction, K’ = 1/K.
- If a reaction is multiplied by n, the new constant is Kⁿ.
- If two reactions are added, K = K₁ × K₂.
- K depends only on temperature - not on the initial concentrations, pressure, or a catalyst.
- A catalyst speeds up forward and backward rates equally, so it only reaches equilibrium faster - it does not change K.
5. Reaction Quotient (Q) and Predicting Direction
The reaction quotient Q has the same form as Kc but uses concentrations at any instant, not just at equilibrium. Comparing Q with K tells you which way the reaction will move.
- Q less than K: too many reactants → reaction proceeds forward (towards products).
- Q = K: system is at equilibrium.
- Q greater than K: too many products → reaction proceeds backward (towards reactants).
Link to thermodynamics: ΔG = ΔG° + RT ln Q, and at equilibrium ΔG = 0, giving ΔG° = −RT ln K.
6. Le Chatelier’s Principle and Factors Affecting Equilibrium
Le Chatelier’s principle: if a system at equilibrium is disturbed by a change in concentration, pressure, or temperature, the equilibrium shifts in the direction that tends to undo the change.
Effect of Each Factor
| Change | Equilibrium shifts | Example (N₂ + 3H₂ ⇌ 2NH₃, exothermic) |
|---|---|---|
| Add reactant / remove product | Forward (towards products) | Adding N₂ or H₂ makes more NH₃ |
| Increase pressure (decrease volume) | Towards fewer gas moles | Shifts forward (4 mol → 2 mol) |
| Increase temperature | Towards endothermic direction | Shifts backward (less NH₃) |
| Add catalyst | No shift | Equilibrium reached faster only |
| Add inert gas at constant volume | No shift | Partial pressures unchanged |
Industrial use: the Haber process (NH₃) uses high pressure and moderate temperature; the Contact process (SO₃) uses similar reasoning.
7. Ionic Equilibrium - Strong and Weak Electrolytes
Ionic equilibrium is the equilibrium established between unionised molecules and their ions in solution.
- Strong electrolytes (NaCl, HCl, NaOH) ionise almost completely - no real equilibrium.
- Weak electrolytes (CH₃COOH, NH₄OH) ionise only partially, setting up a genuine ionic equilibrium.
Ostwald’s dilution law (for a weak electrolyte, degree of dissociation α): Ka = Cα²/(1 − α), which is approximately Cα² when α is small, so α = √(Ka/C). Dilution increases α.
8. Acids and Bases - Three Concepts
| Concept | Acid | Base |
|---|---|---|
| Arrhenius | Gives H⁺ in water | Gives OH⁻ in water |
| Brønsted–Lowry | Proton (H⁺) donor | Proton (H⁺) acceptor |
| Lewis | Electron-pair acceptor | Electron-pair donor |
In the Brønsted–Lowry view, every acid has a conjugate base (formed by losing H⁺) and every base has a conjugate acid. Example: in HCl + H₂O ⇌ H₃O⁺ + Cl⁻, the pairs HCl/Cl⁻ and H₂O/H₃O⁺ are conjugate pairs. A strong acid has a weak conjugate base.
9. Ionization of Water, pH Scale and Kw
Water self-ionises: H₂O ⇌ H⁺ + OH⁻. The ionic product of water is:
Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 298 K
The pH measures acidity: pH = −log[H⁺], and similarly pOH = −log[OH⁻].
- pH + pOH = 14 at 298 K.
- Neutral: pH = 7; Acidic: pH below 7; Basic: pH above 7.
- For a strong acid, [H⁺] = its molarity; for a strong base, [OH⁻] = its molarity.
10. Ionization Constants of Weak Acids and Bases (Ka, Kb)
For a weak acid HA ⇌ H⁺ + A⁻: Ka = [H⁺][A⁻]/[HA]. For a weak base BOH ⇌ B⁺ + OH⁻: Kb = [B⁺][OH⁻]/[BOH].
- A larger Ka (or Kb) means a stronger acid (or base).
- pKa = −log Ka; a smaller pKa means a stronger acid.
- For a conjugate acid–base pair: Ka × Kb = Kw, so pKa + pKb = 14.
- For a weak acid, [H⁺] = √(Ka·C), so pH = ½(pKa − log C).
11. Common Ion Effect and Buffer Solutions
The common ion effect is the suppression of the ionisation of a weak electrolyte by adding a strong electrolyte that shares a common ion. For example, adding CH₃COONa to CH₃COOH lowers the H⁺ concentration.
A buffer solution resists changes in pH on adding small amounts of acid or base.
- Acidic buffer: weak acid + its salt (CH₃COOH + CH₃COONa). pH = pKa + log([salt]/[acid]) (Henderson–Hasselbalch equation).
- Basic buffer: weak base + its salt (NH₄OH + NH₄Cl). pOH = pKb + log([salt]/[base]).
12. Solubility Product (Ksp) and Salt Hydrolysis
For a sparingly soluble salt AₓBᵧ ⇌ xAʸ⁺ + yBˣ⁻, the solubility product is:
Ksp = [Aʸ⁺]ˣ [Bˣ⁻]ʸ
- For AB type (e.g. AgCl) with solubility s: Ksp = s².
- For AB₂ type (e.g. CaF₂) with solubility s: Ksp = 4s³.
- If ionic product is greater than Ksp → precipitation occurs; if less than Ksp → more salt dissolves.
Hydrolysis of Salts
| Salt type | Example | Nature of solution |
|---|---|---|
| Strong acid + strong base | NaCl | Neutral (pH = 7) |
| Strong acid + weak base | NH₄Cl | Acidic (pH below 7) |
| Weak acid + strong base | CH₃COONa | Basic (pH above 7) |
| Weak acid + weak base | CH₃COONH₄ | Depends on Ka vs Kb |
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Weightage in Board & Entrance Exams
| Exam | Typical Weightage | Most-Tested Areas |
|---|---|---|
| CBSE Board (Class 11) | 6–8 marks | Kc/Kp relation, Le Chatelier, pH, buffers, Ksp |
| JEE Main / Advanced | 2–3 questions | Q vs K, ICE-table numericals, Ksp, buffer pH |
| NEET | 2–3 questions | Le Chatelier, pH/pOH, common ion effect, salt hydrolysis |
[TABLE: Question-type split - VSA (1 mark): definitions, conjugate pairs, pH; SA (2–3 marks): Kp–Kc, Le Chatelier shifts, buffer/Henderson; LA (5 marks): Ksp numericals, ionic-equilibrium derivations, pH calculations.]
Important Definitions
| Term | Definition |
|---|---|
| Chemical equilibrium | State where forward and backward reaction rates are equal; concentrations stay constant |
| Equilibrium constant (Kc) | Kc = [products]/[reactants], each raised to its coefficient, at a fixed temperature |
| Kp–Kc relation | Kp = Kc(RT)^Δn, where Δn = gaseous products − gaseous reactants |
| Reaction quotient (Q) | Same form as Kc but at any instant; predicts direction by comparing with K |
| Le Chatelier’s principle | A disturbed equilibrium shifts to oppose (undo) the change imposed |
| Brønsted acid/base | Acid = proton donor; base = proton acceptor |
| Lewis acid/base | Acid = electron-pair acceptor; base = electron-pair donor |
| pH | pH = −log[H⁺]; measures acidity (pH + pOH = 14 at 298 K) |
| Kw | Ionic product of water: [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 298 K |
| Common ion effect | Suppression of a weak electrolyte’s ionisation by adding a common ion |
| Buffer solution | Solution that resists pH change on adding small amounts of acid/base |
| Solubility product (Ksp) | Product of ionic molar concentrations of a saturated sparingly-soluble salt |
Solved Examples
Example 1
For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), find the relation between Kp and Kc.
Answer: Δn = 2 − (1 + 3) = −2. So Kp = Kc(RT)⁻², i.e. Kp = Kc/(RT)².
Example 2
In the reaction H₂ + I₂ ⇌ 2HI, equilibrium concentrations are [H₂] = 0.5 M, [I₂] = 0.5 M, [HI] = 2 M. Find Kc.
Answer: Kc = [HI]²/([H₂][I₂]) = (2)²/(0.5 × 0.5) = 4/0.25 = 16.
Example 3
Calculate the pH of a 0.001 M HCl solution.
Answer: HCl is a strong acid, so [H⁺] = 0.001 = 10⁻³ M. pH = −log(10⁻³) = 3.
Example 4
The Ka of acetic acid is 1.8 × 10⁻⁵. Find the pH of a 0.1 M solution.
Answer: [H⁺] = √(Ka·C) = √(1.8 × 10⁻⁵ × 0.1) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ M. pH = −log(1.34 × 10⁻³), which is approximately 2.87.
Example 5
A buffer is made of 0.2 M CH₃COOH and 0.2 M CH₃COONa (pKa = 4.74). Find its pH.
Answer: pH = pKa + log([salt]/[acid]) = 4.74 + log(0.2/0.2) = 4.74 + 0 = 4.74.
Example 6
The solubility of AgCl is 1.0 × 10⁻⁵ mol/L. Calculate its Ksp.
Answer: AgCl ⇌ Ag⁺ + Cl⁻, so Ksp = s² = (1.0 × 10⁻⁵)² = 1.0 × 10⁻¹⁰.
Important Questions for Board Exams
1-Mark Questions (VSA)
- Why is chemical equilibrium called dynamic?
- Write the conjugate base of H₂SO₄ and the conjugate acid of NH₃.
- What is the value of Kw at 298 K?
- What happens to the pH of water as temperature increases? Justify briefly.
- Does a catalyst change the value of the equilibrium constant? Explain.
2–3-Mark Questions (SA)
- Derive the relation Kp = Kc(RT)^Δn for a gaseous reaction.
- State Le Chatelier’s principle and apply it to the Haber process (N₂ + 3H₂ ⇌ 2NH₃) for pressure and temperature.
- What is a buffer solution? Derive the Henderson–Hasselbalch equation for an acidic buffer.
- Explain the common ion effect with a suitable example and its role in salt analysis.
5-Mark Questions (LA)
- Distinguish between Arrhenius, Brønsted–Lowry, and Lewis concepts of acids and bases with examples.
- Define solubility product. Derive the relation between Ksp and solubility for AB and AB₂ type salts, and explain precipitation using the ionic product.
- Explain salt hydrolysis. Discuss the nature (acidic/basic/neutral) of solutions of NaCl, NH₄Cl, and CH₃COONa with reasons.
Quick Revision Points
- Equilibrium is dynamic: forward rate = backward rate in a closed system at constant T
- Kc = [products]/[reactants] (powers = coefficients); pure solids/liquids excluded
- Kp = Kc(RT)^Δn; Δn = gaseous products − gaseous reactants; if Δn = 0, Kp = Kc
- Reverse reaction: K’ = 1/K; multiply by n: Kⁿ; add reactions: K = K₁ × K₂
- Q below K → forward; Q = K → equilibrium; Q above K → backward; ΔG° = −RT ln K
- Le Chatelier: system shifts to undo the change in concentration/pressure/temperature
- Acid/base: Arrhenius (H⁺/OH⁻), Brønsted (proton donor/acceptor), Lewis (electron-pair acceptor/donor)
- pH = −log[H⁺]; pH + pOH = 14; Kw = 10⁻¹⁴ at 298 K
- Ka × Kb = Kw; pKa + pKb = 14; weak acid [H⁺] = √(Ka·C)
- Buffer: pH = pKa + log([salt]/[acid]) (Henderson–Hasselbalch)
- Ksp: AB type = s²; AB₂ type = 4s³; ionic product above Ksp → precipitation
- Hydrolysis: SA+SB neutral, SA+WB acidic, WA+SB basic
Next Chapter: Chapter 7 - Redox Reactions
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Dynamic Equilibrium & Kc
At equilibrium the forward and backward reactions run at equal rates, so concentrations stop changing.
Use equilibrium concentrations (mol L⁻1); omit pure solids and pure liquids (activity = 1).
- Equilibrium is dynamic, not static — both reactions keep going
- Large Kc (≫1) → products dominate; small Kc (≪1) → reactants dominate
- Kc depends only on temperature, not on concentration, pressure or catalyst
Manipulating the Equilibrium Constant
Change the way you write a reaction and K transforms in a fixed way.
Multiply all coefficients by n → K is raised to the nᵗʰ power.
- Reversing a reaction inverts K
- Scaling coefficients by n raises K to power n
- Adding two reactions multiplies their K values
Connecting Kp and Kc
For gas equilibria, partial-pressure K (Kp) and concentration K (Kc) are linked through the gas law.
Δn = (gas product moles) − (gas reactant moles); R = 0.0821 L atm K⁻1 mol⁻1; T in kelvin.
- Δn = 0 → Kp = Kc (e.g. H2 + I2 ⇌ 2HI)
- Δn > 0 → Kp > Kc; Δn < 0 → Kp < Kc
- For N2 + 3H2 ⇌ 2NH3, Δn = 2 − 4 = −2 → Kp = Kc(RT)⁻2
Reaction Quotient Q
Q has the same form as K but uses the concentrations at any instant, so it tells you which way the reaction moves.
As the reaction proceeds, Q always moves toward K.
- Q < K → too few products → reaction goes forward (→)
- Q = K → system is at equilibrium, no net change
- Q > K → too many products → reaction goes backward (←)
Shifting Equilibrium
Disturb a system at equilibrium and it shifts to partly oppose the change.
Concentration and pressure move the position of equilibrium but never change K.
- Add reactant / remove product → shifts forward
- High pressure favours NH3 in N2 + 3H2 ⇌ 2NH3 (4 → 2 mol); Δn = 0 → pressure has no effect
- Catalyst only reaches equilibrium faster — it does not shift it or change K
Temperature: the Only Thing That Changes K
Treat heat as a reactant or product to predict the shift when temperature changes.
Endothermic (ΔH > 0): raising T shifts forward and K increases.
- Exothermic: heat is a product → raising T shifts backward, K decreases
- Endothermic: heat is a reactant → raising T shifts forward, K increases
- Inert gas at constant V → no effect; at constant P → shifts toward more gas moles
Water, pH and Kw
Water self-ionises, fixing the link between acidity and basicity.
pH = −log[H⁺]; pH + pOH = 14; neutral water has pH 7.
- Brønsted acid donates H⁺, base accepts H⁺; a conjugate pair differs by one H⁺
- Strong acids/bases ionise completely; weak ones set up an equilibrium
- Kw = 10⁻14 holds only at 25 °C — it rises at higher temperature
Ka, pKa and Ostwald’s Law
A weak acid only partly ionises, so its [H⁺] follows a square-root law.
pKₐ = −log Kₐ; smaller pKₐ means a stronger acid.
- Larger Kₐ → stronger acid
- Degree of ionisation α rises on dilution (Ostwald’s dilution law, α ∝ 1/√C)
- Use [H⁺] = √(Kₐ·C) for weak acids, never [H⁺] = C (that is only for strong acids)
Buffers & Salt Hydrolysis
A weak acid plus its salt resists pH change; its pH follows Henderson–Hasselbalch.
When [salt] = [acid], the log term is zero, so pH = pKₐ.
- Acidic buffer = weak acid + its salt; basic buffer = weak base + its salt
- Salt of strong base + weak acid → basic solution; weak base + strong acid → acidic
- Strong base + strong acid (NaCl) → neutral solution
Solubility Product Ksp
For a sparingly soluble salt, the solid stays in equilibrium with its dissolved ions.
Compare the ionic product Qsp with Ksp to predict precipitation.
- Qsp < Ksp → no precipitate; Qsp = Ksp → saturated; Qsp > Ksp → precipitate forms
- Pick the right s-relation by salt type — never use s2 for every salt
- Common ion effect always lowers solubility (equilibrium shifts left)
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Frequently Asked Questions
Chemical equilibrium is the state of a reversible reaction where the forward and backward reactions occur at the same rate, so the concentrations of reactants and products stay constant. It is dynamic, meaning both reactions keep going even though no net change is visible.
For a gaseous equilibrium, Kp = Kc(RT) raised to the power delta-n, where delta-n is the moles of gaseous products minus moles of gaseous reactants, R is 0.0821 L atm per K per mol and T is the temperature in kelvin. When delta-n is zero, Kp equals Kc.
Yes, Equilibrium is part of the Class 11 NCERT Chemistry syllabus and is high-yield for NEET. Questions on Kc and Kp, Le Chatelier’s principle, pH, buffers and Ksp appear frequently and are usually direct and scoring.
Q and K have the same mathematical form, but Q uses the concentrations at any instant while K uses only the equilibrium concentrations. Comparing them gives direction: if Q is less than K the reaction moves forward, if Q is greater than K it moves backward, and if Q equals K the system is already at equilibrium.
Only temperature changes the value of K. Changing concentration, pressure or adding a catalyst can shift the position of equilibrium but never alters K. For an exothermic reaction raising the temperature lowers K, while for an endothermic reaction it raises K.