Chapters

Everything NCERT Class 11 Unit 6 (Equilibrium) says that can be asked, from its start up to and including the factors that affect equilibria (sections 6.1–6.8: physical and chemical equilibria, Kc and Kp, Q, ΔG and K, Le Chatelier), plus what JEE Main adds beyond the book (marked Beyond NCERT). The rest of Unit 6 (ionic equilibrium, from 6.9) is in Ionic Equilibrium. The short note stays for the last week; this is for a proper revision pass.

Formula sheet

Every formula in the chapter on one screen, for the last pass. Details and examples are in the sections below.

Quantity	Formula	Watch
Law of equilibrium	Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ	Equilibrium values; pure solids and liquids left out
Pressure form	Kp = p(C)ᶜ p(D)ᵈ / p(A)ᵃ p(B)ᵇ	Pressures in bar
Kp and Kc	Kp = Kc (RT)^Δn	Δn = gas moles, products − reactants; R = 0.0831 L bar K⁻¹ mol⁻¹
Reverse the equation	K′ = 1/K
Multiply by n	K′ = Kⁿ	Halve it: √K
Add equations	K = K₁ × K₂	Subtract: divide
Partial pressure	p = (n/n(total)) × P	Also p = cRT
Direction	Q < K forward; Q > K backward; Q = K none	Same expression as K, any moment
ΔG and Q	ΔG = ΔG° + RT ln Q	ΔG = 0 at equilibrium
ΔG° and K	ΔG° = −RT ln K = −2.303 RT log K	K = e^(−ΔG°/RT)
Temperature	log(K₂/K₁) = (ΔH°/2.303R)(1/T₁ − 1/T₂)	Beyond NCERT
A ⇌ B + C	Kp = α²P / (1 − α²)	Beyond NCERT; P = total pressure
A ⇌ 2B	Kp = 4α²P / (1 − α²)	Beyond NCERT

0 · What JEE Main asks from this chapter

87 questions in JEE Main 2019–2026, about 1 per paper, mostly medium. Half are numerical-answer questions. Grouped by type (counts are approximate):

Type	Share	Typical question
ICE tables: K from data, equilibrium amounts, α and total pressure	~38%	5 mol PCl₅ + 4 mol Ar, 100 L, 610 K, P = 6 atm: find Kp
Handling K: reverse, multiply, combine; Kp ↔ Kc	~22%	K for SO₃ ⇌ SO₂ + ½O₂ given: K for 2SO₂ + O₂ ⇌ 2SO₃
Le Chatelier: concentration, pressure, inert gas, temperature, catalyst	~15%	Xenon added at constant T and p to PCl₅ ⇌ PCl₃ + Cl₂
ΔG° and K; K with temperature	~13%	N₂O₄ 50% dissociated at 27 °C, 1 atm: ΔG°
Heterogeneous equilibria (solids, liquids)	~7%	5.1 g NH₄SH, 30% decomposed: Kp
Physical equilibria and the nature of equilibrium	~5%	Ice and water at 273 K: pressure doubled

NCERT's problems and exercises come back: exercise 6.14 in 2023, Problem 6.1 in 2024, Problem 6.8's numbers (Kp 85.87, Kc 2.586) in 2026.

1 · Equilibrium in physical processes

At equilibrium two opposite processes run at the same rate, so nothing measurable changes. It is dynamic, not static.

• Solid ⇌ liquid: ice and water in a thermos at 273 K and 1 atm. Their masses stay constant while molecules keep crossing the boundary both ways. Normal melting (freezing) point: the temperature at which solid and liquid are in equilibrium at 1 atm.
• Liquid ⇌ vapour: water in a closed box. The rate of evaporation is constant; condensation speeds up until it equals evaporation. The pressure then stays at the equilibrium vapour pressure, which rises with temperature.
  • At a given T, liquids differ in vapour pressure. Higher vapour pressure means more volatile and a lower boiling point.
  • Normal boiling point: liquid and vapour in equilibrium at 1 atm (1.013 bar); water's is 100 °C. It falls at high altitude.
  • In an open dish the vapour spreads into the room, condensation stays slower than evaporation, and the liquid vanishes. No equilibrium in an open system. The time it takes depends on the liquid, its amount and the temperature.
• Solid ⇌ vapour (sublimation): I₂(s) ⇌ I₂(vapour), violet colour becomes constant; also camphor, NH₄Cl.
• Solid in liquid: a saturated solution has sugar(solution) ⇌ sugar(solid); rate of dissolution = rate of crystallisation. Shown to be dynamic with radioactive sugar: radioactivity appears in both the solid and the solution.
• Gas in liquid: CO₂(g) ⇌ CO₂(solution). Henry's law: the mass of a gas dissolved in a given mass of solvent at any temperature is proportional to the gas's pressure above it. Solubility falls as temperature rises. A soda bottle fizzes when opened and goes flat as the CO₂ pressure falls to its partial pressure in air.

What stays constant at equilibrium (NCERT Table 6.1):

Process	Constant at a given T
H₂O(l) ⇌ H₂O(g)	vapour pressure p(H₂O)
H₂O(s) ⇌ H₂O(l)	melting point (at fixed pressure)
Sugar(s) ⇌ sugar(solution)	concentration of solute (solubility)
CO₂(g) ⇌ CO₂(aq)	[CO₂(aq)] / [CO₂(g)]

General characteristics of physical equilibria:

1. Possible only in a closed system at a given temperature.
2. Opposing processes occur at the same rate: dynamic but stable.
3. All measurable properties stay constant.
4. One parameter has a constant value at a given T (the table above).
5. Its size at any stage shows how far the process has gone.

Example (2023) · Correct statements about physical equilibria: closed system only (yes); same rate both ways (yes); "the value of all its parameters" constant (no: NCERT says one parameter has a constant value, point 4); solubility constant at a given T (yes). Answer 3.

Example (2025) · Ice and water at 273.15 K, 1 atm; pressure doubled at constant T. Ice is less dense than water, so higher pressure favours the liquid (smaller volume) and the ice melts completely. Le Chatelier applied to a physical equilibrium.

2 · Equilibrium in chemical processes

For A + B ⇌ C + D, the forward rate falls as A and B are used up and the reverse rate rises as C and D build up. When the two rates are equal the concentrations stop changing: chemical equilibrium, also dynamic.

At equilibrium the rates are equal and the concentrations constant, not equal to each other: the curves level off. (2025 asked which graph shows this; 2020 asked for the rate–time graph: forward rate falls and reverse rate rises until they meet and stay equal.)

• From either side: H₂ + I₂ ⇌ 2HI reaches the same equilibrium mixture starting from H₂ and I₂ or from HI alone, provided the total numbers of H and I atoms in the volume are the same. NH₃ can likewise be reached from N₂ + H₂ or from NH₃.
• Proof that it is dynamic: Haber's synthesis run once with H₂ and once with D₂ under identical conditions gives the same composition. Mix the two equilibrium mixtures: the amount of ammonia doesn't change, but the mass spectrometer finds NH₃, NH₂D, NHD₂, ND₃ and H₂, HD, D₂. The H and D atoms can only scramble if the forward and reverse reactions keep running.
• Three kinds of extent: reactions that go nearly to completion; reactions that barely start; reactions with comparable amounts of reactants and products at equilibrium. The extent depends on conditions (concentration, temperature).

3 · Law of chemical equilibrium and the equilibrium constant

Guldberg and Waage (1864), the law of mass action ("active mass" was the old name for concentration). For the general reaction aA + bB ⇌ cC + dD:

Equilibrium law: at a given temperature, the product of the products' concentrations, each raised to its stoichiometric coefficient, divided by the same product for the reactants, is constant. The concentrations are equilibrium values in mol L⁻¹.

• NCERT's evidence: six H₂ + I₂ experiments at 731 K, four from H₂ and I₂ and two from HI. [HI]/[H₂][I₂] varied, but [HI]²/[H₂][I₂] was constant. The powers are the coefficients.
• 4NH₃ + 5O₂ ⇌ 4NO + 6H₂O: Kc = [NO]⁴[H₂O]⁶ / [NH₃]⁴[O₂]⁵.

K depends on how the equation is written (NCERT Table 6.4):

Equation	Constant
aA + bB ⇌ cC + dD	Kc
reversed	1/Kc
multiplied by n	Kcⁿ
divided by 2	√Kc
sum of two equations	K₁ × K₂ (Beyond NCERT as a rule; follows from the law)

So always quote K with its balanced equation.

Example (NCERT Problem 6.1, asked in 2024 with new numbers) · N₂ + 3H₂ ⇌ 2NH₃ at 500 K, [N₂] 1.5 × 10⁻², [H₂] 3.0 × 10⁻², [NH₃] 1.2 × 10⁻² M. Kc = (1.2 × 10⁻²)² / [(1.5 × 10⁻²)(3.0 × 10⁻²)³] = 3.55 × 10². With [N₂] 2 × 10⁻², [H₂] 3 × 10⁻², [NH₃] 1.5 × 10⁻² (2024): 417.

Example (NCERT Problem 6.2) · N₂ + O₂ ⇌ 2NO at 800 K: Kc = (2.8 × 10⁻³)² / (3.0 × 10⁻³ × 4.2 × 10⁻³) = 0.622.

Example (2024) · SO₃ ⇌ SO₂ + ½O₂, Kc = 4.9 × 10⁻². For 2SO₂ + O₂ ⇌ 2SO₃: reverse and double, K = 1/(4.9 × 10⁻²)² = 416.

Example (2023) · K₁ for N₂ + 3H₂ ⇌ 2NH₃ = 4 × 10⁵; K₂ for N₂ + O₂ ⇌ 2NO = 1.6 × 10¹²; K₃ for H₂ + ½O₂ ⇌ H₂O = 1.0 × 10⁻¹³. Target 2NH₃ + 5/2 O₂ ⇌ 2NO + 3H₂O = −(1) + (2) + 3 × (3), so K = K₂K₃³/K₁ = 1.6 × 10¹² × 10⁻³⁹ / (4 × 10⁵) = 4 × 10⁻³³.

Example (2019) · S + O₂ ⇌ SO₂, K₁ = 10⁵²; 2S + 3O₂ ⇌ 2SO₃, K₂ = 10¹²⁹. For 2SO₂ + O₂ ⇌ 2SO₃: K₂/K₁² = 10¹²⁹⁻¹⁰⁴ = 10²⁵.

K from rate constants · Beyond NCERT (2021, 2023) · For an elementary reversible reaction, at equilibrium the forward and backward rates are equal, so K = k(forward) / k(backward). (2021: rate = 4.8 × 10⁻⁵[PtCl₄²⁻] − 2.4 × 10⁻³[Pt(H₂O)Cl₃⁻][Cl⁻] gives K = 0.02, so 1/K = 50.)

4 · Homogeneous equilibria and Kp

Homogeneous: all species in one phase. Gases (N₂ + 3H₂ ⇌ 2NH₃) or one solution (CH₃COOC₂H₅ + H₂O ⇌ CH₃COOH + C₂H₅OH; Fe³⁺ + SCN⁻ ⇌ Fe(SCN)²⁺).

For gases, use partial pressures. From pV = nRT, p = (n/V)RT = cRT with c in mol L⁻¹, p in bar and R = 0.0831 bar L mol⁻¹ K⁻¹. Substituting c = p/RT into Kc:

• Δn = 0 (H₂ + I₂ ⇌ 2HI, N₂ + O₂ ⇌ 2NO): Kp = Kc.
• N₂ + 3H₂ ⇌ 2NH₃: Δn = −2, Kp = Kc(RT)⁻².
• Use pressures in bar (the standard state is 1 bar; 1 bar = 10⁵ Pa). R must match the pressure unit: 0.0831 L bar or 0.0821 L atm.

Example (NCERT Problem 6.5) · 2NOCl ⇌ 2NO + Cl₂, Kc = 3.75 × 10⁻⁶ at 1069 K, Δn = 1. Kp = 3.75 × 10⁻⁶ × 0.0831 × 1069 = 0.033.

Example (NCERT exercise 6.10) · 2SO₂ + O₂ ⇌ 2SO₃, Kp = 2.0 × 10¹⁰ bar⁻¹ at 450 K, Δn = −1. Kc = Kp × RT = 2.0 × 10¹⁰ × 37.4 = 7.48 × 10¹¹ L mol⁻¹.

Example (2021, 2024) · N₂O₄ ⇌ 2NO₂, Kc = 20.4 and Kp = 600.1: RT = 29.42, T = 29.42 / 0.0831 = 354 K. Kp = 0.492 atm at 300 K: Kc = 0.492 / 24.6 = 0.02.

Example (2024, 2020) · CO + ½O₂ ⇌ CO₂: Δn = −½, Kp/Kc = 1/√(RT). Fe₂N(s) + 3/2 H₂ ⇌ 2Fe(s) + NH₃: only gases count, Δn = −½, so Kc = Kp(RT)^½.

Example (2026) · xA ⇌ yB at 400 K (RT = 33.2). Kp/Kc = 85.87/2.586 = 33.2 → Δn = +1 → x = 1, y = 2. Kp/Kc = 0.862/28.62 = 1/33.2 → Δn = −1 → x = 2, y = 1.

Units of K: Kc in (mol L⁻¹)^Δn, Kp in bar^Δn. None when Δn = 0 (H₂ + I₂ ⇌ 2HI); N₂O₄ ⇌ 2NO₂ has Kc in mol L⁻¹ and Kp in bar. Measured against standard states (1 bar for a gas, 1 M for a solute), K becomes dimensionless, with different numbers for Kp and Kc.

5 · Heterogeneous equilibria

More than one phase: H₂O(l) ⇌ H₂O(g); Ca(OH)₂(s) + (aq) ⇌ Ca²⁺(aq) + 2OH⁻(aq).

The molar concentration of a pure solid or pure liquid is constant whatever its amount, so it is left out of K. Gases and solutes stay in.

• CaCO₃(s) ⇌ CaO(s) + CO₂(g): Kc = [CO₂], Kp = p(CO₂). At 1100 K, p(CO₂) = 2.0 × 10⁵ Pa, so Kp = 2.00. At a given T the CO₂ pressure over CaCO₃ and CaO is fixed.
• Ni(s) + 4CO(g) ⇌ Ni(CO)₄(g): Kc = [Ni(CO)₄]/[CO]⁴ (nickel purification).
• Ag₂O(s) + 2HNO₃(aq) ⇌ 2AgNO₃(aq) + H₂O(l): Kc = [AgNO₃]²/[HNO₃]².
• The solid must still be present, however little, for the equilibrium to exist.
• Adding more of a solid doesn't shift the equilibrium (2024: adding Fe₂O₃ to the blast-furnace equilibrium changes nothing).

Example (NCERT Problem 6.6) · CO₂(g) + C(s) ⇌ 2CO(g), Kp = 3.0 at 1000 K, p(CO₂) initially 0.48 bar. (2x)²/(0.48 − x) = 3, so 4x² + 3x − 1.44 = 0 and x = 0.33. p(CO) = 0.66 bar, p(CO₂) = 0.15 bar. NCERT's last two lines print both as "pCO₂": the 0.66 bar is CO.

Example (2025) · Same reaction, p(CO₂) = 0.5 atm, total at equilibrium 0.8 atm. 0.5 − x + 2x = 0.8, x = 0.3: p(CO) 0.6, p(CO₂) 0.2. Kp = 0.36/0.2 = 1.8 atm.

Example (2019) · 5.1 g NH₄SH (0.1 mol) in 3.0 L at 600 K, 30% decomposes: 0.03 mol each of NH₃ and H₂S. Each p = 0.03 × 0.082 × 600 / 3 = 0.492 atm. Kp = 0.492² = 0.242 atm². Only the gases count.

Example (2019) · A(s) ⇌ B(g) + C(g), Kp₁ = x; D(s) ⇌ C(g) + E(g), Kp₂ = y, both together. p(C) = p(B) + p(E), and p(B)p(C) + p(C)p(E) = x + y, so p(C)² = x + y. Total = 2p(C) = 2√(x + y).

Example (2026) · CaCO₃ ⇌ CaO + CO₂, Kp₁ = 0.08 atm; C + CO₂ ⇌ 2CO, Kp₂ = 2 atm, together. p(CO₂) = 0.08, so p(CO)² = 2 × 0.08 = 0.16 and p(CO) = 0.4 atm.

6 · Applications of the equilibrium constant

Features of K:

1. It applies only once concentrations have become constant (equilibrium).
2. It doesn't depend on the initial concentrations.
3. It depends on temperature: one value for a given balanced equation at a given T.
4. The reverse reaction's K is 1/K.
5. Multiplying or dividing the equation changes K as in section 3.

Extent of reaction: large K means mostly products. K says nothing about the rate.

Kc	Composition	NCERT examples
> 10³	products predominate; nearly complete	H₂ + O₂ at 500 K, 2.4 × 10⁴⁷; H₂ + Cl₂, 4.0 × 10³¹; H₂ + Br₂, 5.4 × 10¹⁸
10⁻³ to 10³	appreciable amounts of both	H₂ + I₂ ⇌ 2HI at 700 K, 57.0; N₂O₄ ⇌ 2NO₂ at 25 °C, 4.64 × 10⁻³
< 10⁻³	reactants predominate; barely goes	H₂O → H₂ + O₂ at 500 K, 4.1 × 10⁻⁴⁸; N₂ + O₂ ⇌ 2NO at 298 K, 4.8 × 10⁻³¹

Reaction quotient Q: the same expression as K, with the concentrations (Qc) or pressures (Qp) at any moment, not necessarily at equilibrium.

Q always moves towards K.

Example (NCERT) · H₂ + I₂ ⇌ 2HI, Kc = 57.0 at 700 K, with 0.10 M H₂, 0.20 M I₂, 0.40 M HI. Qc = 0.40² / (0.10 × 0.20) = 8.0 < 57.0, so more HI forms.

Example (NCERT Problem 6.7) · 2A ⇌ B + C, Kc = 2 × 10⁻³, all at 3 × 10⁻⁴ M: Qc = 1 > Kc, so it goes backward.

Example (NCERT exercise 6.12) · 1.57 mol N₂, 1.92 mol H₂, 8.13 mol NH₃ in 20 L at 500 K, Kc = 1.7 × 10². Qc = 0.4065² / (0.0785 × 0.096³) = 2.38 × 10³ > Kc: not at equilibrium, net reaction backward.

Calculating equilibrium concentrations (NCERT's steps):

1. Write the balanced equation.
2. Under it, make a table: initial, change, equilibrium. Let x be the amount (mol L⁻¹) of one substance that reacts; get the others from the coefficients.
3. Put the equilibrium values into the K expression; solve for x. Of two roots, keep the one that makes chemical sense (no negative amounts, none above the start).
4. Find each equilibrium concentration.
5. Check by putting them back into K.

Work in moles first if the volume is 1 L or cancels; convert to concentration or partial pressure before using K. When Δn ≠ 0, the volume (or total pressure) matters.

Example (NCERT Problem 6.4) · CO + H₂O ⇌ CO₂ + H₂, Kc = 4.24 at 800 K, 0.10 M each of CO and H₂O. x²/(0.1 − x)² = 4.24. Take the square root: x/(0.1 − x) = 2.06, so x = 0.067 M CO₂ and H₂; CO and H₂O = 0.033 M. (NCERT solves the quadratic; the square root is faster when both sides are perfect squares.)

Example (NCERT Problem 6.9) · 3.00 mol PCl₅ in 1 L at 380 K, Kc = 1.80. x²/(3 − x) = 1.8, so x² + 1.8x − 5.4 = 0 and x = 1.59. [PCl₅] = 1.41 M, [PCl₃] = [Cl₂] = 1.59 M. (2021 used Kc = 1.844: PCl₅ left = 1.40 mol, answer 1400 × 10⁻³.)

Example (NCERT Problem 6.8) · 13.8 g N₂O₄ (0.15 mol) in 1 L at 400 K: p = 0.15 × 0.083 × 400 = 4.98 bar. Total at equilibrium 9.15 bar = 4.98 + x, so x = 4.17. p(N₂O₄) = 0.81, p(NO₂) = 8.34 bar. Kp = 8.34²/0.81 = 85.87; Kc = 85.87 / (0.083 × 400) = 2.6.

Example (NCERT exercise 6.14, asked in 2023) · 1 mol each of H₂O and CO in 10 L at 725 K; 40% of the water reacts. 0.4 mol each of H₂ and CO₂, 0.6 mol each of H₂O and CO. Δn = 0, so the volume cancels: Kc = 0.16/0.36 = 0.44.

Example (2021) · A + B ⇌ 2C, K = 100, all three start at 1 M. Q = 1 < K, forward. (1 + 2x)²/(1 − x)² = 100, so (1 + 2x)/(1 − x) = 10 and x = 0.75. [C] = 2.5 M.

Example (2022) · 4.0 mol Ar and 5.0 mol PCl₅ in 100 L at 610 K; total pressure at equilibrium 6.0 atm. Total moles = PV/RT = 600 / (0.082 × 610) = 12, so 9 + x = 12 and x = 3. PCl₅ 2, PCl₃ 3, Cl₂ 3 mol, each pᵢ = (nᵢ/12) × 6. Kp = 1.5 × 1.5 / 1.0 = 2.25. The argon counts in the total moles and pressure, not in K.

Example (2025) · 37.8 g N₂O₅ (0.35 mol) in 1 L at 500 K: p₀ = 0.35 × 0.082 × 500 = 14.35 bar. 2N₂O₅ ⇌ 2N₂O₄ + O₂: with p(O₂) = x, total = 14.35 + x = 18.65, so x = 4.3. p(N₂O₅) 5.75, p(N₂O₄) 8.6. Kp = 8.6² × 4.3 / 5.75² = 9.62.

Degree of dissociation α · Beyond NCERT

α = fraction of the starting moles that dissociate. From 1 mol at total equilibrium pressure P:

Reaction	Moles at equilibrium	Kp
A ⇌ B + C (PCl₅)	1 − α, α, α; total 1 + α	α²P / (1 − α²)
A ⇌ 2B (N₂O₄)	1 − α, 2α; total 1 + α	4α²P / (1 − α²)
2A ⇌ B + C (2HI)	1 − α, α/2, α/2	α² / 4(1 − α)² (no P)

• For A ⇌ B + C, if α ≪ 1 then α ≈ √(Kp/P): α falls as P rises (2025).
• When K is small, drop α beside 1 and solve the simple power. 2025: H₂O ⇌ H₂ + ½O₂, Kp = 8 × 10⁻³, P = 1 bar. Kp ≈ α(α/2)^½ = α^(3/2)/√2, so α^(3/2) = 0.0113 and α = 0.05.
• From vapour density: α = (D − d) / ((n − 1)d), where D = vapour density before dissociation, d = at equilibrium, n = moles of product from 1 mol.

Example (2021) · Cl₂ ⇌ 2Cl at 1 atm with equal numbers of Cl₂ and Cl: each p = 0.5 atm. Kp = 0.25/0.5 = 0.5 atm.

Example (2021) · AB₂ ⇌ A + 2B, 1 mol in 25 L at 300 K, equilibrium pressure 1.9 atm. p₀ = 1 × 0.08206 × 300 / 25 = 0.985 atm. Total = p₀(1 + 2α) = 1.9, so α = 0.465. Kp = (p₀α)(2p₀α)² / p₀(1 − α) = 4p₀²α³/(1 − α) = 0.73.

Adding more of one species at equilibrium: find K from the first equilibrium, then run a new ICE table from the disturbed amounts.

Example (2026) · X₂ + Y₂ ⇌ 2Z at equilibrium with 3, 3, 9 mol in 1 L: K = 81/9 = 9. Add 10 mol Z: (19 − 2x)²/(3 + x)² = 9, so 19 − 2x = 3(3 + x) and x = 2. Z = 15 mol.

Example (2026) · A ⇌ B with 0.5 M A, 0.375 M B: K = 0.75. Add 0.1 mol A (1 L): (0.375 + x)/(0.6 − x) = 0.75, x = 0.043. A 0.557, B 0.418 M.

7 · Equilibrium constant, Q and Gibbs energy

K doesn't depend on the rate, but it is fixed by the thermodynamics of the reaction.

• ΔG < 0: spontaneous, proceeds forward. ΔG > 0: non-spontaneous; the reverse has ΔG < 0, so products turn back into reactants. ΔG = 0: equilibrium, no free energy left to drive it.

• ΔG° < 0: −ΔG°/RT is positive, K > 1, products predominate.
• ΔG° > 0: K < 1, only a little product forms.

Example (NCERT Problem 6.10) · Phosphorylation of glucose, ΔG° = 13.8 kJ mol⁻¹ at 298 K. ln Kc = −13 800 / (8.314 × 298) = −5.569, so Kc = 3.81 × 10⁻³.

Example (NCERT Problem 6.11) · Sucrose hydrolysis, Kc = 2 × 10¹³ at 300 K. ΔG° = −8.314 × 300 × ln(2 × 10¹³) = −7.64 × 10⁴ J mol⁻¹.

Example (NCERT exercise 6.24) · NO + ½O₂ ⇌ NO₂: ΔG° = ΔfG°(NO₂) − ΔfG°(NO) = 52.0 − 87.0 = −35.0 kJ mol⁻¹. log K = 35 000 / (2.303 × 8.314 × 298) = 6.13, K = 1.36 × 10⁶.

Example (2021) · A ⇌ B at 495 K, ΔG° = −9.478 kJ mol⁻¹: ln K = 9478 / (8.314 × 495) = 2.303, so K = 10. From 22 mmol A: B = 22 × 10/11 = 20 mmol.

Example (2022) · N₂O₄ 50% dissociated at 27 °C, 1 atm: Kp = 4(0.25)(1)/0.75 = 1.33. ΔG° = −8.31 × 300 × 2.3 × 0.1239 = −710 J mol⁻¹.

Example (2022) · 40% of HI decomposes at 300 K. From 1 mol: HI 0.6, H₂ 0.2, I₂ 0.2. The key's answer, 2735 J mol⁻¹, is for HI ⇌ ½H₂ + ½I₂: K = 0.2/0.6 = 1/3, ΔG° = 8.31 × 300 × 2.3 × 0.477 = 2735. For 2HI ⇌ H₂ + I₂ (K = 1/9) it would be twice that. ΔG° depends on how the equation is written.

Example (2019) · X ⇌ Y with ΔG° = 120 − (3/8)T kJ mol⁻¹: ΔG° = 0 at T = 320 K. Below 320 K, ΔG° > 0 and K < 1, so X is the major component (at 315 K, X).

K and temperature (van 't Hoff) · Beyond NCERT (also in Thermodynamics)

Example (2026) · log Kp against 1/T: 3.5, 2.5, 1.5 at 0.05, 0.06, 0.07. Slope = −100 = −ΔH°/2.303R, so ΔH°/R = 230.

8 · Factors affecting equilibria: Le Chatelier's principle

Le Chatelier's principle: a change in any of the factors that fix the equilibrium conditions makes the system change so as to reduce or counteract the effect of that change. It applies to all physical and chemical equilibria.

Only temperature changes K. Concentration, pressure, volume, inert gas and catalyst change Q (or nothing); the system then shifts until Q = K again.

1. Concentration:

• Add a reactant or product: net reaction goes the way that uses it up. Remove one: net reaction goes the way that replaces it.
• The new equilibrium amount of the added substance is less than just after adding, but more than at the start.
• In terms of Q: adding H₂ to H₂ + I₂ ⇌ 2HI makes Qc < Kc, so the forward reaction runs.
• Removing a product keeps Q < K, so the reaction keeps going: NH₃ is liquefied and removed in the Haber process; CO₂ is swept out of the lime kiln (CaCO₃ → CaO + CO₂).

NCERT's experiment: Fe³⁺ (yellow) + SCN⁻ (colourless) ⇌ [Fe(SCN)]²⁺ (deep red).

Added	Effect	Colour
KSCN	more SCN⁻, shifts right	deeper red
Oxalic acid	takes Fe³⁺ as [Fe(C₂O₄)₃]³⁻, shifts left	paler
HgCl₂	takes SCN⁻ as [Hg(SCN)₄]²⁻, shifts left	paler

2. Pressure (by changing the volume):

• Matters only for gas reactions with Δn ≠ 0. Solids and liquids are ignored: their volume hardly changes with pressure.
• Increase pressure (reduce volume): shifts to the side with fewer gas moles. Decrease it: shifts to more gas moles.
• CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g): 4 → 2 gas moles. Halve the volume: every concentration doubles, Qc = Kc/4 < Kc, so it goes forward. K itself is unchanged.
• C(s) + CO₂(g) ⇌ 2CO(g): more gas on the right, so higher pressure pushes it backward.

3. Inert gas:

Added at	Effect	Why
Constant volume	none	partial pressures and concentrations unchanged
Constant pressure	shifts to more gas moles	volume grows, like lowering the pressure · Beyond NCERT

Example (2025) · Xenon added to PCl₅ ⇌ PCl₃ + Cl₂ at constant T and p: the volume grows, the reaction shifts to more moles, so PCl₃ (and Cl₂) increase.

4. Temperature: the only factor that changes K. Raising T favours the endothermic direction.

Reaction	Raise T	K
Exothermic (ΔH < 0)	shifts backward	decreases
Endothermic (ΔH > 0)	shifts forward	increases

• N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92.38 kJ mol⁻¹: low T gives a better yield, but the rate is then too slow, so a catalyst is used.
• NCERT's experiment: 2NO₂ (brown) ⇌ N₂O₄ (colourless), ΔH = −57.2 kJ mol⁻¹. In a freezing mixture the brown fades (N₂O₄ favoured); in hot water (363 K) it deepens.
• Endothermic example: [Co(H₂O)₆]²⁺ (pink) + 4Cl⁻ ⇌ [CoCl₄]²⁻ (blue) + 6H₂O. Blue at room temperature, pink in a freezing mixture. (NCERT prints the pink ion as 3+; cobalt here is +2.)

5. Catalyst:

• Opens a new, lower-energy path. Lowers the activation energy of the forward and reverse reactions by the same amount, so both speed up equally.
• Equilibrium is reached faster; K and the equilibrium composition don't change. A catalyst doesn't appear in the equation or in K.
• With an exceedingly small K, a catalyst is of little help.

Industrial conditions:

Process	Reaction	Conditions
Haber	N₂ + 3H₂ ⇌ 2NH₃, exothermic, Δn = −2	iron catalyst; about 500 °C and 200 atm (high pressure raises the yield)
Contact	2SO₂ + O₂ ⇌ 2SO₃, Kc = 1.7 × 10²⁶	Pt or V₂O₅ catalyst: K is huge but the reaction is very slow

Example (2019) · For 2SO₂ + O₂ ⇌ 2SO₃ (exothermic, Kc huge), the incorrect statement: "K is large, so no catalyst is needed." A large K says nothing about the rate.

Example (2023) · 2SO₂ + O₂ ⇌ 2SO₃, ΔH = −190 kJ. Factors that raise the yield of SO₃: higher pressure, more SO₂, more O₂ = 3. Higher temperature lowers it; a catalyst doesn't change it.

Example (2020) · N₂O₄ ⇌ 2NO₂, ΔH° = +58 kJ. Lower the temperature: shifts towards N₂O₄. Raise the pressure by adding N₂ at constant volume: no change.

Example (NCERT exercise 6.28) · CH₄ + H₂O ⇌ CO + 3H₂, endothermic. Higher pressure: shifts back, Kp unchanged. Higher temperature: shifts forward, Kp increases. Catalyst: neither changes; equilibrium is reached sooner.

9 · Traps

Trap	What goes wrong	Do this
Solids and liquids	Putting CaCO₃(s) or H₂O(l) in K	Only gases and solutes; H₂O counts only as a gas or when not the solvent
Δn in Kp = Kc(RT)^Δn	Counting solid or liquid moles	Gas moles only, products − reactants
R in Kp ↔ Kc	8.314 instead of 0.0821/0.0831	Use R in L atm or L bar to match the pressure
Moles in Kc	Using moles when Δn ≠ 0	Divide by the volume first
Equation as written	Quoting ΔG° or K for the wrong form (2022 HI)	Write the exact equation; halving it halves ΔG°
"K changes with pressure"	Saying K rises when the yield rises	Only temperature changes K
Inert gas	Shifting at constant volume	Constant V: nothing; constant P: towards more gas moles
Catalyst	Saying it raises the yield	Same composition, reached sooner
Quadratic roots	Keeping the root larger than the start	Take the one that keeps every amount positive
Total moles with an inert gas	Leaving Ar out of the mole fraction	Ar is in n(total) and P, not in K
Q vs K direction	Mixing up the inequality	Q < K → forward; Q always moves to K
Same concentrations ≠ equilibrium	Thinking [A] = [B] at equilibrium	Rates are equal; concentrations need not be

10 · Numbers worth knowing

• R = 0.0821 L atm K⁻¹ mol⁻¹ = 0.0831 L bar = 8.314 J K⁻¹ mol⁻¹. 1 atm = 1.013 bar; 1 bar = 10⁵ Pa.
• RT (L atm): 24.6 at 300 K; 24.5 at 298 K; 33.2 at 400 K (L bar, R = 0.083).
• 2.303 RT = 5.71 kJ mol⁻¹ at 298 K; 5.74 at 300 K.
• K > 10³ mostly products; K < 10⁻³ mostly reactants.
• Haber: Fe catalyst, ~500 °C, ~200 atm, ΔH = −92.4 kJ mol⁻¹. Contact: V₂O₅ or Pt.
• 2NO₂ ⇌ N₂O₄: ΔH = −57.2 kJ mol⁻¹.

11 · For the boards only

JEE Main doesn't ask this, but the CBSE paper can.

• Equilibria with haemoglobin carry O₂ from the lungs to the muscles; a similar equilibrium with CO explains why CO is toxic.
• Optimising conditions so that equilibrium favours the product matters in industry and the laboratory (H₂, NH₃, CaO). World ammonia production is about a hundred million tonnes a year, mostly for fertilisers.
• NCERT's student activity: two measuring cylinders, coloured water moved back and forth with two glass tubes until the levels stop changing, while transfer continues. Tubes of different diameters give different final levels. It models a dynamic equilibrium.
• NCERT exercise 6.1 (sealed liquid, volume suddenly increased): the vapour pressure drops at first; evaporation continues at the same rate while condensation slows; at the new equilibrium the vapour pressure returns to its original value, which depends only on T.