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State Board · Class 11 · Chemistry

Lesson 12: Chemical Equilibrium

7 concepts · 56 practice questions with answers · Free to read

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Reversible reactions and dynamic equilibrium

Link to Std 9–10 : In Std 9–10 reactions went one way; here we study reactions that run both ways.

From daily life

People keep going up and down a two-storey building yet the head-count on each floor stays the same: that is dynamic equilibrium.

Simple definition

A reversible reaction runs in both directions (⇌). At equilibrium forward rate = reverse rate and measurable properties (concentration, pressure, colour) stay constant. The reaction does not stop, hence dynamic.

Dynamic equilibriumDynamic equilibrium: rates and concentrationsRatetime →forwardreverse=Concentrationreactantproducttime →At equilibrium both rates are equal; concentrations stay constant

In the figure : Rates and concentrations level off at equilibrium.

Physical equilibriaPhysical equilibria (four types)Solid ⇌ liquidice ⇌ water273 K, 1 atmLiquid ⇌ vapourwater ⇌ vapourclosed vesselSolid ⇌ vapourI₂(s) ⇌ I₂(g)camphorSolute ⇌ solutionsugar(s) ⇌ sugar(aq)saturated solutionA measurable property stays constant (e.g. vapour pressure)

In the figure : Types of physical equilibria.

Rule 1: Features of equilibrium — Reached only in a closed vessel; same state from either side; a catalyst speeds up reaching equilibrium but does not change its position.

  • Easy : H₂ + I₂ ⇌ 2HI — the same mixture whichever side you start from.
  • Medium : N₂O₄ (colourless) ⇌ 2NO₂ (brown): constant colour means equilibrium.
  • Exam-type : In an open vessel CaCO₃ → CaO + CO₂ goes to completion, no equilibrium.

Rule 2: Physical equilibria — Solid ⇌ liquid (ice/water at 273 K); liquid ⇌ vapour (constant vapour pressure); solid ⇌ vapour; solution (saturated).

  • Easy : Soda in a sealed bottle: CO₂(g) ⇌ CO₂(aq).
  • Medium : Fizz on opening — equilibrium is disturbed.
  • Exam-type : In saturated sugar solution the mass of added crystals stays the same.

Rule 3: Chemical equilibrium — Both rates equal, no net change visible. Large K ⇒ more product; small K ⇒ more reactant.

  • Easy : N₂ + O₂ ⇌ 2NO: K tiny (≈ 10⁻³⁰ at 298 K), negligible product.
  • Medium : K > 10³ ⇒ nearly complete.
  • Exam-type : K near 1 ⇒ significant amounts of both.

Why are the rates equal at equilibrium?

  • Step 1: At first reactants are plentiful ⇒ forward rate is high.
  • Step 2: Products build up ⇒ reverse rate rises; reactants fall ⇒ forward rate drops.
  • Step 3: The two rates meet ⇒ concentrations are constant.

CET level (MCQ)

  • CET 1) At equilibrium? 1) reaction stops 2) forward rate = reverse rate 3) reactants are used up 4) products are used up — Answer: 2).
  • CET 2) A catalyst? 1) raises K 2) shifts position 3) speeds up reaching equilibrium 4) raises yield — Answer: 3).

Real-life use: Ammonia and sulphuric acid industries; oxygen transport in blood (Hb + O₂ ⇌ HbO₂).

Remember

⇌ = both ways; equal rates; constant concentrations; closed vessel; catalyst does not shift position.

Common mistake

Writing that concentrations are equal; they are constant, not equal.

CET trap

'Reaction stops at equilibrium' is wrong; it is dynamic.

Shortcut

Memory clue: "rates equal, concentrations constant".

Spot the difference

TypeExampleConstant property
Solid ⇌ liquidice/water273 K
Liquid ⇌ vapourwater/vapourvapour pressure
ChemicalN₂O₄ ⇌ 2NO₂colour

Practice questions

Easy
  1. 1What is a reversible reaction?
  2. 2Why is equilibrium called dynamic?
  3. 3Effect of a catalyst on equilibrium?
Medium
  1. 4How do you recognise equilibrium in N₂O₄ ⇌ 2NO₂?
  2. 5Why is a closed vessel needed?
  3. 6What does a very small K mean?
Challenging
  1. 7State two types of physical equilibrium.
  2. 8Difference of CaCO₃ decomposition in open and closed vessels.
Show answers (try them yourself first)
  1. 1A reaction that runs in both directions.
  2. 2Both rates continue.
  3. 3Only speeds up; position same.
  4. 4When the brown colour stays constant.
  5. 5So that nothing escapes.
  6. 6Negligible product.
  7. 7Solid ⇌ liquid, liquid ⇌ vapour.
  8. 8Closed: equilibrium; open: complete decomposition.

Law of mass action, Kc and Kp

From daily life

In cooking, when salt and chilli are in the right ratio the taste is fixed; at equilibrium the concentration ratio K is fixed.

Simple definition

For aA + bB ⇌ cC + dD, Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ) (in mol/L). For gases Kp uses partial pressures. Kp = Kc (RT)^Δn, Δn = gaseous product mol − reactant mol.

Kc and KpN₂ + 3H₂ ⇌ 2NH₃: Kc and KpKc = [NH₃]² / ([N₂][H₂]³)Kp = (p NH₃)² / (p N₂ · (p H₂)³)Kp = Kc (RT)^ΔnΔn = (gaseous product mol) − (reactant mol) = 2 − 4 = −2Example: Kc = 0.36 at 500 K, R = 0.0821Kp = 0.36 × (0.0821 × 500)⁻² = 2.14 × 10⁻⁴Pure solids and liquids are left out of K

In the figure : Kc and Kp for N₂ + 3H₂ ⇌ 2NH₃.

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Law of mass action, Kc and Kp

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Rule 1: Rules for writing Kc — Pure solids and liquids are omitted; coefficients become powers; K depends only on temperature.

  • Easy : N₂ + 3H₂ ⇌ 2NH₃: Kc = [NH₃]² / ([N₂][H₂]³).
  • Medium : CaCO₃(s) ⇌ CaO(s) + CO₂(g): Kc = [CO₂], Kp = p(CO₂).
  • Exam-type : Unit of Kc for N₂ + 3H₂ ⇌ 2NH₃ is (mol/L)⁻².

Rule 2: Relation between Kp and Kc — Kp = Kc (RT)^Δn, R = 0.0821 L atm K⁻¹ mol⁻¹ (pressure in atm). Δn = 0 ⇒ Kp = Kc.

  • Easy : H₂ + I₂ ⇌ 2HI: Δn = 0 ⇒ Kp = Kc.
  • Medium : Kc = 4, Δn = 1, 300 K: Kp = 4 × 24.63 = 98.5.
  • Exam-type : Kc = 0.36 (500 K), Δn = −2: Kp = 0.36 × (41.05)⁻² = 2.14 × 10⁻⁴.

Rule 3: Manipulating K — Reverse the reaction: K' = 1/K; multiply coefficients by n: K' = Kⁿ; add two reactions: K = K₁ × K₂.

  • Easy : K = 50 for H₂ + I₂ ⇌ 2HI; reverse: 1/50 = 0.02.
  • Medium : ½H₂ + ½I₂ ⇌ HI: K' = √50 = 7.07.
  • Exam-type : A ⇌ B (K₁ = 4), B ⇌ C (K₂ = 5) ⇒ A ⇌ C: K = 20.

Derivation of Kp = Kc(RT)^Δn (ideal gases)

  • Step 1: pV = nRT ⇒ p = (n/V)RT = CRT, so pᵢ = [i]RT.
  • Step 2: Put pᵢ = [i]RT in Kp = (pC)ᶜ(pD)ᵈ / ((pA)ᵃ(pB)ᵇ).
  • Step 3: The power of RT is (c + d) − (a + b) = Δn ⇒ Kp = Kc (RT)^Δn.

CET level (MCQ)

  • CET 1) In which reaction Kp = Kc? 1) N₂ + 3H₂ ⇌ 2NH₃ 2) PCl₅ ⇌ PCl₃ + Cl₂ 3) H₂ + I₂ ⇌ 2HI 4) 2SO₂ + O₂ ⇌ 2SO₃ — Answer: 3). (Δn = 0.)
  • CET 2) If K = 50, K of the reverse reaction? 1) 50 2) 25 3) 0.02 4) 100 — Answer: 3).

Real-life use: Predicting yield in industrial reactions from K.

Remember

Solids/liquids omitted; Kp = Kc(RT)^Δn; reverse ⇒ 1/K; adding ⇒ multiply.

Common mistake

Counting liquids/solids in Δn; use gaseous moles only.

CET trap

Match R to the pressure unit: 0.0821 (atm) or 8.314 (SI).

Shortcut

Memory clue: "products on top, reactants below".

Spot the difference

ItemKcKp
Unitmol/Latm
Used forsolutions/gasesgases only
Depends onTT

Practice questions

Easy
  1. 1Write Kc for 2SO₂ + O₂ ⇌ 2SO₃.
  2. 2When are Kp and Kc equal?
  3. 3Kc for CaCO₃ ⇌ CaO + CO₂.
Medium
  1. 4Kc = 4, Δn = 1, 300 K: find Kp.
  2. 5Why does K need a temperature?
  3. 6A ⇌ B (K = 4) reversed: K?
Challenging
  1. 7K' of ½N₂ + 3/2 H₂ ⇌ NH₃ if K = 0.36?
  2. 8K₁ = 4, K₂ = 5: combined K?
Show answers (try them yourself first)
  1. 1Kc = [SO₃]² / ([SO₂]²[O₂]).
  2. 2When Δn = 0.
  3. 3Kc = [CO₂].
  4. 498.5.
  5. 5K depends on temperature.
  6. 60.25.
  7. 7√0.36 = 0.6.
  8. 820.

Equilibrium calculations and comparing Q with K

From daily life

A shopkeeper looks at the stock-to-customer ratio to decide whether to reorder; similarly Q tells the direction.

Simple definition

Reaction quotient Q has the K expression at any moment. Q < K ⇒ forward; Q > K ⇒ reverse; Q = K ⇒ equilibrium.

Q versus KComparing Q and K: which way will it go?KQ < KQ > Kforward (more product)reverse (more reactant)H₂ + I₂ ⇌ 2HI, Kc = 50[H₂]=[I₂]=1, [HI]=5 ⇒ Q = 25 < 50 ⇒ forward

In the figure : Direction of reaction from Q and K.

Rule 1: ICE table — Initial, Change, Equilibrium — use x for the change and put equilibrium concentrations in K.

  • Easy : H₂ + I₂ ⇌ 2HI: (1 − x), (1 − x), 2x.
  • Medium : K = 50 ⇒ 2x/(1 − x) = √50 = 7.071.
  • Exam-type : x = 7.071/9.071 = 0.7795 ⇒ [HI] = 1.559 M.

Rule 2: Direction from Q — Compute Q and compare it with K.

  • Easy : [H₂] = [I₂] = 1, [HI] = 5 ⇒ Q = 25 < 50 ⇒ forward.
  • Medium : [HI] = 10 ⇒ Q = 100 > 50 ⇒ reverse.
  • Exam-type : [HI] = 7.07 ⇒ Q = 50 ⇒ equilibrium.

Rule 3: Magnitude of K — K > 10³: nearly complete; K < 10⁻³: hardly any; in between: both present.

  • Easy : H₂ + Cl₂ ⇌ 2HCl: K very large.
  • Medium : N₂ + O₂ ⇌ 2NO: K extremely small.
  • Exam-type : K = 50: products favoured.

Calculation: H₂ + I₂ ⇌ 2HI (K = 50) from 1 M each

  • Step 1: K = (2x)² / (1 − x)² = 50.
  • Step 2: Take square roots: 2x/(1 − x) = 7.071.
  • Step 3: x = 0.7795; [H₂] = [I₂] = 0.2205 M, [HI] = 1.559 M.

CET level (MCQ)

  • CET 1) If Q > K? 1) forward 2) reverse 3) equilibrium 4) nothing — Answer: 2).
  • CET 2) H₂ + I₂ ⇌ 2HI (Kc = 50), [H₂] = [I₂] = 1, [HI] = 5: Q? 1) 5 2) 10 3) 25 4) 50 — Answer: 3).

Real-life use: Steering industrial reactions; K for drug binding.

Remember

Q < K forward; Q > K reverse; ICE; the square-root trick.

Common mistake

Mixing up Q and K: Q uses current concentrations, K only equilibrium ones.

CET trap

Take square roots when it is a perfect square; otherwise use the quadratic formula.

Shortcut

Memory clue: "small Q goes forward".

Spot the difference

ComparisonDirection
Q < Kforward
Q = Kequilibrium
Q > Kreverse

Practice questions

Easy
  1. 1Difference between Q and K?
  2. 2Q = 25, K = 50: direction?
  3. 3Expand ICE.
Medium
  1. 4Find Q when [HI] = 10.
  2. 5Find equilibrium concentrations from 1 M each, K = 50.
  3. 6Meaning of K = 10⁻³⁰?
Challenging
  1. 7What if Q = K?
  2. 8When does the square-root trick work?
Show answers (try them yourself first)
  1. 1Q at any moment, K at equilibrium.
  2. 2Forward.
  3. 3Initial, Change, Equilibrium.
  4. 4100.
  5. 5[H₂] = [I₂] = 0.2205 M; [HI] = 1.559 M.
  6. 6Negligible product.
  7. 7At equilibrium.
  8. 8When both sides are perfect squares.

Le Chatelier's principle

From daily life

On a bicycle in a headwind you lean forward; similarly equilibrium shifts to oppose a change.

Simple definition

Principle: if a change (concentration, pressure, temperature) is imposed on an equilibrium it shifts to oppose the change. Only temperature alters K.

Le ChatelierLe Chatelier: N₂ + 3H₂ ⇌ 2NH₃ (ΔH < 0)ChangeShiftKAdd N₂forward →sameRemove NH₃forward →sameRaise Pforward → (4 → 2 mol)sameRaise Treverse ← (exothermic)fallsCatalystno shiftsameOnly temperature changes K

In the figure : Effects on N₂ + 3H₂ ⇌ 2NH₃.

Live figure

Le Chatelier's principle

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Rule 1: Concentration and pressure — Add a substance ⇒ shift toward consuming it; remove ⇒ shift toward making it. Raise pressure ⇒ side with fewer gas moles; Δn = 0 ⇒ no effect.

  • Easy : Add N₂ ⇒ forward.
  • Medium : Raise P ⇒ more NH₃ (4 → 2 mol).
  • Exam-type : H₂ + I₂ ⇌ 2HI: pressure has no effect.

Rule 2: Temperature — Exothermic (ΔH < 0): raise T ⇒ reverse, K falls. Endothermic: raise T ⇒ forward, K rises.

  • Easy : N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92 kJ: high T lowers NH₃.
  • Medium : N₂O₄ ⇌ 2NO₂, ΔH > 0: heating deepens the brown.
  • Exam-type : Ice ⇌ water: raise T ⇒ water.

Rule 3: Catalyst and inert gas — Catalyst: faster, same position. Inert gas at constant volume: no effect; at constant pressure the shift follows Δn.

  • Easy : Fe catalyst does not change K.
  • Medium : Add Ar at constant V ⇒ partial pressures unchanged.
  • Exam-type : Add Ar at constant P ⇒ V rises ⇒ shift to more moles.

Why does higher pressure favour fewer moles?

  • Step 1: Halve the volume ⇒ every concentration doubles.
  • Step 2: In N₂ + 3H₂ ⇌ 2NH₃ the numerator of Q rises ×4, denominator ×16 ⇒ Q = K/4.
  • Step 3: Q < K ⇒ forward, so NH₃ increases.

CET level (MCQ)

  • CET 1) Pressure has no effect on? 1) N₂ + 3H₂ ⇌ 2NH₃ 2) H₂ + I₂ ⇌ 2HI 3) 2SO₂ + O₂ ⇌ 2SO₃ 4) PCl₅ ⇌ PCl₃ + Cl₂ — Answer: 2).
  • CET 2) What changes K? 1) pressure 2) concentration 3) temperature 4) catalyst — Answer: 3).

Real-life use: Higher NH₃ and SO₃ yields; divers' gas mixtures; breathing at altitude (Hb + O₂ ⇌ HbO₂).

Remember

ΔH < 0: raise T ⇒ reverse; raise P ⇒ fewer moles; catalyst ⇒ same position; only T changes K.

Common mistake

Writing that K changes with concentration; K is unchanged, only the position shifts.

CET trap

An inert gas at constant V shifts nothing.

Shortcut

Memory clue: "exothermic: heat it, goes back".

Spot the difference

ChangeShiftK
Concentrationopposessame
Pressurefewer molessame
Temperatureper ΔHchanges

Practice questions

Easy
  1. 1State Le Chatelier's principle.
  2. 2Effect of raising P on N₂ + 3H₂ ⇌ 2NH₃?
  3. 3Does a catalyst change K?
Medium
  1. 4Colour of N₂O₄ ⇌ 2NO₂ on heating?
  2. 5Why no effect of P on H₂ + I₂ ⇌ 2HI?
  3. 6If NH₃ is continually removed?
Challenging
  1. 7Add Ar at constant V: effect?
  2. 8How is Q = K/4 when pressure is doubled?
Show answers (try them yourself first)
  1. 1Equilibrium shifts to oppose the change.
  2. 2More NH₃.
  3. 3No.
  4. 4Darker brown.
  5. 52 mol of gas on both sides.
  6. 6Equilibrium keeps shifting forward.
  7. 7None.
  8. 8Numerator ×4, denominator ×16.

Industrial applications: Haber and Contact processes

From daily life

Fertiliser needs ammonia, soap and batteries need H₂SO₄; both are clever uses of equilibrium.

Simple definition

Haber: N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92 kJ. Contact: 2SO₂ + O₂ ⇌ 2SO₃, ΔH = −196 kJ. Balance high yield with a reasonable rate.

HaberNH₃ yield versus T and P (qualitative)200 atm100 atm50 atmtemperature →%Haber: ≈ 700–773 K, ≈ 200–250 atm, Fe (+ Mo)Contact: V₂O₅, 720 K, 2 bar

In the figure : NH₃ yield against T and P.

Rule 1: Conditions for the Haber process — Pressure ≈ 200–250 atm (fewer moles); T ≈ 700–773 K (compromise); Fe catalyst + Mo promoter; NH₃ is liquefied and removed.

  • Easy : Low T ⇒ higher yield but slow.
  • Medium : High P ⇒ more NH₃.
  • Exam-type : N₂ : H₂ fed in a 1 : 3 ratio.

Rule 2: Conditions for the Contact process — V₂O₅ catalyst, T ≈ 720 K, P ≈ 2 bar; SO₃ is absorbed in H₂SO₄ to give oleum.

  • Easy : SO₃ + H₂SO₄ → H₂S₂O₇ (oleum).
  • Medium : H₂S₂O₇ + H₂O → 2H₂SO₄.
  • Exam-type : Very high P not needed; conversion is already high.

Rule 3: The compromise — ΔH < 0 favours low T, but the rate is then too slow; so moderate T plus a catalyst. K depends only on temperature.

  • Easy : Raising T lowers K (exothermic reaction).
  • Medium : Catalyst raises rate, not yield.
  • Exam-type : Economics: high pressure is costly, so moderate.

Reasoning: how the Haber conditions are chosen

  • Step 1: Δn = −2 ⇒ high pressure pushes forward.
  • Step 2: ΔH < 0 ⇒ low T is good but the reaction is too slow.
  • Step 3: Moderate T (≈ 700 K) + Fe catalyst ⇒ reasonable rate.

CET level (MCQ)

  • CET 1) Catalyst in the Haber process? 1) V₂O₅ 2) Fe 3) Pt 4) Ni — Answer: 2).
  • CET 2) In the Contact process SO₃ is absorbed in? 1) water 2) conc. H₂SO₄ 3) NaOH 4) HCl — Answer: 2).

Real-life use: Fertiliser industry, explosives, battery acid.

Remember

Haber Fe (+Mo), ≈ 773 K, 200–250 atm; Contact V₂O₅, 720 K, 2 bar; oleum.

Common mistake

Absorbing SO₃ directly in water (forms a mist); use oleum.

CET trap

Both are exothermic, yet very low T is not used.

Shortcut

Memory clue: "Haber = iron, Contact = vanadium".

Spot the difference

ItemHaberContact
ProductNH₃SO₃ → H₂SO₄
CatalystFeV₂O₅
T700–773 K720 K
P200–250 atm2 bar

Practice questions

Easy
  1. 1Write the Haber reaction.
  2. 2Why use high pressure?
  3. 3Catalyst of the Contact process?
Medium
  1. 4Why a compromise temperature?
  2. 5What is oleum?
  3. 6Role of Mo?
Challenging
  1. 7Effect of liquefying and removing NH₃?
  2. 8Why is 2 bar enough in the Contact process?
Show answers (try them yourself first)
  1. 1N₂ + 3H₂ ⇌ 2NH₃.
  2. 24 → 2 mol of gas.
  3. 3V₂O₅.
  4. 4Low T is slow, high T lowers K.
  5. 5H₂S₂O₇.
  6. 6Promoter (boosts Fe).
  7. 7Equilibrium shifts forward.
  8. 8Conversion is already high.

Degree of dissociation and equilibrium calculations

From daily life

If 2 of 10 pupils leave the class, 20 % are absent; α is the dissociated fraction in the same way.

Simple definition

α = moles dissociated ÷ initial moles. N₂O₄ ⇌ 2NO₂: Kp = 4α²P/(1 − α²); PCl₅ ⇌ PCl₃ + Cl₂: Kp = α²P/(1 − α²).

Degree of dissociationDegree of dissociation α and KpN₂O₄ ⇌ 2NO₂α = 0.2P = 1 atmKp = 0.1667Kp = 4α²P / (1 − α²)PCl₅ ⇌ PCl₃ + Cl₂α = 0.5P = 2 atmKp = 0.667Kp = α²P / (1 − α²)Raising P lowers α (reaction increases moles)

In the figure : α and Kp of two reactions.

Rule 1: Dissociation of N₂O₄ — From 1 mol N₂O₄: N₂O₄ (1 − α), NO₂ 2α, total 1 + α. Kp = (2α/(1 + α) × P)² / ((1 − α)/(1 + α) × P) = 4α²P/(1 − α²).

  • Easy : α = 0.2, P = 1 atm: Kp = 4 × 0.04 ÷ 0.96 = 0.1667.
  • Medium : Total moles = 1.2.
  • Exam-type : Mole fraction of NO₂ = 0.4 ÷ 1.2 = 1/3.

Rule 2: Dissociation of PCl₅ — PCl₅ (1 − α), PCl₃ α, Cl₂ α; total 1 + α. Kp = α²P/(1 − α²).

  • Easy : α = 0.5, P = 2 atm: Kp = 0.25 × 2 ÷ 0.75 = 0.667.
  • Medium : Total moles = 1.5.
  • Exam-type : Mole fraction of Cl₂ = 0.5 ÷ 1.5 = 1/3.

Rule 3: Effect of pressure — Kp fixed ⇒ α falls as P rises because pressure opposes a reaction that increases moles.

  • Easy : N₂O₄: α ∝ 1/√P for small α.
  • Medium : P ×4 ⇒ α halves.
  • Exam-type : Adding Cl₂ to PCl₅ lowers α.

Derivation: Kp and α for N₂O₄

  • Step 1: N₂O₄ = (1 − α), NO₂ = 2α, total = 1 + α.
  • Step 2: p(NO₂) = 2αP/(1 + α), p(N₂O₄) = (1 − α)P/(1 + α).
  • Step 3: Kp = p(NO₂)²/p(N₂O₄) = 4α²P/((1 + α)(1 − α)) = 4α²P/(1 − α²).

CET level (MCQ)

  • CET 1) α = 0.2, P = 1 atm: Kp of N₂O₄? 1) 0.1667 2) 0.2 3) 0.4 4) 1.0 — Answer: 1).
  • CET 2) Total moles in PCl₅ dissociation (α = 0.5)? 1) 1 2) 1.5 3) 2 4) 2.5 — Answer: 2).

Real-life use: Finding average molar mass of gas mixtures.

Remember

α = dissociated ÷ initial; total = 1 + α; N₂O₄: 4α²P/(1 − α²).

Common mistake

Writing total moles as 1 − α; it is 1 + α.

CET trap

N₂O₄ has the factor 4, PCl₅ does not.

Shortcut

Memory clue: "N₂O₄ has four; PCl₅ has one".

Spot the difference

ReactionKpTotal moles
N₂O₄ ⇌ 2NO₂4α²P/(1 − α²)1 + α
PCl₅ ⇌ PCl₃ + Cl₂α²P/(1 − α²)1 + α

Practice questions

Easy
  1. 1Define α.
  2. 2Write Kp for N₂O₄.
  3. 3α = 0.2, P = 1 atm: Kp?
Medium
  1. 4PCl₅: α = 0.5, P = 2 atm: Kp?
  2. 5Why does α fall as P rises?
  3. 6Why is total moles 1 + α?
Challenging
  1. 7What happens to α if Cl₂ is added to PCl₅?
  2. 8Where does the 4 come from in N₂O₄?
Show answers (try them yourself first)
  1. 1Moles dissociated ÷ initial moles.
  2. 24α²P/(1 − α²).
  3. 30.1667.
  4. 40.667.
  5. 5Pressure opposes more moles.
  6. 61 − α + 2α = 1 + α.
  7. 7α decreases.
  8. 8From (2α)².

Other equilibria and applications

From daily life

Oxygen carried in blood, sugar settling in sweet water — equilibrium is everywhere.

Simple definition

Solubility equilibria, vapour pressure, Henry's law (gas solubility ∝ pressure). K tells how far a reaction proceeds.

Rule 1: Henry's law — S = kH × p; higher pressure raises solubility.

  • Easy : CO₂ in soda under pressure.
  • Medium : Fizz when opened.
  • Exam-type : Divers get 'the bends'.

Rule 2: Solubility and temperature — Gas solubility falls as T rises; most solids dissolve more (endothermic dissolving).

  • Easy : Less oxygen in warm water.
  • Medium : KNO₃ dissolves more in hot water.
  • Exam-type : Fish suffer in a warmed lake.

Rule 3: Yield from K — Large K ⇒ high conversion; small K ⇒ low.

  • Easy : H₂ + Cl₂ ⇌ 2HCl: K ≫ 1.
  • Medium : N₂ + O₂ ⇌ 2NO: K ≪ 1 (only in lightning).
  • Exam-type : Hb + O₂ ⇌ HbO₂: forward in lungs, reverse in tissues.

Reasoning: why is breathing hard at altitude?

  • Step 1: At altitude oxygen partial pressure is lower.
  • Step 2: Hb + O₂ ⇌ HbO₂ shifts back.
  • Step 3: Less HbO₂ ⇒ fatigue.

CET level (MCQ)

  • CET 1) Henry's law relates to? 1) solids 2) gas solubility 3) pH 4) boiling — Answer: 2).
  • CET 2) As T rises, gas solubility? 1) rises 2) falls 3) same 4) zero — Answer: 2).

Real-life use: Soda, blood gases, aquatic life.

Remember

Henry: S ∝ p; heat ⇒ less gas; large K ⇒ products.

Common mistake

Writing that all solubilities rise with T.

CET trap

Henry's law holds only if the gas does not react with the liquid.

Shortcut

Memory clue: "more pressure, more gas".

Spot the difference

ChangeGas solubility
P ↑increases
T ↑decreases

Practice questions

Easy
  1. 1State Henry's law.
  2. 2Why does soda fizz on opening?
  3. 3Why less O₂ in warm water?
Medium
  1. 4Why do divers get the bends?
  2. 5Why is breathing hard at altitude?
  3. 6Meaning of K ≪ 1?
Challenging
  1. 7Which way does Hb + O₂ ⇌ HbO₂ shift in the lungs?
  2. 8How does KNO₃ solubility vary with T?
Show answers (try them yourself first)
  1. 1S ∝ p.
  2. 2Pressure drops and CO₂ escapes.
  3. 3Heating lowers gas solubility.
  4. 4N₂ bubbles when pressure drops.
  5. 5Lower O₂ partial pressure.
  6. 6Negligible product.
  7. 7Forward.
  8. 8Increases.

Quick revision of the whole lesson

ConceptKey formula / fact
Equilibriumrates equal; concentrations constant
Kcproducts ÷ reactants (with powers)
KpKp = Kc(RT)^Δn
QQ < K forward; Q > K reverse
Le Chatelieropposes the change
HaberFe (+Mo), ≈ 773 K, 200–250 atm
ContactV₂O₅, 720 K, 2 bar
αN₂O₄ : 4α²P/(1 − α²)
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🎯 Most important points for the exam

  • Reversible reactions and dynamic equilibrium.
  • Kc, Kp and their relation.
  • Q comparison and ICE calculations.
  • Le Chatelier's principle.
  • Haber and Contact processes.
  • Degree of dissociation α.
  • Henry's law and applications.
Study mantra

understand the idea → remember the formula → solve the example → solve problems yourself.