Chapter 22: Principles Related to Practical Chemistry

Detection of Extra Elements (N, S, Halogens)

Preparation of Lassaigne’s Extract

• Heat small sodium piece in ignition tube till molten.
• Add 50-60 mg organic compound, heat strongly 2-3 min to fuse.
• Cool, break tube in distilled water, crush fused material.
• Boil mixture; sodium salts (NaCN, Na₂S, NaX, NaSCN) dissolve.
• Excess sodium reacts with water forming NaOH; solution is Lassaigne’s extract.
• Lithium and potassium not used due to slow reaction and handling issues.

Detection of Nitrogen

• Nitrogen and carbon form NaCN in Lassaigne’s extract.
• Unreacted sodium reacts with water: 2Na + 2H₂O → 2NaOH + H₂.
• NaCN + FeSO₄ forms sodium ferrocyanide; Fe³⁺ reacts to give Prussian blue (ferric ferrocyanide).
• FeSO₄ + 2NaOH → Fe(OH)₂ + Na₂SO₄
• Fe(OH)₂ + 6NaCN → Na₄[Fe(CN)₆] + 2NaOH
• 4Fe³⁺ + 3Na₄[Fe(CN)₆] → Fe₄[Fe(CN)₆]₃ (Prussian blue)
• Test fails if nitrogen present without carbon (e.g., hydrazine).
• If N and S both present, blood red colour due to ferric thiocyanate (Fe(CNS)₃).
• Excess sodium decomposes sodium thiocyanate: 2Na + NaCNS → Na₂S + NaCN.

Detection of Sulphur

• Sulphur converts to sodium sulphide (Na₂S) in Lassaigne’s extract.
• Na + S → Na₂S (on heating)
• Na₂S + sodium nitroprusside → sodium thionitroprusside (violet colour).
• Na₂S + lead acetate → PbS (black precipitate).

Detection of Halogens

• Beilstein test: Copper wire flame test shows halogens (except fluorine).
• Lassaigne’s test: Halogens form sodium halides; treated with HNO₃ and AgNO₃ gives precipitates:
  • AgCl: white, soluble in NH₄OH (chlorine)
  • AgBr: pale yellow, sparingly soluble (bromine)
  • AgI: yellow, insoluble (iodine)
• Chlorine water test: Lassaigne’s extract + Cl₂ + CS₂ shows orange (Br₂) or violet (I₂) colour.

Solved Examples

• Ex 1: Identify nitrogen in organic compound using Lassaigne’s test.
  • Fuse compound with sodium, prepare extract.
  • Add FeSO₄, observe Prussian blue colour.
• Ex 2: Detect sulphur using lead acetate.
  • Prepare Lassaigne’s extract.
  • Add lead acetate, observe black PbS precipitate.
• Ex 3: Confirm halogen by Lassaigne’s test.
  • Boil extract with HNO₃, add AgNO₃.
  • Observe precipitate colour and solubility.

Practice Set

• Q1: Which element forms NaCN in Lassaigne’s extract?
  • A. Sulphur
  • B. Nitrogen
  • C. Halogen
  • D. Oxygen
• Q2: What colour indicates presence of nitrogen in Lassaigne’s test with FeSO₄?
  • A. Blood red
  • B. Prussian blue
  • C. Violet
  • D. Black
• Q3: Which precipitate confirms iodine presence in Lassaigne’s test?
  • A. White AgCl
  • B. Pale yellow AgBr
  • C. Yellow AgI
  • D. Black PbS

Answer Key

• Q1: B - Nitrogen forms NaCN in Lassaigne’s extract.
• Q2: B - Prussian blue indicates nitrogen presence.
• Q3: C - Yellow AgI precipitate confirms iodine.

Memory Aids

• Lithium & potassium not used in Lassaigne’s test.
• Prussian blue = nitrogen test.
• Blood red = nitrogen + sulphur.
• AgCl white, AgBr pale yellow, AgI yellow for halogens.

Detection of Organic Functional Groups

Alcoholic Group (-OH)

• Sodium metal test: Alcohol + Na → sodium alkoxide + H₂ (effervescence).
• Ester test: Alcohol + carboxylic acid + H₂SO₄ → sweet smelling ester.
• Ceric ammonium nitrate test: Alcohol + ceric ammonium nitrate → pink/red complex.
• Acetyl chloride test: Alcohol + acetyl chloride → ester + HCl; HCl + NH₃ → white fumes (NH₄Cl).

Distinction between Primary, Secondary and Tertiary Alcohols

• Lucas test: Alcohol + Lucas reagent (HCl + ZnCl₂) → alkyl halide (cloudiness).
  • Order of reactivity: Tertiary > Secondary > Primary.
  • Primary alcohols react slowly or require heat.
• Victor Meyer’s test: Series of reactions producing color changes to identify alcohol type.
  • Primary: blood red
  • Secondary: blue
  • Tertiary: no color
• Catalytic dehydrogenation:
  • Primary alcohol → aldehyde + H₂
  • Secondary alcohol → ketone + H₂
  • Tertiary alcohol → alkene + H₂O (dehydration)

Phenolic Group

• Litmus test: Phenols turn blue litmus red; no effervescence with Na₂CO₃.
• Ferric chloride test: Phenol + FeCl₃ → coloured complex (varies with phenol type).
• Liebermann’s nitroso test: Phenol + NaNO₂ + conc. H₂SO₄ → blue colour, changes to red on dilution.

Carbonyl Group (Aldehydes and Ketones)

• 2,4-Dinitrophenyl hydrazine test: Carbonyl + DNP → yellow-orange precipitate.
• Sodium bisulphite test: Aldehydes/methyl ketones + NaHSO₃ → crystalline addition product.
• Tollens’ test: Aldehydes + Tollens’ reagent → silver mirror.
• Fehling’s test: Aldehydes + Fehling’s solution → red Cu₂O precipitate.
• Benedict’s test: Similar to Fehling’s test.
• Schiff’s test: Aldehydes + Schiff’s reagent → pink/red colour.
• Ketone tests:
  • m-Dinitrobenzene test: Violet colour confirms ketone.
  • Sodium nitroprusside test: Ketones + nitroprusside → red complex.

Carboxylic Acid Group

• Litmus test: Turns blue litmus red.
• Ester test: Carboxylic acid + ethyl alcohol + H₂SO₄ → fruity smelling ester.
• Sodium bicarbonate test: Carboxylic acid + NaHCO₃ → brisk effervescence (CO₂).

Amino Group

• Litmus test: Amines turn red litmus blue.
• Solubility: Amines dissolve in mineral acids forming salts.
• Carbylamine test: Primary amines + chloroform + alcoholic KOH → isocyanide (offensive smell).
• Azo-dye test: Primary aromatic amines + nitrous acid + phenol → coloured azo dye.
• Nitrous acid test:
  • Primary amines liberate N₂ gas.
  • Secondary amines form yellow oily nitrosoamines.
  • Tertiary amines form water-soluble salts.
• Hinsberg test:
  • Primary and secondary amines react with benzene sulphonyl chloride to form sulphonamides.
  • Primary amine sulphonamide soluble in NaOH; secondary insoluble.
  • Tertiary amines do not react.

Solved Examples

• Ex 1: Identify primary alcohol using Lucas test.
  • Mix alcohol with Lucas reagent at room temp.
  • No cloudiness indicates primary alcohol.
• Ex 2: Confirm aldehyde by Tollens’ test.
  • Warm compound with Tollens’ reagent.
  • Silver mirror confirms aldehyde.
• Ex 3: Detect amino group by carbylamine test.
  • Mix amine with chloroform and alcoholic KOH.
  • Offensive smell confirms primary amine.

Practice Set

• Q1: Which test gives pink/red colour with alcohol?
  • A. Lucas test
  • B. Ceric ammonium nitrate test
  • C. Tollens’ test
  • D. Fehling’s test
• Q2: Which test distinguishes primary, secondary, tertiary alcohols by cloudiness?
  • A. Victor Meyer’s test
  • B. Lucas test
  • C. Carbylamine test
  • D. Schiff’s test
• Q3: Which test confirms ketones?
  • A. Schiff’s test
  • B. Sodium nitroprusside test
  • C. Tollens’ test
  • D. Carbylamine test

Answer Key

• Q1: B - Ceric ammonium nitrate gives pink/red with alcohols.
• Q2: B - Lucas test shows cloudiness order for alcohol types.
• Q3: B - Sodium nitroprusside test is specific for ketones.

Memory Aids

• Lucas test: Tertiary reacts fastest, primary slowest.
• Tollens’ test positive only for aldehydes.
• Carbylamine test positive only for primary amines.
• Ketones give violet in m-dinitrobenzene test.

Preparation of Organic Compounds

Acetanilide

• Prepared by acetylation of aniline with acetic anhydride in glacial acetic acid.
• Reaction: Aniline + Acetic anhydride → Acetanilide + Acetic acid.
• Boil mixture with reflux, cool in ice water to get crude white precipitate.
• Purify by recrystallisation.
• Zinc added to prevent aniline oxidation and reduce impurities.
• Acetanilide protects amino group in reactions.

Nitration of Acetanilide (p-Nitroacetanilide)

• Mix powdered acetanilide with glacial acetic acid and conc. H₂SO₄, cool below 5°C.
• Add nitrating mixture (conc. HNO₃ + conc. H₂SO₄) below 25°C with stirring.
• Pour into crushed ice to get crude product.
• Recrystallise in ethanol; p-nitroacetanilide crystallises out.
• Direct nitration of aniline not possible due to oxidation; acetylation protects -NH₂.

Aniline Yellow (p-Aminoazobenzene)

• Prepared by coupling benzenediazonium chloride with aniline in acidic medium.
• Forms yellow solid dye.
• Not used as dye due to acid sensitivity.

Iodoform

• Prepared from acetone or ethyl alcohol (better yield from acetone).
• React acetone/ethyl alcohol + KI + NaOH + sodium hypochlorite.
• Yellow precipitate of iodoform (CHI₃) formed, crystallised from spirit.
• Detects CH₃-CH-OH or CH₃-CO- groups.
• Used as antiseptic.

Solved Examples

• Ex 1: Prepare acetanilide.
  • Mix aniline, acetic anhydride, glacial acetic acid.
  • Boil with reflux, cool, filter precipitate.
  • Recrystallise for pure product.
• Ex 2: Prepare p-nitroacetanilide.
  • Mix acetanilide, glacial acetic acid, conc. H₂SO₄, cool.
  • Add nitrating mixture below 25°C.
  • Isolate and recrystallise product.
• Ex 3: Perform iodoform test.
  • Mix acetone/ethyl alcohol with KI, NaOH, sodium hypochlorite.
  • Observe yellow precipitate.

Practice Set

• Q1: Which reagent is used for acetylation of aniline?
  • A. Acetic acid
  • B. Acetic anhydride
  • C. Acetyl chloride
  • D. Both B and C
• Q2: p-Nitroacetanilide is major product due to?
  • A. Ortho directing effect
  • B. Para directing effect
  • C. Meta directing effect
  • D. No directing effect
• Q3: Iodoform test detects?
  • A. Alcohols with CH₃-CH-OH group
  • B. Ketones with CH₃-CO- group
  • C. Both A and B
  • D. None

Answer Key

• Q1: D - Both acetic anhydride and acetyl chloride are used.
• Q2: B - Acetamido group is para directing.
• Q3: C - Iodoform test detects both methyl alcohol and methyl ketone groups.

Memory Aids

• Acetanilide protects -NH₂ during nitration.
• p-Nitroacetanilide is major due to para directing acetamido group.
• Iodoform test positive for CH₃-CH-OH and CH₃-CO- groups.

Volumetric Analysis

Overview

• Measures concentration of a chemical by volume of its solution in a reaction.
• Main process is titration.
• Types: acid-base, redox, iodine, complexometric titrations.

Important Terms

• Analyte (titre): Substance being analyzed.
• Titrant: Substance added to analyte.
• Equivalence point: Reaction completion point; sudden change in indicator color, pH, conductivity, or absorbance.
• Indicator: Compound changing color near equivalence point to signal endpoint.
• Standard solution: Solution with known concentration.
• Standardization: Process to determine reagent concentration using a known reagent.
• Primary standard: Pure reagent for direct standard solution preparation (e.g., oxalic acid, anhydrous Na₂CO₃, Mohr’s salt).
• Secondary standard: Reagent not pure enough for direct standard solution (e.g., KMnO₄, NaOH, KOH).

Law of Chemical Equivalence

• Equivalents of analyte react with equal equivalents of titrant.
• Number of equivalents = Normality (N) × Volume (V in L).
• Normality equation: N₁V₁ = N₂V₂.
• Molarity equation (for dilution): M₁V₁ = M₂V₂.
• Normality = Molarity × valency factor (n).
• General form: M₁V₁n₁ = M₂V₂n₂.

Theory of Titration

• Known volume (usually 20 mL) of one solution in conical flask with indicator.
• Other solution in burette added dropwise with stirring.
• Sudden change in property (color) indicates endpoint.
• Volume of titrant used is noted.

General Calculations

• Apply law of chemical equivalence: N₁V₁ = N₂V₂.
• Calculate unknown normality or molarity.
• Strength (g/L) = Normality × Equivalent weight or Molarity × Molecular mass.
• Percentage purity = (Strength of pure sample / Strength of given sample) × 100.

Solved Examples

Ex 1

• Given volume and normality of titrant, calculate normality of analyte using N₁V₁ = N₂V₂.
• Calculate strength using strength formula.

Practice Set

• Q1: What is the equivalence point in titration?
  A. Start of reaction
  B. Point of color change
  C. Point where reactants are in stoichiometric ratio
  D. Point of maximum volume
  
• Q2: Which is a primary standard?
  A. KMnO₄
  B. NaOH
  C. Oxalic acid
  D. KOH
  
• Q3: Normality is related to molarity by?
  A. Normality = Molarity / n
  B. Normality = Molarity × n
  C. Normality = Molarity + n
  D. Normality = Molarity - n
  

Answer Key

• Q1: C - Equivalence point is where reactants react in stoichiometric ratio.
• Q2: C - Oxalic acid is a primary standard.
• Q3: B - Normality = Molarity × valency factor.

Memory Aids

• Primary standards: Pure, stable, non-hygroscopic (e.g., oxalic acid).
• Normality equation: N₁V₁ = N₂V₂.
• Use indicators that change color near equivalence point.

Acid-Base Titration

Concepts

• Neutralization reaction: H⁺ + OH^- → H₂O.
• End point: Sudden pH change with small titrant addition.
• Indicators: Weak acids/bases changing color at specific pH.
• Examples: Phenolphthalein (colorless to pink), Methyl orange (red to yellow).
• Indicator pK_a ≈ pH at half-equivalence point.
• Indicator choice depends on acid/base strength.

Titration Types

• Strong acid + strong base: Equivalence pH = 7; salt neutral.
• Weak acid + strong base: Buffer before equivalence; pH > 7 at equivalence due to salt hydrolysis.
• Weak base + strong acid: Buffer before equivalence; pH < 7 at equivalence due to salt hydrolysis.
• Weak acid + weak base: No sharp endpoint; pH ≈ 7 at equivalence.
• Diacidic base (Na₂CO₃) + monoprotic acid: Two equivalence points; use phenolphthalein and methyl orange.
• Diprotic acid (oxalic acid) + strong base: Two equivalence points; stepwise neutralization.

Solved Examples

Ex 1

• Titration of strong acid (HCl) with strong base (NaOH): pH at equivalence = 7.
• Reaction: HCl + NaOH → NaCl + H₂O.

Ex 2

• Titration of weak acid (CH₃COOH) with strong base (NaOH): buffer forms before equivalence; pH > 7 at equivalence.

Practice Set

• Q1: pH at equivalence point of strong acid-strong base titration is?
  A. 0
  B. 7
  C. 14
  D. Depends on indicator
  
• Q2: Which indicator is best for weak acid-strong base titration?
  A. Methyl orange
  B. Phenolphthalein
  C. Methyl red
  D. Thymol blue
  
• Q3: In titration of Na₂CO₃ with HCl, number of equivalence points are?
  A. 1
  B. 2
  C. 3
  D. 0
  

Answer Key

• Q1: B - pH = 7 for strong acid-strong base titration.
• Q2: B - Phenolphthalein suitable for weak acid-strong base.
• Q3: B - Two equivalence points due to diacidic base.

Memory Aids

• Strong acid + strong base: pH = 7 at equivalence.
• Weak acid + strong base: pH > 7 at equivalence.
• Weak base + strong acid: pH < 7 at equivalence.
• Use phenolphthalein for weak acid-strong base titrations.

Redox Titrations

Concepts

• Permanganometric titrations use KMnO₄ as oxidizing agent in acidic medium.
• Reducing agents: Mohr’s salt, H₂O₂, oxalic acid, etc.
• KMnO₄ acts as self-indicator: colorless to pink at endpoint.
• Redox reaction: KMnO₄ reduced to Mn²⁺, reducing agent oxidized.
• Equivalent mass of KMnO₄ in acidic medium = 158/5 = 31.6.

Experiment 1: Oxalic Acid vs KMnO₄

• Indicator: KMnO₄ self-indicator.
• End point: Colorless to permanent pink.
• Oxalic acid heated to 60-70°C to speed reaction.
• Reaction: 2MnO₄^- + 16H⁺ + 5C₂O₄^2- → 2Mn²⁺ + 8H₂O + 10CO₂.
• 2 moles KMnO₄ react with 5 moles oxalic acid.

Experiment 2: Mohr’s Salt vs KMnO₄

• Indicator: KMnO₄ self-indicator.
• End point: Colorless to light pink.
• Reaction: MnO₄^- + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.
• 1 mole KMnO₄ reacts with 5 moles Mohr’s salt.

Calculations

• Use normality equation: N₁V₁ = N₂V₂.
• Molarity relation: M_KMnO4V_KMnO4 / M_MohrV_Mohr = 1/5.
• Strength = Normality × Equivalent mass or Molarity × Molecular mass.

Solved Examples

Ex 1

• Calculate normality of KMnO₄ given volume and normality of oxalic acid using N₁V₁ = N₂V₂.

Practice Set

• Q1: KMnO₄ acts as what in permanganometric titrations?
  A. Titrant only
  B. Indicator only
  C. Self-indicator
  D. Reducing agent
  
• Q2: Equivalent mass of KMnO₄ in acidic medium is?
  A. 158
  B. 31.6
  C. 79
  D. 63
  
• Q3: In titration of Mohr’s salt with KMnO₄, mole ratio is?
  A. 1:1
  B. 1:2
  C. 1:5
  D. 5:1
  

Answer Key

• Q1: C - KMnO₄ acts as self-indicator.
• Q2: B - Equivalent mass = 158/5 = 31.6.
• Q3: C - 1 mole KMnO₄ reacts with 5 moles Mohr’s salt.

Memory Aids

• KMnO₄ reduced from MnO₄^- to Mn²⁺ by gaining 5 electrons.
• Oxalic acid reacts with KMnO₄ in 5:2 mole ratio.
• Mohr’s salt reacts with KMnO₄ in 5:1 mole ratio.

Enthalpy of Solution

Concepts

• Enthalpy of solution: Heat change when 1 mole of substance dissolves in excess solvent.
• Heat of solution can be endothermic (+) or exothermic (-).
• Measured using calorimeter (polythene bottle).
• Formula: ΔH = (W + m) × Δt × 4.2 × M / w (J/mol) or without 4.2 in cal/mol.
• W = water equivalent of calorimeter, m = mass of water, Δt = temperature change, M = molecular mass, w = mass of solute.
• Water equivalent: Heat required to raise calorimeter temperature by 1°C equals that for equivalent water mass.

Chemical Principle

• Dissolution involves lattice energy (energy to dissociate solid) and hydration energy (energy released on hydration).
• If lattice energy > hydration energy: endothermic dissolution.
• If lattice energy < hydration energy: exothermic dissolution.
• Example: Anhydrous CuSO₄ dissolution is exothermic; KCl dissolution is endothermic.
• Hydrated salts dissolution is endothermic (e.g., CuSO₄·5H₂O).

Solved Examples

Ex 1

• Calculate enthalpy of solution using temperature change and known masses.

Practice Set

• Q1: Enthalpy of solution is heat change when?
  A. 1 mole solute dissolves
  B. 1 mole solvent evaporates
  C. 1 mole solute reacts
  D. 1 mole solute melts
  
• Q2: If lattice energy > hydration energy, dissolution is?
  A. Exothermic
  B. Endothermic
  C. Neutral
  D. Spontaneous
  
• Q3: Water equivalent of calorimeter is?
  A. Mass of water
  B. Heat capacity of calorimeter
  C. Equivalent heat to raise temperature by 1°C
  D. Heat of solution
  

Answer Key

• Q1: A - Heat change when 1 mole solute dissolves.
• Q2: B - Endothermic if lattice energy > hydration energy.
• Q3: C - Heat required to raise calorimeter temperature by 1°C equals that for equivalent water mass.

Memory Aids

• Endothermic dissolution: lattice energy > hydration energy.
• Exothermic dissolution: hydration energy > lattice energy.
• Hydrated salts usually dissolve endothermically.

Preparation of Lyophilic and Lyophobic Sols

Colloids

• Intermediate state between true solutions and suspensions.
• Particle size: 1-1000 nm.
• Two phases: dispersed phase (solute equivalent) and dispersion medium (solvent equivalent).
• Dispersed phase distributed in dispersion medium.
• Types based on affinity:
• Lyophilic sols: High affinity, stable, reversible, prepared by direct mixing (e.g., gum, gelatin, starch).
• Lyophobic sols: Low affinity, unstable, irreversible, prepared by special methods (e.g., metal sols, hydroxides).

Theory of Preparation

• Lyophilic sols: prepared by mixing dispersed phase with medium (e.g., starch in boiling water).
• Lyophobic sols: prepared by hydrolysis of chlorides (e.g., FeCl₃, AlCl₃).
• Hydrolysis reactions produce insoluble hydroxides and HCl, which is removed by dialysis.
• Arsenious sulphide sol prepared by hydrolysis of As₂O₃ and passing H₂S gas.

Methods of Preparation

• Condensation methods: Combine atomic/molecular particles to form colloidal size aggregates via chemical or physical means.
• Chemical methods: double decomposition, oxidation, reduction, hydrolysis.
• Physical methods: solvent exchange, excessive cooling.
• Dispersion methods: mechanical dispersion, electrical disintegration (Bredig’s arc), peptization.

Solved Examples

Ex 1

• Prepare ferric hydroxide sol by hydrolysis of FeCl₃.
• Prepare arsenious sulphide sol by hydrolysis of As₂O₃ and H₂S gas.

Practice Set

• Q1: Lyophilic sols are?
  A. Unstable
  B. Reversible
  C. Irreversible
  D. Insoluble
  
• Q2: Bredig’s arc method is used for?
  A. Lyophilic sols
  B. Lyophobic sols
  C. True solutions
  D. Suspensions
  
• Q3: Peptization involves?
  A. Precipitate to sol conversion
  B. Sol to precipitate
  C. Mixing two sols
  D. Heating sols
  

Answer Key

• Q1: B - Lyophilic sols are reversible.
• Q2: B - Bredig’s arc prepares lyophobic sols.
• Q3: A - Peptization converts precipitate to sol.

Memory Aids

• Lyophilic = love solvent, stable, reversible.
• Lyophobic = fear solvent, unstable, irreversible.
• Bredig’s arc for metal sols.

Kinetic Study of Iodide Ion with Hydrogen Peroxide

Reaction Rate

• Rate = change in concentration per unit time.
• For A + B → C + D, rate = -Δ[A]/Δt = -Δ[B]/Δt = +Δ[C]/Δt = +Δ[D]/Δt.
• Negative sign: concentration decreases; positive: increases.

Reaction Theory

• H₂O₂ oxidizes I^- to I₂ in acidic medium (slow reaction).
• Reaction: H₂O₂ + 2I^- + 2H⁺ → I₂ + 2H₂O.
• I₂ reacts with sodium thiosulphate (Na₂S₂O₃) to form iodide ions and tetrathionate ions.
• When thiosulphate is consumed, I₂ reacts with starch to form blue complex.
• Time before blue color appears indicates reaction rate.
• Rate increases with KI concentration.

Chemical Principle

• Rate law: Rate = k[X]^a[Y]^b, where a,b are experimental orders.
• Order = a + b; molecularity = x + y (stoichiometric coefficients).
• Rate depends on concentration, temperature, catalyst, surface area, light.

Solved Examples

Ex 1

• Calculate rate from concentration changes over time.

Practice Set

• Q1: Rate of reaction is change in?
  A. Volume
  B. Concentration
  C. Mass
  D. Temperature
  
• Q2: In rate law Rate = k[X]^a[Y]^b, a and b are?
  A. Stoichiometric coefficients
  B. Experimental orders
  C. Rate constants
  D. Concentrations
  
• Q3: Molecularity is?
  A. Sum of experimental orders
  B. Sum of stoichiometric coefficients
  C. Rate constant
  D. Concentration
  

Answer Key

• Q1: B - Rate is change in concentration per unit time.
• Q2: B - a and b are experimentally determined orders.
• Q3: B - Molecularity is sum of stoichiometric coefficients.

Memory Aids

• Rate law exponents are experimental, not always stoichiometric.
• Molecularity is always an integer.
• Higher concentration → faster reaction.