Chapter summaries, must-read concepts, exam tips, tricks and numerical patterns for Solutions, Electrochemistry and Chemical Kinetics.
| Concept | Key Fact | Example / Value |
|---|---|---|
| Types of solutions | Gas in liquid (most common), liquid in liquid, solid in liquid | O₂ in water, ethanol in water, salt in water |
| Mole fraction (χ) | χA = nA / (nA + nB); χA + χB = 1 | Dimensionless |
| Molality (m) | moles of solute / kg of solvent | mol kg⁻¹; independent of temperature |
| Molarity (M) | moles of solute / L of solution | mol L⁻¹; depends on temperature |
| Raoult's Law | pA = χA × p°A ; p_total = χA p°A + χB p°B | Ideal solution — like dissolves in like |
| Positive deviation | A–B < A–A / B–B interactions; p_obs > Raoult | Ethanol + cyclohexane, acetone + CS₂ |
| Negative deviation | A–B > A–A / B–B; p_obs < Raoult | CHCl₃ + acetone, HNO₃ + water |
| Relative lowering of VP | ΔP/P°A = χB = nB/(nA+nB) | Dimensionless colligative property |
| Elevation of BP | ΔTb = Kb × m (× i for electrolyte) | Kb(water) = 0.52 K kg mol⁻¹ |
| Depression of FP | ΔTf = Kf × m (× i for electrolyte) | Kf(water) = 1.86 K kg mol⁻¹ |
| Osmotic pressure | π = iCRT = inRT/V | R = 0.083 L bar mol⁻¹ K⁻¹ |
| van't Hoff factor i | i > 1 for dissociation; i < 1 for association | NaCl: i≈2, MgCl₂: i≈3, acetic acid (assoc): i<1 |
pA = χA × p°A
p°A = vapour pressure of pure A; χA = mole fraction
(p°A − pA) / p°A = χB = nB / (nA + nB)
ΔTb = Kb × m × i
Kb(water) = 0.52 K kg mol⁻¹
ΔTf = Kf × m × i
Kf(water) = 1.86 K kg mol⁻¹
π = iCRT = i(n/V)RT
C = molarity; R = 0.083 L bar mol⁻¹ K⁻¹ or 8.314 J mol⁻¹ K⁻¹
i = 1 + (n − 1)α
n = number of ions; α = degree of dissociation
i = 1 − (1 − 1/n)α
n = number of molecules associating; e.g., acetic acid dimerisation n=2
p = KH × χ
Higher KH → lower solubility of gas
M₂ = (w₂ × Kf × 1000) / (ΔTf × w₁)
w₁ = solvent mass in g; w₂ = solute mass in g
Standard CBSE 5-mark numerical: "2g of X dissolved in 50g water, ΔTf = 0.372°C. Find molar mass."
Step 1: m = ΔTf / Kf = 0.372 / 1.86 = 0.2 mol kg⁻¹
Step 2: moles of X = m × mass of solvent (in kg) = 0.2 × 0.05 = 0.01 mol
Step 3: M = mass / moles = 2 / 0.01 = 200 g mol⁻¹
Single formula: M₂ = (w₂ × Kf × 1000) / (ΔTf × w₁) = (2 × 1.86 × 1000) / (0.372 × 50) = 200 ✓
Given: 0.5M KCl solution, ΔTf = 1.85°C. Kf = 1.86. Find i and α.
| Concept | Key Fact | Value / Formula |
|---|---|---|
| Galvanic cell | Chemical energy → electrical energy. Anode: oxidation (−). Cathode: reduction (+). | Zn−Cu Daniell cell |
| EMF of cell | E°cell = E°cathode − E°anode | E°cell > 0 → spontaneous |
| SHE potential | Standard Hydrogen Electrode = reference electrode | E° = 0.00 V (by convention) |
| Nernst equation | E = E° − (RT/nF) ln Q = E° − (0.0591/n) log Q | At 298K only |
| ΔG° and E°cell | ΔG° = −nFE°cell | F = 96500 C mol⁻¹ |
| E° and K | E° = (0.0591/n) log K at 298K | K = equilibrium constant of cell rxn |
| Conductance (G) | G = 1/R; unit = S (siemens) | Increases with concentration (for strong electrolyte) |
| Molar conductivity (Λm) | Λm = κ × 1000/C (C in mol L⁻¹) | κ = specific conductance (S cm⁻¹) |
| Kohlrausch's Law | Λ°m = ν₊λ°₊ + ν₋λ°₋ (sum of ionic conductivities) | For any electrolyte at infinite dilution |
| Faraday's 1st Law | m = Z × I × t (Z = electrochemical equivalent) | m in grams; I in amperes; t in seconds |
| Faraday's 2nd Law | Same charge → mass deposited ∝ equivalent weight | M/n ratio determines amount |
| Mass from electrolysis | m = (M × I × t) / (n × F) | n = electrons per ion; F = 96500 C |
| Oxidized Form + e⁻ | Reduced Form | E° (V) | |
|---|---|---|---|
| F₂(g) + 2 e⁻ | → | 2 F⁻(aq) | +2.87 |
| H₂O₂(aq) + 2 H⁺(aq) + 2 e⁻ | → | 2 H₂O(l) | +1.78 |
| MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ | → | Mn²⁺(aq) + 4 H₂O(l) | +1.51 |
| Cl₂(g) + 2 e⁻ | → | 2 Cl⁻(aq) | +1.36 |
| Cr₂O₇²⁻(aq) + 14 H⁺(aq) + 6 e⁻ | → | 2 Cr³⁺(aq) + 7 H₂O(l) | +1.33 |
| O₂(g) + 4 H⁺(aq) + 4 e⁻ | → | 2 H₂O(l) | +1.23 |
| Br₂(l) + 2 e⁻ | → | 2 Br⁻(aq) | +1.09 |
| Ag⁺(aq) + e⁻ | → | Ag(s) | +0.80 |
| Fe³⁺(aq) + e⁻ | → | Fe²⁺(aq) | +0.77 |
| O₂(g) + 2 H⁺(aq) + 2 e⁻ | → | H₂O₂(aq) | +0.70 |
| I₂(s) + 2 e⁻ | → | 2 I⁻(aq) | +0.54 |
| O₂(g) + 2 H₂O(l) + 4 e⁻ | → | 4 OH⁻(aq) | +0.40 |
| Cu²⁺(aq) + 2 e⁻ | → | Cu(s) | +0.34 |
| Sn⁴⁺(aq) + 2 e⁻ | → | Sn²⁺(aq) | +0.15 |
| 2 H⁺(aq) + 2 e⁻ | → | H₂(g) SHE | 0.00 |
| Pb²⁺(aq) + 2 e⁻ | → | Pb(s) | −0.13 |
| Ni²⁺(aq) + 2 e⁻ | → | Ni(s) | −0.26 |
| Cd²⁺(aq) + 2 e⁻ | → | Cd(s) | −0.40 |
| Fe²⁺(aq) + 2 e⁻ | → | Fe(s) | −0.45 |
| Zn²⁺(aq) + 2 e⁻ | → | Zn(s) | −0.76 |
| 2 H₂O(l) + 2 e⁻ | → | H₂(g) + 2 OH⁻(aq) | −0.83 |
| Al³⁺(aq) + 3 e⁻ | → | Al(s) | −1.66 |
| Mg²⁺(aq) + 2 e⁻ | → | Mg(s) | −2.37 |
| Na⁺(aq) + e⁻ | → | Na(s) | −2.71 |
| Li⁺(aq) + e⁻ | → | Li(s) | −3.04 |
Reading the table: Top entries (F₂, +2.87V) are strongest oxidizing agents — they gain electrons easily. Bottom entries (Li, −3.04V) are strongest reducing agents — they lose electrons easily. E°cell = E°cathode − E°anode. If E°cell > 0, the reaction is spontaneous.
E = E° − (0.0591/n) × log([products]/[reactants])
n = moles of electrons transferred; use reduction potential for each electrode
ΔG° = −nFE°cell (n in mol, F = 96500 C mol⁻¹, E° in V)
E° = (0.0591/n) × log K at 298K
Λm = (κ × 1000) / C
κ = specific conductance (S cm⁻¹), C = molarity (mol L⁻¹), Λm in S cm² mol⁻¹
Λ°m = ν₊λ°₊ + ν₋λ°₋
Λ°m(CH₃COOH) = Λ°m(HCl) + Λ°m(CH₃COONa) − Λ°m(NaCl)
α = Λm / Λ°m
m = (M × I × t) / (n × F)
M = molar mass (g/mol), I = current (A), t = time (s), n = charge on ion, F = 96500 C
Strong electrolytes (KCl, NaCl): linear decrease — Λ°m found by extrapolation. Weak electrolytes (CH₃COOH): steep rise near zero — Λ°m found via Kohlrausch's Law.
Q: Calculate E_cell for Zn−Cu cell. [Zn²⁺] = 0.001M, [Cu²⁺] = 0.1M. E°(Zn²⁺/Zn) = −0.76V, E°(Cu²⁺/Cu) = +0.34V.
Q: How many grams of Cu are deposited by 0.5A for 30 min from CuSO₄ solution? (M = 63.5, n = 2)
Shortcut check: 96500C deposits 63.5/2 = 31.75g. 900C deposits 31.75 × (900/96500) = 0.296g ✓
| Type | Anode Reaction | Cathode Reaction | EMF / Key Fact |
|---|---|---|---|
| Dry Cell (Leclanche) Primary; non-rechargeable |
Zn → Zn²⁺ + 2e⁻ | MnO₂ + NH₄⁺ + e⁻ → MnO(OH) + NH₃ | ~1.5 V. Zn container = anode. Graphite rod = cathode. NH₄Cl paste = electrolyte. |
| Mercury Cell Primary; constant EMF |
Zn(Hg) + 2OH⁻ → ZnO + H₂O + 2e⁻ | HgO + H₂O + 2e⁻ → Hg + 2OH⁻ | ~1.35 V. Constant EMF throughout life. Used in hearing aids, watches. |
| Lead Storage Battery Secondary; rechargeable |
Pb + SO₄²⁻ → PbSO₄ + 2e⁻ | PbO₂ + 4H⁺ + SO₄²⁻ + 2e⁻ → PbSO₄ + 2H₂O | ~2 V per cell; 12 V battery has 6 cells. Electrolyte: dil. H₂SO₄. On charging, reactions reverse. H₂SO₄ consumed during discharge. |
| Nickel–Cadmium Cell Secondary; rechargeable |
Cd + 2OH⁻ → Cd(OH)₂ + 2e⁻ | NiO(OH) + H₂O + e⁻ → Ni(OH)₂ + OH⁻ | ~1.25 V. Longer life than dry cell. Used in portable electronics. |
| H₂–O₂ Fuel Cell Continuous; not a battery |
H₂ + 2OH⁻ → 2H₂O + 2e⁻ (anode, KOH electrolyte) | O₂ + 2H₂O + 4e⁻ → 4OH⁻ (cathode) | Overall: 2H₂ + O₂ → 2H₂O. ~70% efficient (vs ~40% for combustion engine). No pollution — only water produced. Used in space vehicles. |
Battery: Stores chemical energy; reactants are finite; goes "dead". Fuel cell: Converts fuel energy directly to electrical energy as long as fuel (H₂) is supplied — it is NOT exhausted. Efficiency higher than combustion. Products are just water (eco-friendly). Used in Apollo space missions.
Lead storage battery memory: BOTH electrodes become PbSO₄ during discharge ("both lead to sulfate"). During charging: PbSO₄ at anode → Pb; PbSO₄ at cathode → PbO₂.
| Aspect | Detail |
|---|---|
| Definition | Slow deterioration of a metal by reaction with its environment (oxidation by O₂/moisture) |
| Rusting of iron — anode (oxidation) | Fe → Fe²⁺ + 2e⁻ (at surface pits, impurities) |
| Rusting of iron — cathode (reduction) | O₂ + 2H₂O + 4e⁻ → 4OH⁻ (at grain boundaries) |
| Rust formation | Fe²⁺ + 2OH⁻ → Fe(OH)₂ → further oxidised to Fe(OH)₃ → dehydrates to Fe₂O₃·xH₂O (rust, reddish-brown) |
| Why moist conditions? | Water acts as electrolyte to complete the circuit between anode and cathode regions on the metal surface |
| Why CO₂/acid rain speeds corrosion? | CO₂ + H₂O → H₂CO₃ → increases H⁺ → faster reduction at cathode |
| Prevention — Galvanisation | Coating with Zn. Zn is more active than Fe → Zn corrodes preferentially (sacrificial). Even if Zn coat scratched, Fe still protected. |
| Prevention — Tinning | Coating with Sn (less active than Fe). Only protects while coat is intact — if scratched, Fe corrodes faster (Fe acts as anode, Sn as cathode in galvanic couple). |
| Prevention — Cathodic protection | Connect a more active metal (Mg, Zn) as sacrificial anode to iron structure. Used for underground pipes, ship hulls. |
| Prevention — Coating/Painting | Prevents contact of O₂ and moisture with metal surface. Physical barrier method. |
| Prevention — Alloying | Stainless steel (Fe + Cr + Ni) — Cr forms thin passive oxide layer (Cr₂O₃) that prevents further corrosion. |
| Concept | Key Fact | Formula / Value |
|---|---|---|
| Rate of reaction | Change in concentration per unit time; always positive | r = −(1/a) d[A]/dt = +(1/c) d[C]/dt |
| Rate law | r = k[A]ˣ[B]ʸ; order = x+y (found experimentally) | k = rate constant; depends on T, not concentration |
| Zero order | [A]t = [A]₀ − kt | k units: mol L⁻¹ s⁻¹; t½ = [A]₀/2k |
| First order | ln[A]t = ln[A]₀ − kt | k units: s⁻¹; t½ = 0.693/k (independent of [A]) |
| Second order | 1/[A]t = 1/[A]₀ + kt | k units: L mol⁻¹ s⁻¹; t½ = 1/(k[A]₀) |
| Pseudo first order | Second order but one reactant in large excess → behaves as first order | e.g., hydrolysis of ester in water (water >> ester) |
| Molecularity | Number of molecules reacting in elementary step | Always 1, 2, or 3 (never 0 or fraction) |
| Arrhenius equation | k = A × e^(−Ea/RT) | A = frequency factor; Ea = activation energy (J mol⁻¹) |
| Activation energy | Minimum energy needed for reaction to occur | Higher Ea → slower reaction; catalyst lowers Ea |
| Catalyst effect | Lowers Ea without being consumed. Does NOT change ΔH or K. | Both forward and backward rates increase equally |
r = k[A]ˣ[B]ʸ (overall order = x + y)
k = (2.303/t) × log([A]₀/[A]t)
Equivalent: [A]t = [A]₀ × e^(−kt)
t½ = 0.693/k
Independent of initial concentration — unique property of 1st order
fraction remaining = (1/2)ⁿ after n half-lives
After 1 t½ → 50%; after 2 t½ → 25%; after 3 t½ → 12.5%
k = A × e^(−Ea/RT)
log(k₂/k₁) = (Ea/2.303R) × (1/T₁ − 1/T₂)
R = 8.314 J mol⁻¹ K⁻¹; use K for temperature, not °C
zero: mol L⁻¹ s⁻¹ | first: s⁻¹ | second: L mol⁻¹ s⁻¹
General formula: units = (mol L⁻¹)^(1−n) × s⁻¹
Standard question: Given 3 rows of [A], [B], rate data. Find x, y, k.
| Exp | [A] (mol/L) | [B] (mol/L) | Rate (mol/L/s) |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10⁻⁴ |
| 2 | 0.20 | 0.10 | 4.0 × 10⁻⁴ |
| 3 | 0.10 | 0.20 | 4.0 × 10⁻⁴ |
Q: Rate constant doubles from 300K to 310K. Find Ea.
| Parameter | Zero Order | First Order (for comparison) |
|---|---|---|
| Rate law | rate = k (independent of [A]) | rate = k[A] |
| Integrated rate law | [A] = [A]₀ − kt | ln[A] = ln[A]₀ − kt |
| Half-life (t½) | t½ = [A]₀ / 2k (depends on [A]₀) | t½ = 0.693 / k (independent of [A]₀) |
| Units of k | mol L⁻¹ s⁻¹ (concentration/time) | s⁻¹ |
| Graph: [A] vs t | Straight line, slope = −k | Curve (exponential decay) |
| Examples | Decomposition of NH₃ on Pt surface; decomposition of HI on Au; enzyme-catalysed reactions at high substrate concentration | Radioactive decay; hydrolysis of alkyl halides in dilute alkali |
Zero Order — [A] vs t
Straight line. t½ = [A]₀/2k
t½ decreases with each successive half-life
First Order — [A] vs t
Exponential decay. t½ = 0.693/k
t½ is constant (independent of [A]₀)
| Aspect | Detail |
|---|---|
| Definition | A reaction that is actually of higher order (usually second order overall) but appears to be first order because one reactant is present in large excess (its concentration is effectively constant) |
| Condition | One reactant present in very large excess → its concentration doesn't change appreciably → treated as constant → "pseudo" first order |
| Example 1 — Acid hydrolysis of ethyl acetate | CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH Rate = k[ester][H₂O] but [H₂O] is constant (solvent) → Rate = k′[ester] where k′ = k[H₂O] |
| Example 2 — Inversion of sucrose | C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ + C₆H₁₂O₆ H₂O in excess → pseudo first order. Rate = k′[sucrose]. Studied by measuring optical rotation (sucrose is dextrorotatory; product mixture is levorotatory). |
| Why called "pseudo"? | The reaction is NOT truly first order — it's actually second order. It only behaves as first order under specific conditions (excess of one reactant). The effective rate constant k′ = k × [excess reactant]. |
| Distinguishing from true first order | Vary concentration of the "constant" reactant — if k′ changes, it was pseudo first order. A true first-order rate constant k does not depend on any concentration. |
Zero order: [A] falls linearly with time. t½ is halved every half-life (successive half-lives decrease). k has units of concentration/time. Reaction stops when [A] = 0 (limited catalyst surface). Examples: NH₃/Pt, HI/Au.
Pseudo first order trick: Whenever water is a reactant and the reaction is in aqueous solution, it's a candidate for pseudo first order. Water is the solvent → always in huge excess → its concentration (~55.5 mol/L) is constant.