CIE Syllabus 9701 · AS & A2

A Level Physical Chemistry
Complete Study Guide

Atomic Structure  ·  Stoichiometry  ·  Bonding  ·  States of Matter  ·  Energetics  ·  Electrochemistry  ·  Equilibria  ·  Kinetics

Chapter 1 · AS & A2

Atomic Structure

The atom is the fundamental building block of all matter. Understanding its structure is the foundation for all of chemistry.

Sub-atomic Particles

ParticleSymbolRelative MassRelative ChargeLocation
Protonp1+1Nucleus
Neutronn10Nucleus
Electrone1/1836 ≈ 0−1Shells/orbitals
Proton Number (Atomic Number) Z

The number of protons in the nucleus of an atom. All atoms of the same element have the same proton number.

Mass Number (Nucleon Number) A

The total number of protons plus neutrons in the nucleus. A = Z + N, where N is the number of neutrons.

Nuclide notation
AZX — e.g. 126C means 6 protons, 6 neutrons, 6 electrons (if neutral)

Isotopes

Isotopes

Atoms of the same element with the same number of protons but different numbers of neutrons. They have the same chemical properties but different physical properties (e.g. density, boiling point).

🧪 Isotopes of Hydrogen

  • 11H — Protium (1p, 0n)
  • 21H — Deuterium (1p, 1n)
  • 31H — Tritium (1p, 2n) radioactive

🧪 Isotopes of Chlorine

  • 3517Cl — 75% natural abundance
  • 3717Cl — 25% natural abundance

🧪 Isotopes of Carbon

  • 126C — most abundant
  • 136C — ~1%
  • 146C — radioactive (carbon dating)

Relative Atomic Mass (Ar)

Definition: weighted mean mass relative to 1/12 the mass of a 12C atom
Ar = Σ (isotope mass × % abundance) / 100
Worked Example — Chlorine:
Ar(Cl) = (35 × 75 + 37 × 25) / 100 = (2625 + 925) / 100 = 35.5

Mass Spectrometry AS

⚡ 4 Stages

  • Ionisation: electron bombardment removes electrons → positive ions
  • Acceleration: electric field accelerates ions
  • Deflection: magnetic field deflects — lighter ions deflect more
  • Detection: ions detected; signal gives m/z and abundance

📊 Reading a Mass Spectrum

  • x-axis = mass/charge ratio (m/z)
  • y-axis = relative abundance (%)
  • Each peak = one isotope
  • Tallest peak = most abundant isotope
  • Ar = weighted mean of all peaks

Electron Configuration AS

Sub-shells and Orbitals

Electrons occupy orbitals within sub-shells: s (max 2e), p (max 6e), d (max 10e), f (max 14e). Each orbital holds a maximum of 2 electrons with opposite spins (Pauli exclusion principle).

1s2s 2p3s 3p 3d4s 4p 4d 4f
ElementConfigurationNotes
H (Z=1)1s11 electron
He (Z=2)1s2Noble gas — full shell
Na (Z=11)1s2 2s2 2p6 3s1or [Ne] 3s1
Cl (Z=17)1s2 2s2 2p6 3s2 3p5or [Ne] 3s2 3p5
Fe (Z=26)1s2 2s2 2p6 3s2 3p6 3d6 4s2or [Ar] 3d6 4s2
Cu (Z=29)[Ar] 3d10 4s1Exception: half-filled 4s for stability
Cr (Z=24)[Ar] 3d5 4s1Exception: half-filled d sub-shell
⚠️ Common Exam Errors: When ions form, d-block elements lose 4s electrons FIRST. So Fe2+ = [Ar] 3d6 (NOT [Ar] 3d4 4s2); Fe3+ = [Ar] 3d5.

Ionisation Energies AS

First Ionisation Energy (1st IE)

The energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous unipositive ions under standard conditions.

Equation for 1st IE of element X
X(g) → X+(g) + e   ΔH = 1st IE (kJ mol−1)

📈 Factors Affecting IE

  • Nuclear charge: more protons → higher IE
  • Atomic/ionic radius: larger radius → lower IE (greater distance)
  • Shielding: more inner electrons → lower IE
  • Sub-shell: p electrons shielded by s electrons in same shell

📉 Trends Across Period 3

  • General increase Na → Ar
  • Dip at Al: 3p electron easier to remove than 3s (shielded by 3s)
  • Dip at S: paired electron in 3p repels, easier to remove
Successive IE equations for Na
Na(g) → Na+(g) + e    (1st IE = 496 kJ mol−1)
Na+(g) → Na2+(g) + e    (2nd IE = 4563 kJ mol−1)
Na2+(g) → Na3+(g) + e    (3rd IE = 6913 kJ mol−1)
Identifying group from successive IEs: A large jump between the nth and (n+1)th IE indicates the element is in Group n. For Na, the huge jump between 1st and 2nd IE confirms it is in Group 1.
Chapter 2 · AS

Atoms, Molecules & Stoichiometry

Stoichiometry is the quantitative study of chemical reactions — using moles, formulae and equations to calculate what reacts and what forms.

The Mole Concept

The Mole

One mole is the amount of substance containing 6.022 × 1023 particles (the Avogadro constant, L or NA). This number was defined relative to exactly 12 g of 12C.

🔢 Key Formulae

n = m / M

n = moles; m = mass (g); M = molar mass (g mol−1)

n = V / 22.4   (gas at STP)
n = V / 24.0   (gas at RTP, 25°C, 1 atm)
n = c × V   (V in dm3)

c = concentration (mol dm−3)

📐 Molar Mass Examples

  • H2O: M = 2(1) + 16 = 18 g mol−1
  • NaCl: M = 23 + 35.5 = 58.5 g mol−1
  • H2SO4: M = 2 + 32 + 4(16) = 98 g mol−1
  • CaCO3: M = 40 + 12 + 3(16) = 100 g mol−1
  • Na2CO3: M = 2(23)+12+3(16) = 106 g mol−1

Empirical & Molecular Formulae

Empirical Formula

The simplest whole-number ratio of atoms of each element in a compound. E.g. glucose C6H12O6 has empirical formula CH2O.

Molecular Formula

The actual number of atoms of each element in one molecule. Molecular formula = n × empirical formula. n = Mr(molecule) / Mr(empirical unit).

Method to find empirical formula: 1. Convert % by mass → grams  →  2. Divide by Ar to get moles  →  3. Divide by smallest mole value  →  4. Round to nearest whole number ratio.
Example: Compound contains 40.0% C, 6.67% H, 53.3% O
C: 40.0/12 = 3.33   H: 6.67/1 = 6.67   O: 53.3/16 = 3.33
Ratio = 1 : 2 : 1 → Empirical formula: CH2O

Balancing Equations

⚠️ State Symbols Must Be Included: (s) = solid, (l) = liquid, (g) = gas, (aq) = aqueous solution
ReactionBalanced Equation
Combustion of methaneCH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Neutralisation (HCl/NaOH)HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Combustion of ethanolC2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
Iron + dilute H2SO4Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g)
CaCO3 decompositionCaCO3(s) → CaO(s) + CO2(g)
Haber processN2(g) + 3H2(g) ⇌ 2NH3(g)
Contact process2SO2(g) + O2(g) ⇌ 2SO3(g)

Percentage Yield & Atom Economy

📊 Percentage Yield

% yield = (actual yield / theoretical yield) × 100

Theoretical yield is calculated from stoichiometry assuming complete reaction.

♻️ Atom Economy

% atom economy = (Mr of desired product / sum of Mr of all products) × 100

Higher atom economy = greener, less waste. Addition reactions have 100% atom economy.

Concentration & Titrations

Concentration — converting units
c (mol dm−3) = n / V(dm3)
1 dm3 = 1000 cm3, so V(dm3) = V(cm3) / 1000
Titration Example: 25.00 cm3 of NaOH is neutralised by 22.50 cm3 of 0.100 mol dm−3 HCl.
n(HCl) = 0.100 × 22.50/1000 = 0.00225 mol
Ratio HCl : NaOH = 1:1 → n(NaOH) = 0.00225 mol
c(NaOH) = 0.00225 / (25.00/1000) = 0.0900 mol dm−3

Ideal Gas Law

Ideal Gas Equation
pV = nRT

📏 Units

  • p = pressure in Pa (pascals)
  • V = volume in m3
  • n = moles
  • R = 8.314 J K−1 mol−1
  • T = temperature in K (= °C + 273)

🔢 Conversions

  • 1 dm3 = 10−3 m3
  • 1 cm3 = 10−6 m3
  • 1 atm = 101 325 Pa
  • STP: 273 K, 100 kPa
  • RTP: 298 K, 100 kPa
Chapter 3 · AS & A2

Chemical Bonding

Bonding determines the structure and properties of all substances — from the salt in food to the steel in buildings.

Types of Chemical Bond

🔵 Ionic Bonding

Electrostatic attraction between oppositely charged ions formed by transfer of electrons from metal to non-metal.

  • Strong forces → high melting points
  • Conduct electricity when molten or in solution
  • Example: Na+Cl, Mg2+O2−

🟢 Covalent Bonding

Sharing of a pair of electrons between two atoms, each contributing one electron (or both from one atom — dative bond).

  • Single: 1 shared pair — H–H, H–Cl
  • Double: 2 shared pairs — O=O, C=O
  • Triple: 3 shared pairs — N≡N

⚪ Metallic Bonding

Lattice of positive metal ions surrounded by a 'sea' of delocalised electrons. Electrostatic attraction between ions and electrons.

  • Good conductors (delocalised electrons)
  • Malleable and ductile
  • Generally high melting points

Lewis (dot-and-cross) Structures

MoleculeFormulaBonding pairsLone pairs on central atom
WaterH2O22
AmmoniaNH331
MethaneCH440
Carbon dioxideCO22 double bonds0
Sulfur hexafluorideSF660
Phosphorus pentachloridePCl550

VSEPR Theory & Shapes

VSEPR — Valence Shell Electron Pair Repulsion

Electron pairs around a central atom arrange themselves to minimise repulsion. Lone pairs repel more than bonding pairs: lone-lone > lone-bond > bond-bond.

Bonding pairsLone pairsShapeBond angleExample
20Linear180°BeCl2, CO2
30Trigonal planar120°BF3, SO3
40Tetrahedral109.5°CH4, SiCl4
31Trigonal pyramidal107°NH3
22Bent / V-shaped104.5°H2O
50Trigonal bipyramidal90° & 120°PCl5
60Octahedral90°SF6
⚠️ Note on SO2: SO2 has 2 bonding regions + 1 lone pair → bent shape, bond angle ~119° (not 120° due to lone pair repulsion).

Electronegativity & Polarity

Electronegativity

The ability of an atom to attract the shared pair of electrons in a covalent bond towards itself. Increases across a period (more protons) and up a group (smaller atomic radius, less shielding).

Pauling Scale highlights: F = 4.0 (most electronegative), O = 3.5, N = 3.0, Cl = 3.0, H = 2.1, C = 2.5. If Δelectronegativity > ~1.7 → ionic; 0.5–1.7 → polar covalent; < 0.5 → non-polar covalent.

➕➖ Polar vs Non-Polar Molecules

  • CCl4: polar bonds but symmetric tetrahedral → dipoles cancel → non-polar
  • CHCl3: asymmetric → net dipole → polar
  • CO2: linear, dipoles cancel → non-polar
  • H2O: bent, dipoles don't cancel → polar

🔗 Dative Covalent (Co-ordinate) Bond

Both electrons in the shared pair come from the same atom. Once formed, identical to a covalent bond.

  • NH4+: lone pair from NH3 → H+
  • H3O+: lone pair from H2O → H+
  • Al2Cl6: Cl lone pair → Al
  • CO: one of three bonds is dative

Intermolecular Forces

🌊 van der Waals / London Forces

Temporary dipoles induced by instantaneous uneven electron distribution. Present in ALL molecules.

  • Strength increases with: more electrons / larger surface area
  • Weakest intermolecular force
  • Explains why I2 has higher bp than F2

↔️ Permanent Dipole–Dipole

Electrostatic attraction between δ+ of one polar molecule and δ− of another.

  • Stronger than London forces
  • Example: HCl, SO2, CH2Cl2
  • Explains higher bp of HCl vs noble gases

🔵 Hydrogen Bonding

Strong dipole–dipole interaction between H bonded to N, O, or F and a lone pair on another N, O, or F atom.

  • H2O: up to 4 H-bonds per molecule
  • Explains anomalously high bp of H2O (100°C vs predicted −80°C)
  • Explains high bp of NH3, HF, alcohols
  • Explains ice being less dense than water
Chapter 4 · AS & A2

States of Matter

The physical state of a substance and its bulk properties are determined by the nature and strength of the forces between its particles.

Giant Structures vs Simple Molecular

🏗️ Giant Ionic Lattice

  • Regular 3D array of oppositely charged ions
  • High melting points (strong electrostatic forces)
  • Brittle — layers shift, like charges repel
  • Conducts when molten or aqueous (not solid)
  • Example: NaCl (6:6 coordination), CsCl (8:8)

🔷 Giant Covalent (Macromolecular)

  • Diamond: each C bonded to 4 C tetrahedrally; very hard, very high m.p., non-conductor
  • Graphite: layers of hexagonal C rings; delocalised electrons within layers → conductor; soft (layers slide)
  • SiO2: giant covalent, high m.p.
  • Si: semiconductor

🧊 Simple Molecular

  • Discrete molecules held by weak intermolecular forces
  • Low melting and boiling points
  • Non-conductors (no ions or free electrons)
  • Examples: H2O, CO2, I2, CH4, S8

⚙️ Metallic Lattice

  • Close-packed metal cations + delocalised electron sea
  • Good conductors of heat and electricity
  • Malleable, ductile, lustrous
  • m.p. increases with: charge on ion, number of delocalised electrons, smaller ionic radius
  • Example: Na < Mg < Al (increasing ionic charge and fewer shells)

Allotropes of Carbon A2

PropertyDiamondGraphiteBuckminster-
fullerene C60
Graphene
StructureGiant covalentGiant covalentSimple molecular2D layer
Bonding4 C–C bonds3 C–C + 1 deloc πCovalent within3 C–C + 1 deloc
Melting pointVery high (~3550°C)Very high (~3675°C)Low (sublimes ~600°C)Very high
Electrical conductionNoYesNo (poor)Yes
HardnessVery hardSoft (slippery)SoftExtremely strong

Gas Laws & Kinetic Theory

💨 Kinetic Molecular Theory

  • Gas particles in constant, random motion
  • Particles have negligible volume
  • No intermolecular forces (ideal gas)
  • Elastic collisions (no energy lost)
  • KE ∝ absolute temperature T

📐 Gas Laws Summary

  • Boyle's Law: pV = constant (fixed T, n)
  • Charles' Law: V/T = constant (fixed p, n)
  • Pressure Law: p/T = constant (fixed V, n)
  • Ideal: pV = nRT
⚠️ Real vs Ideal Gases: At high pressure and low temperature, real gases deviate from ideal behaviour. Particles have finite volume (V is larger than ideal) and there ARE intermolecular attractions (p is lower than ideal). Van der Waals equation: (p + an²/V²)(V − nb) = nRT.

Liquid Crystals A2

Liquid Crystals

A state of matter with properties between crystalline solids and isotropic liquids. Molecules are ordered (like a crystal) but can flow (like a liquid). Used in LCD screens. Typically long rod-shaped molecules with polar groups.

Chapter 5 · AS & A2

Chemical Energetics

Thermochemistry allows us to measure and predict the energy changes in chemical reactions, from combustion to dissolution.

Enthalpy Changes — Key Definitions

🔥 Standard Enthalpy of Combustion ΔH°c

Enthalpy change when 1 mole of a substance burns completely in oxygen under standard conditions (298 K, 100 kPa), with all reactants and products in standard states. Always negative (exothermic).

C(graphite,s) + O2(g) → CO2(g)   ΔH°c = −394 kJ mol−1

⚗️ Standard Enthalpy of Formation ΔH°f

Enthalpy change when 1 mole of a compound is formed from its elements in their standard states under standard conditions. ΔH°f of an element = 0 by definition.

2C(graphite,s) + 3H2(g) + ½O2(g) → C2H5OH(l)   ΔH°f = −277 kJ mol−1

💧 Enthalpy of Neutralisation ΔH°neut

Enthalpy change when an acid and base react to form 1 mole of H2O under standard conditions.

H+(aq) + OH(aq) → H2O(l)   ΔH°neut = −57.1 kJ mol−1

🧊 Enthalpy of Atomisation ΔH°at

Enthalpy change to produce 1 mole of gaseous atoms from the element in its standard state. Always endothermic (positive).

Na(s) → Na(g)   ΔH°at = +107 kJ mol−1

Hess's Law

Hess's Law

The enthalpy change of a reaction is independent of the route taken, provided the initial and final conditions are the same. This is a consequence of conservation of energy.

Using formation enthalpies
ΔH°rxn = Σ ΔH°f(products) − Σ ΔH°f(reactants)
Using combustion enthalpies
ΔH°rxn = Σ ΔH°c(reactants) − Σ ΔH°c(products)
Worked Example: Find ΔH°rxn for C(s) + 2H2(g) → CH4(g)
Given: ΔH°c(C) = −394, ΔH°c(H2) = −286, ΔH°c(CH4) = −890 kJ mol−1
ΔH°f(CH4) = [−394 + 2(−286)] − [−890] = −966 + 890 = −76 kJ mol−1

Bond Enthalpy

Mean Bond Enthalpy

The average energy required to break one mole of a particular type of covalent bond in the gaseous state, averaged over many different compounds. Always positive (endothermic — breaking bonds requires energy).

Energy cycle using bond enthalpies
ΔH°rxn = Σ (bonds broken, BE reactants) − Σ (bonds formed, BE products)
BondMean Bond Enthalpy / kJ mol−1
H–H+436
O=O+496
C–H+412
C=O (in CO2)+743
O–H+463
N≡N+944
N–H+388
Cl–Cl+243

Born–Haber Cycles A2

Lattice Enthalpy ΔH°LE

The energy released when one mole of an ionic solid is formed from its gaseous ions under standard conditions. It is always negative (exothermic). More negative = stronger lattice = higher melting point.

Born–Haber cycle for NaCl — applying Hess's Law
ΔH°f(NaCl) = ΔH°at(Na) + ΔH°at(Cl) + IE1(Na) + EA1(Cl) + ΔH°LE(NaCl)
−411 = +107 + (+122) + (+496) + (−349) + ΔH°LE
ΔH°LE = −411 − 107 − 122 − 496 + 349 = −787 kJ mol−1
Enthalpy of hydration ΔH°hyd = energy released when 1 mol of gaseous ions dissolve in water to form 1 mol of aqueous ions.
Enthalpy of solution ΔH°sol = ΔH°LE(dissociation) + ΔH°hyd(cation) + ΔH°hyd(anion)

Entropy & Gibbs Free Energy A2

Entropy (S)

A measure of the disorder or randomness of a system. Units: J K−1 mol−1. Increases when: solids dissolve, gases form, temperature rises, number of moles of gas increases.

Gibbs Free Energy
ΔG = ΔH − TΔS

✅ Spontaneity Conditions

  • Reaction is spontaneous (feasible) when ΔG < 0
  • ΔG = 0 at equilibrium
  • ΔG > 0 → non-spontaneous
  • If ΔH −ve and ΔS +ve → always spontaneous
  • If ΔH +ve and ΔS −ve → never spontaneous

🌡️ Temperature Dependence

  • ΔH −ve, ΔS −ve → spontaneous at low T
  • ΔH +ve, ΔS +ve → spontaneous at high T
  • Crossover temperature: T = ΔH / ΔS
  • Example: CaCO3(s) → CaO(s) + CO2(g)
    ΔH = +178 kJ mol−1, ΔS = +165 J K−1 mol−1
    T = 178000/165 ≈ 1079 K
Chapter 6 · AS & A2

Electrochemistry

Electrochemistry links chemical change with electrical energy — from batteries to electrolysis of aluminium.

Oxidation & Reduction

⬆️ OIL RIG Mnemonic

  • Oxidation Is Loss (of electrons)
  • Reduction Is Gain (of electrons)
  • Oxidising agent = accepts electrons (gets reduced)
  • Reducing agent = donates electrons (gets oxidised)

🔢 Oxidation State Rules

  • Free element = 0 (e.g. Fe, O2, Na)
  • Monatomic ion = charge (e.g. Fe2+ = +2)
  • O in compounds = −2 (except peroxides = −1, OF2 = +2)
  • H in compounds = +1 (except metal hydrides = −1)
  • Sum of OS in neutral molecule = 0
  • Sum of OS in ion = charge of ion
Oxidation State Examples:
Cr in K2Cr2O7: 2(+1) + 2x + 7(−2) = 0 → 2x = 12 → x = +6
Mn in MnO4: x + 4(−2) = −1 → x = +7
S in H2SO4: 2(+1) + x + 4(−2) = 0 → x = +6

Half-Equations & Full Redox Equations

Balancing in acidic solution — add H+ and H2O
MnO4(aq) + 8H+(aq) + 5e → Mn2+(aq) + 4H2O(l)   (reduction)
Fe2+(aq) → Fe3+(aq) + e   (oxidation) × 5
────────────────────────────────────────────────────
MnO4(aq) + 8H+(aq) + 5Fe2+(aq) → Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)

Electrochemical Cells A2

Standard Electrode Potential E°

The EMF of a half-cell connected to a standard hydrogen electrode (SHE) under standard conditions (298 K, 100 kPa, 1 mol dm−3 solutions). The SHE has E° = 0.00 V by definition.

EMF of electrochemical cell
cell = E°cathode (reduction) − E°anode (oxidation)
= E°right − E°left
Half-equation (Reduction)E° / V
F2(g) + 2e → 2F(aq)+2.87
MnO4(aq) + 8H+(aq) + 5e → Mn2+(aq) + 4H2O(l)+1.51
Cl2(g) + 2e → 2Cl(aq)+1.36
Cu2+(aq) + 2e → Cu(s)+0.34
2H+(aq) + 2e → H2(g)0.00
Fe2+(aq) + 2e → Fe(s)−0.44
Zn2+(aq) + 2e → Zn(s)−0.76
Li+(aq) + e → Li(s)−3.04
Predicting feasibility: A reaction is feasible (spontaneous) when E°cell > 0. The more positive the E°cell, the greater the tendency to react. Example: Cu2+ + Fe → Cu + Fe2+: E°cell = +0.34 − (−0.44) = +0.78 V → feasible.
⚠️ Limitations: E° predictions assume standard conditions. Non-standard concentrations, temperatures, and kinetic barriers may prevent a thermodynamically feasible reaction from occurring.

The Nernst Equation A2

Standard electrode potentials E° are measured at 298 K with all ion concentrations at exactly 1 mol dm−3. In real systems, concentrations differ from standard, so the actual electrode potential E changes. The Nernst equation corrects E° for non-standard conditions.

The Nernst Equation — for a half-cell at non-standard conditions
E = E° − (RT / nF) × ln Q

At 298 K, substituting R = 8.314 J K−1 mol−1, T = 298 K, F = 96 485 C mol−1:

E = E° − (0.02569 / n) × ln Q

Or equivalently using log base 10:

E = E° − (0.0592 / n) × log10 Q    (at 298 K)

🔑 What Each Symbol Means

  • E — actual electrode potential under non-standard conditions (V)
  • — standard electrode potential (V)
  • R — gas constant = 8.314 J K−1 mol−1
  • T — absolute temperature (K); always use K, not °C
  • n — number of moles of electrons transferred in the half-equation
  • F — Faraday constant = 96 485 C mol−1
  • Q — reaction quotient = ratio of products to reactants at the current concentrations (same form as K, but using actual concentrations)
  • ln Q — natural logarithm of Q; use log10 Q with the 0.0592/n form

📐 Writing the Reaction Quotient Q

Q has the same form as the equilibrium constant K, using actual (non-standard) concentrations or partial pressures instead of equilibrium values.

  • For: Cu2+(aq) + 2e → Cu(s)
    Q = 1 / [Cu2+]  (solid Cu omitted)
  • For: Zn2+(aq) + 2e → Zn(s)
    Q = 1 / [Zn2+]
  • For: Fe3+(aq) + e → Fe2+(aq)
    Q = [Fe2+] / [Fe3+]
  • For: MnO4 + 8H+ + 5e → Mn2+ + 4H2O
    Q = [Mn2+] / ([MnO4][H+]8)
Worked Example 1 — Effect of concentration on E:
Calculate E for Cu2+(aq) + 2e → Cu(s) when [Cu2+] = 0.010 mol dm−3. E° = +0.34 V, n = 2.

Q = 1/[Cu2+] = 1/0.010 = 100
E = 0.34 − (0.0592/2) × log10(100)
E = 0.34 − (0.0296) × 2
E = 0.34 − 0.0592 = +0.281 V

Interpretation: Lowering [Cu2+] below 1 mol dm−3 makes the Cu2+/Cu electrode less positive — the system has less tendency to be reduced.
Worked Example 2 — Full cell with non-standard concentrations:
Cell: Zn(s) | Zn2+(aq, 0.001 mol dm−3) ‖ Cu2+(aq, 2.0 mol dm−3) | Cu(s)
E°(Cu2+/Cu) = +0.34 V; E°(Zn2+/Zn) = −0.76 V; n = 2 for both.

E(Cu) = 0.34 − (0.0592/2) × log10(1/2.0) = 0.34 − (0.0296)(−0.301) = 0.34 + 0.0089 = +0.349 V
E(Zn) = −0.76 − (0.0592/2) × log10(1/0.001) = −0.76 − (0.0296)(3) = −0.76 − 0.0888 = −0.849 V
Ecell = E(cathode) − E(anode) = +0.349 − (−0.849) = +1.198 V

Compare with E°cell = 0.34 − (−0.76) = +1.10 V — the non-standard conditions increase the cell EMF.

📊 Effect of Concentration on E — Key Rules

  • Increasing [oxidised form] (e.g. more Cu2+) → Q decreases → E increases (more positive) → greater tendency to be reduced
  • Decreasing [oxidised form] → Q increases → E decreases
  • Increasing [reduced form] (e.g. more Fe2+) → Q increases → E decreases
  • At equilibrium: Ecell = 0 and Q = K; this links E° to K: ln K = nFE°/RT

🔗 Nernst Equation & Equilibrium Constant

At equilibrium, Q = K and Ecell = 0. Substituting into the Nernst equation for the full cell:

0 = E°cell − (RT / nF) × ln K

∴  ln K = nFE°cell / RT

At 298 K:  log10 K = nE°cell / 0.0592

This means a large positive E°cell → large K → reaction strongly favours products. A negative E°cell → K < 1 → reactants favoured.

🌡️ Effect of Temperature on E

  • The Nernst equation contains T explicitly: E = E° − (RT/nF) ln Q
  • Higher T → larger (RT/nF) factor → greater correction from E°
  • For an exothermic cell reaction: increasing T lowers E°cell (Le Chatelier — backward shift)
  • For an endothermic cell reaction: increasing T raises E°cell
  • Fuel cells and batteries lose efficiency at low T because k is also affected
⚠️ Common Exam Mistakes with the Nernst Equation:
  • Using °C instead of K — always convert: T(K) = T(°C) + 273
  • Using the wrong value of n — n is the number of electrons in the balanced half-equation, NOT a stoichiometric coefficient
  • Forgetting that pure solids and pure liquids are omitted from Q (just as in K expressions)
  • Confusing Q (reaction quotient, at any point) with K (at equilibrium)
  • Applying the 0.0592/n shortcut at temperatures other than 298 K — this value is only valid at exactly 25°C

Electrolysis A2

⚡ Electrolysis of Molten NaCl

  • Cathode (negative): Na+(l) + e → Na(l)
  • Anode (positive): 2Cl(l) → Cl2(g) + 2e
  • Overall: 2NaCl(l) → 2Na(l) + Cl2(g)

⚡ Electrolysis of Aqueous NaCl

  • Cathode: 2H+(aq) + 2e → H2(g) (H+ from water preferentially discharged)
  • Anode: 2Cl(aq) → Cl2(g) + 2e
  • Solution becomes NaOH(aq) — chlor-alkali process

⚡ Electrolysis of CuSO4(aq)

  • Cathode: Cu2+(aq) + 2e → Cu(s)
  • Anode (inert): 2H2O(l) → O2(g) + 4H+(aq) + 4e
  • Anode (Cu): Cu(s) → Cu2+(aq) + 2e (used in electroplating/purification)

Faraday's Laws

Faraday's Laws of Electrolysis
Q = I × t    (charge = current × time)
n(e) = Q / F    (F = 96 485 C mol−1)
n(substance) = n(e) / number of electrons in half-equation
Chapter 7 · AS & A2

Equilibria

Many reactions are reversible and reach a state of dynamic equilibrium. Understanding equilibria is key to industrial processes like the Haber process.

Dynamic Equilibrium

Dynamic Equilibrium

In a closed system, when the forward and reverse reactions occur at the same rate, so concentrations of reactants and products remain constant (but not necessarily equal). The system is dynamic because both reactions still occur.

General equilibrium
aA(g) + bB(g) ⇌ cC(g) + dD(g)

Equilibrium Constants Kc and Kp

Expression for Kc (concentration equilibrium constant)
Kc = [C]c[D]d / [A]a[B]b
Expression for Kp (partial pressure equilibrium constant)
Kp = (pC)c × (pD)d / (pA)a × (pB)b
Relationship Kp and Kc
Kp = Kc(RT)Δn   where Δn = moles of gaseous products − moles of gaseous reactants
Key rules for K: Only gaseous and aqueous species are included (not solids or pure liquids). K has no units if written in terms of activities. In practice, units depend on the powers in the expression.
ReactionKc ExpressionUnits of Kc
N2(g) + 3H2(g) ⇌ 2NH3(g)[NH3]² / ([N2][H2]³)mol−2 dm6
H2(g) + I2(g) ⇌ 2HI(g)[HI]² / ([H2][I2])dimensionless
2SO2(g) + O2(g) ⇌ 2SO3(g)[SO3]² / ([SO2]²[O2])mol−1 dm3

Le Chatelier's Principle

Le Chatelier's Principle

When a system in dynamic equilibrium is subjected to a change, the position of equilibrium shifts in the direction that opposes that change.

ChangeEffect on positionEffect on K
Increase concentration of reactantShifts right (→)No change
Decrease concentration of reactantShifts left (←)No change
Increase pressure (↑p)Shifts to side with fewer moles of gasNo change
Decrease pressure (↓p)Shifts to side with more moles of gasNo change
Increase temperature (exothermic rxn)Shifts left (↓K)K decreases
Increase temperature (endothermic rxn)Shifts right (↑K)K increases
Add catalystNo shift (equilibrium reached faster)No change

Industrial Applications

🏭 Haber Process (NH3)

N2(g) + 3H2(g) ⇌ 2NH3(g)   ΔH = −92 kJ mol−1
  • Conditions: 450°C, 200 atm, Fe catalyst
  • High pressure favours NH3 (fewer moles of gas), but very expensive
  • Lower T favours yield (exothermic) but rate too slow → compromise at 450°C
  • Catalyst doesn't affect K but speeds equilibrium

🏭 Contact Process (H2SO4)

2SO2(g) + O2(g) ⇌ 2SO3(g)   ΔH = −196 kJ mol−1
  • Conditions: 450°C, 1–2 atm, V2O5 catalyst
  • SO3 + H2SO4 → H2S2O7 (oleum) → then diluted
  • Conversion ~99.5%

Acid–Base Equilibria A2

Brønsted–Lowry Theory

An acid is a proton (H+) donor; a base is a proton acceptor. Conjugate acid–base pairs differ by one H+.

Water autoionisation
H2O(l) ⇌ H+(aq) + OH(aq)    Kw = [H+][OH] = 1.00 × 10−14 mol2 dm−6 at 25°C
pH definitions
pH = −log10[H+]
[H+] = 10−pH
pOH = −log10[OH]
pH + pOH = 14 (at 25°C)
pKa = −log10 Ka

🟡 Weak Acid Calculations

HA(aq) ⇌ H+(aq) + A(aq)
Ka = [H+][A] / [HA]

Assumption: [H+] = [A] and [HA]eq ≈ [HA]initial (valid when Ka << c)

[H+] = √(Ka × c) → pH = ½(pKa − log c)

📊 Buffer Solutions

Resist pH change on addition of small amounts of acid or alkali. Made from weak acid + its conjugate base (or excess weak acid + strong base).

Henderson–Hasselbalch:
pH = pKa + log([A]/[HA])

Example: CH3COOH / CH3COONa+

Chapter 8 · AS & A2

Reaction Kinetics

Kinetics studies how fast reactions proceed and what factors affect the rate — essential knowledge for everything from food preservation to industrial chemistry.

Rates of Reaction

Rate of Reaction

The change in concentration of a reactant or product per unit time. Units: mol dm−3 s−1.

Rate expression
rate = −Δ[A] / Δt   or   rate = +Δ[P] / Δt

Factors Affecting Rate

🌡️ Temperature

Increasing temperature increases rate because:

  • Particles have greater kinetic energy
  • More particles have energy ≥ activation energy (Ea)
  • Higher frequency of successful collisions
  • Rule of thumb: ~10°C rise ≈ doubles rate

💊 Concentration

  • Higher concentration → more particles per unit volume
  • More frequent collisions
  • Greater rate
  • For gases: higher pressure = higher concentration

🪨 Surface Area

  • Smaller particles → greater surface area
  • More exposed particles → more collisions
  • Increases rate
  • Example: flour dust explosions, powdered vs lump zinc

🧪 Catalyst

  • Provides alternative reaction pathway with lower Ea
  • More particles have energy ≥ Ea
  • Unchanged at end of reaction
  • Does NOT shift equilibrium position but equilibrium reached faster

Collision Theory & Maxwell–Boltzmann Distribution

Activation Energy (Ea)

The minimum energy that colliding particles must possess for a reaction to occur. Even if particles collide, reaction only occurs if they have energy ≥ Ea AND the correct orientation.

Kinetic energy → Number of molecules → Ea T1 T2 > T1

Maxwell–Boltzmann distribution: at higher temperature (T2), more molecules have energy ≥ Ea

Rate Equations A2

General rate equation for: A + B → products
rate = k[A]m[B]n

📊 Reaction Orders

  • Zero order (m=0): rate unaffected by [A]; graph [A] vs t is straight line
  • First order (m=1): rate ∝ [A]; [A] vs t is exponential decay
  • Second order (m=2): rate ∝ [A]²; steeper curve
  • Overall order = m + n
  • Determined experimentally — NOT from stoichiometry!

⏱️ Half-Life (t½)

  • Time for concentration to halve
  • First order: t½ = constant (independent of [A])
  • t½ = 0.693 / k (first order)
  • Radioactive decay is always first order
  • Second order: t½ increases as [A] decreases
Finding order from initial rate data: Double [A] and see effect on rate. If rate × 2 → 1st order; rate × 4 → 2nd order; rate unchanged → 0th order.

Rate Constant k & Arrhenius Equation A2

Arrhenius Equation
k = A · e−Ea/RT
ln k = ln A − Ea/RT
log k = log A − Ea / (2.303RT)

Plot ln k vs 1/T → straight line, gradient = −Ea/R

📐 Units of k

  • 0th order: mol dm−3 s−1
  • 1st order: s−1
  • 2nd order: mol−1 dm3 s−1
  • 3rd order: mol−2 dm6 s−1

🔬 Rate-Determining Step

  • The slowest step in a multi-step mechanism
  • Determines the rate equation
  • Only species in or before the RDS appear in the rate equation
  • Example: SN1 vs SN2 mechanisms
🏆 Final Challenge

20-Question Quiz

Test your knowledge across all 8 chapters. Select the best answer for each question.

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