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
Particle
Symbol
Relative Mass
Relative Charge
Location
Proton
p
1
+1
Nucleus
Neutron
n
1
0
Nucleus
Electron
e
1/1836 ≈ 0
−1
Shells/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
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
Element
Configuration
Notes
H (Z=1)
1s1
1 electron
He (Z=2)
1s2
Noble gas — full shell
Na (Z=11)
1s2 2s2 2p6 3s1
or [Ne] 3s1
Cl (Z=17)
1s2 2s2 2p6 3s2 3p5
or [Ne] 3s2 3p5
Fe (Z=26)
1s2 2s2 2p6 3s2 3p6 3d6 4s2
or [Ar] 3d6 4s2
Cu (Z=29)
[Ar] 3d10 4s1
Exception: half-filled 4s for stability
Cr (Z=24)
[Ar] 3d5 4s1
Exception: 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.
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
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
Molecule
Formula
Bonding pairs
Lone pairs on central atom
Water
H2O
2
2
Ammonia
NH3
3
1
Methane
CH4
4
0
Carbon dioxide
CO2
2 double bonds
0
Sulfur hexafluoride
SF6
6
0
Phosphorus pentachloride
PCl5
5
0
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 pairs
Lone pairs
Shape
Bond angle
Example
2
0
Linear
180°
BeCl2, CO2
3
0
Trigonal planar
120°
BF3, SO3
4
0
Tetrahedral
109.5°
CH4, SiCl4
3
1
Trigonal pyramidal
107°
NH3
2
2
Bent / V-shaped
104.5°
H2O
5
0
Trigonal bipyramidal
90° & 120°
PCl5
6
0
Octahedral
90°
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.
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
Property
Diamond
Graphite
Buckminster- fullerene C60
Graphene
Structure
Giant covalent
Giant covalent
Simple molecular
2D layer
Bonding
4 C–C bonds
3 C–C + 1 deloc π
Covalent within
3 C–C + 1 deloc
Melting point
Very high (~3550°C)
Very high (~3675°C)
Low (sublimes ~600°C)
Very high
Electrical conduction
No
Yes
No (poor)
Yes
Hardness
Very hard
Soft (slippery)
Soft
Extremely 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).
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.
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.
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)
Bond
Mean 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.
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.
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
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.
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)
E° — 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.
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.
Reaction
Kc Expression
Units 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.
Change
Effect on position
Effect on K
Increase concentration of reactant
Shifts right (→)
No change
Decrease concentration of reactant
Shifts left (←)
No change
Increase pressure (↑p)
Shifts to side with fewer moles of gas
No change
Decrease pressure (↓p)
Shifts to side with more moles of gas
No change
Increase temperature (exothermic rxn)
Shifts left (↓K)
K decreases
Increase temperature (endothermic rxn)
Shifts right (↑K)
K increases
Add catalyst
No 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+.
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 / CH3COO−Na+
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.
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.