Edexcel Chemistry — Book 2

Rate = how fast a reactant disappears or a product appears, per unit time

rate = Δ[ ] / Δt units: mol dm⁻³ s⁻¹

Measure by: gas volume, mass loss, colour, pH, conductivity, or sampling + titration

Faster reaction = bigger rate, not shorter time on its own

Rate equation: rate = k[A]ᵐ[B]ⁿ m, n = orders; m+n = overall order

Zero (0): change [ ] → no change in rate

First (1): double [ ] → rate doubles

Second (2): double [ ] → rate × 4

Orders come from experiment, NOT from the balanced equation (unless single-step)

Find them by initial-rate method: change one [ ], keep others fixed, watch rate

Or by graph shape (see diagram) or by constant half-life (first order signature)

Units of k depend on overall order:

0 order → mol dm⁻³ s⁻¹ • 1 order → s⁻¹

2 order → dm³ mol⁻¹ s⁻¹ • 3 order → dm⁶ mol⁻² s⁻¹

Trick: units of rate ÷ units of all [ ]ᵐ⁺ⁿ = units of k

Mechanism = the steps the reaction actually goes through

The slowest step = rate-determining step (RDS)

Only species in or before the RDS appear in the rate equation

Catalyst: lowers Eₐ via a different pathway — does NOT change ΔH or position of equilibrium

Arrhenius: k = A·e^(−Ea/RT) → ln k = −Ea/R · 1/T + ln A

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

So: Ea = − gradient × R R = 8.314 J K⁻¹ mol⁻¹

Temperature must be in Kelvin everywhere it appears

Memory Tip: Zero / First / Second order = line, curve, steeper curve

First order gives constant half-life — useful clue on conc-time graphs

"Powers in rate equation come from experiment, not from coefficients"

Arrhenius gradient is ALWAYS negative — Eₐ comes out positive after × −R

Common Mistake: saying rate = time taken — rate is per unit time, so faster reaction = bigger rate

Reading orders straight off the balanced equation — only valid for single-step reactions

Forgetting T must be Kelvin in Arrhenius — using °C gives nonsense

Including intermediates in the rate equation — replace them using earlier fast steps

Entropy (S) = how spread out the particles and their energy are

More spread = higher S. Solid < liquid < gas (by a lot)

Heat it up, melt it, vaporise it, mix it, make more moles of gas → S goes up

Units: J K⁻¹ mol⁻¹ (small numbers — be careful with kJ vs J)

A reaction happens if ΔS_total > 0 (it spreads out the universe a bit more)

ΔS_total = ΔS_system + ΔS_surroundings

Surroundings care about heat: ΔS_surr = − ΔH / T (T in K)

Exothermic → ΔS_surr positive (releases heat → universe more spread out)

Lattice energy: gaseous ions → solid, per mole

Na⁺(g) + Cl⁻(g) → NaCl(s) (always exothermic — strongly negative)

More negative when ions are smaller & more charged (stronger attraction)

MgO has a much bigger lattice energy than NaCl (charges 2+/2− vs 1+/1−)

Theoretical lattice energy: pure ionic model

Experimental (Born–Haber): real value

Big difference → bond has covalent character

Cation small + highly charged + anion big = more covalent (polarised)

Dissolving: ΔH_sol = ΔH_lattice diss. + Σ ΔH_hyd

ΔH_hydration: gaseous ion → aqueous ion (always negative — water attracts ions)

Even if ΔH_sol is slightly positive, salt can still dissolve thanks to entropy

Na⁺ pulls water O-side; Cl⁻ pulls water H-side (ion-dipole)

Memory Tip: S: solid < liquid < gas — gas is always the messiest

ΔS_surr = −ΔH/T — exothermic reactions help the surroundings

Small + highly charged ions = the most exothermic lattice energy

Check ΔS_total, not just ΔH — entropy can rescue an endothermic process

Common Mistake: mixing kJ and J — ΔH usually in kJ mol⁻¹, S in J K⁻¹ mol⁻¹. Convert before adding.

Using °C in −ΔH/T — must be Kelvin

Drawing lattice formation arrow upwards — it goes downwards (releases energy)

Saying "entropy = mess" — call it "particles and energy spreading out"

For aA + bB ⇌ cC + dD : Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ

Products on top, reactants on bottom ("top is products")

Ignore pure solids and pure liquids — they don't change

Big Kc → equilibrium lies right; small Kc → lies left

For gases use partial pressures: Kp = p_C^c · p_D^d / p_A^a · p_B^b

mole fraction = moles of gas ÷ total moles of gas

partial pressure = mole fraction × total pressure

Use equilibrium moles (not starting moles) every time

Only temperature changes the value of K

Exothermic forward → heat shifts left, K decreases

Endothermic forward → heat shifts right, K increases

Catalyst, pressure, concentration → may shift position but K stays the same

Units of K depend on the expression — count powers of (mol dm⁻³) or (Pa)

If powers cancel completely (e.g. H₂ + I₂ ⇌ 2HI), K has no units

ICE table: Initial, Change, Equilibrium — fill column by column

Memory Tip: "Top is products, bottom is reactants" — write Kc straight off the equation

Only T moves K. Everything else just shifts the position.

Pure solid/liquid → leave out of Kc. Aqueous and gas → in.

For Kp: count moles of GAS, not total moles

Common Mistake: using starting moles instead of equilibrium moles when calculating Kp / Kc

Saying "a catalyst increases K" — it does not. It speeds both directions equally.

Including solids in the expression (e.g. CaCO₃) — only species whose conc/pressure can vary go in

Forgetting the units — they depend on the specific expression each time

Brønsted acid = proton donor; Brønsted base = proton acceptor

Strong acid: fully dissociates (HCl → H⁺ + Cl⁻)

Weak acid: partial dissociation (CH₃COOH ⇌ H⁺ + CH₃COO⁻)

Strong ≠ concentrated. Strength = how much it dissociates; concentration = mol per dm³

pH = − log[H⁺] [H⁺] = 10^(−pH)

Water: Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ at 298 K

Pure water: [H⁺] = [OH⁻] = 1.00 × 10⁻⁷ → pH = 7

pH < 0 and pH > 14 are allowed for very concentrated solutions

Weak acid: HA ⇌ H⁺ + A⁻ Ka = [H⁺][A⁻] / [HA]

Approximation: assume [H⁺] = [A⁻] and [HA] ≈ initial concentration

So [H⁺] ≈ √(Ka × [HA]) then pH = −log[H⁺]

pKa = − log Ka — lower pKa = stronger acid (lower number wins)

Pick indicator by where the vertical jump sits:

Strong/strong → either methyl orange or phenolphthalein

Weak acid + strong base → phenolphthalein (jump in alkaline)

Strong acid + weak base → methyl orange (jump in acidic)

Buffer = weak acid + its salt (or weak base + its salt)

Example: CH₃COOH + CH₃COONa. Resists pH change when small amounts of acid/alkali added

Henderson form: [H⁺] = Ka × [HA] / [A⁻]

At half-neutralisation, [HA] = [A⁻] so pH = pKa — useful exam trick

Memory Tip: Lower pKa = stronger acid (lower number wins)

pH = pKa at the half-neutralisation point — quick way to read Ka off a curve

Indicator rule: choose by where the vertical jump is, not by what neutral means

Buffer makes the change smaller, not zero

Common Mistake: confusing strong (fully dissociated) with concentrated (lots of it)

Forgetting to take the square root in [H⁺] ≈ √(Ka × [HA])

Picking phenolphthalein for strong acid + weak base — jump is in the acidic range

Saying "buffer fixes pH" — it minimises the change, never stops it completely

Chiral C = 4 different groups on one carbon

Two non-superimposable mirror images = enantiomers

They rotate plane-polarised light in opposite directions — same melting point, same boiling point

A C in a C=C double bond can't be chiral — only 3 groups attached

Carbonyl group = C=O. Aldehyde RCHO (end of chain). Ketone RCOR' (middle).

Aldehydes oxidise easily → carboxylic acid. Ketones don't oxidise easily.

Tollens' (silver mirror): aldehyde only. Fehling's (brick-red ppt): aldehyde only.

Aldehyde gives a SILVER mirror — KETOne keeps QUIET.

Carbonyl C is δ+ (O pulls electron density). Nucleophiles attack there.

HCN reaction: CN⁻ attacks C=O → hydroxynitrile (gains 1 carbon)

Two steps: CN⁻ adds → O⁻ intermediate → grabs H⁺ from solution

New chiral C can form → racemic mix (50/50 enantiomers)

Carboxylic acid (−COOH): weak acid, partially dissociates

Acid + carbonate → salt + CO₂ + H₂O (fizz test — distinguishes from phenol)

Acid + alcohol ⇌ ester + water (conc H₂SO₄ catalyst — reversible)

Acyl chlorides RCOCl: very reactive — vigorous with water/alcohols/amines

Spectroscopy ID:

IR — C=O ~1700 cm⁻¹; broad O–H of COOH 2500–3300

¹H NMR — number of peaks = number of H environments; integration = ratio of Hs

Mass spec — M⁺ peak = molecular mass; base peak = tallest fragment

Memory Tip: Aldehyde → silver mirror (Tollens). Ketone → no reaction.

Nucleophile attacks δ+ carbon (C=O carbon) — every time

HCN adds 1 C — handy for "increase the chain length by one" questions

IR shortcut: broad 2500–3300 = COOH; sharp 1700 = C=O

Common Mistake: writing HCN as the attacker — it's CN⁻ (curly arrow from lone pair)

Saying a C=C carbon can be chiral — only sp³ C with 4 different groups counts

Forgetting that the HCN reaction usually gives a racemic mixture

Mixing up aldehyde and ketone tests — only aldehydes give Tollens / Fehling positives

Standard electrode potential E° = how much a half-cell wants to be reduced

Standard conditions: 298 K, 1 mol dm⁻³, 100 kPa

Reference: standard hydrogen electrode E° = 0.00 V

Bigger (more positive) E° = stronger oxidising agent (eats electrons)

Ecell = E°(right) − E°(left) or more positive E° − more negative E°

Positive Ecell → reaction is feasible as written

Negative Ecell → not feasible in that direction (try reverse)

Feasibility ≠ "happens quickly" — kinetics may still be slow

Standard means standard. Change [ ] or T → values change.

Increase [reactant] of a half-cell → its E° becomes more positive (Le Chatelier on the half-equation)

Apparatus must always have a salt bridge — KCl or KNO₃ — to keep both halves neutral

Redox titration: MnO₄⁻ + Fe²⁺ — purple → colourless endpoint (no indicator needed)

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Iodine–thiosulfate: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻

Add starch near endpoint → blue-black → colourless

Memory Tip: "Bigger E° eats electrons" — more +ve E° = better oxidising agent

Wires carry ELECTRONS, salt bridge carries IONS — keep that straight

Ecell positive = "yes it can happen" (thermodynamically). Doesn't say how fast.

MnO₄⁻ titration is self-indicating — first permanent pink = end

Common Mistake: saying electrons flow through the salt bridge — they don't

Confusing "feasible" (Ecell > 0) with "fast" — they're different things

Forgetting to acidify the MnO₄⁻ titration — only works with excess H⁺

Adding starch too early to an iodine titration — it traps I₂ and gives a false endpoint

Transition metal = forms at least one ion with a partially filled d sub-shell

Four signature jobs: colour • catalysis • variable oxidation • complex ions

Common: Fe²⁺/Fe³⁺, Cu⁺/Cu²⁺, Mn²⁺/MnO₄⁻, Cr³⁺/Cr₂O₇²⁻

Sc and Zn are NOT transition (Sc only Sc³⁺ has empty d; Zn²⁺ has full d¹⁰)

Ligand = lone-pair donor that bonds to the metal (coordinate bond)

Monodentate: H₂O, NH₃, Cl⁻, OH⁻, CN⁻

Multidentate (chelate): EDTA⁴⁻ (6 bonds), ethanedioate (2 bonds)

Coordination number = number of coordinate bonds to metal

Shapes by coordination number:

2 → linear • 4 → tetrahedral or square planar • 6 → octahedral

Isomerism: cis-trans (e.g. cis-platin used as anticancer drug)

Octahedral complexes with 3 bidentate ligands show optical isomerism

Colour: ligands split d-orbitals into two energy sets

Visible light supplies ΔE → an electron jumps up

You see the complementary colour of what was absorbed (absorb red → see green)

Change metal / oxidation state / ligand / shape → colour changes

Catalysts work because they can change oxidation state easily

Heterogeneous (different phase): Fe in Haber, V₂O₅ in Contact, Ni in hydrogenation

Homogeneous (same phase): Fe²⁺ in S₂O₈²⁻ + I⁻ reaction

Mn²⁺ autocatalysis: a product (Mn²⁺) catalyses its own formation in MnO₄⁻ + C₂O₄²⁻

Memory Tip: four T-metal jobs: colour, catalysis, variable oxidation, complex ions

Coord 2 / 4 / 6 → linear / (tet or sq-planar) / octahedral

Cu(H₂O)₆²⁺ pale blue → CuCl₄²⁻ yellow-green (octahedral → tetrahedral)

"Absorb red, see green" — colour seen is the complement of the colour absorbed

Common Mistake: saying tetrahedral and square planar are the same — same coord # (4), different shapes

Calling Sc/Zn transition metals — they don't form ions with partially filled d sub-shell

Forgetting that ligand substitution often changes both shape AND colour

Writing H₂O ligand as just "O" — show the whole molecule (or use abbreviation aq)

Benzene C₆H₆ — 6 carbons in a flat ring with 6 delocalised π electrons

All C–C bonds the same length (between single and double)

Extra stability from delocalisation = "aromatic stabilisation"

Skeletal symbol: hexagon with a circle inside — not 3 fixed double bonds

Benzene reacts by electrophilic substitution (keeps the aromatic ring)

Alkenes react by addition — losing C=C is fine

Substitution lets benzene keep its 6π stability — that's why it prefers it

Slogan: benzene substitutes, alkenes add

Nitration: conc HNO₃ + conc H₂SO₄ → NO₂⁺ attacks ring

Product: nitrobenzene

Halogenation: Br₂ / FeBr₃ (halogen carrier) → bromobenzene + HBr

Mechanism step 1: FeBr₃ generates Br⁺ from Br₂

Phenol = benzene with –OH attached. Oxygen lone pair feeds into the ring → more reactive

Phenol + Br₂(aq) → 2,4,6-tribromophenol (no halogen carrier needed)

Observation: bromine water decolourises + white precipitate forms

Phenol is weakly acidic — reacts with NaOH (but not Na₂CO₃ — distinguishes from COOH)

Memory Tip: Benzene substitutes, alkenes add

Phenol > benzene reactivity because the OH lone pair joins the π cloud

Distinguish phenol vs carboxylic acid: phenol reacts with NaOH but not Na₂CO₃

Always show the electrophile (NO₂⁺, Br⁺) — not just HNO₃ or Br₂ — in the mechanism

Common Mistake: drawing benzene as 3 fixed C=C when explaining stability — use the circle / delocalised cloud

Writing the electrophile as HNO₃ — it's NO₂⁺ (made by H₂SO₄ protonating HNO₃)

Forgetting the halogen carrier (AlCl₃ / FeBr₃) for benzene halogenation

Saying benzene undergoes addition — it doesn't (too stable to break the π system)

Amine = nitrogen with a lone pair → behaves as a base (accepts H⁺)

Primary RNH₂, secondary R₂NH, tertiary R₃N (count C bonded to N)

Make by: NH₃ + halogenoalkane (excess NH₃) • reduce nitrile (LiAlH₄) • reduce nitrobenzene → phenylamine

Reducing a nitrile adds one extra C to the chain — useful for "lengthen" routes

Basicity = how available the N lone pair is

Alkyl groups push electrons in → more basic (RNH₂ > NH₃)

Phenylamine less basic — lone pair delocalises INTO the benzene ring

Order: secondary > primary > NH₃ > phenylamine (rough order in water)

Amide R–C(=O)–NH₂. Made from acyl chloride + NH₃ (or amine) → amide + HCl

Acyl chlorides are very reactive: vigorous + steam at room T

Acyl chloride reaction spider: water → acid; alcohol → ester; NH₃ → amide; amine → N-substituted amide

Hydrolyse an amide with acid or base → carboxylic acid + amine (slow with hot reflux)

Amino acid = molecule with both –NH₂ and –COOH on the same C

Acts as acid AND base (amphoteric)

At the right pH (isoelectric pH) it exists as a zwitterion: ⁺H₃N–CHR–COO⁻

Isoelectric pH = pH where the amino acid has zero net charge — does not move in electrophoresis

Amino acids join via condensation — peptide bond −CO−NH− + water out

Many peptides → protein

Primary = sequence; Secondary = α-helix / β-sheet; Tertiary = 3D fold; Quaternary = multiple chains

Hydrolysis (acid + reflux) breaks peptide bonds back to amino acids

Memory Tip: amino acid charge by pH: Low → +, iso → ±, High → −

Phenylamine is the LEAST basic amine — lone pair delocalised into ring

Condensation MAKES peptide bonds. Hydrolysis BREAKS them.

Nitrile → amine reduction adds 1 carbon — handy for chain-extension routes

Common Mistake: saying phenylamine is more basic because it has a benzene ring — it's actually less basic

Drawing the zwitterion as neutral H₂N–CHR–COOH at neutral pH — it's ⁺H₃N–CHR–COO⁻

Confusing amide (CONH₂) with amine (NH₂) — amide has the C=O too

Writing "hydrolysis makes peptide bonds" — it breaks them, condensation makes them

Synthesis = plan a route from start material → product using known reactions

Always quote reagents AND conditions (e.g. KOH in ethanol, hot, reflux)

Aim for high atom economy, few steps, and easy purification

Chain length-changers: CN⁻ (+1 C, makes nitrile); reduction of nitrile keeps the +1 C

Functional-group tests (quick recall):

Alkene → Br₂(aq) decolourises • Aldehyde → Tollens/Fehling • Phenol → Br₂(aq) white ppt

Carboxylic acid → Na₂CO₃ fizzes (CO₂ gas) • Halogenoalkane → NaOH/warm, acidify, AgNO₃

Alcohol → acidified K₂Cr₂O₇ orange → green (1° and 2° only)

Practical techniques you must name:

Reflux — heat without losing volatile substances

Distillation — separate by boiling point

Separating funnel — separate immiscible layers

Recrystallisation — purify solid; melting-point test confirms purity

Identifying unknowns — order of attack:

1) IR for functional groups • 2) Mass spec for Mᵣ and fragments

3) ¹H NMR for H environments + ratios + splitting (n+1 rule)

4) ¹³C NMR for number of C environments

Chromatography: Rf = distance spot / distance solvent

Memory Tip: chain extension by 1 C → use CN⁻ then reduce (nitrile → amine route)

Distinguish phenol vs acid: acid fizzes with Na₂CO₃, phenol doesn't

Distinguish aldehyde vs ketone: Tollens silver mirror for aldehyde only

Always quote conditions — NaOH(aq)/warm gives substitution; NaOH in ethanol/hot gives elimination

Common Mistake: writing reagent without conditions — examiners want both

Skipping the silver-mirror step before going to carboxylic acid in a multi-step route

Using K₂Cr₂O₇ to "oxidise" a tertiary alcohol — it doesn't react

Forgetting that benzene needs a halogen carrier (alkenes don't)

CP9a — Iodine + propanone (acid catalysed)

Aim: find order with respect to I₂, propanone and H⁺ by sampling + titration

Quench samples with NaHCO₃ to stop the acid catalysis; titrate remaining I₂ vs thiosulfate

Plot [I₂] vs time → straight line = zero order in I₂

CP9b — Iodine clock

Time to first appearance of blue-black (starch + I₂) for different concentrations

Initial rate ∝ 1 / time — repeat at varied [reactant]

Common timing error: blink before the colour change → use white tile + uniform light

CP10 — Activation energy

Measure rate at several temperatures (same [ ] each time)

Calculate k for each, plot ln k vs 1/T → straight line

gradient × −R = Ea (Ea in J mol⁻¹ — divide by 1000 to get kJ mol⁻¹)

CP11 — Ka of a weak acid

Measure pH of known concentration of weak acid → find [H⁺] = 10⁻ᵖᴴ

Ka = [H⁺]² / [HA] (assumes [H⁺] = [A⁻] and [HA] ≈ initial)

Calibrate pH meter with buffer of pH 4 (and 7) at the same temperature

CP12 — Electrochemical cells

Build two half-cells; connect via salt bridge (KNO₃ or KCl)

High-resistance voltmeter reads E_cell (≈ no current drawn)

Compare measured value to E° = E°(right) − E°(left) from data book

CP13a — Fe²⁺ vs acidified KMnO₄

Self-indicating: first permanent pale pink = endpoint

Acidify with dilute H₂SO₄ (not HCl — Cl⁻ reduces MnO₄⁻)

Mole ratio: MnO₄⁻ : Fe²⁺ = 1 : 5

CP13b — Iodine / thiosulfate titration

Add starch only near the endpoint (when solution is straw-yellow) — else it traps I₂

Endpoint: blue-black → colourless

Mole ratio: I₂ : S₂O₃²⁻ = 1 : 2

CP14 — Preparing a transition-metal complex

Dissolve salt in minimum hot water → add ligand source (NH₃ or Cl⁻)

Crystallise on cooling → filter under reduced pressure → wash → dry

Calculate % yield from limiting reagent moles → expected product mass

CP15 — Identifying an organic / inorganic unknown

Build a logical test sequence: physical state → solubility → cation/anion tests → functional groups

Record observations as table — colour, gas, precipitate, smell (waft, don't sniff)

Use specific tests early (Tollens, Br₂, Na₂CO₃) — broad ones late

CP16 — Preparation of aspirin

2-hydroxybenzoic acid + ethanoic anhydride (conc H₂SO₄ catalyst) → aspirin + ethanoic acid

Reflux → cool → crystallise → recrystallise from hot water → dry

Check purity by melting point (pure aspirin mp 135 °C — sharp range)