IB Chemistry SL — All Flashcards

A pure substance made of **only one type of atom**; it cannot be broken down chemically.

**Two or more different atoms chemically bonded** together in a fixed ratio.

Two or more substances **physically combined but not chemically bonded**, in any ratio.

Only by **chemical** means (a reaction) — not by physical methods.

By **physical** means (e.g. filtration, distillation), because nothing is bonded.

Homogeneous = **uniform** (e.g. solution, air); heterogeneous = **not uniform**, parts visible (e.g. sand + iron).

Element = one type of atom; compound = different atoms **chemically bonded** in a fixed ratio.

A single element or compound — it has a **sharp, fixed** melting and boiling point.

A pure substance melts **sharply**; impurities **lower** it and spread it over a **range**.

A **mixture** (an alloy of copper and zinc) — the metals are not chemically bonded.

**Solid**, **liquid** and **gas** — they differ in how close the particles are and how freely they move.

**Packed** close in fixed positions; they only **vibrate**. A solid has a fixed shape and volume.

**Touching** but not fixed; they **slide** past each other. A liquid has fixed volume but takes the container's shape.

**Far apart**, moving **fast and randomly**. A gas fills its container and is easily compressed.

A model treating matter as **small particles in constant random motion**, with attractive forces between them that weaken as they spread apart.

The **average kinetic energy** of the particles — hotter means the particles move faster on average.

**Add 273.15** (≈ 273): T(K) = θ(°C) + 273.15.

**0 K** (about −273 °C) — the lowest possible temperature, where particle motion is at a minimum.

Gas particles are **far apart** with large gaps to close up; liquid particles are already **touching** with little space.

The added energy is used to **overcome the forces** between particles, not to speed them up, so the average kinetic energy (temperature) stays the same.

Its particles are **not held in fixed positions**, so they **slide** and flow to fit the container.

They **gain kinetic energy** and **vibrate more**, until they have enough energy to break free of their fixed positions and the solid melts.

Their components are **not chemically bonded**, so they keep their own properties and can be separated by **physical** methods.

An **insoluble solid** from a liquid — the solid is too large to pass through the **filter paper** (uses particle size).

A **dissolved (soluble) solid** from its solution — the **solvent boils off**, leaving the solid behind.

Liquids (or a liquid from a dissolved solid) by their difference in **boiling point**.

The dissolved components of a mixture by their difference in **solubility / attraction** to the paper.

R_{f} = distance moved by **spot** ÷ distance moved by **solvent** — a ratio with **no units**, between 0 and 1.

The component is **more soluble** in the solvent, so it was carried **further** up the paper.

Add water to dissolve the salt, then **filter** — the sand stays as the residue.

Use a **magnet** — iron is **magnetic**, sand and salt are not.

**Crystallisation** — dissolve in hot solvent, cool to form pure crystals, then filter them off.

**No** — the spot cannot move further than the solvent front, so 0 < R_{f} < 1.

Size → **filtration**; boiling point → **distillation**; solubility → **crystallisation / chromatography**.

Together in the tiny, dense central **nucleus** of the atom.

Moving around the nucleus in **shells** (energy levels); this region is mostly empty space.

Relative mass **1**, relative charge **+1**.

Relative mass **1**, relative charge **0** (neutral).

Relative mass ≈ **1/1836** (negligible), relative charge **−1**.

The number of **protons** in the nucleus; it defines the element.

The number of **protons + neutrons** (nucleons) in the nucleus.

**neutrons = A − Z** (mass number − atomic number).

**electrons = protons = Z** — the + and − charges balance.

Adjust electrons by the charge: **electrons = Z − charge** (lose e⁻ for +, gain e⁻ for −).

Top = **mass number A**, bottom = **atomic number Z**, X = element symbol.

Only the **electron** count; the protons and neutrons stay the same.

Atoms of the **same element** with the **same number of protons** but **different numbers of neutrons**.

The number of **protons** in an atom — it defines which element it is.

The total number of **protons + neutrons** in an atom.

The same atomic number Z, but a **different mass number A** (because they have different numbers of neutrons).

They have the **same number of electrons** and the **same electron arrangement**, and chemistry is controlled by the electrons.

**Physical** properties that depend on mass — e.g. **density** and rate of **diffusion** — because of the different number of neutrons.

Neutrons = **A − Z** (mass number minus atomic number).

37 − 17 = **20 neutrons**.

An isotope with an **unstable nucleus** that decays and gives out radiation; chemically it behaves like the stable isotope.

**Carbon-14** for dating, **cobalt-60** for radiotherapy/sterilising, or **iodine-131** for treating the thyroid.

No — neutrons have **no charge** and do not affect the electrons, so **bonding and reactions are unchanged**.

The **weighted average** mass of an element's isotopes, relative to one-twelfth of a carbon-12 atom. It has **no units**.

Because it averages **isotopes of different masses**, weighted by their **abundance** (e.g. Cl = 35.5).

It separates the atoms/ions of a sample by **mass**, producing a **mass spectrum**.

**m/z** (mass-to-charge ratio) on the x-axis; **relative abundance** on the y-axis.

For singly-charged ions, the **mass of that isotope**.

The **relative abundance** of that isotope — the taller the peak, the more common the isotope.

$A_{r} = \dfrac{\sum(\text{mass} \times \%\,\text{abundance})}{100}$ — weight each isotope mass by its abundance, sum, divide by 100.

**Three** peaks — one peak per isotope.

Divide the weighted sum by the **total of the abundances** instead of by 100.

It must lie **between** the lightest and heaviest isotope masses, closest to the **most abundant** one.

No — chlorine atoms are ³⁵Cl or ³⁷Cl; 35.5 is the **weighted average** (75% ³⁵Cl, 25% ³⁷Cl).

A tiny **packet of light energy**; its energy is given by E = hf (higher frequency → more energy).

A **fixed, allowed energy** an electron can have in an atom; energy levels are **discrete (quantised)**.

Continuous = an **unbroken rainbow** (all wavelengths). Line = a few **discrete bright lines** on black, from an excited element.

An excited electron **falls** from a higher to a lower energy level, emitting a photon of fixed energy (one line per allowed jump).

That the electron's energy levels are **discrete (quantised)** — fixed lines mean only fixed energy gaps are allowed.

The spectral lines get **closer together** toward **high frequency/energy**, because the energy levels bunch up at higher n.

The **biggest energy gap** — an electron falling **to n = 1** (the ground state).

$c = \lambda f$ — speed of light = wavelength × frequency, so **high f means short λ**.

$E = hf$ — photon energy = Planck's constant × frequency (higher f → higher E).

**Radio < red < violet** in frequency, so radio is lowest energy and violet is highest.

The lines merge; the electron gains just enough energy to **leave the atom** — this gives the **ionisation energy**.

The major 'shell' of an atom (n = 1, 2, 3, …); higher n means **higher energy** and **further** from the nucleus.

A subdivision of a main level, labelled **s, p, d, f**, differing slightly in energy (s < p < d < f).

A region around the nucleus that can hold up to **2 electrons**.

A **sphere** centred on the nucleus.

A **dumbbell** — two lobes pointing in opposite directions through the nucleus.

s = **1**, p = **3**, d = **5**, f = **7** orbitals.

s = **2**, p = **6**, d = **10**, f = **14** (2 electrons per orbital).

**2n²** — so n = 1 → 2, n = 2 → 8, n = 3 → 18, n = 4 → 32.

**4s** fills first — it is slightly lower in energy than 3d.

Electrons occupy orbitals of a sublevel **singly** (parallel spins) before any pairing up.

**s < p < d < f** (s is lowest, f is highest).

**3s, 3p and 3d** (max 18 electrons).

Electrons fill the **lowest-energy** sub-shell available first (build up: 1s, 2s, 2p, 3s, …).

Each orbital holds **at most 2 electrons**, and they must have **opposite spins**.

Within a sub-shell, electrons occupy orbitals **singly with parallel spins** before any pairing occurs.

1s, 2s, 2p, 3s, 3p, **4s, 3d**, 4p — note **4s fills before 3d**.

**s = 2**, **p = 6**, **d = 10** (each orbital holds 2).

1s² 2s² 2p⁶ 3s² 3p⁴.

Replace the inner electrons with the **previous noble gas** in [ ], then list the outer electrons — e.g. Ca = [Ar] 4s².

Start from the atom and **remove electrons from the highest main shell (largest n) first** — for transition metals, **4s before 3d**.

**[Ar] 3d⁶** — the two **4s** electrons are removed first, not the 3d.

**Add** the gained electrons to the next available sub-shell — e.g. O²⁻ = 1s² 2s² 2p⁶.

A **half-full** 3d⁵ sub-shell is extra stable, so one 4s electron promotes to 3d.

A **full** 3d¹⁰ sub-shell is extra stable, so one 4s electron promotes to 3d.

The amount of substance containing **6.02 × 10²³** particles (Avogadro's constant, N_{A}).

N_{A} = **6.02 × 10²³ mol⁻¹** — the number of particles in one mole.

The mass of **one mole** of a substance, in **g mol⁻¹**; numerically equal to the relative atomic/formula mass.

$n = \dfrac{m}{M}$ — amount (mol) = mass (g) ÷ molar mass (g mol⁻¹).

$N = n\,N_{A}$ — number of particles = amount (mol) × Avogadro's constant.

Rearrange to $m = nM$ — multiply the amount (mol) by the molar mass.

$M = \dfrac{m}{n}$ — divide the mass by the amount in moles.

2 mol of O atoms = **1.20 × 10²⁴** atoms (each CO_{2} has 2 oxygens).

**g mol⁻¹** (grams per mole).

Forgetting to scale by the **number of that atom or ion in the formula** (e.g. 2 Cl⁻ per MgCl_{2}).

The **simplest whole-number ratio** of the atoms of each element in a compound.

The **actual number** of atoms of each element in one molecule of the compound.

The molecular formula is a **whole-number multiple** of the empirical formula (molecular = empirical × x).

**CH_{2}O** — divide every subscript by 6 to get the simplest ratio.

Treat % as g per 100 g → divide each by A_{r} (n = m/M) → divide by the **smallest** → round / scale to whole numbers.

**n(C) = n(CO_{2})** — every carbon atom ends up in one CO_{2}.

**n(H) = 2 × n(H_{2}O)** — each water molecule contains two H atoms.

By **difference**: subtract the masses of C and H from the sample mass, then divide the leftover by 16.00.

$x = \dfrac{M_{r}}{\text{empirical formula mass}}$, then multiply every subscript by x.

Multiply the **whole ratio** by 2 (for .5) or 3 (for .33) to clear it into whole numbers.

Atoms combine in whole-**number** ratios, which only show up once masses are turned into **moles** (÷ A_{r}).

An **empirical** formula — ionic compounds have no separate molecules, so the formula is the simplest ratio.

How much **solute** is dissolved in a given volume of **solution** — usually in **mol dm⁻³**.

$n = CV$ — amount (mol) = concentration (mol dm⁻³) × volume (**dm³**).

Rearrange to $C = \dfrac{n}{V}$ — divide the amount in moles by the volume in dm³.

**Divide by 1000** (1 dm³ = 1000 cm³). E.g. 250 cm³ = 0.250 dm³.

**Multiply by the molar mass M**: g dm⁻³ = mol dm⁻³ × M.

$C_{1}V_{1} = C_{2}V_{2}$ — the amount of solute is unchanged when you add solvent.

Diluting only adds solvent, so the **moles of solute (n = CV) stay constant**.

**1 mg dm⁻³** (1 part per million) — used for very dilute solutions.

A solution of **precisely known concentration**, made up in a **volumetric flask**.

Forgetting to convert the **volume from cm³ to dm³** (÷ 1000) before using n = CV.

The **total** final volume. Water added = V_{2} − V_{1}.

**Dissolve** the weighed solid → **transfer** to a volumetric flask (rinse beaker in) → **make up** to the mark → **invert** to mix.

At **constant temperature** (and amount), the pressure of a gas is **inversely proportional** to its volume: $P_{1}V_{1} = P_{2}V_{2}$.

Pressure is **directly proportional** to the **kelvin** temperature: $\dfrac{P_{1}}{T_{1}} = \dfrac{P_{2}}{T_{2}}$.

$\dfrac{P_{1}V_{1}}{T_{1}} = \dfrac{P_{2}V_{2}}{T_{2}}$ — with T in **kelvin**. It is given in the data booklet.

**T/K = θ/°C + 273** — always do this before a gas-law calculation.

The particles have **no volume** of their own and there are **no forces** between them.

At **high temperature** and **low pressure** — particles are far apart and fast-moving.

At **low temperature** and **high pressure** — particle volume and intermolecular forces become significant.

The pressure **halves** (Boyle's law: P ∝ 1/V).

Only the **kelvin** scale starts at true zero (0 K), so only it gives the correct proportionality; °C would give wrong ratios.

Because P ∝ kelvin T — at 0 K the particles would stop and the pressure would be **zero**.

The **temperature** and the **amount** of gas; only P and V change.

Use the combined gas law: $P_{2} = P_{1}\times\dfrac{V_{1}}{V_{2}}\times\dfrac{T_{2}}{T_{1}}$ (T in kelvin).

$PV = nRT$ — links pressure, volume, amount and temperature of an ideal gas.

**Standard temperature and pressure**: 273 K (0 °C) and 100 kPa.

V_{m} = **22.7 dm³ mol⁻¹** — the volume of one mole of any ideal gas at STP.

$n = \dfrac{V}{V_{m}}$ — divide the volume (in dm³) by 22.7.

$V = n\,V_{m}$ — multiply the amount (mol) by 22.7 dm³ mol⁻¹.

**Pa** (pressure), **m³** (volume) and **K** (temperature), because R = 8.31 J K⁻¹ mol⁻¹ is in SI units.

R = **8.31 J K⁻¹ mol⁻¹** (given in the data booklet).

**Multiply by 1000** — e.g. 101 kPa = 1.01 × 10⁵ Pa.

**Divide by 1000** — e.g. 24.0 dm³ = 0.0240 m³.

**Add 273** — e.g. 25 °C = 298 K.

**At STP** use V_{m} = 22.7; at **any other conditions** use PV = nRT with SI units.

Find n from PV = nRT, then $M = \dfrac{m}{n}$ using the sample mass.

An atom (or group of atoms) with an overall **charge** because it has **lost or gained electrons**.

A **positive** ion, formed when an atom **loses** electrons (more protons than electrons).

A **negative** ion, formed when an atom **gains** electrons (more electrons than protons).

**Cations** — metals **lose** their outer electrons to form **positive** ions.

**Anions** — non-metals **gain** electrons to form **negative** ions.

To reach a **full outer shell** — the stable **noble-gas configuration** of a Group 18 atom.

**1+, 2+, 3+** — these metals lose 1, 2 or 3 outer electrons.

**3−, 2−, 1−** — these non-metals gain 3, 2 or 1 electrons.

The **electrostatic attraction between oppositely charged ions** (a cation and an anion).

Group 17, so it **gains 1** electron → a **1−** ion (reaching 2, 8, 8).

**Na⁺** (sodium loses 1 e⁻) and **Cl⁻** (chlorine gains 1 e⁻).

Ionic = **attraction between charged ions** (electrons transferred); covalent = a **shared pair** of electrons.

A **positively** charged ion (a metal, or NH_{4}⁺).

A **negatively** charged ion (a non-metal, or a polyatomic ion).

A charged group of **bonded atoms** that acts as a single unit (e.g. SO_{4}²⁻, NO_{3}⁻).

The ratio of ions is chosen so the **total positive and negative charges cancel** to zero.

Write each ion with its charge, **cross over** the charge sizes as subscripts, then **simplify** to the smallest whole-number ratio.

**MgCl_{2}** — Mg²⁺ needs two Cl⁻ to balance.

**Al_{2}O_{3}** — crossover of Al³⁺ and O²⁻ (6+ balances 6−).

When a **polyatomic ion appears two or more times**, e.g. Ca(NO_{3})_{2}, (NH_{4})_{2}SO_{4}.

**SO_{4}²⁻** — a 2− polyatomic ion.

**NH_{4}⁺** — a 1+ polyatomic cation.

Cation name first, then the non-metal anion ending in **-ide** (e.g. magnesium ox**ide**).

**Ca_{3}N_{2}** — Ca²⁺ and N³⁻ crossover (6+ balances 6−).

A regular, repeating **3-D array** of oppositely charged ions, with each ion surrounded by ions of the opposite charge.

**Strong electrostatic forces of attraction** between the oppositely charged ions (this is the ionic bond).

Many **strong electrostatic attractions** between the ions must be broken, which needs a **large amount of energy**.

**Higher ionic charge** and **smaller ionic radius** — both increase the electrostatic attraction.

When **molten** or **dissolved in water (aqueous)** — the ions are then **free to move**. Not as a solid.

The ions are held in **fixed positions** in the lattice, so no charged particles are free to move.

A force makes layers **shift**, bringing **like charges** next to each other; they **repel** and split the crystal.

Water is **polar**: its δ⁻ oxygen attracts cations and δ⁺ hydrogens attract anions, pulling ions out of the lattice (hydration).

Solid = ions **fixed**, does **not** conduct. Molten = lattice broken, ions **free to move**, **conducts**.

High melting point + does **not** conduct as a solid + **conducts when molten/aqueous** = ionic.

Mg^{2+} and O^{2−} carry **higher charges** than Na^{+} and Cl^{−}, so the electrostatic attraction is much stronger.

A **shared pair of electrons** between two (usually non-metal) atoms.

A **non-bonding** pair of electrons that stays on one atom (drawn as two dots).

A **bonding pair** (one shared pair of electrons).

Atoms tend to gain a full outer shell of **8 electrons** by sharing (or transferring) electrons.

Number of **shared pairs**: 1, 2, 3 — bond order 1, 2, 3. Higher order → shorter, stronger.

O=C=O — **two double bonds**, two lone pairs on each oxygen, none on carbon.

N≡N — a **triple bond** with **one lone pair on each** nitrogen.

**BF_{3}** (boron has 6 electrons) and **BeCl_{2}** (beryllium has 4) — electron-deficient.

Count valence electrons → least electronegative atom central → single bonds → complete outer octets → multiple bonds if the centre is short.

**One** (three bonding pairs to H, one lone pair).

**V**alence **S**hell **E**lectron **P**air **R**epulsion.

Any group of electrons around the central atom — a single/double/triple **bond (each = 1 domain)** or a **lone pair**.

**Linear**, 180° (e.g. CO_{2}, HCN).

**Trigonal planar**, 120° (e.g. BF_{3}).

**Tetrahedral**, 109.5° (e.g. CH_{4}).

**Trigonal pyramidal**, ~107° (e.g. NH_{3}).

**Bent**, ~104.5° (e.g. H_{2}O).

Lone pairs repel **more** than bonding pairs, so they **reduce** the bond angle.

Each double bond is **one** electron domain; 2 domains, 0 lone pairs → linear, 180°.

CH_{4} (109.5°) > NH_{3} (107°) > H_{2}O (104.5°) — angle falls as lone pairs increase.

A measure of how strongly an atom **attracts the shared (bonding) electrons** in a covalent bond.

A **difference in electronegativity** between the two atoms — the electrons are pulled towards the more electronegative atom.

The **more electronegative** atom (it gets a bigger share of the electrons); the less electronegative atom is **δ+**.

Δχ = 0 → **pure covalent**; small Δχ → **polar covalent** (δ+/δ−); large Δχ → **ionic**.

The small separation of charge (δ+ → δ−) along a polar bond; drawn as an **arrow** pointing to the δ− atom.

When the molecule is **symmetrical**, so the bond dipoles **cancel** (e.g. CO_{2}, CCl_{4}, BF_{3}).

It is **linear** — the two equal C=O dipoles point in opposite directions and **cancel**.

It is **bent** (lone pairs on O), so the two O–H dipoles **do not cancel** and give a net dipole.

Yes — it is **trigonal pyramidal** (a lone pair on N), so the N–H dipoles do not cancel; NH_{3} is polar.

(1) the bonds are **polar** (electronegativity difference) and (2) the **shape** means the dipoles **do not cancel**.

No — both atoms are identical, so Δχ = 0; the bond is **non-polar** and there is no dipole.

A continuous lattice of atoms joined by **covalent bonds** in every direction — there are **no separate small molecules**.

Melting requires breaking **many strong covalent bonds**, which needs a large amount of energy.

Different structural forms of the **same element** — e.g. diamond and graphite are both pure carbon.

To **four** other carbons in a rigid **3-D tetrahedral** network.

Its **rigid 3-D framework** of strong covalent bonds cannot be pushed out of shape.

All **four** outer electrons of each carbon are used in bonds, so there are **no delocalised electrons** to carry charge.

To **three** others in flat **layers**; the **fourth** electron is **delocalised**.

The **delocalised electrons** between the layers are free to move and carry charge.

**Weak forces** between the layers let the **layers slide** over each other (the covalent bonds within a layer stay strong).

**Diamond**, **graphite** (carbon allotropes), **silicon (Si)** and **silicon dioxide (SiO_{2})**.

Giant covalent → break **strong covalent bonds**; molecular → only overcome **weak intermolecular forces**.

Diamond uses all 4 electrons in bonds (**no** delocalised e⁻ → no conduction); graphite has **1 delocalised** e⁻ per carbon (conducts).

A force of attraction **between** separate molecules — much weaker than the covalent bonds **inside** a molecule.

The strength of its **intermolecular forces** — stronger IMFs need more energy, so a **higher** boiling point.

**London (dispersion) < dipole–dipole < hydrogen bonding.**

Forces from **temporary, instantaneous dipoles**; present between **all** molecules and the **only** force in non-polar ones.

**More electrons** (a larger, more polarisable molecule) — so they increase **down a group** and with molecular size.

When it is **polar** — it has a **permanent dipole** (δ+ and δ− ends) from an electronegativity difference.

The **strongest** IMF: a very δ+ H bonded to **N, O or F** is attracted to a lone pair on the N, O or F of a neighbour.

Hydrogen bonded directly to **N, O or F** ('H bonds to NOF').

NH_{3} has **hydrogen bonding** (H on N); PH_{3} has only weaker dipole–dipole/London forces.

Larger molecules have **more electrons → stronger London forces → higher boiling point**.

**No** — boiling only **separates the molecules** by overcoming intermolecular forces; the covalent bonds stay intact.

N, O and F are very electronegative, so the H is very δ+ and the attraction to a lone pair is especially strong.

The electrostatic attraction between a lattice of **positive metal cations** and a **sea of delocalised electrons**.

Electrons that are **not fixed to one atom** — free to move throughout the whole lattice.

The **delocalised electrons are free to move**, so they carry charge through the metal (solid or molten).

The bonding is **non-directional**, so layers of cations can **slide** over each other while the electron sea keeps them bonded.

A lot of energy is needed to overcome the **strong attraction** between the cations and the delocalised electron sea.

Sliding an ionic lattice brings **like charges** together → they repel and crack; a metal's non-directional sea has no like-charge layer, so it bends.

**Higher cation charge** and **smaller cation radius** (and more delocalised electrons).

Mg²⁺ has a **higher charge**, donates **two** electrons (denser sea) and is **smaller** than Na⁺.

It **weakens** — the cation radius **increases**, so the electron sea sits further from the nucleus.

The **delocalised electrons** (the cations stay fixed) — unlike a molten ionic compound, where the **ions** move.

The mobile **delocalised electrons** transfer kinetic energy quickly through the lattice.

That ionic, covalent and metallic bonding are the three **extremes** of one **continuum** — real compounds sit in between.

**Metallic** (bottom-left), **covalent** (bottom-right) and **ionic** (top).

How strongly an atom **attracts a shared pair of electrons**; values are in the data booklet.

Average the two electronegativities: $\chi_{avg} = \dfrac{\chi_A + \chi_B}{2}$ — it sets the **horizontal** position.

Take the difference: $\Delta\chi = |\chi_A - \chi_B|$ — it sets the **vertical** (ionic) position.

Electrons are essentially **transferred** → the bonding is **ionic** (high up the triangle).

Electrons are **shared** between similar non-metals → **covalent** (bottom-right corner).

A sea of delocalised electrons among metal atoms → **metallic** (bottom-left corner).

NaCl → **ionic** (top, large Δχ); Cl_{2} → **covalent** (bottom-right); Na → **metallic** (bottom-left).

It uses the **actual χ values**, so it correctly classifies polar-covalent metal compounds like BeCl_{2}.

Ionic = electrons **transferred** (large Δχ); covalent = electrons **shared** (small Δχ).

A **mixture** of a metal with one or more other elements (it is **not** a compound — no fixed ratio).

Its **different-sized atoms disrupt the regular layers**, so the layers **cannot slide** over each other as easily.

Yes — they keep **metallic bonding** (a sea of delocalised electrons); they are just **harder** than the pure metal.

**Brass** = copper + zinc; **steel** = iron + carbon (also bronze = copper + tin).

A **small molecule** that joins to many others to form a **polymer** (a giant molecule).

A long-chain molecule made by joining many **alkene monomers** (with **C=C**), with **no atoms lost**.

The **double bond opens up** — one bond becomes a single bond, the other joins to the next monomer.

The part of the polymer chain that **repeats**; get it by **opening the C=C** and drawing a bond out of each end.

**Monomer** has the **C=C double bond**; **repeat unit** has a **single** C–C bond with a bond out of each end.

Take **one repeating unit** and **put the C=C double bond back** between the two carbons.

**Ethene, CH_{2}=CH_{2}** — the repeat unit is –CH_{2}–CH_{2}–.

It is **chemically unreactive (inert)** and waterproof, so it resists corrosion — useful for packaging and containers.

By **increasing atomic number** (number of protons), not by relative atomic mass.

A horizontal **row**; the period number equals the highest occupied **main energy level (n)**.

A vertical **column**; elements in a group have the **same number of outer (valence) electrons**.

The **sublevel** that the outermost electrons are filling (s, p, d or f).

Groups **1 and 2** (plus H and He) — outer electrons fill the **s** sublevel.

Groups **13–18** — outer electrons fill the **p** sublevel.

The **centre** of the table (groups 3–12) — the **transition metals**, filling the d sublevel.

The **two detached rows** at the bottom — the **lanthanides and actinides**, filling the f sublevel.

Name the **sublevel the outermost electron enters** (e.g. …3p⁵ → p-block; …3d⁶ → d-block).

**Period** number = n of the outer shell; **group** number = number of outer electrons (group 17 → 7).

The **s-block** — its next electron would enter the **8s** sublevel (group 1, period 8).

**Nuclear charge** (proton pull) and **shielding/distance** (inner shells + extra shells).

The energy needed to remove one mole of electrons from one mole of **gaseous** atoms: X(g) → X⁺(g) + e⁻.

**Half** the distance between the nuclei of two bonded atoms — a measure of atom size.

How strongly an atom attracts a **bonding pair** of electrons (Pauling scale).

The energy change when one mole of gaseous atoms **gains** an electron: X(g) + e⁻ → X⁻(g).

**Decreases** — greater nuclear charge with similar shielding pulls the outer shell in.

**Increases** — each element has an extra electron shell.

**Increases** across a period (stronger pull); **decreases** down a group (further out, more shielded).

**Increases** across a period, **decreases** down a group (fluorine is the most electronegative).

A cation is **smaller** than its atom (it often loses a whole shell).

An anion is **larger** than its atom (extra electron–electron repulsion spreads the shell out).

Compare **nuclear charge**, compare **shielding/distance**, then state the **net effect** (held more/less tightly).

The same number of **outer (valence) electrons**, so they react in similar ways.

It **increases** — the outer electron is further out and more shielded, so it is **lost more easily**.

It **decreases** — the atom is bigger, so an incoming electron is **harder to gain**.

K is lower in group 1: **bigger atom + more shielding** → outer electron lost more easily.

F is smaller with less shielding, so it **gains** an electron more easily.

Able to act as **both an acid and a base** — reacts with acids **and** alkalis (e.g. Al_{2}O_{3}).

It **decreases** — elements change from **metallic** (Na) to **non-metallic** (Cl, Ar).

**Basic → amphoteric → acidic** left to right (Na_{2}O/MgO basic, Al_{2}O_{3} amphoteric, SO_{3} acidic).

**Basic** (e.g. Na_{2}O, MgO). Non-metal oxides are **acidic** (e.g. SO_{3}, P_{4}O_{10}).

**Caesium + fluorine** — lowest (most reactive) metal + top (most reactive) halogen.

Li < Na < K < Rb < Cs (increases down).

F > Cl > Br > I (decreases down).

The chemistry of **carbon compounds**.

A family of organic compounds with the **same general formula** and **functional group**, each member differing by **CH_{2}**.

The reactive atom or group of atoms that gives a series its **characteristic chemistry** (e.g. C=C, –OH).

Same **general formula**; differ by **CH_{2}**; **gradual change** in physical properties; **similar chemical** properties.

**C_{n}H_{2n+2}** (saturated — only single C–C bonds).

**C_{n}H_{2n}** (unsaturated — one C=C double bond).

**C_{n}H_{2n+1}OH** (functional group –OH).

**Saturated** = only single C–C bonds (max H); **unsaturated** = at least one **C=C** double bond (fewer H).

Longer chains are bigger/heavier, so **intermolecular forces** are stronger → **higher boiling point**.

**Two** fewer hydrogens in the alkene — the C=C double bond replaces two C–H bonds.

**Ethene, C_{2}H_{4}** (alkenes start at n = 2).

The **simplest whole-number ratio** of the atoms in a compound (e.g. CH_{2}O for glucose).

The **actual number** of each type of atom in one molecule (e.g. C_{6}H_{12}O_{6} for glucose).

A diagram showing **every atom and every bond** in the molecule.

Atoms written **grouped in a line** with the bonds implied (e.g. CH_{3}CH_{2}OH).

Only the **carbon skeleton** drawn as lines; carbons are corners/ends and **H on carbon is implied**; functional groups are shown.

A **carbon** atom (each with enough H to make four bonds, not drawn).

Divide **every** subscript by their **highest common factor** (e.g. C_{6}H_{12}O_{6} ÷ 6 = CH_{2}O).

Molecules with the **same molecular formula** but a **different arrangement** of atoms (different connectivity).

Chain **branching**, **position** of a group, or different **functional group / class**.

Keep the **same molecular formula** but **change the connectivity** — never just rotate or flip the original.

No — CH_{2}O is an **empirical** formula; C_{2}H_{4}O_{2} (ethanoic acid) is one **molecular** formula with that ratio.

**No** — it is the same molecule; an isomer must have a genuinely different structure.

The **reactive atom or group of atoms** that defines an organic molecule's class and chemistry.

A family of compounds with the **same functional group** and the same general formula.

Saturated = only single C–C bonds (alkane); unsaturated = has a **C=C** (or C≡C) bond (alkene).

**–OH** (hydroxyl); name ends in **-ol** (e.g. propan-1-ol).

**–COOH** (carboxyl); name ends in **-oic acid** (e.g. propanoic acid).

Both have C=O. Aldehyde = carbonyl at the **end** (-al); ketone = carbonyl in the **middle** (-one).

An alkane with a **halogen** (–F, –Cl, –Br, –I) in place of an H; named with a prefix (chloro-, bromo-…).

**-ene**, because it contains a **C=C** double bond (e.g. propene).

**Stem** (number of carbons) + **suffix** (functional group) + **locant** (where the group is).

1 = meth-, 2 = eth-, 3 = prop-, 4 = but-.

The **number** in a name showing the position of the functional group on the chain (e.g. the 2 in but-2-ene).

Give the functional group the **lowest possible locant**.

Its m/z value is the **relative molecular mass (Mr)** — M⁺ is the peak at the **highest** m/z.

The **mass lost** (M⁺ − fragment) shows which **group broke off** (e.g. loss of 15 = CH_{3}).

Loss of a **CH_{3}** (methyl) group.

Loss of an **OH** group.

The **functional group**, from a characteristic absorption **wavenumber** (cm⁻¹) given in the data booklet.

An **O–H** group (an alcohol). A carboxylic acid O–H is even broader, ~2500–3000.

A **C=O** (carbonyl) — aldehyde, ketone, acid or ester.

The **number of peaks = number of different hydrogen environments** in the molecule.

**Three** — the CH_{3}, CH_{2} and OH hydrogens are three different environments.

By **symmetry** the two CH_{3} groups are equivalent, so all six H are in one environment.

**MS** (Mr + fragments), **IR** (functional group), **¹H NMR** (number of H environments) — used together.

In the **data booklet** — you read the wavenumbers off, you don't memorise them.

The **heat energy** released or absorbed by a reaction at **constant pressure** (units: kJ mol⁻¹).

A reaction that **releases** energy to the surroundings, so they get **hotter**; **ΔH is negative**.

A reaction that **absorbs** energy from the surroundings, so they get **colder**; **ΔH is positive**.

Exothermic → **ΔH < 0** (negative); endothermic → **ΔH > 0** (positive).

**Endothermic** — energy must be **put in** to break a bond.

**Exothermic** — energy is **released** when a new bond forms.

When **making** the new bonds releases **more** energy than **breaking** the old bonds absorbed (net energy out).

The **minimum** energy reactants need to react — the reactant level up to the **peak** of the profile.

It is the energy gap between the **reactant** and **product** levels (down for exothermic, up for endothermic).

**Exothermic** products are **lower** in energy and so **more stable** than the reactants.

**Endothermic** — energy is absorbed from the surroundings, so **ΔH is positive**.

**Combustion** and **neutralisation** (also respiration) — they release energy.

Measuring the **temperature change** of a known mass of water (or solution) to find the heat transferred by a reaction.

The energy needed to raise **1 g** of a substance by **1 K** (1 °C). For water, **c = 4.18 J g⁻¹ K⁻¹**.

$Q = mc\Delta T$ — heat (J) = mass (g) × specific heat capacity × temperature change.

ΔT = **T_{final} − T_{initial}**. A change of 1 °C equals a change of 1 K, so the number is the same.

$\Delta H = -\dfrac{Q}{n}$ — divide Q (in kJ) by the amount that reacted, and add the sign.

**Exothermic** — heat released to the water — so **ΔH is negative**.

**Endothermic** — heat absorbed from the water — so **ΔH is positive**.

The mass of **water** (the substance heated), **not** the mass of fuel or reactant.

Q from $mc\Delta T$ is in **joules**; enthalpy changes are quoted in **kJ mol⁻¹**, so divide by 1000.

**Heat loss** to the surroundings and apparatus — so the measured ΔH is **less exothermic** than the true value.

All the heat goes to the **water**, and the **specific heat capacity** (and density) of the solution equals that of water.

ΔT → **Q = mcΔT** → ÷1000 for kJ → **÷ n** for per mole → add the **sign**.

The energy needed to **break one mole** of a particular bond in the **gaseous** state (always a positive value).

**Endothermic** — breaking a bond always **costs** (absorbs) energy.

**Exothermic** — forming a bond always **releases** energy.

$\Delta H = \Sigma(\text{bonds broken}) - \Sigma(\text{bonds made})$.

The reaction is **exothermic** — more energy was released making bonds than was used breaking them.

The reaction is **endothermic** — breaking bonds cost more energy than was released making them.

A bond (e.g. C–H) exists in many molecules with slightly different strengths, so the booklet gives an **average**; ΔH is therefore an **estimate**.

Only when **all species are gaseous**, because bond enthalpy is defined for the gaseous state.

Only the bonds that **break or form** — unchanged bonds (spectator bonds) cancel out.

**Higher** — a larger bond enthalpy means a stronger bond that needs more energy to break.

Because the bond enthalpies are **averages**, so the calculated ΔH is only an **estimate**.

The total enthalpy change of a reaction is the **same** whatever route is taken, because ΔH depends only on the initial and final states.

A property that depends only on the **current state** of the system, not on the path taken to reach it (enthalpy is one).

Because enthalpy is a **state function**, so ΔH is **path-independent** — you can add up the steps of an alternative route.

Its **sign is reversed** (the magnitude stays the same).

ΔH is **doubled** — multiply ΔH by the same factor as the equation.

$\Delta H^{\ominus} = \Sigma\,\Delta H_{f}^{\ominus}(\text{products}) - \Sigma\,\Delta H_{f}^{\ominus}(\text{reactants})$.

**Zero** by definition (e.g. O_{2}(g), C(s) graphite).

**With** an arrow → **add** its ΔH; **against** it (reverse) → **subtract** its ΔH (flip the sign).

Forgetting to **multiply** a step by the number of moles in the target equation.

To find a ΔH that **cannot be measured directly** (e.g. the reaction is too slow or has side reactions).

The ΔHf equation **is** a Hess cycle — going down to the elements (reverse ΔHf of reactants) and up to the products (ΔHf of products).

The enthalpy change when **1 mol** of a compound forms from its **elements in their standard states** (100 kPa, stated T).

The enthalpy change when **1 mol** of a substance is **completely burned in oxygen** under standard conditions; always **negative**.

**Zero** — e.g. O_{2}(g), N_{2}(g), C(graphite); there is nothing to form.

$\Delta H^{\ominus} = \sum \Delta H_{f}^{\ominus}(\text{products}) - \sum \Delta H_{f}^{\ominus}(\text{reactants})$.

$\Delta H^{\ominus} = \sum \Delta H_{c}^{\ominus}(\text{reactants}) - \sum \Delta H_{c}^{\ominus}(\text{products})$.

Both reactants and products burn down to the **same products** (CO_{2} + H_{2}O), so the Hess cycle runs the other way → **reactants − products**.

A pressure of **100 kPa** and a stated temperature (usually **298 K**), with all substances in their standard states.

Enthalpy is a **state function** — ΔH depends only on the start and end states, so a 'paper' Hess route gives the same answer as experiment.

Forgetting to multiply each value by its **stoichiometric coefficient** (e.g. 2 H_{2}O) or forgetting an **element is zero**.

**Negative** (exothermic) — a quick check that you used the correct rule.

**kJ mol⁻¹** (kilojoules per mole).

A substance that releases useful **energy** when it is **burned** (combusted) in oxygen.

The reaction of a fuel with **oxygen** that releases energy as heat; it is always **exothermic** (ΔH < 0).

**Carbon dioxide (CO_{2}) and water (H_{2}O)** only — with maximum energy released.

**Carbon monoxide (CO) and/or carbon (soot)** plus water — less energy is released.

When there is a **limited supply of oxygen**, so the carbon is not fully oxidised.

CO is a **toxic** gas that binds to haemoglobin, stopping the blood from carrying oxygen.

Energy released **per unit mass** of fuel (e.g. **kJ g⁻¹**) — matters when weight is important.

Energy released **per unit volume** of fuel (e.g. **kJ cm⁻³**) — matters when storage space is important.

Fossil fuels (coal, oil, gas) are **non-renewable**; biofuels (e.g. ethanol, biodiesel) are **renewable**.

The crop **absorbs CO_{2}** as it grows, roughly balancing the CO_{2} released when the fuel is burned.

They release carbon that was **locked away for millions of years**, adding **new** CO_{2} to the atmosphere.

**Ethanol** (from fermented sugar cane/corn) or **biodiesel** (from plant oils).

An equation with the **same number of each kind of atom** on both sides — atoms are conserved.

The study of the **whole-number ratios** in which substances react and are formed, read from a balanced equation.

The ratio of the **coefficients** in a balanced equation — how many moles of one substance react with or form another.

Only the **coefficients** (the big numbers in front) — **never** a subscript inside a formula.

Changing a subscript changes the **substance** itself (e.g. H_{2}O → H_{2}O_{2}), so it no longer describes the same reaction.

**(s)** solid, **(l)** pure liquid, **(g)** gas, **(aq)** aqueous (dissolved in water).

**Aqueous** — the substance is **dissolved in water** (different from a pure liquid, (l)).

The ratio N_{2} : H_{2} : NH_{3} is **1 : 3 : 2** — 1 mol N_{2} reacts with 3 mol H_{2} to make 2 mol NH_{3}.

Balance **C first, then H, then O last** (oxygen appears in more than one product), then reduce to smallest whole numbers.

**CH_{4} + 2 O_{2} → CO_{2} + 2 H_{2}O** — smallest whole-number coefficients.

The C : CO_{2} ratio is 1 : 1, so **0.5 mol** of CO_{2}.

Changing a **formula** (subscript) instead of a coefficient, or forgetting the **state symbols** when asked.

The reactant that **runs out first** — it controls (limits) the amount of product that can form.

The reactant **left over** once the limiting reactant has been used up.

The **maximum** amount (or mass) of product, calculated from the **limiting** reactant.

Convert each reactant mass to **moles**, divide each by its **coefficient**, and the **smallest** result is limiting.

It compares the reactants fairly against the **mole ratio** in the balanced equation, so you can see which runs out first.

Always the **limiting** reactant — never the one in excess.

$n = \dfrac{m}{M}$ — convert every mass to moles before using the mole ratio.

Balanced equation → mass to **moles** (n = m/M) → scale by the **mole ratio** → moles back to **mass** (m = nM).

From the **coefficients** of the balanced equation (e.g. N_{2} + 3H_{2} → 2NH_{3} is 1 : 3 : 2).

Working out the product from the reactant in **excess**, or forgetting the **mole ratio** when coefficients are not 1 : 1.

**A** (0.3 mol runs out first); B is in excess by 0.2 mol.

$\%\text{ yield} = \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100$ — how much product you actually obtained versus the maximum predicted by the equation.

The amount of product predicted from the balanced equation if the **limiting reactant** reacted completely.

The amount of product you really obtain — always **less** than theoretical, due to side reactions, reversible reactions and losses.

Side reactions, reversible reactions not going to completion, and losses during separation/purification.

$\%\text{ AE} = \dfrac{M(\text{desired product})}{M(\text{all reactants})} \times 100$ — the fraction of reactant atoms ending up in the wanted product.

Yield = **how much product you made**; atom economy = **how little reactant mass you wasted** as by-products. They are independent.

**Addition** reactions — all reactants combine into a single product, so there are no by-products.

Sum the molar masses of **all** reactants, each multiplied by its **coefficient** in the balanced equation.

Fewer atoms wasted as by-products → less raw material used and less waste to treat → more **sustainable and economical**.

No — actual yield cannot beat the theoretical maximum. A value over 100% signals an error (e.g. impure/wet product).

$\text{actual} = \dfrac{\%\text{ yield}}{100} \times \text{theoretical}$.

Putting only **one** reactant (or forgetting coefficients) on the bottom — you must sum **every** reactant's molar mass.

At the **same temperature and pressure**, equal **volumes** of gases contain equal numbers of **moles** — so the volume ratio equals the coefficient ratio.

Because at fixed T and P volume is **proportional to amount**, so the balanced **coefficients** give the **volume ratio** — no moles needed.

**22.7 dm³ mol⁻¹** at STP (273 K, 100 kPa) — given in the data booklet.

$n = \dfrac{V}{V_{m}}$ with $V_{m} = 22.7$ dm³ mol⁻¹ (volume in dm³).

Multiply the amount by the molar volume: $V = n\,V_{m} = n \times 22.7$ dm³.

**1000 cm³** — divide a cm³ value by 1000 before using the molar volume 22.7 dm³ mol⁻¹.

**2 volumes** of NH_{3} (the volume ratio matches the 1 : 3 : 2 coefficients).

Subtract the volume that **reacted** (from the coefficient ratio) from the volume **supplied**.

**No** — only **gases** contribute; liquids and solids (like condensed water) add zero volume.

Forgetting to **subtract the gas that reacted** when asked for the volume remaining, or counting **liquid** products as gas.

**273 K and 100 kPa** (standard temperature and pressure).

A precise technique to find an **unknown concentration** by reacting it with a **standard solution** to the **end point** (an indicator colour change).

A solution of **precisely known concentration**, made up in a **volumetric flask**.

Delivers a **fixed, exact** volume of the solution being analysed (e.g. 25.0 cm³).

Delivers the **variable** volume of titrant (the **titre**), read to ±0.05 cm³.

$n = CV$ — amount (mol) = concentration (mol dm⁻³) × volume (**dm³**). Given in the data booklet.

**(1)** n = CV on the known reagent → mol. **(2)** Cross by the **mole ratio**. **(3)** C = n/V (or M = m/n) on the unknown.

The volume must be in **dm³** — divide a cm³ titre by **1000** first.

Titres that **agree** (typically within 0.10 cm³). Only the concordant titres are **averaged** — a rough trial is ignored.

Add a **known excess** of a reagent, let it react, then titrate the **leftover** excess. Amount reacted = **added − leftover**.

When the reaction is **slow** or the sample is an **insoluble solid** (e.g. a carbonate), making a direct titration impractical.

**2 : 1** — sulfuric acid is diprotic, so it needs **two** moles of NaOH per mole of acid.

Forgetting the **mole ratio** from the balanced equation, or leaving a volume in **cm³** instead of dm³.

The **change in concentration** of a reactant or product **per unit time**.

**mol dm⁻³ s⁻¹** — a concentration (mol dm⁻³) divided by a time (s).

It is the **gradient** (steepness) of the curve — the tangent at a point gives the instantaneous rate.

The **reactant concentration is highest** at t = 0, so effective collisions are most frequent and the curve is **steepest**.

**Average** = total change ÷ total time (slope of the **chord**); **instantaneous** = slope of the **tangent** at one moment.

Particles must **collide** to react, but only **effective** collisions (enough energy + correct orientation) lead to a reaction.

Energy **≥ the activation energy Eₐ**, AND the particles collide in the **correct orientation**.

The **minimum energy** that colliding particles must have for a reaction to occur.

Reactants are **used up**, so their concentration falls and effective collisions become **less frequent**; rate drops to zero when reactants run out.

Measure the **volume of gas** collected vs time, or the **mass lost** vs time.

Use a **colorimeter** to measure the **light absorbed** as it changes with time.

The rate at t = 0 — the **slope of the tangent drawn at the start** of a concentration–time graph (the steepest point).

Energy **≥ the activation energy (E_{a})** AND the **correct orientation**.

The **minimum** energy a colliding pair of particles must have for a reaction to occur.

**Concentration, pressure, surface area, temperature** and a **catalyst**.

They put more particles in the reaction space, so collisions are **more frequent** (the energy per collision is unchanged).

Particles move faster (collisions **more frequent**) AND the distribution shifts right so a **greater fraction** have energy ≥ E_{a} — the second effect is the main one.

A substance that speeds up a reaction by providing an **alternative pathway of lower E_{a}**, and is **not used up** itself.

**No** — the reactant and product energy levels are unchanged, so ΔH is the same.

How the **kinetic energies** of particles are **spread out**; only those to the right of E_{a} can react.

**Lower and shifted to the right** (broader/flatter), but with the **same area** underneath.

The **fraction of particles** with enough energy to react (energy ≥ E_{a}).

The curve is **unchanged**; the **E_{a} line moves left**, so a larger fraction lies to the right of it.

The reaction goes **faster**, AND the solid is **recovered unchanged** (same mass/nature) at the end.

A reaction that can go in **both directions** — reactants can form products and products can re-form reactants. Shown with the **⇌** symbol.

The state, in a **closed system**, where the **forward and reverse reactions occur at equal rates**, so the **concentrations of reactants and products stay constant**.

Because **both** the forward and reverse reactions are **still happening** — the reaction has **not** stopped; the opposite changes simply cancel out.

**No** — they are **constant** (unchanging), but generally **not equal** to one another.

The **rate of the forward reaction = the rate of the reverse reaction**.

So that **nothing is added or escapes** (no reactant/product/heat leaves); an open system could never settle to constant concentrations.

**Colour (absorbance)**, **pressure** (for gases) and **pH** — all remain constant because the concentrations are constant.

**How far** a reaction has gone — the **relative amounts** of reactants and products. To the **right** = mostly products; to the **left** = mostly reactants.

Measure a **macroscopic property** over time (e.g. colour intensity); when it **levels off to a constant value**, equilibrium has been reached.

That the reaction has **stopped** — in fact both reactions continue (it is **dynamic**); the concentrations are merely constant.

The **forward rate falls** and the **reverse rate rises** until they **meet (become equal)** — that point is equilibrium.

Both the reactant and product curves **level off** (become flat) and stay constant — at **different** values.

If a system at equilibrium is disturbed, the position shifts in the direction that **opposes** (partly cancels) the change.

…towards the **products** (right) — the system uses up the added reactant.

…towards the **products** (right) — the system replaces the lost product.

The position shifts to the side with **fewer moles of gas**, to reduce the pressure.

Changing the pressure causes **no shift** in the position.

The position shifts in the **endothermic** direction (it absorbs the added heat).

A change in **temperature** — concentration, pressure and a catalyst leave K_{c} unchanged.

**No shift** in position and **no change** in K_{c}; it just reaches equilibrium **sooner** (speeds up both directions equally).

K_{c} **decreases** (the position shifts towards the reactants).

**Count the moles of gas** on each side; the position shifts towards the side with **fewer** gas moles when pressure rises.

Write **heat** as a species (exo: products + heat; endo: reactants + heat), then treat adding heat like adding that species.

**Endothermic** — adding heat favours the heat-absorbing direction.

The **fixed ratio** of product to reactant concentrations at equilibrium, at a given temperature — it shows **how far** a reaction goes.

**Products over reactants**, each concentration **raised to the power of its balancing coefficient**; use [ ] for equilibrium concentration in mol dm⁻³.

$K_{c} = \dfrac{[\text{NH}_{3}]^{2}}{[\text{N}_{2}][\text{H}_{2}]^{3}}$ — the 2 and 3 become powers.

**Products are favoured** — the equilibrium lies to the **right** (mostly products).

**Reactants are favoured** — the equilibrium lies to the **left** (mostly reactants).

The **reciprocal**: K_{reverse} = **1 / K_{forward}**.

A change in **temperature** — concentration, pressure and a catalyst all leave K_{c} unchanged.

K_{c} **increases** (the position shifts towards products).

K_{c} **decreases** (the position shifts towards reactants).

Convert each **amount (mol)** to a **concentration (mol dm⁻³)** using **c = n/V**, then substitute.

Pure **solids** and pure **liquids** — only gases and dissolved (aqueous) species appear.

**No** — K_{c} describes the **extent** (how far), not the **rate** (how fast).

A **proton (H⁺) donor**.

A **proton (H⁺) acceptor**.

A **hydrogen ion, H⁺** — a hydrogen atom that has lost its electron.

Two species that differ by **exactly one H⁺** (an acid and the base left after it donates).

**Remove** one H⁺ from the acid (e.g. HCl → Cl⁻).

**Add** one H⁺ to the base (e.g. NH_{3} → NH_{4}^{+}).

A species that can **both donate and accept** a proton (e.g. H_{2}O, HCO_{3}⁻).

**HSO_{4}⁻** (remove one H⁺ — not SO_{4}^{2-}, which is two H⁺ away).

**H_{3}O^{+}** (the oxonium / hydronium ion).

**H_{2}O** and **HCO_{3}⁻** — both can donate or accept a proton.

**HCl** — it donates the proton; water is the base.

An acid can only **donate** H⁺ if a base is there to **accept** it — every proton transfer has both.

A measure of acidity based on hydrogen-ion concentration: $\text{pH} = -\log_{10}[\text{H}^{+}]$.

$\text{pH} = -\log_{10}[\text{H}^{+}]$ — given in the data booklet.

$[\text{H}^{+}] = 10^{-\text{pH}}$ — the rearranged given equation.

The ionic product of water, $K_{w} = [\text{H}^{+}][\text{OH}^{-}] = 1.0\times10^{-14}$ at 25 °C.

pH < 7 acidic · pH = 7 neutral · pH > 7 basic (alkaline), at 25 °C.

[H_{+}] changes by a factor of **10** (pH is a log scale).

Strong = **fully** dissociated into ions; weak = only **partially** dissociated.

No — strength is the **degree of dissociation**; concentration is the amount dissolved.

Single arrow, e.g. $\text{HCl} \rightarrow \text{H}^{+} + \text{Cl}^{-}$ (full dissociation).

Equilibrium arrows, e.g. $\text{CH}_{3}\text{COOH} \rightleftharpoons \text{H}^{+} + \text{CH}_{3}\text{COO}^{-}$ (partial).

Strong acid has a **lower pH**, **higher conductivity** and a **faster** reaction (more H_{+} ions).

It is fully dissociated, so it gives a **higher [H_{+}]**, and a higher [H_{+}] means a lower pH.

The reaction of an **acid with a base** to give a **salt and water**; the H⁺ and OH⁻ cancel out.

The ionic compound formed when the **H⁺** of an acid is replaced by a **metal ion** (or NH_{4}⁺).

**salt + hydrogen** (e.g. Mg + 2HCl → MgCl_{2} + H_{2}).

**salt + water** (neutralisation; the base is a metal oxide or hydroxide).

**salt + water + carbon dioxide** (e.g. 2HCl + CaCO_{3} → CaCl_{2} + H_{2}O + CO_{2}).

A **chloride** (e.g. NaCl, MgCl_{2}).

A **sulfate** (e.g. Na_{2}SO_{4}, MgSO_{4}).

A **nitrate** (e.g. NaNO_{3}, Ca(NO_{3})_{2}).

**Hydrogen** gives a squeaky **'pop'** with a lit splint.

**Carbon dioxide** turns **limewater milky** (cloudy).

It is **diprotic** — it provides **two H⁺**, so it neutralises two 1+ bases: 2NaOH + H_{2}SO_{4} → Na_{2}SO_{4} + 2H_{2}O.

Balance the **ionic charges** (e.g. Mg²⁺ with Cl⁻ → MgCl_{2}), then balance the whole equation.

A number tracking how many electrons an atom has **gained or lost** relative to the free element — the charge it would have if all bonds were ionic.

Always **0** (e.g. Na, O_{2}, Cl_{2}, S_{8}).

**Equal to its charge** (e.g. Mg²⁺ is +2, Cl⁻ is −1).

Oxygen is **−2**; hydrogen is **+1** — except peroxides (O is −1) and metal hydrides (H is −1).

They add up to the **total charge**: 0 for a neutral compound, the ion charge for a polyatomic ion.

An **increase** in oxidation state — the atom has **lost** electrons (OIL).

A **decrease** in oxidation state — the atom has **gained** electrons (RIG).

**O**xidation **I**s **L**oss, **R**eduction **I**s **G**ain (of electrons).

The species that **takes** electrons and is itself **reduced** (its oxidation state goes down), e.g. O_{2}, Cl_{2}.

The species that **gives** electrons and is itself **oxidised** (its oxidation state goes up), e.g. a reactive metal.

An atom's oxidation state **changes** during the reaction — so electrons have been transferred.

**+6** — four O at −2 (= −8) plus S must equal the ion charge −2, so S = +6.

An equation showing **just the oxidation or just the reduction** part of a redox reaction, with the electrons (e⁻) included.

On the **right** (product) side — oxidation is **loss** of electrons.

On the **left** (reactant) side — reduction is **gain** of electrons.

**O**xidation **I**s **L**oss, **R**eduction **I**s **G**ain (of electrons).

Balance the **atoms** first, then add **electrons** to the more positive side so the **charge** balances.

$\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^{-}$ — an oxidation (loses 2e⁻).

$\text{Cu}^{2+} + 2e^{-} \rightarrow \text{Cu}$ — a reduction (gains 2e⁻).

**Multiply** so both transfer the same number of electrons, then **add** them and **cancel** the e⁻.

So the **electrons lost equal the electrons gained** — they must cancel exactly in the overall equation.

Both the **atoms** and the **total charge** must balance, with **no electrons** left over.

$\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$ — both halves transfer 2e⁻, which cancel.

A ranking of metals from **most reactive** (top) to **least reactive** (bottom), by how readily they lose electrons.

When a **more reactive** metal pushes a **less reactive** metal out of a solution of its ions.

A **more reactive** metal **displaces** a less reactive metal from a solution of its ions.

Electrons are **transferred**: the metal is **oxidised** (loses e⁻) and the metal ion is **reduced** (gains e⁻).

**O**xidation **I**s **L**oss, **R**eduction **I**s **G**ain — of electrons.

Oxidation: Zn → Zn²⁺ + 2e⁻; Reduction: Cu²⁺ + 2e⁻ → Cu.

Metals **above hydrogen** in the series → give a **salt + hydrogen gas**. Cu, Ag (below H) do not.

**Salt + hydrogen** (e.g. Mg + 2HCl → MgCl_{2} + H_{2}).

The most reactive ones (K, Na, Ca) → **metal hydroxide + hydrogen** (e.g. 2Na + 2H_{2}O → 2NaOH + H_{2}).

Add each metal to the **other's salt solution**; the metal that reacts (displaces) is the **more reactive**.

A **reducing agent** — it donates electrons (and is itself oxidised).

A **colour change** of the solution and a **deposit** of the displaced metal on the added metal.

A device that links a **redox reaction** to a flow of electrons through a wire — either making electricity (voltaic) or driven by it (electrolytic).

A cell in which a **spontaneous** redox reaction converts **chemical energy into electrical energy** (a battery).

A cell in which an external power supply drives a **non-spontaneous** reaction — **electrical energy into chemical energy** (electrolysis).

**Oxidation** (loss of electrons) — remember **AN OX**.

**Reduction** (gain of electrons) — remember **RED CAT**.

Anode = **negative (−)**, cathode = **positive (+)**.

Anode = **positive (+)**, cathode = **negative (−)** — the opposite of a voltaic cell.

Always from the **anode to the cathode** (in both cell types).

Completes the circuit and keeps each half-cell **neutral**: anions move toward the anode, cations toward the cathode.

Voltaic = **spontaneous**, makes electricity; electrolytic = **driven** by a supply, uses electricity.

$\text{Ag}^{+}(aq) + e^{-} \rightarrow \text{Ag}(s)$ — reduction (gain of one electron).

A species with an **unpaired electron**, written with a dot (e.g. Cl•, •CH_{3}); very reactive.

A bond breaks **evenly** — **one electron goes to each** atom, forming two **radicals**.

A bond breaks **unevenly** — **both electrons go to one** atom, forming **ions** (a cation and an anion).

**Homolytic** fission makes radicals; **heterolytic** fission makes ions.

An alkane reacts with a halogen in **UV light**, replacing an H atom with a halogen atom, via a radical chain.

**UV light** breaks the halogen molecule by **homolytic** fission, e.g. Cl_{2} → 2 Cl•.

A radical reacts to give a product **and a new radical**, so the chain continues (radical count unchanged).

Cl• + CH_{4} → •CH_{3} + HCl, then •CH_{3} + Cl_{2} → CH_{3}Cl + Cl•.

**Two radicals combine** into one molecule, removing radicals and **stopping** the chain (e.g. •CH_{3} + Cl• → CH_{3}Cl).

It supplies the energy to break the halogen bond **homolytically** and create the first radicals.

Each propagation step **regenerates** a radical, so one initiation triggers many cycles (and a mixture of products).

With a **dot (•)** next to it, showing the single unpaired electron (e.g. Cl•).

An **electron-pair donor** — it has a lone pair and is attracted to a δ+ (electron-poor) carbon. Examples: OH⁻, CN⁻, NH_{3}.

The C–halogen bond is **polar**: the more electronegative halogen pulls the bonding electrons, leaving carbon slightly positive (**δ+**).

The movement of a **pair of electrons** — the tail is at the electrons that move, the head is where the pair ends up.

Arrow 1: the nucleophile's **lone pair → δ+ carbon** (new bond). Arrow 2: the **C–X bond → halogen**, which leaves as X⁻.

The atom/ion that departs **with the bonding pair** — here the **halide ion, X⁻** (e.g. Br⁻, Cl⁻).

An **alcohol** (the –halogen is replaced by –OH), plus a halide ion.

**Warm** (gentle heat) and **aqueous** sodium or potassium hydroxide.

A reaction in which **one group replaces another** on the carbon skeleton, which is otherwise unchanged.

Nucleophile = electron-pair **donor** (attacks δ+); electrophile = electron-pair **acceptor** (attacks δ−). Opposites.

**C–I** — it is the **weakest** bond, so it breaks most easily. C–F is strongest, so the fluoroalkane is slowest.

A **nitrile** (–CN) — and the chain gains one carbon atom.

An **amine** (–NH_{2}), using excess ammonia.

An **electron-pair acceptor** — an electron-poor (often positive) species attracted to an electron-rich centre such as a C=C. Examples: Br_{2}, HBr, H⁺.

It is **electron-rich** (two shared pairs / exposed π electrons), so it readily donates electrons and attacks electrophiles.

Two molecules join to form **one** product; the C=C opens to a single bond and the reactant adds across it. **Nothing leaves**.

Saturated = only **single** bonds (alkane); unsaturated = has a **C=C** (alkene) so more atoms can be **added**.

Arrow 1: the **C=C π electrons → a bromine** (new bond). Arrow 2: the **Br–Br bond → the other bromine**, which leaves as Br⁻.

**1,2-dibromoethane, CH_{2}BrCH_{2}Br** — one bromine adds to each carbon.

**Bromoethane, CH_{3}CH_{2}Br** — H and Br add across the C=C.

**Ethanol, CH_{3}CH_{2}OH** — water adds across the C=C with an H_{3}PO_{4} catalyst at high T and P.

**Ethane, CH_{3}CH_{3}** (saturated) — hydrogen adds across the C=C with a Ni catalyst.

Add **bromine water**: an **alkene decolourises** it (orange → colourless); an **alkane** gives **no change**.

For an unsymmetrical alkene + HX, the **H adds to the carbon that already has more hydrogens**, giving the **major** product.

**2-bromopropane, CH_{3}CHBrCH_{3}** — H goes to the CH_{2} end, Br to the middle carbon (Markovnikov).