Free flashcards · International Baccalaureate IB Diploma Programme

Chemistry flashcards

181 cards covering every topic of the syllabus (First assessment 2025). Study one topic or mix them all, turn each card, and mark yourself honestly.

Study mode needs JavaScript. Every card is listed below, so you can still cover the answers and test yourself.

All 181 cards

For a quick read-through or printing. Cover the right-hand side to test yourself.

Topic 1: Structure 1 – Models of the particulate nature of matter

Element, compound and mixture: Element: one type of atom only. Compound: atoms of different elements chemically bonded in a fixed ratio; separated only by a chemical reaction. Mixture: not chemically bonded, no fixed ratio; separable by physical methods.
Homogeneous vs heterogeneous mixture: Homogeneous: uniform composition, one phase (vinegar, bronze, filtered air). Heterogeneous: non-uniform, two or more phases (muddy water).
Recrystallization: method: Dissolve the impure solid in the minimum of hot solvent; filter hot to remove insoluble impurities; cool so the product crystallizes (soluble impurities stay in solution); filter, wash with a little cold solvent, dry.
Solvent front 9.0 cm, spot 2.7 cm from the base line. Rf?: Rf = 2.7 ÷ 9.0 = 0.30. Rf always lies between 0 and 1; both distances are measured from the base line.
Evaporation vs boiling: Evaporation: at the surface only, below the boiling point; the most energetic particles escape, so the liquid cools. Boiling: throughout the liquid, at a fixed temperature for a given external pressure.
What does kelvin temperature measure? Convert −18 °C to kelvin.: The average kinetic energy of the particles (double the kelvin temperature, double the average kinetic energy). −18 + 273.15 = 255.15 K.
Why is a heating curve flat during melting and boiling?: The energy supplied overcomes the attractions between particles, so the average kinetic energy and temperature stay constant while two states coexist.
Relative charge, relative mass and location of the three subatomic particles: Proton: +1, 1, nucleus. Neutron: 0, 1, nucleus. Electron: −1, negligible (about 1/1836), outside the nucleus.
Protons, neutrons and electrons in ¹¹⁹₅₀Sn²⁺: 50 protons, 119 − 50 = 69 neutrons, 50 − 2 = 48 electrons.
Why do isotopes have the same chemical properties but different physical properties?: Same number of protons, so the same number and arrangement of electrons, which decide reactions. Different numbers of neutrons, so different masses: density, melting and boiling points and diffusion rate differ.
Rhenium is 37.4% ¹⁸⁵Re (184.95) and 62.6% ¹⁸⁷Re (186.96). Ar?: (184.95 × 37.4 + 186.96 × 62.6) ÷ 100 = 186.21
Isotopes of masses 64.00 and 66.00 give Ar = 64.70. Percentage abundances?: 64.00x + 66.00(1 − x) = 64.70, so 66.00 − 2.00x = 64.70 and x = 0.650. 65.0% of mass 64.00 and 35.0% of mass 66.00.
Reading a mass spectrum of an element: Each peak is one isotope (m/z = isotope mass for 1+ ions); peak height is relative abundance. Ar is the weighted mean of the m/z values. (HL only)
Hydrogen emission spectrum: which transitions give ultraviolet, visible and infrared lines?: Falls to n = 1: ultraviolet. To n = 2: visible. To n = 3: infrared. In each series the lines converge at higher frequency, because the levels get closer together as n increases.
Orbitals per sublevel, and the maximum electrons in main level n: s 1, p 3, d 5, f 7 orbitals. Main level n holds n² orbitals and up to 2n² electrons. Each orbital holds two electrons of opposite spin.
Aufbau principle, Pauli exclusion principle, Hund’s rule: Aufbau: fill the lowest-energy orbital first. Pauli: at most two electrons per orbital, with opposite spins. Hund: occupy equal-energy orbitals singly with parallel spins before pairing.
Configurations of chromium and copper: Cr: [Ar] 3d⁵ 4s¹ (half-filled 3d). Cu: [Ar] 3d¹⁰ 4s¹ (filled 3d). Each gives a lower overall energy than the expected form.
Filling order vs loss order: 4s fills before 3d but empties before 3d.
Configuration of Ni²⁺ (Z = 28), and its unpaired electrons: Ni: [Ar] 3d⁸ 4s². Ni²⁺: [Ar] 3d⁸ (4s emptied first), with 2 unpaired electrons.
Lithium’s convergence limit is at 1.30 × 10¹⁵ Hz (h = 6.63 × 10⁻³⁴ J s). First ionization energy?: E = hf = 6.63 × 10⁻³⁴ × 1.30 × 10¹⁵ = 8.62 × 10⁻¹⁹ J. × 6.02 × 10²³ = 5.19 × 10⁵ J mol⁻¹ = 519 kJ mol⁻¹. (HL only)
Successive ionization energies of boron: 801, 2427, 3660, 25 026 kJ mol⁻¹. Group?: The biggest increase comes after the third electron, so boron has three outer electrons: group 13. Look for the biggest increase, not the biggest value. (HL only)
The mole and the Avogadro constant: The mole is the SI unit for amount of substance; one mole contains NA elementary entities. NA = 6.02 × 10²³ mol⁻¹.
Ar and Mr versus molar mass M: The same number, but Ar and Mr are ratios with no units, and M is the mass of one mole in g mol⁻¹. n = m ÷ M.
How many sodium ions are in 0.0200 mol of Na₂CO₃?: 2 × 0.0200 = 0.0400 mol Na⁺; 0.0400 × 6.02 × 10²³ = 2.41 × 10²² ions.
5.55 g of CaCl₂ (M = 110.98 g mol⁻¹) is made up to 500.0 cm³. [CaCl₂] and [Cl⁻]?: n = 5.55 ÷ 110.98 = 0.0500 mol. [CaCl₂] = 0.0500 ÷ 0.5000 = 0.100 mol dm⁻³; [Cl⁻] = 0.200 mol dm⁻³.
Empirical formula: what to do with a mole ratio ending in .5, .33 or .25: .5: multiply by 2. .33 or .67: multiply by 3. .25 or .75: multiply by 4. Molecular formula multiple = M ÷ empirical formula mass.
Avogadro's law: At equal temperature and pressure, equal volumes of gases contain equal numbers of molecules, so gas volume ratios equal mole ratios.
Dilution: The amount of solute stays the same, so C₁V₁ = C₂V₂. Concentration uses the final volume of solution.
Assumptions of the ideal gas model: Particles move constantly and randomly; their own volume is negligible; there are no intermolecular forces; collisions are elastic.
When and why do real gases deviate from ideal behaviour?: High pressure: the particles’ own volume is no longer negligible. Low temperature: intermolecular attractions are no longer negligible. Most ideal at high T, low P, with small non-polar particles.
Molar volume of an ideal gas at STP: 22.7 dm³ mol⁻¹ at 273.15 K and 100 kPa. Under other conditions Vm = RT ÷ P.
PV = nRT: the units to use: P in Pa, V in m³, T in K, R = 8.31 J K⁻¹ mol⁻¹. kPa × 10³ = Pa; dm³ × 10⁻³ = m³; cm³ × 10⁻⁶ = m³.
A gas at 100 kPa and 20.0 °C is compressed to half its volume; the pressure becomes 230 kPa. New temperature?: T₂ = T₁ × P₂V₂ ÷ (P₁V₁) = 293.15 × (230 × 0.5) ÷ (100 × 1) = 337 K (64.0 °C).
Volume against temperature in °C for a fixed mass of gas at constant pressure: A straight line that does not pass through the origin; extrapolated to V = 0 it meets the temperature axis near −273 °C (absolute zero). V is proportional to T only in kelvin.

Topic 2: Structure 2 – Models of bonding and structure

Ionic bond: Electrostatic attraction between oppositely charged ions.
Structure of an ionic compound: Giant 3D lattice; each ion surrounded by oppositely charged ions; attractions act in all directions.
Why is the formula of an ionic compound an empirical formula?: Because a crystal has no fixed number of ions and no discrete molecules.
Formulas of rubidium nitride and iron(III) carbonate: Rb⁺ and N³⁻: Rb₃N. Fe³⁺ and CO₃²⁻: Fe₂(CO₃)₃.
Why does CaS have a larger lattice enthalpy than K₂S?: Both contain S²⁻, but Ca²⁺ has a higher charge and is smaller than K⁺, so the ion centres are closer and the electrostatic attraction is stronger. Charge usually dominates.
Electrical conductivity: molten vs aqueous vs solid ionic compound: The first two conduct, the last does not. Ions move; electrons do not.
The seven polyatomic ions to know: Ammonium NH₄⁺, hydroxide OH⁻, nitrate NO₃⁻, hydrogencarbonate HCO₃⁻, carbonate CO₃²⁻, sulfate SO₄²⁻, phosphate PO₄³⁻.
Covalent bond, and coordination bond: Covalent: two nuclei both attracting the same shared electron pair. Coordination: a covalent bond where both shared electrons come from the same atom (H₃O⁺, NH₄⁺, Al₂Cl₆, CO). Once formed it is identical to the others.
Shapes and approximate angles of BF₃, NH₃ and OF₂: BF₃: 3 domains, 0 lone pairs, trigonal planar, 120°. NH₃: 4 domains, 1 lone pair, trigonal pyramidal, about 107°. OF₂: 4 domains, 2 lone pairs, bent, about 104.5°.
Single, double and triple bonds between the same two atoms: length and strength: More shared pairs means more electron density between the nuclei: triple is shortest and strongest, single is longest and weakest.
Why does SiF₄ have polar bonds but no net dipole?: It is tetrahedral with identical outer atoms, so the bond dipoles cancel. Lone pairs on the central atom (pyramidal, bent) or different outer atoms usually prevent cancelling.
Intermolecular forces in order of strength (similar molar mass): London (always present) < dipole-induced dipole < dipole–dipole < hydrogen bond (H on N, O or F plus a lone pair on N, O or F).
Diamond, graphite, graphene and C₆₀: bonds per carbon and conductivity: Diamond: 4, does not conduct. Graphite: 3 in layers, conducts along the layers. Graphene: 3 in one sheet, conducts very well. C₆₀: 3 in a cage, molecular, conducts poorly.
Resonance: Two or more valid Lewis formulas differ only in where π electrons sit. The real species is one resonance hybrid; its delocalized bonds are equal in length, between single and double (e.g. 1.5 for three pairs over two positions). (HL only)
Formal charge: formula and how to choose the preferred Lewis formula: FC = V − (L + ½B). Prefer formal charges closest to zero; if tied, put the negative formal charge on the more electronegative atom. (HL only)
σ vs π bonds: Head-on vs sideways; on the axis vs either side of it. A single bond is 1 σ; a double bond 1 σ + 1 π; a triple bond 1 σ + 2 π. (HL only)
Hybridization for 4, 3 and 2 electron domains: 4 domains: sp³, tetrahedral, 109.5°. 3: sp², trigonal planar, 120° (one p left for a π bond). 2: sp, linear, 180° (two p left). Count domains, not atoms. (HL only)
Molecular geometry of XeF₄ and ClF₃: XeF₄: 6 domains, 2 lone pairs opposite each other, square planar. ClF₃: 5 domains, 2 lone pairs equatorial, T-shaped. (HL only)
Metallic bond: The electrostatic attraction holding a lattice of cations to the delocalized electrons around them. The bonding is non-directional.
How the metallic model explains electrical and thermal conductivity: Electrical: under a potential difference the delocalized electrons drift, carrying charge; the cations stay put. Thermal: mobile electrons gain kinetic energy at the hot end and carry it to cooler regions.
Why are metals malleable and ductile?: Layers of cations slide over each other and the delocalized electrons move with them, so the attraction is kept and the metal reshapes instead of breaking.
What makes a metallic bond stronger?: A higher cation charge (more delocalized electrons per atom) and a smaller ion radius. Example: melting point rises Na → Mg → Al.
Why does strontium melt higher than rubidium?: Sr gives Sr²⁺ and two delocalized electrons per atom; Rb gives Rb⁺ and one. Sr²⁺ is also smaller, so the attraction is stronger and more energy is needed (about 1050 K vs about 312 K).
Why do transition elements have high melting points and conduct well?: 3d as well as 4s electrons can be delocalized, giving greater electron density and higher cation charge with small ions: very strong metallic bonding and more mobile charge carriers. (HL only)
The bonding triangle: what the axes show: Vertical: electronegativity difference, Δχ. Horizontal: average electronegativity, (χA + χB) ÷ 2. Top: ionic; bottom left: metallic; bottom right: covalent.
Place NaBr on the bonding triangle (χ: Na 0.93, Br 2.96): Δχ = 2.03; average = 1.945. Ionic region: high melting point, conducts when molten.
Why is an alloy harder than the pure metal?: Atoms of a different size distort the layers of the lattice, so the layers resist sliding. The delocalized electrons remain, so the alloy still conducts.
Why are most plastics insulators that soften on heating?: No mobile ions or delocalized electrons, so no conduction. Heating overcomes only the intermolecular forces between chains, so they are mouldable (cross-linked ones do not soften).
Repeating unit of the addition polymer from 3-chloropropene, CH₂=CH–CH₂Cl: –[–CH₂–CH(CH₂Cl)–]–ₙ. The C=C becomes C–C; CH₂Cl is a side chain. Addition polymerization has 100% atom economy.
Polyesters vs polyamides: Polyester: –COOH + –OH, ester link –COO–. Polyamide: –COOH + –NH₂, amide link –CONH–. Both are condensation polymers that release water at each link. (HL only)
How many water molecules are released when N monomer units form a condensation polymer chain?: N − 1 links, so N − 1 water molecules; the same number is needed to hydrolyse it completely. Proteins, polysaccharides and nucleic acids all form this way. (HL only)

Topic 3: Structure 3 – Classification of matter

Period and group from an electron configuration: Period = outer occupied main level (n). Group = number of valence electrons for groups 1 and 2; group number minus 10 for groups 13 to 18. The block shows which sublevel is filling.
First ionization energy: The energy, per mole, to take the outermost electron from each atom of a gaseous element in its ground state, giving gaseous 1+ ions.
Electron affinity: The energy change, per mole, as each gaseous atom picks up one electron to become a gaseous 1− ion.
Periodic trends: the explanation across a period: More protons, same main level, similar shielding → outer electrons pulled harder.
Periodic trends: the explanation down a group: Extra main level, more shielding → outer electrons further away and held less tightly.
Order the radii of N³⁻, O²⁻, F⁻, Na⁺, Mg²⁺: N³⁻ > O²⁻ > F⁻ > Na⁺ > Mg²⁺. All have 10 electrons; more protons pull them in more tightly.
Ocean acidification: Extra CO₂ dissolves → H₂CO₃ ⇌ H⁺ + HCO₃⁻ → pH falls; H⁺ removes CO₃²⁻ as HCO₃⁻, so less carbonate is available for shells.
Oxidation state of iodine in HIO₃: (+1) + x + 3(−2) = 0, so x = +5. Write the sign first (+5); 5+ would be an ion charge.
Why does first ionization energy dip from group 2 to 13 and from group 15 to 16?: 2 → 13: the electron removed is in a p sublevel, higher in energy than s. 15 → 16: the electron removed is paired, and repulsion raises its energy. (HL only)
Typical transition element properties, and why zinc is excluded: Variable oxidation states, high melting points, magnetic and catalytic properties, coloured compounds, complex ions. Caused by an incomplete d-sublevel; Zn is 3d¹⁰ in the atom and in Zn²⁺. (HL only)
Why are transition element complexes coloured?: Ligands split the d orbitals into two sets; an electron absorbs visible light matching the gap and is promoted. The colour seen is the complementary colour. (HL only)
Homologous series: Same functional group and general formula; successive members differ by CH₂; similar chemistry and a gradual trend in physical properties.
Effect of branching on boiling point: Branching lowers the boiling point: less surface contact, weaker London forces.
Effect of the functional group on boiling point: For similar numbers of electrons, hydrogen bonding (–OH, –COOH) > dipole–dipole (C=O, C–X) > London forces only (alkanes).
Name ClCH₂CH₂CH₂CH(CH₃)₂ and CH₂=CHCOCH₂CH₃: 1-chloro-4-methylpentane (lowest locants {1,4}). Pent-1-en-3-one (C=O is C3 either way, so the C=C decides).
Alcohol vs amine classification (primary, secondary, tertiary): Count around the carbon vs around the nitrogen. So (CH₃)₃CNH₂ is a primary amine.
Isomer vs homologue: Isomers share a molecular formula; homologues differ by CH₂.
When does a compound show cis-trans isomerism?: When rotation is restricted (a C=C, or a C₃ or C₄ ring) and each of the two carbons carries two different groups. (HL only)
Chiral carbon, enantiomers and a racemic mixture: A chiral carbon has four different groups. Enantiomers are non-superimposable mirror images that rotate plane-polarized light by equal angles in opposite directions. A racemic mixture has equal amounts: no net rotation. (HL only)
IR: wavenumbers for C=O, alcohol O–H and carboxylic acid O–H: C=O: 1700–1750 cm⁻¹. Alcohol O–H: broad, 3200–3600 cm⁻¹. Acid O–H: very broad, 2500–3000 cm⁻¹. (HL only)
Which vibrations are IR active?: Only those that change the dipole moment. N₂ and O₂ stretches and the CO₂ symmetric stretch are not; the CO₂ asymmetric stretch and bend are. (HL only)
¹H NMR: what the number of signals, integration and splitting tell you: Signals: number of hydrogen environments. Integration: ratio of H in each. Splitting: n H on adjacent carbons give n + 1 peaks (ethyl: quartet + triplet). (HL only)
Mass spectra of butan-1-ol and butan-2-ol (both M = 74): what distinguishes them?: A strong peak at m/z 31 (CH₂OH⁺) points to OH on the end carbon: butan-1-ol. A strong peak at 45 (CH₃CHOH⁺) points to OH on carbon 2: butan-2-ol. (HL only)

Topic 4: Reactivity 1 – What drives chemical reactions?

Heat vs temperature: Energy on the move (J) vs a measure of mean particle kinetic energy (K). Heat depends on the amount of substance; temperature does not.
The thermometer rises during a reaction in solution. Exothermic or endothermic, and the sign of ΔH?: The thermometer reads the surroundings. They gained energy, so the system lost it: exothermic, ΔH negative.
Standard enthalpy change, ΔH⦵: ΔH under standard conditions (100 kPa; values usually for 298 K) with every substance in its standard state, in kJ mol⁻¹.
0.0150 mol of limiting reactant raises 40.0 g of solution by 4.3 K. ΔH?: Q = mcΔT = 40.0 × 4.18 × 4.3 = 719 J = 0.719 kJ. ΔH = −Q ÷ n = −0.719 ÷ 0.0150 = −47.9 kJ mol⁻¹.
Why is an experimental ΔH usually less negative than the data value?: Energy is lost to the air and bench; energy warms the cup or can; combustion is incomplete; fuel evaporates; products are not in standard states.
ΔH vs Ea on an energy profile: Gap between reactant and product levels vs gap from reactants up to the peak.
Bond enthalpy: The energy absorbed when one mole of a particular covalent bond is broken in the gas phase; always positive. Average bond enthalpies are means over a range of compounds.
CH₂=CHCH₃ + Cl₂ → CH₂ClCHClCH₃. C=C 614, Cl–Cl 242, C–C 346, C–Cl 324 kJ mol⁻¹. ΔH?: Broken: 614 + 242 = 856. Formed: 346 + 2 × 324 = 994. ΔH = Σ(broken) − Σ(formed) = 856 − 994 = −138 kJ mol⁻¹.
Why is a ΔH from bond enthalpies only an estimate?: The values are averages over many molecules, and they apply to gases, so intermolecular forces in liquids or solids are ignored.
Hess's law: ΔH for a reaction is the same whatever pathway links the initial and final states (conservation of energy). Reverse an equation: change the sign of ΔH; scale it: scale ΔH.
Standard enthalpy of formation, and its value for an element: ΔH for making one mole of a substance from its elements, every species in its standard state. It is zero for an element in its standard state. (HL only)
Calculating ΔH⦵ from formation data vs combustion data: Formation: Σ ΔHf⦵(products) − Σ ΔHf⦵(reactants). Combustion: Σ ΔHc⦵(reactants) − Σ ΔHc⦵(products). (HL only)
Born–Haber: Li(s) → Li(g) +159; IE₁ +520; ½I₂ → I(g) +107; EA −295; lattice enthalpy +759 kJ mol⁻¹. ΔHf⦵ of LiI?: Elements to gaseous ions: 159 + 520 + 107 − 295 = +491. ΔHf⦵ = 491 − 759 = −268 kJ mol⁻¹ (lattice enthalpy here is for separating the solid). (HL only)
Complete vs incomplete combustion of an organic fuel: Complete: excess oxygen; CO₂ and H₂O only; blue, clean flame. Incomplete: limited oxygen; carbon ends up as CO and/or C (soot); yellow, sooty flame; less energy per mole.
Balanced equation for the complete combustion of propan-2-ol, C₃H₇OH: 2C₃H₇OH + 9O₂ → 6CO₂ + 8H₂O (3CO₂ + 4H₂O need 10 O atoms; subtract the 1 in the alcohol: 4½O₂ per molecule).
Why is carbon monoxide dangerous?: It attaches to haemoglobin more tightly than oxygen, so the blood carries less oxygen; it has no colour or smell.
Coal, crude oil and natural gas: energy per gram and CO₂ per unit of energy: Natural gas (mostly methane): highest energy per gram, lowest CO₂ per unit energy. Coal (mostly carbon): lowest energy per gram, highest CO₂ per unit energy. Crude oil in between.
How do greenhouse gases warm the lower atmosphere?: The ground absorbs visible light and emits infrared. CO₂, H₂O and CH₄ absorb the infrared (their bonds vibrate) and re-emit it randomly, so part of it returns to the ground.
Why is a biofuel rarely fully carbon neutral?: Its carbon was fixed recently by photosynthesis (6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂), but farming, fertiliser, processing and transport usually burn fossil fuels.
Hydrogen fuel cell half-equations with an acidic electrolyte: Anode (negative, oxidation): H₂ → 2H⁺ + 2e⁻. Cathode (positive, reduction): O₂ + 4H⁺ + 4e⁻ → 2H₂O. Overall 2H₂ + O₂ → 2H₂O.
Entropy, and the standard entropy of an element: A measure of how dispersed the matter and energy of a system are (J K⁻¹ mol⁻¹). S⦵ is positive for every pure substance above 0 K, elements included; a perfect crystal at 0 K has zero entropy. (HL only)
When is ΔS of the system positive?: When the moles of gas increase (strongest effect); when a solid melts, a liquid boils or a solid sublimes; when a solid dissolves or substances mix; when a substance is heated. (HL only)
Gibbs energy equation and its units: ΔG⦵ = ΔH⦵ − TΔS⦵, with ΔH and ΔG in kJ mol⁻¹, ΔS divided by 1000 to kJ K⁻¹ mol⁻¹, T in K. Spontaneous when ΔG is negative. (HL only)
Spontaneity from the signs of ΔH and ΔS: ΔH −, ΔS +: at all temperatures. ΔH +, ΔS −: never. ΔH −, ΔS −: below T = ΔH/ΔS. ΔH +, ΔS +: above T = ΔH/ΔS. (HL only)
ΔH = +65.0 kJ mol⁻¹, ΔS = +160 J K⁻¹ mol⁻¹. When is the change spontaneous?: T = 65.0 ÷ 0.160 = 406 K. Both positive, so spontaneous above 406 K. (HL only)
Link between ΔG⦵ and K: ΔG⦵ = −RT lnK (ΔG⦵ in J mol⁻¹, R = 8.31 J K⁻¹ mol⁻¹). Negative ΔG⦵ gives K > 1, products favoured; positive gives K < 1. At equilibrium ΔG = 0. (HL only)
ΔG for a mixture that is not at equilibrium: ΔG = ΔG⦵ + RT lnQ. If Q < K, ΔG is negative and the forward reaction proceeds, even if ΔG⦵ is positive. (HL only)

Topic 5: Reactivity 2 – How much, how fast and how far?

Limiting reactant: The reactant that is used up first. It caps the amount of product.
Theoretical yield: The amount of product calculated from the limiting reactant, assuming complete reaction and no losses.
Atom economy: Molar mass of the desired product as a percentage of the total molar mass of all reactants in the balanced equation; a green chemistry measure of efficiency.
4Li + O₂ → 2Li₂O with 0.50 mol Li and 0.15 mol O₂. Which is limiting, and how much Li₂O forms?: Divide by coefficients: Li 0.50 ÷ 4 = 0.125; O₂ 0.15 ÷ 1 = 0.15. Lithium is limiting. Li₂O = 0.50 × 2/4 = 0.25 mol; 0.025 mol O₂ is left over.
2C₃H₆(g) + 9O₂(g) → 6CO₂(g) + 6H₂O(l). Volumes for 10 cm³ of propene?: Oxygen needed: 10 × 9/2 = 45 cm³. CO₂ made: 10 × 6/2 = 30 cm³. Liquid water takes up no gas volume.
Percentage yield vs atom economy: Yield depends on how the reaction is carried out (experimental ÷ theoretical × 100). Atom economy is fixed by the equation, so better technique cannot change it.
Rate of reaction, and the instantaneous rate: Rate: how fast the concentration of a named reactant or product changes per unit time (mol dm⁻³ s⁻¹). Instantaneous rate: gradient of the tangent to the curve at that moment.
The two conditions for a successful collision: Energy at least equal to the activation energy, Ea, and the proper orientation (collision geometry).
Which factors change collision frequency only, and which change the fraction with E ≥ Ea?: Concentration, pressure, surface area: frequency only. Temperature: the fraction with E ≥ Ea (main effect), and frequency slightly. Catalyst: the fraction, because Ea is lower.
Maxwell–Boltzmann: temperature vs catalyst: Temperature changes the curve; a catalyst moves the Ea line.
Kinetic vs thermodynamic stability: A high Ea makes a reaction slow; a positive or negative ΔH says nothing about speed. A high Ea lets a fuel be stored in air without igniting.
An endothermic reaction has Ea = 140 kJ mol⁻¹ and ΔH = +35 kJ mol⁻¹. Ea of the reverse reaction?: Ea(reverse) = 140 − 35 = 105 kJ mol⁻¹.
How are the orders in a rate equation found?: Only by experiment, never from the coefficients of the balanced equation. rate = k[A]ᵐ[B]ⁿ; overall order = m + n. (HL only)
Mechanism: L + M → LM (slow); LM + M → LM₂ (fast). Rate equation?: rate = k[L][M]. Only one M is in the rate-determining (slow) step, so it is first order in M even though the overall equation has 2M. LM is an intermediate. (HL only)
Intermediate vs transition state: Intermediate: made in one step and used in a later one; a trough on the profile; can be detected. Transition state: the highest-energy arrangement in one step; a peak; cannot be isolated. (HL only)
Units of k for zero, first, second and third order overall: Zero: mol dm⁻³ s⁻¹. First: s⁻¹. Second: mol⁻¹ dm³ s⁻¹. Third: mol⁻² dm⁶ s⁻¹. (HL only)
Arrhenius plot: what is plotted, and how are Ea and A found?: ln k against 1/T (K). Gradient = −Ea/R, so Ea = −gradient × 8.31 J mol⁻¹. The y-intercept is ln A. (HL only)
Features of a dynamic equilibrium: Closed system; forward and backward rates equal and not zero; concentrations constant (not necessarily equal); observable properties constant; it can be reached from either side.
Equilibrium constant expression for 2N₂O(g) + 3O₂(g) ⇌ 4NO₂(g): K = [NO₂]⁴ ÷ ([N₂O]²[O₂]³). Products on top; coefficients as powers.
K for the forward reaction is 0.25. K for the reverse at the same temperature?: K(reverse) = 1 ÷ K(forward) = 4.0.
Effect of heating on K for an exothermic forward reaction: K gets smaller and there is less product. Only temperature changes K; concentration and pressure move the position but leave K the same.
Effect of a catalyst on an equilibrium: No shift in position and no change in K: it increases the forward and backward rates by the same factor, so equilibrium is reached faster.
Q compared with K: which way does the reaction go?: Q < K: forward, making more products. Q = K: at equilibrium. Q > K: backward. Compare only at the same temperature. (HL only)
0.50 mol SO₂Cl₂ in 2.0 dm³ gives 0.10 mol Cl₂ at equilibrium (SO₂Cl₂ ⇌ SO₂ + Cl₂). K?: Equilibrium: SO₂Cl₂ 0.40 mol, SO₂ 0.10, Cl₂ 0.10. Concentrations 0.20, 0.050, 0.050 mol dm⁻³. K = (0.050 × 0.050) ÷ 0.20 = 0.0125. (HL only)
ΔG⦵ = +10.0 kJ mol⁻¹ at 400 K. K?: lnK = −10 000 ÷ (8.31 × 400) = −3.01, so K = 0.0494. Positive ΔG⦵ and K < 1: reactants favoured. (HL only)

Topic 6: Reactivity 3 – What are the mechanisms of chemical change?

Brønsted–Lowry acid and base, and a conjugate pair: Acid: proton (H⁺) donor. Base: proton acceptor. Conjugate pair: two species whose formulas differ by a single H⁺, e.g. H₂PO₄⁻ / HPO₄²⁻.
Amphiprotic vs amphoteric: Amphiprotic: can donate or accept a proton (H₂O, HSO₄⁻). Amphoteric: a substance, such as Al₂O₃, that reacts with both acids and bases.
pH of 0.0040 mol dm⁻³ NaOH at 298 K: [H⁺] = Kw ÷ [OH⁻] = 1.0 × 10⁻¹⁴ ÷ 0.0040 = 2.5 × 10⁻¹² mol dm⁻³. pH = 11.60
Strong vs weak acids, with examples: Strong: completely ionized (HCl, HBr, HI, HNO₃, H₂SO₄). Weak: partially ionized, an equilibrium (carboxylic acids, H₂CO₃, HF). Strong is not the same as concentrated.
Products of an acid with a carbonate, and with ammonia: Carbonate (or hydrogencarbonate): salt + water + CO₂. Ammonia: an ammonium salt only, e.g. HNO₃ + NH₃ → NH₄NO₃ (no water).
Why is pure water above 298 K not pH 7, and is it still neutral?: Water’s ionization is endothermic, so Kw rises with temperature and the pH falls below 7. It is still neutral because [H⁺] = [OH⁻].
Strong acid titrated with strong base: shape of the pH curve: Starts low, rises slowly while acid is in excess, then rises almost vertically (about pH 3 to 11) around the equivalence point at pH 7 (298 K), then levels off.
Relationship between Ka and Kb for a conjugate pair: Ka × Kb = Kw, so pKa + pKb = 14.00 at 298 K. A weaker acid has a stronger conjugate base. (HL only)
pH of 0.0800 mol dm⁻³ weak acid with pKa 4.87: [H⁺] = √(Ka × c) = √(10⁻⁴·⁸⁷ × 0.0800) = 1.04 × 10⁻³ mol dm⁻³. pH = 2.98 (HL only)
Is NH₄Br solution acidic, neutral or basic? Why?: Acidic: it comes from a strong acid and a weak base, and NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺. (HL only)
pH of an acidic buffer, and its pH when [HA] = [A⁻]: pH = pKa + log₁₀([A⁻]/[HA]). Equal concentrations give pH = pKa. Dilution leaves the pH nearly unchanged but lowers the capacity. (HL only)
How is an acid–base indicator chosen?: It changes colour near pH = pKa(HInd). Its range must sit inside the vertical section of the pH curve around the equivalence pH. Weak acid with weak base: no indicator is suitable. (HL only)
Oxidation and reduction (four definitions): Oxidation: electrons lost, oxidation state rises, oxygen gained, hydrogen lost. Reduction: electrons gained, oxidation state falls, oxygen lost, hydrogen gained.
Cu²⁺ + Fe → Cu + Fe²⁺: oxidizing agent and reducing agent: Fe 0 → +2 is oxidized, so Fe is the reducing agent. Cu +2 → 0 is reduced, so Cu²⁺ is the oxidizing agent.
Oxidation state of sulfur in HSO₃⁻: +1 + x + 3(−2) = −1, so x = +4.
Half-equation for manganate(VII) being reduced in acid: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Charge: −1 + 8 − 5 = +2 on each side.
Electrode signs in voltaic and electrolytic cells: Oxidation at the anode and reduction at the cathode in both. Voltaic: anode negative, cathode positive. Electrolytic: anode positive, cathode negative.
What does the salt bridge do in a voltaic cell?: Anions move towards the anode half-cell and cations towards the cathode half-cell, keeping each solution electrically neutral and completing the circuit. Electrons travel only through the wire.
Oxidizing a primary alcohol: aldehyde vs carboxylic acid: Aldehyde: distil it off as it forms (lower boiling point, no hydrogen bonding between its molecules). Carboxylic acid: reflux with excess oxidizing agent. A tertiary alcohol does not react.
Standard hydrogen electrode: H⁺(aq) + e⁻ ⇌ ½H₂(g): platinum electrode, H₂ at 100 kPa, [H⁺] = 1.00 mol dm⁻³, 298 K. E⦵ = 0.00 V by convention. (HL only)
Mg²⁺/Mg −2.37 V and Sn²⁺/Sn −0.14 V. E⦵cell and ΔG⦵ (F = 96 500 C mol⁻¹)?: E⦵cell = E⦵(cathode) − E⦵(anode) = −0.14 − (−2.37) = +2.23 V. ΔG⦵ = −nFE⦵cell = −2 × 96 500 × 2.23 = −430 kJ mol⁻¹. (HL only)
Electrolysis of concentrated vs dilute NaCl(aq) with inert electrodes: Both: hydrogen at the cathode. Concentrated: chlorine at the anode (NaOH forms). Dilute: mainly oxygen at the anode. (HL only)
Electroplating: what is the anode, the cathode and the electrolyte?: Cathode: the object to be plated (Mⁿ⁺ + ne⁻ → M). Anode: the plating metal (M → Mⁿ⁺ + ne⁻). Electrolyte: a solution of the plating metal’s ions. (HL only)
Radical, and homolytic fission: Radical: a species with an unpaired electron (Cl•, CH₃•). Homolytic fission: a bond splits evenly, one electron to each atom, giving two radicals (X₂ → 2X• with UV light or heat).
How is homolytic fission shown with curly arrows?: Two single-barbed (fish-hook) arrows, each from the centre of the bond to one atom. Each shows one electron moving.
Why do alkanes react only by radical substitution?: C–C and C–H bonds are strong (high Ea) and essentially non-polar, so there is no δ+ or δ− site for nucleophiles or electrophiles.
Mechanism for an alkane RH with halogen X₂: Initiation: X₂ → 2X• (UV). Propagation: X• + RH → R• + HX; R• + X₂ → RX + X•. Termination: X• + X• → X₂; R• + X• → RX; R• + R• → R–R.
Why does radical substitution give a mixture of products?: Further substitution (RX still has C–H bonds), hydrogens in different environments can all be replaced, and termination R• + R• gives a longer alkane.
How many monochloro structural isomers does heptane give?: Four: heptane has four hydrogen environments (C1/C7, C2/C6, C3/C5, C4), giving 1-, 2-, 3- and 4-chloroheptane.
Why does an alkane not decolourize bromine water in the dark?: Without UV light (or heat) Br₂ does not undergo homolytic fission, so no radicals form, and the non-polar alkane offers nothing for bromine to attack.
Nucleophile: A species that supplies both electrons of a new covalent bond. It can be neutral (H₂O, NH₃) or negative (OH⁻, CN⁻).
Electrophile: A species that takes in both electrons of a new covalent bond. It can be neutral (HBr, Br₂) or positive (H⁺, NO₂⁺).
Heterolytic fission: Bond breaking in which one fragment keeps the whole bonding pair, giving an anion and a cation; one double-barbed arrow. Example: (CH₃)₂CHBr → (CH₃)₂CH⁺ + Br⁻.
1-bromobutane + CN⁻: product: CH₃CH₂CH₂CH₂CN + Br⁻: pentanenitrile, because the CN carbon counts in the chain. Nucleophilic substitution; Br⁻ is the leaving group.
Electrophilic addition to alkenes: reagents and products: X₂ → dihalogenoalkane. HX → halogenoalkane. Steam with an acid catalyst → alcohol. Example: but-2-ene + Br₂ → 2,3-dibromobutane.
Lewis acid and Lewis base: Lewis acid: electron-pair acceptor (an electrophile). Lewis base: electron-pair donor (a nucleophile). In SnCl₄ + 2Cl⁻ → SnCl₆²⁻, Cl⁻ is the base. (HL only)
SN1 vs SN2: SN2 (primary): one step, rate = k[RX][Nu], attack opposite the leaving group with inversion. SN1 (tertiary): two steps via a carbocation, rate = k[RX], not stereospecific. Secondary can react by both. (HL only)
[RX] is halved and [Nu] tripled. How does the rate change for SN2 and for SN1?: SN2: × 0.5 × 3 = × 1.5. SN1: × 0.5, because only [RX] is in its rate equation. (HL only)
Why does rate follow RI > RBr > RCl in nucleophilic substitution?: C–I is the longest and weakest bond, and I⁻ is the best leaving group (largest, most stable ion). C–Cl is the most polar but the slowest. (HL only)
Major product of HBr with an unsymmetrical alkene: The product from the more stable carbocation (tertiary > secondary > primary, from the positive inductive effect of alkyl groups). 2-methylhex-1-ene gives mainly 2-bromo-2-methylhexane. (HL only)
Oxidation state of iron in [Fe(H₂O)₅(OH)]²⁺: x + 5(0) + (−1) = +2, so x = +3. Neutral ligands: H₂O, NH₃; anionic ligands: Cl⁻, OH⁻, CN⁻ (−1 each). (HL only)

Flashcards by topic

Written for International Baccalaureate Diploma Programme Chemistry, (First assessment 2025).

Cards by Marlbridge Academic Team; 20 restate definitions from the Chemistry revision notes. This deck was reviewed by Nouman Ahmed on 2026-11-05. Report an error.

All Chemistry notes, practice and topics