Flashcards · YCH11 · A Level Topic 4

Unit 4 – Rates, Equilibria and Further Organic Chemistry (A Level) flashcards

48 cards on A Level Topic 4, subtopics 11 to 15. Pick a subtopic or study them all.

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11 Kinetics

Rate equation and order of reaction: rate = k[A]ᵐ[B]ⁿ, with orders found by experiment (usually 0, 1 or 2). Order with respect to A is m; overall order is m + n.
Units of k for zero, first and second order overall: Zero: mol dm⁻³ s⁻¹. First: s⁻¹. Second: mol⁻¹ dm³ s⁻¹.
rate = k[A][B]; rate 3.0 × 10⁻⁴ mol dm⁻³ s⁻¹, [A] = 0.20, [B] = 0.050 mol dm⁻³. k?: k = 3.0 × 10⁻⁴ ÷ (0.20 × 0.050) = 0.030 mol⁻¹ dm³ s⁻¹
Concentration-time and rate-concentration graphs for zero and first order: Zero order: straight line with negative gradient; rate-concentration horizontal. First order: curve with a constant half-life; rate-concentration a straight line through the origin.
SN1 vs SN2 for halogenoalkane hydrolysis: SN1 (tertiary): rate = k[RX]; slow ionisation to a carbocation, then fast attack by OH⁻. SN2 (primary): rate = k[RX][OH⁻]; one step via a transition state, OH⁻ attacking opposite the halogen.
Intermediate vs catalyst in a mechanism: Intermediate: made in one step, used in a later one. Catalyst: used in one step, regenerated later.
Finding Ea from an Arrhenius plot: ln k = −(Ea/R)(1/T) + ln A. Plot ln k against 1/T (T in K); gradient = −Ea/R, so Ea = −gradient × 8.31 J mol⁻¹.
Iodination of propanone (Core Practical 9a): what does zero order in iodine tell you?: Iodine is not in the rate-determining step; it reacts after it. rate = k[CH₃COCH₃][H⁺]; H⁺ is a catalyst (regenerated).
How does a solid (heterogeneous) catalyst work?: Reactant gases adsorb weakly onto active sites; their bonds weaken and they meet in a good orientation, so Ea falls. Products desorb and free the sites.

12 Entropy and Energetics

Entropy, and the order of entropy for the three states: A measure of disorder: the random dispersal of molecules and of energy quanta between them. S(solid) < S(liquid) < S(gas); a perfect crystal at 0 K has zero entropy.
ΔSsurroundings and ΔStotal: ΔSsurroundings = −ΔH ÷ T (ΔH in J mol⁻¹, T in K). ΔStotal = ΔSsystem + ΔSsurroundings. A change is feasible when ΔStotal > 0.
ΔH = +52 kJ mol⁻¹, ΔSsystem = +130 J mol⁻¹ K⁻¹. When is the reaction feasible?: Changeover T = ΔH ÷ ΔSsystem = 52 000 ÷ 130 = 400 K. Both positive, so feasible above 400 K.
Thermodynamic vs kinetic stability: Thermodynamically stable: ΔStotal for the change is negative, so it is not feasible. Kinetically stable: feasible, but the activation energy is so high the rate is negligible (petrol in air).
Lattice energy (as defined in this specification): The enthalpy change when gaseous ions come together to form 1 mol of ionic solid, e.g. Ca²⁺(g) + 2I⁻(g) → CaI₂(s). Always negative.
First electron affinity, and the sign of the second: First: the enthalpy change when 1 mol of gaseous atoms each gain one electron to form 1 mol of gaseous 1− ions; usually negative. The second electron affinity is always positive.
Born-Haber: ΔfH(LiCl) −409; ΔatH(Li) +159; IE(Li) +520; ΔatH(Cl) +122; EA(Cl) −349 kJ mol⁻¹. Lattice energy?: Steps to gaseous ions: 159 + 520 + 122 − 349 = +452. Lattice energy = −409 − 452 = −861 kJ mol⁻¹.
Enthalpy change of solution from lattice and hydration enthalpies: ΔsolH = −(lattice energy) + Σ ΔhydH(ions). For CaI₂: 2074 + (−1616) + 2(−297) = −136 kJ mol⁻¹.
What does an experimental lattice energy much more exothermic than the theoretical value show?: Partial covalent bonding: the cation polarises the anion, drawing electron density between the nuclei. Biggest for a small, highly charged cation with a large anion.

13 Chemical Equilibria

Homogeneous vs heterogeneous equilibrium: Homogeneous: every species in one phase. Heterogeneous: species in more than one phase. Solids never appear in K expressions.
Kc for N₂(g) + 3H₂(g) ⇌ 2NH₃(g), with units: Kc = [NH₃]² ÷ ([N₂][H₂]³). Units dm⁶ mol⁻².
Mole fraction and partial pressure: x(A) = moles of A ÷ total moles of gas. p(A) = x(A) × total pressure. The partial pressures add up to the total pressure.
C₂H₄ + H₂ ⇌ C₂H₆: p(C₂H₄) 0.40, p(H₂) 0.25, p(C₂H₆) 1.10 atm. Kp?: Kp = 1.10 ÷ (0.40 × 0.25) = 11.0 atm⁻¹
Which changes alter the value of K?: Only temperature. Concentration and pressure move the position of equilibrium; a catalyst changes neither.
Effect of raising the temperature on K for an exothermic forward reaction: K gets smaller, and the equilibrium moves left so the composition matches the new K. (K changes first.)
Link between ΔStotal and K: ΔStotal = R ln K (R = 8.31 J K⁻¹ mol⁻¹). A large positive ΔStotal means a very large K.

14 Acid-base Equilibria

Brønsted–Lowry acid and base, and a conjugate pair: Acid: proton (H⁺) donor. Base: proton acceptor. A conjugate pair is two species that differ by one H⁺, e.g. HSO₄⁻/SO₄²⁻.
Kw vs pKw: Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ and pKw = 14.00, both at 298 K only.
Ka vs pKa: Larger Ka means stronger acid; smaller pKa means stronger acid.
pH of 0.110 mol dm⁻³ weak acid with pKa 4.13: Ka = 10⁻⁴·¹³ = 7.41 × 10⁻⁵. [H⁺] = √(7.41 × 10⁻⁵ × 0.110) = 2.86 × 10⁻³. pH = 2.54
pH of 0.0215 mol dm⁻³ KOH at 298 K: [H⁺] = Kw ÷ [OH⁻] = 1.00 × 10⁻¹⁴ ÷ 0.0215 = 4.65 × 10⁻¹³. pH = 12.33
Buffer solution: A solution whose pH changes very little when a small amount of acid or alkali is added to it.
How does an acidic buffer (HA and A⁻) resist added acid and added alkali?: Added H⁺: H⁺ + A⁻ → HA. Added OH⁻: OH⁻ + HA → A⁻ + H₂O. HA and A⁻ are present in large amounts, so the ratio [HA]/[A⁻] and so [H⁺] change only slightly.
Half-neutralisation point: Half the equivalence volume; here [HA] = [A⁻] and pH = pKa.
Choosing an indicator for weak acid–strong base and strong acid–weak base titrations: Its whole colour-change range must lie in the steep section. Weak acid + strong base: phenolphthalein (about 8.2–10.0). Strong acid + weak base: methyl orange (about 3.1–4.4). Weak + weak: no sharp change; use a pH meter.
How is blood kept close to pH 7.4?: Mainly the H₂CO₃ ⇌ H⁺ + HCO₃⁻ buffer. Extra acid: H⁺ + HCO₃⁻ → H₂CO₃, removed as CO₂ through the lungs.

15 Organic Chemistry: Carbonyls, Carboxylic Acids and Chirality

Chiral centre, and optical isomers: A chiral centre is a carbon atom carrying four different atoms or groups. Optical isomers (enantiomers) are non-superimposable mirror images.
Optical activity, and a racemic mixture: A single optical isomer rotates the plane of polarisation of plane-polarised monochromatic light; enantiomers rotate it by equal angles in opposite directions. A racemic mixture (equal amounts of both) shows no net rotation.
How optical activity distinguishes SN1 from SN2: SN1: planar carbocation attacked equally from either face → racemic product, zero rotation. SN2: one-sided attack with inversion → a single enantiomer, non-zero rotation.
Tests that distinguish an aldehyde from a ketone: Warm with Tollens’ reagent: aldehyde gives a silver mirror. Warm with Fehling’s or Benedict’s: aldehyde gives a red precipitate (Cu₂O); a ketone stays blue. 2,4-DNPH gives a yellow-orange precipitate with both.
Adding HCN to a carbonyl: mechanism type, why KCN is added, and the stereochemistry: Nucleophilic addition. HCN is a weak acid, so KCN supplies enough CN⁻. CN⁻ attacks the planar C=O equally from both sides, so a chiral product forms as a racemic mixture.
Reducing aldehydes, ketones and carboxylic acids: LiAlH₄ in dry ether. Aldehyde → primary alcohol (2[H]); ketone → secondary alcohol (2[H]); carboxylic acid → primary alcohol (4[H]).
Iodoform test: reagents, result and what it shows: Iodine with NaOH(aq), warm: pale yellow precipitate of CHI₃. Shows a CH₃CO– group (ethanal and methyl ketones), or CH₃CH(OH)–.
Acyl chloride reactions with water, alcohols and ammonia: Water → carboxylic acid + HCl. Alcohol → ester + HCl. Concentrated ammonia (2 mol) → amide + NH₄Cl. Fast at room temperature and not reversible.
Ester hydrolysis: acidic vs alkaline: Both heated under reflux. Acidic (dilute HCl or H₂SO₄): carboxylic acid + alcohol; reversible. Alkaline (NaOH(aq)): carboxylate salt + alcohol; goes to completion.
Terylene: Ethane-1,2-diol + benzene-1,4-dicarboxylic acid. Repeat unit –OC–C₆H₄–COO–CH₂CH₂–O–.
¹³C NMR: what does the number of peaks tell you?: The number of different carbon environments, not the number of carbons. Symmetry reduces the count.
¹H NMR: what do peak area and splitting tell you?: Relative area: the number of H atoms in that environment. Splitting (n + 1 rule): n non-equivalent H atoms on adjacent carbons give n + 1 lines. Equivalent H do not split each other; O–H is usually a singlet.
HPLC vs GC: Both are column chromatography that separate by retention time. HPLC: liquid mobile phase under high pressure; solid particles packed in the column. GC: unreactive carrier gas; liquid stationary phase on a solid support. Linked to MS for identification.