Flashcards · YCH11 · AS Topic 2

Unit 2 – Energetics, Group Chemistry, Halogenoalkanes and Alcohols (AS Level) flashcards

44 cards on AS Topic 2, subtopics 6 to 10. Pick a subtopic or study them all.

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6 Energetics

Standard conditions (energetics): 100 kPa and a specified temperature, usually 298 K. Substances in their standard states. Symbol ⦵.
Exothermic vs endothermic (sign of ΔH): Exothermic: ΔH negative; surroundings get hotter. Endothermic: ΔH positive; surroundings get colder.
Standard enthalpy change of formation: The enthalpy change when 1 mol of a compound is formed from its elements, under standard conditions with all substances in their standard states, e.g. Na(s) + ½Cl₂(g) → NaCl(s). It is zero for an element.
Standard enthalpy change of neutralisation: The enthalpy change when an acid and a base react to form 1 mol of water, under standard conditions with all substances in their standard states.
25.0 cm³ of solution warms by 4.4 °C when 0.00500 mol of a metal reacts (c = 4.18 J g⁻¹ °C⁻¹). ΔH?: Energy = 25.0 × 4.18 × 4.4 = 459.8 J = 0.4598 kJ. ΔH = −0.4598 ÷ 0.00500 = −92 kJ mol⁻¹ (negative because the temperature rose).
Hess's law, and ΔrH from formation data: The overall enthalpy change is the same whatever route is taken, if the start and end conditions match. ΔrH = ΣΔfH(products) − ΣΔfH(reactants).
Bond enthalpy vs mean bond enthalpy: One specific bond in one molecule vs an average over many.
Why is a measured enthalpy change of combustion usually less exothermic than the data value?: Heat is lost to the air and the can, combustion is incomplete, fuel evaporates, and conditions are not standard.

7 Intermolecular Forces

London force: An attraction between an instantaneous dipole in one molecule and the dipole it induces in a neighbouring molecule.
Hydrogen bond: An attraction between a δ+ hydrogen atom bonded to N, O or F and a lone pair on an N, O or F atom of another molecule. The X–H···Y arrangement is close to linear.
Why does water boil higher than HF and NH₃?: Water can form the most hydrogen bonds per molecule (2 H and 2 lone pairs). NH₃ has only one lone pair; HF has only one H. It is the number of hydrogen bonds, not the strength of each.
Why is ice less dense than liquid water?: In ice each molecule is hydrogen bonded to four others in a tetrahedral arrangement, an open structure with empty space. On melting some hydrogen bonds break and the molecules move closer.
Why does 2,2-dimethylpropane boil lower than pentane?: Same number of electrons, but the branched molecule is compact with less surface contact, so the London forces are weaker.
Why does boiling temperature rise from HCl to HI?: More electrons, so stronger London forces, which outweigh the falling permanent dipole.
Why do halogenoalkanes not dissolve well in water?: They are polar but cannot hydrogen bond with water, so the energy needed to break water’s hydrogen bonds is not repaid.

8 Redox Chemistry and Groups 1, 2 and 7

Oxidation number rules: the exceptions for hydrogen and oxygen: H is +1, but −1 in metal hydrides (NaH, CaH₂). O is −2, but −1 in peroxides (H₂O₂, Na₂O₂) and positive with F.
Oxidation number of Se in H₂SeO₄: 2(+1) + x + 4(−2) = 0, so x = +6.
Oxidising agent versus oxidised: The oxidising agent is the species that gets reduced.
Disproportionation: Atoms of one element, all in the same species, are both oxidised and reduced. Example: Cl₂ + 2NaOH → NaCl + NaClO + H₂O (Cl 0 → −1 and 0 → +1).
Why does first ionisation energy fall down Groups 1 and 2?: An extra shell each time places the outer electron further out, with more inner-shell shielding; these outweigh the larger nuclear charge, so the attraction is weaker.
Solubility trends down Group 2 for hydroxides and sulfates: Hydroxides become more soluble (Mg(OH)₂ least, Ba(OH)₂ most). Sulfates become less soluble (MgSO₄ most, BaSO₄ least).
Why do Group 2 carbonates become more thermally stable down the group?: The cation gets larger, so it is less polarising: it distorts the carbonate ion’s electron cloud less and weakens the C–O bonds less. Order: MgCO₃ < CaCO₃ < SrCO₃ < BaCO₃.
What does heating sodium nitrate give, and what does a Group 2 nitrate give?: NaNO₃ (and KNO₃): the nitrite plus O₂ only. Group 2 nitrates (and LiNO₃): the oxide + NO₂ + O₂.
Chlorine with cold, dilute NaOH vs hot NaOH: Cold: Cl₂ + 2NaOH → NaCl + NaClO + H₂O (chlorate(I), +1). Hot: 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O (chlorate(V), +5).
Solid NaCl, NaBr and NaI with concentrated sulfuric acid: what forms?: NaCl: HCl only (steamy fumes); not redox. NaBr: HBr, then Br₂ + SO₂ (S +6 → +4). NaI: HI, then I₂ with SO₂, S and H₂S. Reducing power HCl < HBr < HI.
Silver halide precipitates and their behaviour with ammonia: AgCl white, dissolves in dilute NH₃. AgBr cream, dissolves in concentrated NH₃ only. AgI yellow, does not dissolve.
Colour of iodine in water and in cyclohexane: Brown in water; violet in cyclohexane (a non-polar solvent). Bromine is orange to red-brown in cyclohexane.

9 Introduction to Kinetics and Equilibria

Activation energy: The minimum energy colliding particles need for reaction to occur.
Maxwell-Boltzmann distribution: shape, and the effect of a higher temperature: Starts at zero, one asymmetric peak, long tail that never touches the energy axis; area = number of molecules. Higher T: peak lower and further right, same area, much more area beyond Ea.
Main reason a higher temperature increases the rate: A greater proportion of collisions have energy ≥ Ea, so successful collisions are more frequent. Collision frequency rises only slightly.
How does a catalyst appear on a Maxwell-Boltzmann diagram and on a reaction profile?: Maxwell-Boltzmann: the curve is unchanged; only the Ea mark moves left. Profile: a lower route, usually two humps with an intermediate in the dip; ΔH unchanged.
Dynamic equilibrium: the two statements needed: Forward and backward rates are equal, and the concentrations of reactants and products do not change. It needs a closed system.
Effect of a catalyst on the position of equilibrium: None: it speeds up the forward and backward reactions equally, so equilibrium is reached sooner but the yield is unchanged.
Run A takes 50 s and run B takes 20 s. How many times faster is B?: Rate ∝ 1/t: (1/20) ÷ (1/50) = 2.5 times faster.

10 Organic Chemistry: Halogenoalkanes, Alcohols and Spectra

Nucleophile: Donates a lone pair of electrons to form a new covalent bond.
Halogenoalkane with KOH: aqueous vs ethanolic: KOH(aq), heat under reflux: OH⁻ acts as a nucleophile; substitution gives an alcohol. KOH in ethanol, heat under reflux: OH⁻ acts as a base; elimination gives an alkene.
Halogenoalkane + KCN in ethanol: product: A nitrile with one more carbon atom (bromoethane → propanenitrile, CH₃CH₂CN). Heat under reflux.
Why does hydrolysis rate follow R–I > R–Br > R–Cl?: Bond enthalpy, not polarity: C–I is the weakest bond, so it breaks most easily. C–Cl is the most polar but reacts slowest.
Oxidising a primary alcohol to an aldehyde vs to a carboxylic acid: Both: K₂Cr₂O₇ / dilute H₂SO₄, orange → green. Aldehyde: distil off as it forms. Acid: excess oxidising agent, heat under reflux. A tertiary alcohol does not react (stays orange).
Test for an –OH group with PCl₅: Steamy fumes of HCl, using a dry sample: ROH + PCl₅ → RCl + POCl₃ + HCl. Positive for alcohols, carboxylic acids and water.
Why is 50% (not fully concentrated) sulfuric acid used with KBr to brominate an alcohol?: Fully concentrated sulfuric acid oxidises the HBr to bromine, wasting the HBr.
Molecular ion peak: Found at the largest m/z value, if you ignore the tiny M+1 peak from carbon-13. Its m/z gives Mr.
Mass spectrum M⁺ patterns for one Cl and one Br: One Cl: two peaks 2 apart, heights about 3 : 1. One Br: two peaks 2 apart, heights about 1 : 1.
IR: how do the O–H absorptions of an alcohol and a carboxylic acid differ?: Alcohol: broad, about 3550–3230 cm⁻¹. Acid: very broad, about 3300–2500 cm⁻¹, always with a strong C=O near 1700 cm⁻¹. Hydrogen bonding makes O–H bands broad.