2016 C1 Promotional Exam Answers

9 2016 HCI C1 H2 Chemistry Promotional Exam / Paper 1 Answers 1 2 3 4 5 6 7 8 C C B B A B B C 9 10 11 12 13 14 15 16 A B B A A C D C 17 18 19 20 D D D C

13 2016 HCI C1 H2 Chemistry Promotional Exam / Paper 2 Answers: 1 (a) (i) Water: Hydrogen Bonding CCl4: Dispersion Forces [½ each] (ii) Water Tetrachloromethane Hydrocarbon Chain Permanent dipole- induced dipole interactions / dispersion forces Dispersion forces Carboxylic acid functional group Hydrogen bonds Permanent dipole- induced dipole interactions / dispersion forces [½ each] (ii) The energy released in forming new interactions between water and caprylic acid is insufficient to compensate for the energy required to break the existing interactions. However, the energy released is sufficient to compensate the energy required when in CCl4. [1] (b) (i) When Br2 approaches the alkene, the electron cloud on Br2 is distorted to form a temporary dipole moment that causes it to be electron deficient. [1] ICl is a polar molecule, the more electronegative Cl causes I to be electron deficient. [1] (ii) The reaction is kept in the dark to prevent the possibility of free radical substitution. [1] (iii) Mechanism for electrophilic addition. Award 1 mark per step. Deduct ½ per mistake (e.g. no charge separation, missing/ wrongly drawn arrows, missing slow step, wrong intermediate) (c) The electrons in the benzene ring are delocalized, hence there is resonance stabilization. If benzene undergoes addition, the stabilization due to resonance is lost and it destabilizes the compound. [1] (d) Step 1: CH3Cl, AlCl3 or FeCl3, heat Step 2: KMnO4. Dilute H2SO4, heat [1 mark per step]

14 2016 HCI C1 H2 Chemistry Promotional Exam / Paper 2 2 (a) (i) [1] (ii) [1] (iii) CnH2nOn [1] (iv) C12H22O11 [1] (b) (i) Circles laevorotatory. The final glucose/fructose mixture is laevorotatory (Afinal is negative). As glucose is dextrorotatory, fructose must be laevorotatory for this to be possible. [1] (half mark for answers which only point out that the observed rotation is becoming less and less positive) (ii) H+ is a catalyst and overall remains unchanged and not used up (or is regenerated), hence its concentration is constant. [1] (iii) Finding half life = 82 min [1 – subtract 0.5 if no construction lines] As the half-life of sucrose is constant at 82 min, the order with respect to sucrose is one (x=1). [1] (iv) 1 2 1ln2 ln2 0.00845min82effk t  [1] for working+value, [1] for units (v) Initial rate = 0.00845 × 0.580 ÷ 2 = 0.00245 mol dm−3 min−1 [1] (vi) The temperature of the solutions could be kept lower than 30 °C. [1] The concentration of HCl used could be lower than 1 mol dm−3. [1] * *

15 2016 HCI C1 H2 Chemistry Promotional Exam / Paper 2 3 (a) (i) There are no/negligible intermolecular forces of attraction between the gas particles. OR The volume occupied by the gas particles is negligible compared to the volume of the container/ the gas molecules have zero volume. OR Collisions between the gas particles are elastic / no loss of kinetic energy on collision. Any of the above two points [2] (ii) As temperature increases, the gas molecules possess more kinetic energy [1] and collide more frequently OR more forcefully with the walls of the container. [1] (iii) PV = (m/Mr)RT Choose correct point on graph (any point from B to C) [1] Using point B, m = (0.8 x 101325 x 1.25 x 10–3 x 32.0) / (8.31 x 90) = 4.34 g [1] (iv) 3O2 (g)  2O3 (g) before reaction / atm 1.2 0 after reaction / atm 1.2 – 3x 2x total pressure = 1.2 – 3x + 2x = 1.2 – x = 1.14  x = 0.06 [1] partial pressure of oxygen = 1.2 – 0.06 x 3 = 1.02 atm [1] (b) (i) Given 3/2 O2 (g)  O3 (g) H = +143 kJ mol–1 So standard enthalpy change of Reaction 1 = +143 x 2 = +286 kJ mol–1 [1] Entropy change for Reaction 1 is negative because the number of moles of gas molecules decreases after the reaction, so there are fewer ways of distributing the molecules and their energies. [1] Since HO is positive and SO is negative, and GO = HO – TSO (some link between first 2 points and GO), GO is positive at all temperatures, and hence it is non- spontaneous at all temperatures under standard pressure. [1] (ii) The lightning provides an external input of energy to drive the non-spontaneous reaction. [1] (iii) Since GO is positive, K < 1, so the equilibrium position will lie more on the left OR the equilibrium is reactant-favoured OR the extent of reaction at equilibrium is less than half complete [1] (i) Reactions 1 and 4 may be slower because they involve breaking of covalent bonds and higher Ea is likely involved. [1] (ii) Cl + O3  ClO + O2 ClO + O  Cl + O2 [1] elementary steps add up to give overall eqn for reaction (3) [1] Cl is regenerated, ClO is produced in one step and used up in the next

2016 HCI C1 H2 Chemistry Promotional Exam / Paper 3 (Answers) 1 2016 HCI C1 H2 Promotional Examination Answers (Paper 3) 1(a) (i) [½] for each correct structure (ignore repeats) [Total: 2] (ii) would be formed via the most stable carbon radical intermediate. The tertiary radical formed is stabilised through the electron-donating effect of three alkyl groups (vs. two or one for the other radicals). [1] (iii) Name: Free Radical Substitution Initiation Propagation Termination [1] name of mechanism [2] mechanism (negative marking): deduct ½ for each type of mistake (e.g. forgot initiation/propagation/termination, unpaired electron not drawn correctly/missing, omitted steps, etc.) Propose not to penalise if they select incorrect product, since mark for (ii) is already lost, and this qn is to test FRS mechanism. [Total:3]

2016 HCI C1 H2 Chemistry Promotional Exam / Paper 3 (Answers) 2 (b) Test Observations A B Add a little acidified potassium manganate(VII) to a sample of each bottle, and place in a heated water bath. Purple KMnO4 decolourises, effervescence seen Purple KMnO4 decolourises, no effervescence seen [1] correct description of a chemical test that would successfully distinguish A and B [1] correct observations for each compound [Total: 2] (c) (i) ηH+ reacted with the sample = ηH+ total – ηH+ left unreacted ηH+ reacted with the sample = 35.00 1000 × 2.00 − 19.65 1000 × 1.00 = 5.035 × 10−2 ηH+ reacted with the sample = 5.04 × 10−2 mol (3 s.f.) [1] (ii) ηCO2 produced = 16.40 22700 = 7.225 × 10−4 mol Y2(CO3)3•3H2O + 6HCl → 2YCl3 + 3CO2 + 6H2O ---------------- (equation not necessary) So, ηH+ reacted with the carbonate = 7.225 × 10−4 × 6 3 = 1.4449 × 10−3 mol So, ηH+ reacted with the carbonate = 2.408 × 10−4 × 6 3 = 1.45 × 10−3 mol (3 s.f.) [1] (iii) Y H O mass in 100 g /g 63.5 2.14 34.36 amount / mol 63.5 88.9 = 0.714 2.14 1.0 = 2.14 34.36 16.0 = 2.148 mole ratio 1 3 3 Therefore, Q is yttrium(III) hydroxide, Y(OH)3 Y2O3 + 3H2O → 2Y(OH)3 [1] for correct formula of Q from calculations [1] for correct equation [Total: 2]

2016 HCI C1 H2 Chemistry Promotional Exam / Paper 3 (Answers) 3 (iv) Y(OH3) + 3HCl → YCl3 + 3H2O Whether from the hydroxide or oxide, each mole of “Y” requires 3 times the no. of moles of HCl to react completely. (Hence it doesn’t matter what proportion of the oxide has converted to hydroxide ⇒ reacting ratio with HCl is still the same!) One way of calculating (some steps can be skipped/combined): ηH+ reacted with the yttrium oxide and hydroxide = 5.035 × 10−2 − 1.445 × 10−3 ηH+ reacted with the yttrium oxide and hydroxide = 4.891 × 10−2 mol ∴ ηY2O3 in original sample = 1 6 ×