SAJC 2022 P2 ANS

1 [TURN OVER ST ANDREW’S JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS HIGHER 2 CANDIDATE NAME CLASS 2 1 S CHEMISTRY Paper 2 Structured Questions Candidates answer on the Question Paper. Additional Materials: Data Booklet 9729/02 31 August 2022 2 hours READ THESE INSTRUCTIONS FIRST Write your name and class on all the work that you hand in. Write in dark blue or black pen. You may use a HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions in the spaces provided on the Question Paper. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of XX printed pages (including this cover page). For Examiner’s Use Q1 12 Q2 14 Q3 14 Q4 10 Q5 25 Total 75

2 [TURN OVER 1 Hydrazine, N2H4, is a colourless liquid with an ammonia-like odour. It is an important precursor in the pharmaceuticals industry. (a) Hydrazine exists as a liquid while ammonia exists as a gas at room temperature and pressure. State two reasons to explain this difference in physical state. [2] Both ammonia and hydrazine are polar simple covalent molecules, capable of forming intermolecular hydrogen bonds. Hydrazine has more hydrogen bonding sites and therefore, it is able to form more extensive intermolecular hydrogen bonds. At the same time, it has more electrons / larger electron cloud size and this leads to stronger intermolecular instantaneous dipole -induced dipole interactions . Hence, both of these require more energy to overcome. (b) The Kb values of hydrazine, ethylamine, and phenylamine are shown in Table 1.1. Table 1.1 base Kb / mol dm–3 Hydrazine 1.7 x 10–6 (for Kb1) Ethylamine 4.5 x 10–4 Phenylamine 7.4 x 10–10 (i) Explain what is meant by the term Bronsted-Lowry base. [1] A Bronsted-Lowry base refers to a proton acceptor. (ii) Explain the relative magnitudes of the Kb values in Table 1.1. [2] (most basic) ethylamine > hydrazine > phenylamine Unlike hydrazine, ethylamine has an electron donating alkyl/ethyl group which makes the lone pair of electrons on N more available to accept H+. Phenylamine is weaker base than hydrazine as the lone pair of electrons on N is delocalised into the benzene ring , making it less available to accept H+. (iii) The Kb values of diethylamine and triethylamine are shown in Table 1.2.

3 [TURN OVER Table 1.2 base Kb / mol dm–3 Diethylamine 6.9 x 10–4 Triethylamine 6.5 x 10–5 Suggest why the Kb value of triethylamine is significantly smaller than the Kb values of ethylamine and diethylamine. [1] Triethylamine is a tertiary amine and the presence of (one) more R/alkyl groups will result in steric hindrance, thus leading to a smaller extent of base dissociation in aqueous / less likely to accept a H+. (c) The Wolff-Kishner reaction is a valuable synthetic method to convert carbonyl compounds into alkanes. This is done by reacting a carbonyl compound with excess hydrazine in the presence of potassium hydroxide. O RR + RR + +N2 H2ON2H4 KOH (aq), heat (i) Suggest a simple chemical test to monitor the completion of the Wolff -Kishner reaction. [2] Add 2,4-dinitrophenylhydrazine / 2,4-DNPH to the reaction mixture (and warm) If the reaction was complete, there will be no orange ppt. (ii) Propan-1-ol can be synthesised from propene by the following 3-step route that incorporates the Wolff-Kishner reaction.

4 [TURN OVER CH3CH=CH2 CH3CH2CH2OH step 1 step 2 N2H4, KOH (aq) heat Suggest the structures of intermediate products A and B and state the reagents and conditions for each step. Reagents and conditions Step 1: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Step 2: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . [4] A: Br OH B: Br O Reagents and conditions Step 1: Br2(aq) Step 2: K2Cr2O7(aq) or KMnO4(aq), H2SO4(aq), heat [Total: 12] A B (C3H5OBr)

5 [TURN OVER 2 Organic matter is known to decay under low oxygen conditions, such as in swamps. The sulfate- reducing bacteria present in the organic matter will reduce the various sulfates into hydrogen sulfide, H 2S. Some of the hydrogen sulfide will react with Fe 2+ present in swamp to produce insoluble FeS, which is responsible for the brown colour of sludge in the swamp. reaction 1 H2S(g) + aq ⇌ 2H+(aq) + S2–(aq) reaction 2 Fe2+(aq) + S2–(aq) ⇌ FeS(s) Hꝋppt (a) (i) In a saturated solution of hydrogen sulfide, [H +]2[S2–] is 1.0 × 10–23 mol3 dm–9. Calculate the maximum concentration of sulfide ions present in the swamp, given that the pH of swamp water is 6.8. [1] [H+] = 10–6.8 = 1.5848 X 10-7 mol dm–3 [S2–] = 1 × 10-23 [1.5848 × 10-7]2 = 3.98 × 10–10 mol dm–3 (ii) Hence, calculate the minimum concentration of Fe 2+ in the swamp required for the precipitation of FeS. (Ksp of FeS = 4.9 × 10–18 mol2 dm–6) [1] Ecf from (a)(i) Ksp = [Fe2+][S2–] [Fe2+] = 4.9 × 10-18 3.981 × 10-10 = 1.23  10–8 mol dm–3 (iii) Gꝋppt, can be determined by using the following expression, where R is the molar gas constant and T is the temperature measured in K. Gꝋppt = 2.303RT lg Ksp Using the Ksp in (a)(ii), calculate Gꝋppt for the precipitation of FeS. Express your answer in kJ molꟷ1. [2] Gꝋppt = 2.303 × 8.31 × 298 × lg(4.9 × 10–18) = – 98719 J mol–1

6 [TURN OVER = – 98.7 kJ mol–1 (3 sf) (iv) Predict how the brown colour intensity of sludge will change when pH decreases. Explain your answer. [2] When pH decreases, [H+] increases. By Le Chatelier’s Principle, position of equilibrium 1 will shift left, reducing [S2ꟷ]. This will in turn cause position of equilibrium 2 to shift left, resulting in less FeS solid a nd hence, the brown colour intensity will decrease. (b) Using data from Table 2 below, together with relevant data from the Data Booklet, draw an energy cycle and calculate Hꝋppt for reaction 2. Table 2 standard enthalpy change of formation of FeS(s) –102 kJ mol–1 standard enthalpy change of atomisation of Fe(s) +415 kJ mol–1 standard enthalpy change of atomisation of S(s) +279 kJ mol–1 sum of first and second electron affinity of sulfur +337 kJ mol–1 standard enthalpy change of hydration of Fe2+(g) –1970 kJ mol–1 standard enthalpy change of hydration of S2–(g) –1372 kJ mol–1 [4]

7 [TURN OVER OR

8 [TURN OVER Applying Hess’ Law, −102 = 415 + 279 + 762 + 1560 + 337−1970 − 1372 + Hꝋppt(FeS) Hꝋppt (FeS) = − 113 kJ mol−1 (3 s.f.) (c) (i) Use your answers in (a)(iii) and (b), calculate the Spptꝋ for the formation of FeS precipitate. [1] ecf from (a)(iii) and (b) ∆Gꝋ = ∆Hꝋ – T∆Sꝋ –98.7 = (−113) − 298∆Sꝋ ∆Sꝋ = – 0.04798 kJ mol–1 K–1 ≈ – 0.0480 kJ mol–1 K–1 (to 3 sf) (ii) Hence, explain the significance of the sign of Spptꝋ in (c)(i). [1] Entropy change is negative because the degree of disorderliness decreases / less disordered owing to a decrease in the number of aqueous particles during precipitation / aqueous speci es are regularly arranged in crystal lattice structure. (d) Although hydrogen sulfide and water molecules have the same shape, they have slightly different bond angles. State and explain which species has a larger bond angle. [2] In H2O, oxygen has a greater electronegativity than sulfur in H2S. Hence, the bond pair of electrons are more strongly attracted to oxygen