SNGS 2024 Chem Prelim P2 Answers

2024 Prelim Chem Paper 2 1(a) SO2 or NO2 Poly(ethene) is a macromolecule whereby the boiling point will be higher than those with simple covalent structure. SO2 or NO2 has simple covalent structure with molecular mass greater than 32 so their boiling point will be higher than oxygen due to the stronger intermolecular forces of attraction. 1 1 1 (b) NH3 1 (c) ZnO 1 (d) CO /SO2 1 (e) NH3 1 (f) K2O 1 2(a) H+ any suitable indicator eg moist blue litmus turns red OR Add magnesium , lighted splint extinguished with a pop sound. OR Add carbonate, white precipitate formed in limewater. 1 1 (b) Potassium hydroxide/potassium carbonate and phosphoric acid Titration 1 1 (c) Add acidified aqueous silver nitrate White ppt observed Aqueous lead(II) nitrate max 1m 1 1 (d) Mole of Na+ = (156/1000)/23 OR Mole of Na+ = (48/1000)/23 = 0.00678 mol = 0.00209 mol AND Mole of Cl- = (127/1000)/35.5 OR Mole of Cl- = (39/1000)/35.5 = 0.00358 mol = 0.00110 mol For every mole of NaCl, no of mole of Na+ = No of mol of Cl-. Since there are more mole of Na+ than Cl-, there are other compounds of sodium. 1 1 (e) Energy Shape of graph [1] Label axes with units and include correct reactant and product [1] Enthalpy change and activation energy [1] Level diagram 2m max Endo 1m max 3

(f) (i)yeast, 370C, darkness, no air 0.5 ea (ii) C2H5OH + 3O2 → 2CO2 + 3H2O and C6H12O6 → 2C2H5OH + 2CO2 6CO2 + 6H2O → C6H12O6 + 6O2 not balance -1m 1 1 3(a) Stage IV 1 (b) No, a catalyst provides an alternative pathway with lower activation energy but the enthalpy change remains unchanged as the energy of reactant and product remain the same. 1 1 (c) Stage I: S is oxidised as its oxidation state increases from 0 to +4 in SO2 and O2 is reduced as its oxidation state decreases from 0 to -2 in SO2 OR Stage II: SO2 is oxidised as the oxidation state of S increases from +4 in SO2 to +6 in SO3 and O2 is reduced as its oxidation state decreases from 0 in O2 to -2 in SO3 1 1 (d) Stage II is a reversible reaction. *Some of the sulfur trioxide decompose to sulfur dioxide *at the same time/immediately 1 (e) 2S+ 3O2 + 2H2O → 2H2SO4 Not balanced : 0m 1 (f) (i) NO2 Nitrogen and oxygen in air react in the high temperature at the combustion engine OR lightning NO/oxides of nitrogen [0] 1 1 (ii) Zn + 2H+ → Zn2+ + H2 OR CaCO3 + 2H+ → Ca2+ + CO2 + H2O 1 (iii) CaCO3 1 (iv) It causes respiratory problems 1 4(a) Cl H H H H I I I I I Cl – C – C – C – C – C – O – H I I I I I H H H H H *OH must be at the first C Change the position of Cl 1 (b) Steam High temperature, high pressure, phosphoric (V) acid as catalyst Chlorine under UV light 1 1

(c) Aqueous bromine: pent-1-ene decolourises the reddish-brown bromine spontaneously whereas 1,2-dichloropentanol has no visible change. OR warm with acidified potassium manganate(VII) – 1,2-dichloropentanol will decolourise the purple KMnO4 but pent-1-ene shows no visible change Reagent – 1 mark Observation – 1 mark ea 3 5(a) Statement 1: AgCl /PbCl2 are insoluble in water. Statement 3: Chlorine is a stronger oxidising agent than bromine cos it can gain electrons more readily than bromine. 1,1 1,1 (b) (i) 2Fe + 3Cl2 → 2FeCl3 not balanced [0m] 1 (ii) Cl2 gain electrons to become chloride ion 1 (iii) addition of aqueous sodium hydroxide/aqueous ammonia , a green ppt is form Wrong reagent 0m 1 (iv) addition of acidified potassium manganate (VII) Iron (II) chloride will decolourise the KMnO4 but not iron(III) chloride. 1 1 6(a) (i) In Fig 6.1, there are two ions/pacticles/peaks which have different m/z values of 6 and 7[1] This shows that these two ions have different number of neutrons.[1] 2 relative abundance 0m 2 (ii) Ar of Lithium = 3.75 100x6 + 96.25 100 x 7 = 6.96 1 (b) There are3 possible combinations as follows: 2 atoms of chlorine -35 [1] 1 atoms of chlorine –Cl-35 and Cl-37[1] 2 atoms of chlorine -37 [1] 3 (c) (i) Methane has a relative molecular mass of 16 The largest m/z value is 16 [1] 1 (ii) CH+ 1 (iii) CH3Cl+ [1] m/z value is 52[1] 2 7(a) No visible change.[1] Hydroethanoic does not dissociate/ionize in methylbenzene to form H+[1] 2 (b) the stronger the acid as the higher tendency for acid to produce H+ ions and the lower the pH value. [1] Hence with more mobile ions, conductivity will increase.[1] 2 (c) Hydroxypentanoic produces less effervescence per unit time during the reaction / rate of effervescence is slower. [1] Less H+ per unit volume / lower concentration of H+ [1] Frequency of collision between reactant particles is lower 3

Frequency of effective collision is lower [1] Speed of reaction is lower (d) (i) 1 (ii) Economical It can be more expensive than disposal in landfills or incineration as it involves transporting the waste to the processing plant; sorting and cleaning of the waste; carrying out physical or chemical processes. 1 (iii) cracking 1 8(a) (i) Zn→ Zn2+ +2e 1 (ii) Cu2++2e → Cu 1 (b) (i) Electrode Y is the cathode./negative electrode[1]Silver ions will be selectively discharged/reduces to form silver.[1] 2 (ii) Agee with James. The same metals are used as electrodes. The voltage will be the same 0.42V[1]. The voltage will be negative as the direction of electron flow is different.[1] Disagee with Megan. Magnesium is more reactive than zinc.[1] The further apart the reactivity of metal, the higher the voltage[1] 4 (c) number of moles of Cu2+ used = 0.5 x 1.0 = 0.5 mol number of moles of Cu deposited = 19.2 ÷ 64 = 0.3 mol calc of mole for both – 1m number of moles of Cu2+ remaining = 0.5 – 0.3 = 0.2 mol concentration of Cu2+ remaining = 0.2mol ÷ 0.5dm3 = 0.4 mol/dm3 [1] 2 9(a) C, A, B, D [1] The more reactive the metal, the higher the thermal stability of the nitrates, [1] thus the higher the temperature /more energy needed to decompose its nitrate.[1] 3 (b) - Add metal E to aqueous nitrate of D ( or any metal the student identify as the most reactive ); [1] - If solid metal D is seen to be deposited/coated on E/ E gets smaller;[1] - E displaces D, therefore the most reactive. [1] 3 (c) - E is Mg/Al [0.5] extract by electrolysis [0.5] of its molten oxide [1]/ - E is Zn[0.5], extract by heating [0.5] the oxide with C or CO [1]/ - E is Fe[0.5], extract by heating ]0.5] the oxide with C/CO/H2 [1] 2 (d) Mole of Mg(NO3)2 = 200000/[24 +2(14+16x3)] = 1351 mol Mole of NO2 = 2702 mol [1] Mass of NO2 = 2702 x 46 x 0.9= 111863 g[1] 2