NJC Prelim P2_Solutions

1 NJC H2 Chemistry Prelim Paper 2 Suggested Answers 1 (a) Due to high temperature in the car engine, N2 and O2 from the air can react to form NO2. NO2 can be removed from the exhaust gas with the use of a catalytic converter, it can be reduced by CO to form harmless N2. (b) (i) (ii) NO2, being a radical, is very reactive and reacts with other gases in the air and gets destroyed more readily, hence a shorter atmospheric residence time (c) (i) ΔHo = +9.2 – 2(+33.2) = – 57.2 kJ mol‒1 ΔSo = +304 – 2(+240) = – 176 J mol‒1 K‒1 (ii) ΔHo has a negative sign since the dimerization is a bond formation process, hence heat is given out / reaction is exothermic. ΔSo has a negative sign since the dimerization results in fewer gas particles, hence there is a decrease in the disorderliness of the system. (iii) At equilibrium, ΔG = 0 ΔHo – TΔSo = 0 – 57.2 – T(– 176/1000) = 0 T = 325 K (iv) When N2O4 liquefies, ΔG = 0 ΔHvap – TΔSvap = 0 ΔHvap = TΔSvap = 294 x 88 = +25.9 kJ mol‒1 (iv) Using Hess’s law,  =  –  For ΔHrxn = – 57.2 – +25.9 = – 83.1 kJ mol‒1 For ΔSrxn = – 176 – 88 = – 264 J mol‒1 K‒1 2 (a) (i) H2O2 oxidises the I‒, to aqueous I2, so a brown solution / black solid would be obtained. H2O2 + 2I‒ + 2H+ → I2 + 2H2O

2 (ii) EC𝑙2/C𝑙–0 = +1.36 V EH2O2/H2O 0 = +1.77 V Eθcell = +0.41 V > 0 Since Eθcell is positive, reaction is spontaneous, H 2O2 can oxidise chloride to chlorine while it itself is reduced to H2O. Hence, the oxidation of iodide may not be complete. (iii) hexane / cyclohexane The aqueous layer will decrease in brown intensity and the colourless organic layer will turn purple / violet. (iv) Organic solvent is flammable, and can cause a fire to break out. With higher temperature, iodine may sublime and escape. (v) Excess aqueous NH3 should be added. AgCl is soluble in excess NH 3(aq) but Ag I is not. If there was significant amount of silver chloride in the precipitate obtained, most of the precipitate dissolved upon adding excess aqueous NH3 (b) (i) Amount of thioanisole used = 0.9×1.06 124.1 = 0.007687 mol Amount of NaIO4 used = 107×15.40 1000⁄ 214.0 = 0.0077 mol  NaIO4 reacts with thioanisole in a 1:1 mol ratio. (ii) C6H5SCH3 + NaIO4  C6H5SOCH3 + NaIO3 (iii) (c) As a reducing agent, H2O2 will be oxidized. Quote both eqns and Eo (In acidic) O2 + 2H+ + 2e– ⇌ H2O2 Eo = +0.68 V (In alkaline) O2 + H2O + 2e– ⇌ HO2– + OH– Eo = –0.08 V Due to a more negative E o value, HO2– is more likely to be oxidised than H 2O2. H2O2 is a better reducing agent in alkaline condition. 3 (a)

3 (b) Reagents : Dilute KMnO4, NaOH(aq) Conditions : cold (c) Structure of carbocation intermediate The carbocation ion would be formed at C 2 as it will form a more highly substituted carbocat