TJC H2 CHEM P3 STUDENTS ANS Prelim

11 Essay Answers: 1 (a) (i) PV = nRT m / V =  = P  Mr RT  density of NO 2 = 1.01  105  46.0 8.31  298  = 1880 g m3 (to 3 s.f)  = 1.88  103 g cm3 (ii) NO2(g) + N2O(g)  3NO(g)  H r = H  f (products) – H f (reactants) = 3(+90.4) – (+33.9 + 81.6) = +155.7 kJmol -1  At 800 K, Gr = H r – TS r = +155.7 – 800(+171.4  103) = +18.6 kJ mol1  Since Gr > 0 at 800 K, the reaction is not spontaneous. (iii)  OR [Correct number of dots and crosses for N and 2 O atoms.]  Linear molecule (b)  For AgCl to precipitate , [Ag +] [Cl] > 1.8  1010 (1.8  102)  [Cl] > 1.8  1010 [Cl] > 1.8  108 mol dm3  For PbCl 2 to precipitate, [Pb2+] [Cl]2 > 1.7  105 (2.0  102)  [Cl]2 > 1.7  105 [Cl] > 2.9  102 mol dm3  AgCl will precipitate out from the solution first as it requires a much smaller concentration of Cl. (c) (i)  X+(g)  X2+(g) + e  Going down the X + ions of Group II, both the nuclear charge and screening effect increase. As the increase in screening effect outweighs the increase in nuclear charge, the effective nuclear charge of the X + cations decreases down the group.  Hence moving down the group, the valence electron in the X + cation becomes further away from the nucleus and is less strongly attracted, thus needing less energy for its removal. N N O       N N O    

12 (ii)  Mg 2+ ion has a higher charge density than Ba 2+ ion and hence is more polarising. Mg2+ ions polarise the large NO3  ions to a greater extent and hence causes the N – O bonds to be weakened. Thus Mg(NO 3)2 is less thermally stable and decomposes at a lower temperature than Ba(NO3)2.  Equations: Mg(NO3)2  MgO + ½ O2 + 2NO2 Ba(NO 3)2  BaO + ½ O2 + 2NO2 (iii)  M is magnesium.  MgCO 3 undergoes decomposition when heated. Since it forms CO2(g) which escapes when it is formed, there is a loss in mass during heating. MgCO 3(s)  MgO(s) + CO2(g)  When X which is MgO is added to water, it reacts to form insoluble Mg(OH) 2. As the concentration of aqueous OH  is low, the resulting solution is only weakly alkaline. MgO + H 2O  Mg(OH)2  Adding H 2SO4(aq) to the above mixture produces MgSO 4 which is soluble in water. Hence no precipitate is observed.  1 mark for the two equations for decomposition of MgCO 3 and reaction of MgO with water. 2 (a) NaOH(s) + aq  NaOH(aq) ∆Hsol = ∆Hf of NaOH (aq) – ∆Hf of NaOH (s) = –470 – (–425) ● = – 45.0 kJ mol1 Number of moles of NaOH = 10.0 40 = 0.250 mol ● Heat evolved = 45 x 0.25 = 11.3 kJ ● Temperature increase = Q 11.3 1000 mc (250)(4.18)  = 10.8 C (b) (i) [H+] = √Ka.c = √10-8.35  0.1 = 2.11 x 106 mol dm3 ● pH = –lg[H+] = 4.68 (ii) ● Benzoic acid is a stronger acid as it has a lower pK a value. ● For phenoxide ion formed from the dissociation of 3-nitrophenol, the lone pair of electrons on oxygen is delocalised into the benzene ring and the nitro group being electron withdrawing further disperses the negative charge on the phenoxide ion resulting in a stable anion formed. However benzoic acid is a stronger acid as COO- is more stable than phenoxide ion due to the delocalisation of negative charge over two electronegative oxygen atoms, resulting in a much more stabilised ion as the negative charge is better dispersed and is more likely to release H+. (iii) ● Phenolpthalein and bromothymol blue are both weak acids. Small amounts are used so that it will not affect accuracy of the end-point titre value. (If larger amounts are used, a larger titre value will be obtained)

13 (iv) ● A: pH = 4.20 ● B: Volume of titre = 7.20 cm3 (v) (first endpoint with bromothymol blue) ● concentration of benzoic acid = (14.40/1000 x 1) / (25.0/1000) = 0.576 mol dm3 (second endpoint with phenolphthalein) ● concentration of 3-nitrophenol = (7.60/1000 x 1) / (25.0/1000) = 0.304 mol dm3 (c) ●●● N CH3 CH3 NO2 NH2 conc H2SO4 + conc HNO3 60oC Sn/conc HCl heat followed by NaOH CH 3I heat (d) (i) ● Buffers are solutions that resist changes in pH upon additions of small amounts of acid or alkali. (ii) ● H 3N–CH2–COO (iii) ● When small amounts of acid is present, H2C +NH3 CO2 - +H 3O+ H2C +NH3 CO2H +H 2O H + is removed thus there is negligible change in pH. ● When small amounts of base is present, H2C NH3 + CO2 - +O H - H2C NH2 CO2- +H 2O OH - is removed thus there is negligible change in pH. (iv) ● 2CH 4 + 2H2O + NH3  H2NCH2COOH + 5H2 (v) ● Large amounts of energy is required to break the strong covalent bonds. OR Energy is required to overcome activation energy for the reaction OR Energy is required to break the covalent bonds to form free radicals (also can be accepted) 3. (a)  Solution formed is orange (pH~3-4).  A lCl3 undergoes both hydration and hydrolysis as A l3+ has a high charge density, hence a high polarizing power. It draws electrons away from its surrounding water molecules and weakens the O-H bond. AlCl3(s) + 6H2O(l)  [Al (H2O)6]3+(aq) + 3Cl-(aq) [Al(H2O)6]3+(aq)  [Al (H2O)5OH]2+(aq) + H+(aq)  1 mark for both equations +

14 (b) (i)  No, because A l2Cl6 is not electron-deficient and is unable to accept a lone pair to form a dative bond. (ii)  Step 1: Generation of electrophile A lCl3 + CH3COCl +COCH3 + AlCl4  (c) (i)  SiC l4 is a (simple) covalent chloride that undergoes hydrolysis in water. Hydrolysis occurs due to energetically accessible vacant 3d-orbitals available for dative bonding with water. (ii)  SiC l4(l) + 2H2O(l) → SiO2(s) + 4HCl(g)  The standard entropy change is positive because 4 moles of gas is being produced in the reaction, increasing the degree of randomness / disorderliness of the system. (iii)  SiC l4 + 4CH3OH → Si(OCH3)4 + 4HCl  Steamy white fumes will be observed. (d)  Since there are 5 σ bonds ( 5 electron pairs) around P, there should be 5 hybrid orbitals. Since there is only 1 3s orbital and 3 3p orbitals, the fifth orbital has to be a 3d orbital. Hence type of hybridization is sp3d. (e) (i)  Y liberates white fumes with PCl5, therefore it contains –OH group.  Y forms orange ppt with Brady’s reagent, therefore it contains a carbonyl group. Y does not react with Tollen’s reagent, therefore it can only be a ketone.  Y forms a yellow ppt in the tri-iodomethane test, hence it contains either CH3CO-R or CH3CH(OH)-R where R is H or an alkyl chain.  Y is oxidised to form a carboxylic acid, hence it must contain a primary alcohol. Therefore, Y must be  CH 3COCH2CH2OH (ii)  His choice of chemical test was inappropriate because the PC l5 would have undergone hydrolysis to form phosphoric and hydrochloric acid.  PC l5 + 4H2O  H3PO4 + 5HCl 4. (a) (i) ● Anode : Li  Li+ + e- ● Cathode : Cu 2O + H2O + 2e-  2Cu + 2OH- Step 3: fast + COCH3 H + AlCl3 + AlCl4 - COCH3 + HCl + +COCH3 Step 2: slow + COCH3 H    

15 (ii) /LiLiE  = -3.04V E cell = Ered – Eoxid +2.30 = O/CuCu2 E – (-3.04) O/CuCu2 E = 2.30 – 3.04 = -0.74V Assumptions made: ● 298K (25 oC) & Concentration of aqu