SAJC H2 CHEM P3 ANS Prelim

1 [Turn Over SUGGESTED ANSWERS SAJC PRELIM 2014 PAPER 3 (H2 CHEMISTRY) 1 (a) [3] (b) (i) Kp = (ii) PCl5 PC l3 C l2 I / mol Z 0 0 C / mol -0.4 z +0.4 z +0.4 z E / mol 0.6 z 0.4 z 0.4 z Total moles = 1.4 z PPCl5 = (0.6 z / 1.4 z) x 5 = 2.143 atm PPCl3 = PCl2 = (0.4 z / 1.4 z) x 5 = 1.429 atm Kp = 0.953 atm (iii) When temperature increases, the equilibrium will favour the endothermic reaction to absorb the excess heat / to decrease the temperature. Hence, rateforward increases more than ratebackward. Hence, Kp increases. [5] (c) (i) The energy released from forming ion-dipole interaction between aluminium oxide and water is insufficient to overcome ionic bonds in aluminium oxide. (ii) Aluminium oxide is amphoteric. Aqueous solution of PCl5 is acidic. PCl5 + 4 H2O  H3PO4 + 5 HCl Al2O3 + 6HCl 2AlCl3 + 3H2O [4]

2 [Turn Over (d) [4] (e) Between F and G Add Tollen’s reagent to both compounds and warm. No silver mirror seen for compound F, while silver mirror is seen for G. OR Add neutral FeCl 3 F forms a violet solution (complex) whilst G forms no violet solution. OR Add Br2 (aq) F orange will turn to colourless with white ppt and G will remain orange. OR Add 2, 4-DNPH. Orange ppt formed for G and no orange ppt formed for F. Between F and H Add 2, 4-DNPH. Orange ppt formed for H and no orange ppt formed for F. OR Add hot aq H 2SO4 K2Cr2O7 F turns from orange to green whilst H remains orange. [4] [Total: 20]

3 [Turn Over 2. (a) (i) 5 Ca(s) + 3 P(s) + 13/2 O2 (g) + ½ H2 (g)  Ca5(PO4)3OH (s) (ii) 5(178.2) + 5(590+1150) – 230 – 3(1913) + x = - 12969 x = -16 591 kJ mol-1 (iii) 2Ca5(PO4)3OH  3Ca3(PO4)2 + CaO + H2O Moles of hydroxyapatite = 9 / 502.5 = 0.01791 mol Moles of calcium oxide = 8.96 x 10-3 mol Mass of calcium oxide = 8.96 x 10-3 x (40.1+16) = 0.502 g (iv) Quote: Ca2+: 0.099 nm and Al3+: 0.050 Al3+ ion has a higher charge density, and higher polarising power than Ca 2+. Thus, there is more distortion of the electron cloud of the anion, hence requiring less energy to break. [8] (b) (i) Given that 1 kg of fruits can contain 7mg of acephate, 1g of fruit = 7 x 10-6 g Hence, 50 g of fruit = 3.5 x 10-4 g of acephate (ii) Nucleophilic substitution / Condensation

4 [Turn Over (b) (iii) (iv) Alkene, (secondary or tertiary) amide, secondary alcohol OR ether [6] (c) (i) (ii) [6] [Total: 20 marks]

5 [Turn Over 3 (a) (i) A buffer solution is a solution whose pH remains almost unchanged when a small amount of H+ or OH- is added to it. (ii) Cysteine Lysine Serine COOH2N CH2SH H (iv) AND [8] (b) (i) Cysteine can form disulfide linkages which are strong covalent bonds which are hard to break. (ii) 2 -CH2SH  -CH2S-SCH2- + 2H+ + 2e- H2O2 + 2H+ + 2e-  2H2O [3] (c) (i) H2O2 O2 + 2H+ + 2e- 2e- + 2H+ + ClO- Cl- + H2O Overall eqn: H2O2 + ClO-  O2 + H2O + Cl- (ii) White ppt is AgCl. AgCl + 2NH3  [Ag(NH3)2]+ + Cl- AgCl Ag+ + Cl- (Eqm 1) The decrease in [Ag+] due to formation of complex shifts eqm 1 to the right. Also, decrease in [Ag+] also decreases I.P. of AgCl which becomes lower than Ksp of AgCl, hence ppt dissolves.

6 [Turn Over (iii) ClO- can undergo disproportionation at higher temperatures to form ClO3 -. (iv) For CuCl: [Cu+][Cl‒] = 1.2 x 10‒6 [Cl‒] = (1.2 x 10‒6) / 0.04 = 3.00 x 10‒5 mol dm‒3 For AgCl @ the point where IP CuCl = Ksp CuCl Hence, [Ag+][Cl‒] = 1.8 x 10‒10 [Ag+] = 6.00 x 10‒6 mol dm‒3 No of moles of AgCl formed = 250 / 1000 x 0.05 – 250 / 1000 x 6.00 x 10‒6 = 0.012499 mol Total moles of Cl- needed = Cl- in AgCl + Cl- in saturated solution = 0.012499 + 250 / 1000 x 3.00 x 10-5 = 0.012507 mol Maximum mass of NaCl that can be added = 0.012507 x (35.5 + 23) = 0.732g [9] [Total: 20] 4 (a) (i) Product are H+ and O2 Overall equation (ii) Q = (0.25) x (24 x 60 x 60) = 21, 600 C From the above half equation, 1 mole of 1-phenyl-2-aminopropane requires 8F (772 000C). Mol of 1-phenyl-2-aminopropane formed by 21,600 C = 0.0280 mol Mass of 1-phenyl-2-aminopropane = 0.0280 x [ 9 (12) + 13 (1) + 14 (1) ] = 0.280 x 135 = 3.78 g [4] (b) (i)

7 [Turn Over (ii) Moles of HCl = moles of 1-phenyl-2-aminopropane in 25 cm3 = 1.37 x 10-3. moles of 1-phenyl-2-aminopropane in 250 cm3 = 1.37 x 10-2 mol [1-phenyl-2-aminopropane] in g dm-3 = (4 x 1.37 x 10-2 x 135) = 7.40 g dm-3 (iii) Mass in 250 cm3 = 1.85 g percentage purity of the 1-phenyl-2-aminopropane crystals = 1.85 / 3.78 x 100 = 49.0 % [4] (c) (i) (ii) Bidentate ligand can form 2 dative bonds to a central atom or ion. (iii) When 1,4-diaminobutane is added, the colour of the complex changes from blue to violet as there is a ligand exchange of water by 1,4-diaminobutane. The energy absorbed for the excitation of d electrons changes as different ligands are bonded to Cu2+. (iv) Cu H 2O 2 2+ H2N NH2 2 Cu 3 2+ H2N NH2 OR [5] (d) (i)

8 [Turn Over (ii) [2] (e) (i) (ii) Pyruvic is a simple covalent molecule whilst alanine is a zwitterion with giant ionic lattice structure. Intermolecular forces between the pyruvic acid molecules is hydrogen bonding while in alanine, it’s held by strong electrostatic forces of attraction between oppositely charged ions. More energy is needed to break the stronger ionic bonds in alanine than the hydrogen bond in the acid. (iii) [5] 5 (a) (i) [H+] = (10-3.25 x 0.025)1/2 = 3.7494 x 10-3 mol dm-3 pH = 2.43 (ii) (ii) [8]

9 [Turn Over (b) Observations Deductions A dissolves in aqueous HNO3. A contains amine or phenylamine group. Neutralisation B reacts with cold KMnO4 to form A. Mild oxidation. B contains C=C. A is a diol. Brown ppt is MnO2. A forms yellow ppt on addition of ethanolic silver nitrate. Nucleophilic substitution. A contains iodoalkane. Yellow ppt is AgI A decolourises aqueous bromine and white ppt is formed. B contains 3 Br atoms. Electrophilic substitution. A is phenylamine. A produces 0.125 mol of gas when sodium metal is added. Gas is H2. Redox reaction. 1 mol of A produces 1 mol of H2. Two –OH group present in A. D is produced when hot acidified K2Cr2O7 is added to A. Oxidation. Primary or secondary alcohol present in A. Carboxylic acid or ketone in D D reacts with 2,4-DNPH but not with Tollen’s. D contains ketone. Secondary alcohol present in A. (no double award) D undergoes condensation. A and D both form yellow ppt with aqueous alkaline iodine. A contains and D contains . E has a RCOO- Yellow ppt is CHI3. Oxidation. A: B:

10 [Turn Over C: D: E: [12] [Total: 20] END OF PAPER