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