PJC_H2 Chemistry_2008 JC2 H2 Chemistry Prelims Paper 3 Suggested Answers

Pioneer Junior College H2 Chemistry Preliminary Examination Paper 3 Suggested Answers 1 (a) (i) Using pV = nRT (38.9 x 101000)(2.0 x 10 -3) = ntotal x 8.31 x (273 + 30) n total = 3.12 mol SO2Cl2(g) ⇌ SO2(g) + Cl2(g) Eqm amount / mol 2 – x x x At equilibrium, n total = 2 – x + x + x = (2 + x) mol x = 1.12 mol n(SO 2Cl2) = 2 – 1.12 = 0.88 mol n(SO 2) = n(Cl2) = 1.12 mol (ii) [SO 2Cl2] = ½(0.88) = 0.44 mol dm-3 [ S O 2] = [Cl2] = ½(1.12) = 0.56 mol dm-3 K c = ) 44 . 0 ( ) 56 . 0 (2 = 0.713 mol dm-3 (iii) Le Chatelier’s principle states that when a system in equilibrium is subjected to a change, the system will react in such a manner to counteract the change so as to re-establish the equilibrium. (iv) SO2Cl2(g) ⇌ SO2(g) + Cl2(g) ΔH negative By Le Chatelier’s pr inciple, an increase in temperature favours an endothermic reaction. Since the forw ard reaction is exothermic, the position of equilibrium shifts to t he left, resulting in an increase in [SO2Cl2] and a decrease in [SO2] and [Cl2]. Thus, the value of Kc will decrease. (b) Sodium chloride has a high melting point whereas silicon tetrachloride has a low melting point. Sodium chloride is an ioni c compound with giant ionic lattice structure. Strong electrostatic attractions exist between sodium and chloride ions , hence large amount of energy is required to weak the strong ionic bonds during melt ing. Silicon tetrachloride is a covalent compound with simple molecular structure . Weak van der waals forces (induced dipole – induced dipole) exist between silicon tetrachloride molecules. Lesser ener gy is required to overcome these intermolecular forces of attractions during melting. (or difference in electrical conductivity)

2 Sodium chloride gives sodium and chlori de ions on dissolution in water. Both sodium ion (having low polarisi ng power) and chloride ions (anion of a strong acid) do not undergo hydrolysis thus a solution of sodium chloride is neutral. Silicon tetrachloride, on the contrary, undergoes hydrolysis to give an acidic solution. NaCl (s) + aq → Na+(aq) + Cl-(aq) SiCl 4(l) + 2H2O(l) → SiO2(s) + 4HCl(aq) (c) (i) CH3CH2OH CH3CH2Cl CH3CH2CN CH3CH2CH2NH2 I II III Step I: PCl5 Step II: KCN in ethanol, heat Step III: LiAlH4, dry ether (ii) CH3CH=CH2 CH3CH(OH)CH2Br CH3COCH2Br CH3COCH2OH I II III Step I: Br2(aq) Step II: KMnO4(aq), H2SO4(aq), heat Step III: NaOH(aq), heat 2 (a) (i) Standard enthalpy change of neutralisation is the heat evolved when 1 mole of water is formed when an acid reacts with an alkali under the standard conditions of 298 K and 1 atm. (ii) Ethanoic acid is weak