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 acid whic h undergoes partial dissociation. Some of the energy liberated from the neutralisation is used for the dissociation of the weak acid unlike HCl and HNO 3 which dissociate completely. (iii) CH3CH2NH2 + HCl → CH3CH2NH3 +Cl- Heat evolved = mC ΔT = (35.0 + 45.0) x 4.2 x 5.2 = 1747 J = 1.747 kJ No of moles of salt pr oduced = 35.0/1000 x 1.0 = 0.0350 mol

3 Δ Hreaction = -1.747.2/ 0.035 = - 49.9 kJ mol-1 (b) (i) CH3CH2NH2 + H2O ⇌ CH3CH2NH3 + + OH- [ ]eqm / mol dm-3 0.10 – x x x Kb = ] NH CH CH [ ] OH ][ NH CH CH [ 2 2 3 3 2 3 −+ 4.5 x 10-4 = x 10 . 0 x2 − Since CH3CH2NH2 is a weak base, it undergoes partial dissociation. Hence, 0.10 >> x. Thus, 0.10 – x ≈ 0.10 x = [OH-] = 6.71 x 10-3 mol dm-3 pH = 14 – pOH = 14 - [-lg(6.71 x 10 -3)] = 11.8 (ii) A solution that is equally effectiv e in resisting changes in pH is a buffer at its maximum buffer capac ity, i.e. a solution in which [CH3CH2NH2] = [CH3CH2NH3 +]. This occurs when half of the amount of CH3CH2NH2 is converted to the salt. Volume of HCl required for complete neutralisation = 20.00 cm3 Volume of HCl required to neutralise half the amount of CH 3CH2NH2 = 10.00 cm3. At the maximum buffer capacity, pOH = pKb pOH = -lg(4.5 x 10-4) = 3.35 pH = 14 – 3.35 = 10.7 (iii) Vol of HCl / cm3 pH 10 20 <7 10.7 11.8 Equivalence point Max. Buffer capacity (10, 10.7)

4 (c) phenylamine < 4-methylphenylamine < ethylamine Ethylamine is the most basic as it contains one electron donating, ethyl group. This makes the lone pair of electrons on N more available to accept a proton. Phenylamine is the least basic as it contains an amine group bonded directly to the benzene ring. As a result, the lone pair of electrons on the N can delocalised into the ring, making the lone pair of electrons on N less available for coordination wit h proton. 4-methylphenylamine is more basic than phenylamine because it contains –CH 3 group which is electron donating which reduces the ex tent of the delocalisation of the lone pair of electrons on N. (d) (i) CH 3COCl, room temperature (ii) Phenol is not a good nucleophile to react with ethanoyl chloride to form the ester. [Note: To form the ester, phenol is usually treated with NaOH(aq) to form sodium phenoxide. Phenoxide ion is a better nucleophile.] (iii) NaOH(aq) at room temperature Product: O - CH3COHN Na + NaOH(aq), heat Products: H2N O - Na + CH3CO2 -Na+ 3 (a) (i) The order of reaction with respect to a reactant is the power to which the concentration of that reactant is raised in the experimentally determined rate equation. Half-life , t ½, is the time taken for the reactant concentration to decrease to half of its original value.

5 (ii) Let rate= k[H2O2(aq)]m[ I-(aq)]n Compare experiments 1 & 2, keeping [H2O2(aq)] constant nm nm 4 4 ) 050 . 0 ( ) 020 . 0 ( k ) 040 . 0 ( ) 020 . 0 ( k 10 x 5 . 1 10 x 2 . 1=− − (or use inspection method) n = 1 Rate of reaction is 1st order with respect to I-(aq) or rate α[I-(aq)]. Compare experiments 1 & 3, keeping [I-(aq)] constant nm nm 4 4 ) 040 . 0 ( ) 050 . 0 ( k ) 040 . 0 ( ) 020 . 0 ( k 10 x 0 . 3 10 x 2 . 1=− − (or use comparing method) m = 1 Rate of reaction is 1st order with respect to H2O2(aq) or rate α [H2O2] rate= k[H 2O2(aq)][ I-(aq)] (iii) Using experiment 1, rate= k[H 2O2(aq)][I-(aq)] 1.2 x 10 -4 = k(0.020)(0.040) k = 0.15 mol-1 dm3 min-1 (iv) For experiment 4 and 5, since [ I-(aq)] >> [H 2O2(aq)], [ I-(aq)] is approximately constant. Thus, rate = k’[H 2O2(aq)] (a pseudo first order reacti on) where k’ = k[I-(aq)] t1/2 = ] [ k 2 ln ' k 2 ln −= = I t1/2 of H2O2 in experiment 4 = 9.2 min (for [I-(aq)] = 0.500 mol dm-3) t 1/2 of H2O2 in experiment 5 = 4.6 min (for [I-(aq)] = 1.00 mol dm-3) (b) (i) The reaction can be quenched by: • adding NaOH/NaHCO3/Na2CO3 to remove the H+(aq) • sudden cooling of the reaction mixture • sudden dilution through the addition of large volume of water any of the above method (ii) I 2(aq) + 2S2O3 2-(aq) → 2I-(aq) + S4O6 2-(aq)

6 (c) (i) On descending the group, the atomic size of the halo gen increases, the number of electrons that can be polarised increases. The strength of the van der Waals’ forces (induced dipole – induced dipole interactions) increases (or the effects of a stronger induced dipole – induced dipole interaction outweigh the effect of the permanent dipole – permanent dipole interaction) from HC l to H I. More energy is required to overcome