RI H2 CHEM P2 ANS Prelim

© Raffles Institution 9647 / 2011 H2 Chemistry Pape r 2 1 RAFFLES INSTITUTION 2011 PRELIMINARY EXAMINATIONS H2 CHEMISTRY – PAPER 2 (Suggested Solutions) 1 (a) Ba 2+ (aq) + SO 4 2– (aq)  → BaSO 4(s) (b) Insoluble barium carbonate and barium sulfite may co -precipitate with barium sulfate . Concentrated hydrochloric acid reacts with any ba rium carbonate and barium sulfite present to form soluble barium chloride , and hence, this ensures that only barium sulfate is pre cipitated during the experiment. (May accept full equations with state symbols involving acid with CO 3 2– and SO 3 2– .) (c) Assumption: MC l2 is soluble in water. 1 Weigh accurately about 0.5 0 g of the impure solid metal sulfate in a clean and dry weighing bottle . (Accept: 0.50 – 1.00 g.) 2 Using a measuring cylinder , add 50 cm 3 of deionised water to dissolve the sulfate completely in a 500 cm 3 beaker . Stir and add more deionised water to dissolve, if necessary. Using a measuring cylinder, add 2 cm 3 of concentrated hydrochloric acid . Stir and warm the mixture (to remove the impurities carbonate and sulfite). 3 Using a 100 cm 3 measuring cylinder , measure 10 cm 3 of the aqueous barium chloride and pour it into the sulfate solution. Stir the resulting mixture with a glass rod. 4 Continue adding aqueous barium chloride dropwise, w hile stirring , until no more white precipitate forms from the further addition of aqueous barium chloride. 5 Filter the mixture using a filter funnel which has been lined with fi lter paper which was pre-weighed . Record the mass of the filter paper. 6 Using a dropper , add aqueous barium chloride to the filtrate . If no precipitate forms, discard the filtrate. If more pr ecipitate forms, repeat steps 4 to 6 . 7 Rinse the white residue with some deionised water and dry the precipitate using an infra-red lamp until a constant mass is obtained. 8 Weigh the dried precipitate with the filter paper and subtract the pre- weighed mass in Step 5 from it to find the mass of the dried BaSO 4.

© Raffles Institution 9647 / 2011 H2 Chemistry Pape r 2 2 (d) MSO 4 + BaC l2  → MC l2 + BaSO 4 amt of MSO 4 in 0.892 g = amt of BaSO 4 obtained = 1.049 137 32.1 4 16.0 + + × = 0.004500 mol molar mass of M = ×0.804 0.892 0.004500 – 32.1 – 4(16.0) = 63.5 g mol –1 The metal is copper. Therefore, the metal sulfate is CuSO 4. 2 (a) (i) To quench the reaction at the specified times by rapidly lowering the temperature of the reaction mixture. (ii) Br 2 may undergo disproportionation in the presence of NaOH(aq) and it would not be possible to determine the [Br 2] accurately . (iii) 2I – + Br 2  → 2Br – + I2 (iv) At the end point, the yellow solution of I2 decolourises . Hence there is no need for any indicator. (b) By measuring the change in the colour intensity of the orange Br 2(aq) . (The change in colour intensity over time gives the rate of reaction.) May also accept changes in electrical conductivity / pH of solution . (c) (i) y–axis: [Br 2(aq)] / mol dm – 3 & x–axis: time / s Properly plotted curves for Expts 1, 2 and 3. (See next page.)

© Raffles Institution 9647 / 2011 H2 Chemistry Pape r 2 3 (ii) From Expt 1, initial rate = – [(0.020 – 0.050) / (2500 – 0)] = 1.20 × 10 –5 mol dm –3 s–1 From Expt 2, initial rate = – [(0.070 – 0.100) / (2500 – 0)] = 1.20 × 10 –6 mol dm –3 s–1 Compare Expt 1 and Expt 2. When [Br 2] is doubled , there is no change in the initial rate ⇒ reaction is zero order w.r.t. Br 2. ⇒ a = 0 From Expt 3, initial rate = – [(0.030 – 0.050) / (4100 – 0)] = 4.88 × 10 –6 mol dm –3 s–1

© Raffles Institution 9647 / 2011 H2 Chemistry Pape r 2 4 Compare Expt 1 and Expt 3. When [CH 3COCH 2Br] is increased 2.5 times, the initial rate increases 1.20 × 10 –5 / 4.88 × 10 –6 = 2.5 times. ⇒ reaction is zero order w.r.t. Br 2. ⇒ b = 1 OR Compare Expt 1 and Expt 3. From rate = k [Br 2] a [CH 3COCH 2Br] b [H +] 1.20 × 10 –5 = (0.050) b 4.88 × 10 –6 (0.020) b ⇒ b = 1 (iii) mol – 1 dm 3 s– 1 (d) (i) CH 3COCH 2Br + OH – → CH 3COCHBr – + H 2O (ii) Comparing CH 3COCH 2 – and CH 3COCHBr – , the presence of the electron-withdrawing –Br group helps to disperse th e negative charge and stabilises CH 3COCHBr – , hence successive brominations take place more readily in basic solution. Also accepted: The electron-withdrawing –Br group causes the H on the C bearing the Br in CH 3COCH 2Br to be more acidic . (e) (i) Homogeneous equilibrium refers to an equilibrium in which all the substances involved are in the same phase . (ii) 1 The rate of bromination will increase since increasing the total pressure increases the concentration of the gaseous reactant s CH 3COCH 3 and Br 2 and the reactant molecules are closer together. The frequency of collisions between reactant molecu les increases. The frequency of effective collisions also increases . Hence bromination rate increases. 2 The position of equilibrium will be unaffected since an equal number of gaseous molecules is formed in both the forward and backward reactions.

© Raffles Institution 9647 / 2011 H2 Chemistry Pape r 2 5 3 (a) (i) Lattice energy is the amount of energy released when one mole of a pure solid ionic compound is formed from its constituent gaseous ions . (ii) Lattice energy depends directly on the product of t he charges of the ions and inversely proportional to the inter-ionic distance . Down Group II , only the cationic size increases and hence, the magnitude of the lattice energy decreases. (b) (i) 2Sr(NO 3)2(s)  → 2SrO(s) + 4NO 2(g) + O2(g) (ii) Down the group, ionic radius increases, charge density decreases, polarising power decreases, thus decreasing distortion of the electron cloud of the NO 3 − anion. N− O bond in M(NO 3)2 is less weakened and less easily broken. Thermal stability increases down the group. (c) (i) Mg(OH) 2(s) ⇌ Mg 2+ (aq) + 2OH – (aq) K sp = [Mg 2+ ][OH – ]2 = (1.5 × 10 –4 )(2 × 1.5 × 10 –4 )2 = 1.35 x 10 –11 mol 3 dm –9 (ii) Let the solubility of Mg(OH) 2 in NaOH be s mol dm – 3. [OH – ] from NaOH = (10 / 1000 × 1.00) / 1.010 = 0.00990 mol dm –3 Hence, Ksp = s(2 s + 0.00990) 2 Since NaOH is a strong base, the contribution to [OH – ] from Mg(OH) 2, a weak base, is negligible. s = [Mg 2+ ] = 1.37 × 10 –7 mol dm –3 amt of Mg in the ppt = (1.5 × 10 –4 – 1.37 × 10 –7 ) × 1.01 = 1.499 × 10 –4 mol mass of Mg(OH) 2 precipitated = (1.499 × 10 –4 ) [24.3 + 2(17.0)] = 8.74 × 10 –3 g (d) (i) ∆ Hrxn M2+ (aq) + 2e – M(s) ∆ H atm ∆ Hhyd [M 2+ (g)] IE 1 + IE 2 M2+ (g) + 2e – M(g)

© Raffles Institution 9647 / 2011 H2 Chemistry Pape r 2 6 (ii) Hence, Eo ∝ ∆ Hatm + IE 1 + IE 2 + ∆ Hhyd All 3 magnitudes of the enthalpy changes decrease down the group. ∆∆ ∆∆ Hatm + IE 1 + IE 2 is endothermic while ∆∆ ∆∆ Hhyd is exothermic. Hence, adding up these results in insignificant cha nges to Eo down the group. (e) Since Ca 2+ has a smaller ionic radius than Sr 2+ , it has a higher charge density and is more extensively hydrated by water molecules. This produces more drag and hence its ionic speed is low er than expected (in this case, it is identical with that of Sr 2+ ). 4 (a) P has a lower vapour pre