HCI H2 CHEM P2 ANSWERS Prelim

Hwa Chong Institution 2011 Prelim Paper 2 Answers 1(a) (b) (i) 100 cm 3 (accept 10 cm 3 to 200 cm 3) (ii) The maximum mass that can dissolve in 100 cm 3 of water = (6.95 / 1000) x 100 x 53.5 = 37.2 g (c) Assuming a temperature change of 5 oC to be measured and no heat loss to surroundings, nsalt x 15 000 = 100 x 4.3 x 5 ⇒ nsalt = 0.143 mol minimum mass to use = 0.143 x 53.5 = 7.65 g (d) 1. Weigh accurately 8.00 g of ammonium chloride in a weighing bottle, using a weighing balance. 2. Using a 100 cm 3 measuring cylinder, add 100 cm 3 of water into a dry styrofoam cup calorimeter. Support the styrofoam cup on a 250 cm 3 beaker. 3. Stir the water gently using the thermometer. 4. Start the stopwatch. 5. Record the temperature of the water in the styro foam cup using a 0.1 oC thermometer, at 30 s intervals ( accept 1 min ) until 2.5 min. 6. At exactly 3 min ( accept 1–5 min ), tip the ammonium chloride into the water. Do not measure the temperature at this time. 7. Stir the solution gently and record the temperat ure of the solution at 3.5 min. Continue to stir and record the temperature at 30 s intervals until solution returns to room temperature. 8. Reweigh the weighing bottle. thermometer lid styrofoam cup calorimeter 250 cm 3 beaker ammonium chloride and water

Table with headings: Mass of weighing bottle and salt / g m 1 Mass of weighing bottle after experiment / g m 2 Time / min Temperature of solution / oC 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 . (e) (f) ∆ H solution (in J mol –1 ) = +v × 4.3 × T m 53.5       (g) The salt with the more endothermic enthalpy change of solution will be the more effective ingredient in the cold pack. OR The salt that gives the fastest drop in temperatur e will be the more effective ingredient.

2 (a) (i) Examples of possible answers : Category of reaction Equation from Group II or Group VII Decomposition 2Mg(NO 3)2 (s) /barb2right 2MgO(s) + 4NO 2(g) + O 2(g) 2HC l(g) /barb2right H2(g) + C l2(g) Single displacement Ca (s) + 2H 2O(l) /barb2right Ca(OH) 2(aq) + H 2(g) Cl2(g) + 2KI(aq) /barb2right 2KC l(aq) + I2(aq) Double displacement Na 2SO 4(aq) + Ba(NO 3)2(aq) /barb2rightBaSO 4(s) + 2NaNO 3(aq) NaCl(aq) + AgNO 3(aq) /barb2right AgC l(s) + NaNO 3(aq) (ii) Type of reaction : redox Equation : 4K I(aq) + 2CuSO 4(aq) /barb2right 2K 2SO 4(aq) + 2Cu I(s) + I2(aq) (iii) NaBr(s) + H 2SO 4(l) /barb2right HBr(g) + NaHSO 4(s) 2HBr(g) + H 2SO 4(l) /barb2right Br 2(g) + SO 2(g) + 2H 2O(l) (b) (i) Ka = ])([ ])()(][ [ 2 62 52 + ++ OHMg OH OHMg H (ii) I. The solubility equilibrium Mg 2+ (aq) + 2OH _(aq) Mg(OH) 2(s) keeps [OH -] constant (pH is constant at 9). II. The Mg 2+ has been completely precipitated / reacted, and the excess (drop of) NaOH increases the [OH -] / pH drastically. (iii) Initial [Mg 2+ ] = ½ x 0.50 0 .20 00 . 1x = 0.20 mol dm -3 Ka = ][20 . 0 ][ 2 + + − H H where [H +] = 10 -6 mol dm -3 = 5.00 x 10 -12 mol dm -3 Note that the [H +] in the denominator is negligible. (iv) I. pOH = 14 – 9 = 5 [OH – ] = 10 -5 mol dm -3 II. At half-completion, [Mg 2+ ] = 1000 60 20 . 01000 50 2 1 xx = 0.0833 mol dm -3

Hence, K sp = [Mg 2+ ][OH – ]2 = 0.0833 x (10 -5)2 = 8.33 x 10 -12 mol 3 dm -9 (c) (i) A complex is the combination of a central metal atom / cation by dative bonds to a number of ligands Coordination no. = 6 (ii) Colour of solution turns (from pink) to bright blue (with KSCN) and then pink / decolourises (with edta). Ligand exchange occurs with H 2O replaced successively by SCN - and edta. This is allowed by the high lg K stab of the SCN - complex; and subsequently by the higher lg K stab of the edta complex. (accept if student states SC N – is stronger ligand than H 2O and edta is stronger ligand than SCN – ; no need reference to K stab ) (iii) With the replacement of H 2O ligands by NH 3, the E/ring2 for the Co 3+ / Co 2+ system becomes less positive/decreases (literature values : E/ring2 changes from +1.82 V to +0.10 V). lg K stab for [Co(NH 3)6]3+(aq) is much higher than for [Co(NH 3)6]2+ (aq) ; hence the reduction process is not favoured. (accept: NH 3 ligand stabilises Co 3+ much more than Co 2+ ) (iv) From the formula of [Co(SCN) 4]2-, the mole ratio of Co 2+ to SCN – = 1 : 4. The sketch has an inverted V shape and shows a maxi mum absorbance at volume of CoC l2 = 2.0 cm 3. (v) The stated equilibrium is endothermic, i.e. ∆ H is positive.

3(a) Energy kJ mol-1 0 2Fe(s) + 3/2 O2(g) Fe2O3 (s) ∆∆ ∆∆ Hf [Fe2O3 (s)] 2Fe(g) + 3/2 O2(g) ∆∆ ∆∆ Hatm[Fe (s)]2x 2x (1st IE + 2nd IE + 3rd IE) 2Fe3+(g) + 3/2 O2(g) + 6e 3/2 x BE (O=O) 2Fe3+(g) + 3O(g) + 6e 2Fe3+(g) + 3O-(g) + 3e 3x EA1[O] 2Fe3+(g) + 3O2-(g) L.E. By Hess’ Law, /uni2206Hf = 2(414) + 2(762 + 1560 + 2960) + 3/2 (496) + 3 (-141) + 3 (844) + L.E. L.E. = - 1.51 x 10 4 kJ mol –1 (b) • As the CN – ligand approach the iron ion, the degenerate d-orbitals split into two energy levels, giving rise to a ∆ E, that corresponds to the visible light region of the electromagnetic spectrum. • In a partially filled d-orbitals, an electron from the lower energy level d-orbital will absorb energy corresponding to ∆ E and get promoted up to the vacant higher energy level d- orbital. • The color of the iron compounds is the complementary color to the one absorbed. (c) [R] FeO 4 2-(aq) + 8H +(aq) + 3e ⇌ Fe 3+ (aq) + 4H 2O( l) +2.00 V [O] Cr 2O7 2- + 14H + + 6e ⇌ 2Cr 3+ + 7H 2O +1.33 V E/ring2 cell = +2.00 – (+1.33) = +0.67 V > 0 Overall equation: 2FeO 4 2- + 2Cr 3+ + 2H + /barb2right 2Fe 3+ + Cr 2O7 2- + H 2O Observation: The resultant solution should be yellowish-orange

(d)(i) (ii) (e) Electrophilic substitution [1] HNO 3 + 2H 2SO 4 → NO 2 + + 2HSO 4 – + H 3O+ (not required)

4 (a) Phosphorus burns vigorously in O 2 to give a bright flame , forming a white solid . 4P(s) + 3O 2 (g) → P 4O6 (s) or 4P(s) + 5O 2 (g) → P 4O10 (s) Accept P 4 in these equations (b) P4O6 (s) + 6HC l (g) → 2H 3PO 3 (l) + 2PC l3 (g) 4 H 3PO 3 (l) + 6 HC l (g) Let x be the ∆ Hrxn for the reaction of PC l3 with steam By Hess' Law, +177 = +729 – 2 x x = +276 kJ mol –1 (c) Energy released in bond formation for structure B = 322 + 544 + 2(335) + 2(460) = 2456 kJ mol –1 Energy released in bond formation for structure C = 3(335) + 3 (460) = 2385 kJ mol –1 Structure B is more likely to be formed as more energy is rele ased during bond formation. (accept if student only compares f orming P=O and P–H with forming P–O and O–H bonds) (d)(i) (ii) The electron withdrawing effect of oxygen in P=O in H 3PO 4 is spread out over three OH groups as compared to two OH groups in H 3PO 3. +6 H 2O +6 H 2O (+177) (+729) (2x)

5(a)(i) (ii) (iii) C l2 + 2NaSCN → 2NaC l + (SCN) 2 (b)(i) (ii) Delocalization of electrons in the thiocyanat e radical allows an isomer, the isothiocyanate radical, S=C=N /circle6, to be formed, resulting in such a side product. (c)(i) (ii) (iii) Cyanide can act as a ligand