RI H2 Chem 2013 Prelim P2 Soln

1 © Raffles Institution 2013 9647/02/S/13 Raffles Institution 2013 H2 Chemistry Preliminary Examinations Paper 2 (Suggested Solutions) 1 (a) C6H12(l) + 9O2(g)  6CO2(g) + 6H2O(l) (b)(i) Assuming no heat loss, heat evolved from combustion = heat gained by calorimeter set-up – [cis–hex–3–ene] × amt of cis–hex–3–ene = C × temperature rise – (–3733)(0.20 / 84.0) = C(5.0) C = 1.7776 = 1.78 kJ K–1 (3 s.f.) (b)(ii) Assuming no heat loss, heat evolved from combustion = heat gained by calorimeter set-up – [trans–hex–3–ene] × amt of trans–hex–3–ene = C × temperature rise – [trans–hex–3–ene] × (0.22 / 84.0) = (1.7776)(5.4) [trans–hex–3–ene] = –3665 = –3670 kJ mol–1 (3 s.f.) (b)(iii) = [cis–hex–3–ene] – [trans–hex–3–ene] = –3733 – (–3665) = –68.0 kJ mol–1 (c) Experimental set-up 1. Fill up one spirit lamp with cis–hex–3–ene (to about 80% capacity), taking care to avoid getting any cis–hex–3–ene on the outside surface of the lamp. 2. Set up the apparatus as shown above, such that the beaker is about 2-3 cm from the tip of the wick. cap beaker spirit lamp containing alkene to be burnt 203

2 © Raffles Institution 2013 9647/02/S/13 Calibration of calorimeter set-up 3. Using a 250 cm3 beaker, measure out 200 cm3 of water. 4. Using an electronic weighing balance, w eigh the mass of the spirit lamp filled with cis–hex–3–ene. Record this mass. 5. Using the thermometer, read the temperature of the water and record it. 6. Use the lighter to l ight the lamp and place it under the beaker of water , centering the flame under the bottom of the beaker and ensuring the flame is at an approximately constant distance below it. 7. Using the thermometer, carefully and gently stir the water. 8. When the temperature has risen to about 5 °C, blow out (extinguish) the flame. 9. Continue to stir the water, recording the highest temperature reached. 10. While the lamp is cooling, clean the soot off from the bottom of the beaker. 11. Re-weigh the spirit lamp when it has cooled to room temperature. Record this mass. 12. Repeat Steps 3 to 11, using the same lamp after cleaning it. (For a more accurate estimate of the heat capacity.) Determining mass and temperature changes using trans–hex–3–ene 13. Repeat Step 1 by filling the other spirit lamp with trans–hex–3–ene. 14. Repeat Steps 3 to 11, using this new spirit lamp, recording the relevant masses and temperatures. 15. For a more accurate estimate of the data, Step 14 can be repeated. The value for the standard enthalpy change of combustion of trans–hex–3–ene can then be found by similar calculations as shown in (b). (d) Hexenes are inflammable liquids and are hence a fire hazard. Any of the following: Containers of the hexenes should be covered and stored away, when not in use; proper disposal of hexenes; the wick should be lit away from any nearby hexenes. Or Hexenes are volatile and hence a health hazard, causing respiratory problems. Experiment to be carried out in fume cupboard to avoid inhaling toxic vapour of hexene. 204

3 © Raffles Institution 2013 9647/02/S/13 2 (a) (b) sp3 ; 107 (c) (d) pV = nRT = (m/M)RT (101 × 103)(10 × 10–3) = (m / 20)(8.31)(150 + 273) m = 5.75 g (e)(i) Let the initial amt of N2H4 be a. N2H4(g) ⇌ N2(g) + 2H2(g) initial amt / mol a 0 0 eqm amt / mol a(1) a 2a Total amt of gases present at eqm = a(1) + a + 2a = a(1+2) mol average Mr = 1 12     Mr(N2H4) + 12   Mr(N2) + 2 12   Mr(H2) 20 = 1 12     (32.0) + 12   (28.0) + 2 12   (2.0)  = 0.30 N N  × ××  × × ×    H H H H  × N2 pV nRT p ideal gas N2H4 H2 1.0 205

4 © Raffles Institution 2013 9647/02/S/13 (e)(ii) 2 4 2 4 1 1 0.3 (1) 0.4375atm(1 2 ) (1 0.6) N H N H T Tp p p      = 0.438 atm 22 0.3 (1) 0.1875atm(1 2 ) (1 0.6) N N T Tp p p     = 0.188 atm 22 2 2(0.3) (1) 0.375atm(1 2 ) (1 0.6) H H T Tp p p     22 24 2 2 22(0.1875)(0.375) 6.03 10 atm0.4375 NH P NH ppK p     (e)(iii)  will decrease. 206

5 © Raffles Institution 2013 9647/02/S/13 3 (a) When pH increases, the concentration of H+ decreases. By Le Chatel er’ Pr n ple, the position of equilibrium shifts to the right to increase the concentration of H+, causing the fraction of HF to decrease. (b) Since = – = 0.5, ⇒ [HF]eqm = [ ˉ]eqm ⇒ buffer solution at maximum buffering capacity. pH = pKa + [F ]lg [HF]  ⇒ 3.2 = pKa + lg 1 ⇒ Ka = 10–3.2 = 6.31 x 10–4 mol dm–3 (c)(i) a = ½ (c)(ii) The addition of H+ au e the p t n f the equ l br um (2) t h ft r ght, redu ng [ ˉ] n solution. This, in turn, causes the position of equilibrium (1) to shift right, increasing [Ca 2+] in solution. (c)(iii) From the graph, at pH 3.0, fraction of HF = 0.61 and fraction of F– = 0.39 [F (aq)] 0.39 0.639[HF(aq)] 0.61   At pH 3.0, [F–] = 4 4 3(0.639)(4.63 10 ) 2.96 10 mol dm     (c)(iv) [Ca2+] = ½ ( [ ] + [ ˉ] ) (c)(v) From (iii) and (iv), [Ca2+] = 0.5 x (4.63 + 2.96) x 10–4 = 3.795 x 10–4 mol dm–3 Ksp = [Ca2+][F–]2 = (3.795 x 10–4)(2.96 x 10–4)2 = 3.33 x 10–11 mol3 dm–9 207

6 © Raffles Institution 2013 9647/02/S/13 4 (a) Ca(OH)2(aq) + CO2(g)  CaCO3(s) + H2O(l) (b)(i) MgCO3 (b)(ii) As shown by plot (II), MgCO3 decomposes more readily than BaCO3. Mg2+ is smaller than Ba2+ and has a higher charge density and polarising power. Hence it distorts the electron cloud of the CO 3 2– anion, and weakens the carbon- oxygen bond within the anion to a greater extent , resulting in MgCO3 requiring less energy for thermal decomposition and thus it decomposes more readily. (b)(iii) Mg2+: 1s2 2s2 2p6 Mg2+ has one less occupied principal quantum shell than that of its parent atom Mg. (c) (d) Let the Ar of Z be a. 3ZCO3  Z3O4 (i.e. 3 moles of ZCO3 decomposes to give 1 mole of Z3O4) amt of ZCO3 = 3 × amt of Z3O4 1.0 / (a + 12.0 + 3 × 16.0) = 3 × 0.667 / (3a + 4 × 16.0) 1 / (a + 60) = 2.001 / (3a + 64) 3a + 64 = 2.001a + 120.06 a = 56.1 (value is closest to 55.8) Z is Fe. mass of solid / g time 1.0 (II) 0.5 (I) (III) 0.667 0 208

7 © Raffles Institution 2013 9647/02/S/13 5 (a) BE(C–Cl) = +340 kJ mol–1 BE(C–I) = +240 kJ mol–1 The energy given out by the C –I bond formed is insufficient to compensate for the energy needed to break the C –Cl bond / the C–I bond formed is weaker than the C –Cl bond broken, hence H is endothermic. Since G  H > 0, the reaction is not feasible / does not go to completion. (b)(i) Note: The above diagram shows what it actually looks like in 3 -dimensional space i.e. Na + is surrounded by six propanone molecules. Only one Na+ with one propanone molecule is needed for this answer. (b)(ii) NaCl is precipitated out as a solid. By Le Chatel er’ Pr n ple, [NaCl] is low and position of equilibrium shifts to the right towards completion. (c)(i) Mechanism: Nucleophilic Substitution (SN1) + + + + + 209

8 © Raffles Institution 2013 9647/02/S/13 (c)(ii) The C6H5CH2 + carbocation is resonance stabilised as the  electron cloud of the benzene ring can overlap with the empty p orbital of the carbocation. For C 6H5CH2CH2I, the carbocation is not resonance -stabilised. Hence, th