MJC H2 Chem 2013 Prelim P2 Soln

©chemistry@meridian jc 1  (a) (i) 140 × 103 × (0.00412) = (100) (4.18) (T) The temperature change of 1.38C is too small and (experimental) results would be very inaccurate. (ii) Hence, write a detailed plan to ve rify the enthalpy change of reaction between magnesium and sulfuric acid, H1. Your plan should include the following:  Justification of the mass of Mg turnings to be used  Tabulation of raw data  Brief description of how the data can be used to determine the enthalpy change of reaction [6] Justification of mass to be used 140 × 103 × (m/24.3) = (100) (4.18) (7) m = 0.508 g (any suitable mass to achieve a T of 7 C to 10 C) Procedure 1) Using a 100 cm3 measuring cylinder, measure out 100 cm 3 of sulfuric acid into the styrofoam cup supported in a 250 cm3 beaker. 2) Measure and record the initial temperature of the solution. 3) Weigh accurately 0.510 g of Mg turnings in a clean dry weighing bottle using a mass balance. 4) Carefully transfer the Mg turnings into the cup, stir. 5) Measure and record the highest temperature reached. 6) Re-weigh the weighing bottle which may contain some residual Mg turnings; record the actual mass of solid sample used. Tabulation of Data Mass of weighing bottle and Mg turnings / g Mass of weighing bottle + residual solid after transfer / g Mass of Mg turnings used / g M Initial temperature of solution / C Highest temperature of solution / C Change in temperature, T / C Treatment of Data IH1I × M / 24.3 = 100 × 4.18 × T / 1000 H1 = (0.418)(24.3)(T) /M kJ mol-1 2013 MJC H2 Chemistry Paper 2 Suggested Answers 916

2 ©chemistry@meridian jc (b)(i) Is the energy change when 1 mole of MgO (s) is formed from its elements, Mg (s) and O2 (g) under standard conditions (ii) Mg (s) + ½ O2 (g) MgO (s) MgSO 4 (aq) + H2 (g) + ½O2 (g) MgSO4 (aq) + H2O (l) Hf = 602 kJ mol1 2(a) Step I: Nucleophilic Addition ; Step II: Hydrolysis (b)   (c) Acts as a solvent (d)(i) (Strong) covalent bonds in organic mole cules or the strong C-I bond must be broken. (ii) No of moles of pentan–2–one = 6 x 0.814 /86 = 0.05679 No of moles of Mg = 1.50 / 24.3 = 0.0617 Grignard reagent is in excess. + H2SO4 (aq) H1 + H2SO4 (aq) H2 H3   Hf  917

3 ©chemistry@meridian jc (f) (i) w a t e r Mg(OH)I /MgCl/MgI   (ii) saturated sodium hydrogencarbonate HCl (iii) sodium thiosulfat e Iodine (iv) saturated sodium ch loride Water (g) Transfer the diethylether layer into a distilling flask fitted with an air condenser and carry out distillation by boiling over temperature range from 141 oC to 145 oC. 3 (ai) 1 [Cl-] = 31.00 10 10 1000  = 0.100 mol dm-3 [Ag+] = [] spK Cl = 102.02 10 0.100  = 2.02 x 10-9 mol dm-3 2 Kc = -9 2 3 0.10 2.02? 0 譡NH ] = 1.50 x 107 [NH 3] at equlibrium = 1.82 mol dm3 No. of mol of NH 3 for complexation = = 2.00 x 103 Total mol of NH3 = 0.0182 + 2.00 x 10 3 [NH 3]total = 0.0202 10 1000 = 2.02 mol dm3 Organic layer Aqueous layer 918

4 ©chemistry@meridian jc (ii) 1 ∆G ppt = = 2.303 x 8.31 x 298 x lg 3.01 x 10–12 = --65.7 kJ mol-1 2 Precipitation is feasible hence, Ag2CrO4 is insoluble in water (bi) [Ag+] (0.5) = 2.02 x 103 [Ag+] = 4.04 x 1010 mol dm3 (ii) E = 0.80 + 0.059 lg 4.04 x 10 10 = +0.25 V (iii) KCl (aq) dissociates in water to form oppositely charged ions to maintain electrical neutrality or KCl (aq) is acting like a salt bridge since it dissociates in water to form oppositely charged ions. (ci) NaBr or sodium bromide (ii) 2 HBr + H2SO4  Br2 + SO2 + 2H2O or conc H3PO4 is not an oxidising agent (iii) Disproportionation : 6 OH + 3 Br2  5 Br- + BrO3  + 3 H2O (iv) Bond energies are 431, 366 and 299 kJ mol1 respectively. Bond strength of H–X decreases hence bond dissociation energy decreases . Hence, thermal stability of the HX decreases down the group. 4 (ai) The dispersant can form hydrogen bonds with water molecules. The hydrophobic groups form weak Van der Waal s’ forces of attraction with the oil droplets. (ii) Hydrogen Bonding + + + + + +     919

5 ©chemistry@meridian jc (bi) For reaction II: The electron withdrawing C=O group on the dienophile causes it to be more electron- deficient, thus making the dienophile a stronger electrophile. Hence, rxn is faster. Fo r reaction I: The electron donating alkyl group on the dienophile causes it to be less electron-deficient,thus making the dineophile a weaker electrophile. (ii (ci) Order of reaction wrt CH2=C(CH3)CH=CH2 (or isoprene) is 1. (ii) Rate = k[CH2=C(CH3)CH=CH2][Cl2] (iii) Product F (iv) Step 1: Cl Cl Cl-+Cl Step 2: slow fast + - compound F + 920

6 ©chemistry@meridian jc (d) Step 1: Add dilute H2SO4, KMnO4, heat to each compound separately Step 2: Add aq I2, (excess) NaOH, heat to the reaction products Observations: Yellow precip itate will be formed for compound E but not F 5(ai) Dilute or aq HCl (or dilute H2SO4) (ii) Br2 in CCl4 (or aq. Br2) , r.t.p. (iii) H2 with Ni catalyst at high temperature and pressure (b) CH3Cl (c) For acyl bromide. the electron defici ent carbonyl carbon in –COBr is bonded to two electronegative atoms.Hence, –C OBr group is most susceptible to nucleophilic attack, thus it is most reactive. For aryl bromide, the p-orbital of the bromine atom interacts with the  electron cloud of the benzene ring.This strengthens the C-Br bond t hus, rendering nucleophile substitution difficult hence least reactive. (d) Add dilute HC l to each compound and heat .Distill the products and add aqueous Na 2CO3 to each mixture. Gas evolved forms a white ppt with Ca(OH)2. 6(a) [Cr(H2O)6]3+ + 3 OH– Cr(OH)3 + 6 H2O or Cr(H2O)6]3+ + 3 OH– Cr(OH)3(H2O)3 + 3 H2O [Cr(H2O)6]3+ + 6 OH– [Cr(OH)6]3– + 6 H2O or Cr(OH)3(H2O)3 + 3 OH– [Cr(OH)6]3– + 3 H2O (b) Cr2O3 + 3 C + 3 Cl2  2 CrCl3 + 3 CO (c) 6 (d) [Cr(H2O)6]2+ 921

7 ©chemistry@meridian jc (ei)  Ligand exchange (ii) (f)  Cr3+ has partially filled 3d orbitals. The d orbitals are split into two groups via d-d splitting. d-d transition occurs where d e’s are promoted to the higher d orbital During the transition, the d elec tron absorbs wavelength from the yellow region of light and emits the remaining wavelength which appears as purple color. (g) 4.90 = n(n+2) ; n = -6 (rejected) or n=4 High Spin State for Cr2+     ___ ___ ___ ___ ___   (hi) It is harder/more difficult to reduce the Cr 3+ by adding an electron to an anion/negatively charged complex. (ii) Accept :Eo = < -0.41V 922