AJC H2 CHEM P3 ANS

ANDERSON JUNIOR COLLEGE 2009 PRELIMINARY EXAMINATION HIGHER 2 CHEMISTRY PAPER 3 SOLUTIONS AJC 2009 Prelim H2 Chem Paper 3 Solutions 1 1 (a) (i) CH3COCH3 (l) + 4O2 (g)  3CO2 (g) + 3H2O (l) (ii) Heat absorbed by water = 0.70 x 0.050 x 1790 = 62.65 kJ Let m be the maximum mass of water brought to boiling. Heat absorbed by water = mc∆T = m(4.20)(100-15) = 357m J 357m = 62 650 m = 175 g (iii) Heat loss to surroundings. (iv) Since the forward reaction is accompanied by a decrease in number of moles of gaseous particles (n = -1) and there is less disorder (or disorder decreases) which results in a negative entropy change. (v) ΔG = ΔH - TΔS Since ΔS = negative and ΔH = negative, a high temperature will give rise to an increase in  TΔS  such that  TΔS  >  ΔH  , ΔG becomes positive and therefore reaction is not feasible. (b) (i) ΔH at (propanone) CH 3COCH3(l) 3C(g) + 6H(g) + O(g) 3(+715) - 248 6(+218) +249 3C(s) + 3H 2(g) + ½ O2(g) By Hess’ Law, ΔH  at (propanone) = + 3950 kJ mol-1 (ii) ΔHat (propanone) = 2BE(C-C) + 6BE(C-H) + BE(C=O) = 2(+350) + 6(+410) + (+740) = + 3900 kJ mol-1 (iii) ΔH vaporisation (propanone) = + 3950 – 3900 = + 50 kJ mol-1

(c) 2 (a) (i) Nucleophilic substitution C NH2 H ClNC H C NH2 H H ClNC NC C NH2 H H + Cltransition state C---C bond partially formed C---Cl bond partially broken (ii) Reaction II : LiAlH4 in dry ether (or anhydrous) OR H2(g) over Pt catalyst Reaction III : (limiting) Cl2, presence of uv light (or heat) Reaction IV : KOH in ethanol (or alcoholic KOH), reflux CH3COCH3, CH3CH2COONH4, C4H9CHO, CH CH =CHCH (OH) To each sample, add Br2(aq) in the dark AJC 2009 Prelim H2 Chem Paper 3 Solutions 2 CH3CH2=CHCH2(OH) White fu CH3COCH3,CH3CH2COONH4, C H CHO No white fumes observed CH3CH2COONH4 CH3COCH3, C4H9CHO, To each sample, add aqueous NaOH and warm No gas evolved To each sample, add Tollen’s reagent and warm C4H9CHO Silver mirror CH3CH2COCH2CH3 No silver mirror add 2,4-dinitrophenylhydrazine Orange ppt Pungent gas evolved. Turns moist red litmus blue. Orange solution decolorizes Solution remains orange

(iii) H2NC HC H 2NH2 Cl (b) E Heat with NaOH (aq) and acidify with excess dilute HNO3. Add AgNO3 (aq). chloromethylamine : white ppt and E : cream ppt F Add Br2 (aq) F : decolourisation of brown Br2 (aq). (c) The p-orbital on C l can overlap with the  electron cloud of the aromatic ring, making the C–Cl bond stronger and less easily cleaved. (d) (i) Since 1 mol CH3N=NCH3  1 mol CH3CH3  1 mol N2 = 600 - x + x + x = 600 - x (Shown) (ii) 210 1350 1800 At any time t, if x cm3 of CH3N=NCH3 has reacted, then x cm3 of CH3CH3 and x cm3 of N2 would be formed. Hence volume of CH3N=NCH3 at any time t, V = initial total volume of gas – volume of CH3N=NCH3 reacted + volume of CH3CH3 formed + volume of N2 formed V Time / s 0 420 860 Vol of N2 / cm3 0 6 10 15 18 19 0 2 4 6 8 10 12 14 16 18 20 0 500 1000 1500 2000 time/s Volume of N2 /cm3 Constant half-life at 390 s  1st order reaction (i k -3 ii) = ln 2 / t1/2 = ln 2 / 390 = 1.65 x 10 AJC 2009 Prelim H2 Chem Paper 3 Solutions 3 s-1

AJC 2009 Prelim H2 Chem Paper 3 Solutions 4 < (e) As shown on the diagram, when temperature increases, the number of reactant molecules with energy greater or equal to the activation energy, Ea, will increase. This results in an increase in the frequency of effective collisions and the rate of reaction increases. 3 (a) (i) 2Ni(OH)2 + Cd(OH)2  2NiO(OH) + Cd + 2H2O (ii) No. of moles of Cd(OH)2 = 3.65 / (112 + 17 + 17) = 0.025 mol No. of moles of electrons = 0.025 x 2 = 0.05 mol Total amount of charge = 0.05 x 96500 = 4825 C Time taken = 4825 / 3 = 1610 s (26.8 min) (iii) nickel electrode (anode) : OH  O2 + 2H2O + 4e cadmium electrode (cathode) : 2H2O + 2e  H2 + 2OH (b) (i) The carbon electrode coated with MnO2 is the cathode, as the oxidation number of Mn changes from +4 in MnO2 to +3 in MnO(OH). (ii) E red = 1.60  0.76 = + 0.84 V (iii) No change in E cell as the concentration of solid zinc remains the same (iv) More environmentally friendly as cadmium is highly toxic (or no leakage / cheaper) (c) Colour in Mn compound is due to d-d transition. In the presence of ligands, the partially filled 3d orbitals split into two levels with a small energy gap within the visible light energy. An electron from the lower energy level d–orbital can absorb energy from visible light and move to a vacant higher energy d–orbital. The colour of the manganese compound is the complementary colour to the one absorbed. Zn 2+ has d10 configuration, no vacant d orbitals for d-d transition. (d) (i) KMnO4 (ii) Disproportionation (iii) No. of moles of MnO2 = 0.174 / (54.9 + 16 x 2) = 0.002 mol No. of moles of compound H = 40/1000 x 0.5 x 1/5 = 0.004 mol Represents no. of molecules with energy greater or equal to E Number of molecules with energy Ea Energ y (T ) a 1 T > T1 2 Represents no. o f molecules with energy greater or equal to E (T ) a 2 T2 T1

(iv) Let the oxidation number of X be n. Mn(n) + (n – 4) e–  Mn(IV) 0.002 0.002(n-4) 0.002 No of moles of electrons gained = 0.002 (n – 4) Mn(n)  Mn(VII) + (7-n)e– 0.004 0.004 0.004 (7 – n) No. of moles of electrons lost = 0.004 (7 – n) 4 (a) [H+] = 10-6.5 = 3.16 x 10-7 mol dm-3 10-3.86 = acid] lactic [ ) 10 x (3.162 -7  [lactic acid] = 7.24 x 10-10 mol dm-3 (b) (i) pH of buffer solution = pKa + lg ] a [ [salt] cid = 3.86 + lg 0.10 0.200 = 4.16 (ii) no. of moles of H+ added = 10 . 01000 10  = 1.00 x 10-3 mol [lactate]new = 100010.0 1.0 10 x 1.00 - 0.200-3  = 0.197 mol dm-3 [lactic acid]new = 100010.0 1.0 10 x 1.00 0.10-3   = 0.100 mol dm-3 new pH of buffer solution = pKa + lg ] a [ [salt] cid = 3.86 + lg 0.100 0.197 = 4.15 (c) In order of increasing pH: No. of moles of electrons gained = No. of moles of electrons lost 0.002 (n – 4) = 0.004 (7 – n) n = +6 Formula of G = K MnO2 4 OH OH O Cl OH Cl O OH OH O Has lowest pH / most acidic as it hydrolyses readily in water to produce acidic HCl. Presence of electron withdrawing Cl atom disperses the negative charge on the oxygen atom of OH O O Cl , thus stabilising it. AJC 2009 Prelim H2 Chem Paper 3 Solutions 5

(d) (i) dilute HNO3, room temperature (ii) Electrophilic substitution (iii) J K L , M OH COCl NO2 O-Na+ COO- Na+ NH2 OH C NO2 O NH NH2 (e) (i) When the first trace of Fe(OH)2 is formed, [Fe2+] [OH-]2  Ksp (Fe(OH)2) (0.010) [OH-]2  1.6  10-14 [OH-] = 1.265  10-6 mol dm-3 pH = 14 +