TJC H2 Chem 2013 Prelim P3 Soln

9647/03/TJC Prelim 2013 1 9647 TJC H2 Chemistry Prelim P3 Answers: 1 (a)  M(NO 3)2(s) → MO(s) + 2NO2(g) + ½O2(g)  Charge density of the cation decreases down the group as charge of cation is constant and ionic radius increases.  Polarising power of the cation decreases resulting in less distortion of the electron cloud of the NO3 - ion, and the N-O covalent bond is weakened to a smaller extent. Therefore, ease of thermal decomposition decreases down the group. (b) (i)  Mg 2+(aq) + 2OH-(aq) → Mg(OH) 2(s) A  A l3+(aq) + 3OH-(aq) → A l(OH)3(s)  A l(OH)3 + OH-(aq) → A l(OH)4 -(aq) B (ii)  When carbon dioxide is passed into Q, the following reaction occurs. CO 2(g) + OH-(aq) → HCO 3 - (aq)  The concentration of OH - decreases, causing the position of the equilibrium below to shift to the left according to Le Chatelier’s principle. Thus Al(OH)3 (white solid C) is precipitated. A l(OH)3 + OH-(aq) A l(OH)4 -(aq) Or  Carbon dioxide is an acidic gas, and there will be an acid/base reaction with the Al(OH)4 - complex to form salt and water.  CO 2(g) + Al(OH)4 -(aq) → HCO 3 -(aq) + Al(OH)3(s) C or 2CO2(g) + 2Al(OH)4 -(aq) → 2HCO 3 -(aq) + Al2O3(s) + 3H2O(l) C (iii) Residue A is Mg(OH)2.  Mass of Mg = 24.3/58.3 x 0.18 = 0.0750 g % by mass Mg = 0.075/1.75 x 100% = 4.29 % After heating to constant mass, the residue is Al2O3(s).  Mass of A l = 54/102 x 3.16 = 1.67 g % by mass A l = 1.67/1.75 x 100% = 95.4% (c) (i)  (ii) 2O3 3O 2 1381

9647/03/TJC Prelim 2013 2 At t = 0 s, vol. of O3 = 38 cm3 and vol. of O2 = 762 cm3 At t = 1500 s, let vol. of O3 = (38 – y) cm3 and vol. of O2 = (3y/2 + 762) cm3 (38 – y) + (3y/2 + 762) = 818 y = 36.0  Volume of O3 = 38 – 36 = 2.00 cm3 (iii) Plot of Volume of O3 vs time. (correct labeling of axes, plotting of points)  Both half-lives are constant = 400 s (indicate at least 2 t 1/2 on graph)  Therefore, the reaction is first order with respect to ozone. (iv) k = ln2 / t1/2 k = 1.73 x 10-3 s-1 Rate = k[O3] 1382

9647/03/TJC Prelim 2013 3 (v)  A catalyst provides alternative pathway of lower activation energy, which is the minimum amount of energy the particles must collide with for reaction to take place.  The number of molecules having energy greater than or equal to activation energy increases, resulting in an increase in frequency of effective collisions and the rate of decomposition.  [Total: 20] 2 (a) (i)  The primary structure of a protein refers to the number and sequence of amino acids in the polypeptide chains.  The amino acids are bonded by peptide bonds in the polypeptide chain. (ii)  The -helix takes the shape of an extended spiral spring that is held in place by hydrogen bonds between the -C=O of each amino acid the –NH of the amino acid 4 amino acids away.  The hydrophobic R groups of the amino acid point out of the helix and are perpendicular to the main axis of the helix. 1383

9647/03/TJC Prelim 2013 4  Drawing of alpha helix with hydrogen bond: (iii)  The formation of a disulphide bond requires 2 cysteine residues within the polypeptide chain but the -sub-unit only has one. (iv)  When the pH is decreased, the histidine residue can be protonated. This causes ionic bonds to form between the positively charged –R group of the histidine residue and negatively charged –R groups on other amino acid residues in the polypeptide chain.  This results in a change in the conformation and folding of the sub-units which decreases the affinity that haemoglobin has on oxygen. (v)  In the lungs, the concentration of O 2 is high. By Le Chatelier’s principle, the position of equilibrium for the reaction shift to the right causing oxygen to bind to haemoglobin and increasing the concentration of oxyhaemoglobin.  Oxyhaemoglobin travels to the tissues in the body where the concentration of O2 is low. By Le Chatelier’s principle the position of equilibrium for the reaction shift left, releasing oxygen to the surrounding tissues. (b) (i)  –COOH group in lysine is a carboxylic acid group and thus has the lowest pKa value of 2.15. The –CO 2H group can deprotonate easily as the carboxylate ion is stable as the negative charge can be delocalised over the 2 O atoms.  –NH 3 + group on the -carbon is a stronger acid than the –NH 3 + group at the end of the carbon chain. The –NH 3 + group on the  -carbon is closer to the electron withdrawing –COOH group and experiences a stronger inductive effect, the N–H bond is more polarised and a proton is lost more readily.  Hence the –NH 2 group on the  -carbon has a pK a of 9.16 and the –NH 2 group at the end of the carbon chain, being the strongest base, has a pK a of 10.67. Hydrogen Bond 1384

9647/03/TJC Prelim 2013 5 (ii)  Isoelectric point occurs at pH = 9.92 (i.e. 2 10.67)(9.16 )  (c) (i) Concentration of O2 in solution = 769 1 0.2 = 2.60 x 10–4 mol dm–3 Concentration of CO in solution = 1052 1 0.03 = 2.85 x 10–5 mol dm–3 (ii) Let initial concentration of Hb(O2) = n mol dm-3 Hb(O2)(aq) + CO(aq) Hb(CO)(aq) + O2(aq) Initial conc. / mol dm–3 n 0 Change in conc. / mol dm –3 – 24 23n + 24 23n Equil conc. / mol dm–3 24 n 24 23n ● Kc = )][CO][Hb(O ][Hb(CO)][O 2 2 =   5 - 4 - 10 2.8524 n 10 2.6024 23n           ● = 210 (iii) At toxic levels of carbon monoxide, )] [Hb(O [Hb(CO)] 2 = 90 10 Kc = )][CO][Hb(O ][Hb(CO)][O 2 2 210 = (90)[CO] ) 10(10)(2.60 -4  1385

9647/03/TJC Prelim 2013 6  [CO] = 1.38 x 10 -7 mol dm–3 PCO = 1052 x (1.38 x 10 –7) = 1.45 x 10 –4 atm  Percentage of CO in air to cause toxicity = 1001 0.000145  = 0.0145 % [Total: 20] 3 (a) (i) Anode: C2H5OH + 3H2O → 2CO2 + 12H+ + 12e- Cathode: 4H+ + O2 + 4e- → 2H2O Overall: C2H5OH + 3O2 → 2CO2 + 3H2O ● (ii) ● E θ O2/H2O = +1.23 V Eθ cell = E θ O2/H2O - Eθ CO2/C2H5OH 1.61 = 1.23 - E θ CO2/C2H5OH ● E θ CO2/C2H5OH = -0.38 V (iii) ● Ethanol is more energy-efficient than methanol. Or ● Ethanol can be obtained in great quantity (or easily obtained) from a fermentation process. (b) (i) Number of moles of copper deposited = 0.5 mol S i n c e C u ≡ 2 mol of e - Number of moles of electrons passed = 1 mol Since CO 2 ≡ 6 mol of e - Number of moles of CO2 produced = 1/6 mol Volume of CO2 produced = 1/6 x 24 000 = 4000 cm 3 (ii) Since number of moles of electrons passed = 1 mol Amount of charge passed = 96500 c Q = I t 96500 = I x 8 x 60 x 60 I = 3.35 A (c) (i) Amount of carbon dioxide = 2300/44 = 52.27 mol (ii) PV = nRT P = nRT/V = (52.27 x 8.31 x 298)/(2.5 x 10 -3) = 5.18 x 10 7 Pa ● ● ● ● ● ● ● 1386

9647/03/TJC Prelim 2013 7 (iii) ● The calculation in (ii) assumes that CO 2 behaves ideally i.e. there are no forces of attraction between the molecules. ● However, due to the van der Waals’ forces of attraction between the CO 2 molecules, this resulted in the molecules colliding against the cylinder wall with a smaller force. As a result, the pressure of the gas is smaller than expected. (d) (i)  D has both O and N atoms but is neutral  D is n