JJC H2 Chem 2013 Prelim P3 Soln

© Jurong Junior College 9647/03/PRELIM/2013 2013 JC2 H2 Chemistry Preliminary Examination Paper 3 Suggested Answers 1 (a) (i) Pb(s) + 4HNO3(aq)  Pb(NO3)2(aq) + 2NO2(g) + 2H2O(l) (ii) Al3+ left in solution X exists as Al(H2O)6 3+. Due to high charge density of A l3+, it hydrolyses in water to give a weakly acidic solution. Al3+ polarises the electron cloud of H2O molecule attached to it which will weaken and break its OH bond, releasing H+. Al(H2O)6 3+  Al(H2O)5(OH)2+ + H+ (iii) Since Al2(SO4)3  2Al(H2O)6 3+, [Al(H2O)6 3+] in solution X =  602 0.10060 + 40 = 0. 120 mol dm3 [H+] = Kacid ac? = ..  50120 14 10 = 1.30  103 mol dm3 (b) (i) Hf(NO2(g)) = 0.5 [(+183) + (116)] = +33.5 kJ mol1 (ii) Hr = 3(+33.5) + 1 2 (256) + 1 2 (183) = 137 kJ mol1 (c) (i) (ii) HNO3 can accept H + due to the availability of the lone pair of electrons on O atom of –OH group. After accepting H + from H 2SO4, the unstable H 2NO3 + intermediate formed quickly breaks down to NO2 + and H2O. O N O O H Hr 3NO2(g) + H2O(l)  2HNO3(aq) + NO(g) 3 2 N2(g) + 3O2(g) + H2O(l) 1 2 (+183) 1 2 (256) 3(+33.5) 630

2 © Jurong Junior College 9647/03/PRELIM/2013 1 (d) OR O + HNO 3 O NO2 +H 2O Sn, conc HCl, heat under reflux, followed by NaOH(aq) (e) (i) A weak acid dissociates partially in water. HCN  H+ + CN (ii) At high [H+], equilibrium position of HCN  H + + CN  shifts left, resulting in low [CN ]which slows down step 1. At low [H +], equilibrium position of HCN  H+ + CN shifts right, resulting in low [HCN] which slows down step 2. (iii) I: condensation II: reduction (iv) 631

3 © Jurong Junior College 9647/03/PRELIM/2013 2 (a) (i) Since O 2– has a smaller radius, it is less polarisable than C l–. Hence, Fe 2O3 is more thermally stable/ decomposes at a higher temperature. (ii) Fe2O3 has giant ionic structure while FeCl3 has simple covalent structure. Much larger amount of energy is required to overcome the strong ionic bonds between Fe 3+ and O 2– as compared to that required to overcome the weak van der Waals’ forces between FeCl3 molecules. (b) (i) D : CuCl E: Cu(H2O)6 2+ (ii) Cu(I) in D has a d 10 configuration/ completely filled d–orbitals and hence, electron transition between d–orbitals is not possible. Thus, CuCl is white in colour. Cu(II) in E has a d 9 configuration/ partially filled d–orbital and hence, electrons transition between d–orbitals is possible. In Cu(II) complex ion, the presence of ligands causes the five 3d orbitals to split into 2 sets of different energies. The difference in energies between the 2 sets of 3d orbitals is relatively small such that radiation from the visible regi on region of the electromagnetic spectrum when an electron moves from a d–orbital of lower energy to another partially–filled d–orbital of higher energy. Hence, the Cu( II) compounds are coloured and the colour observed is the complement of the colours absorbed. (c) (i) Electrophilic substitution (ii) Step 3: acidified K2Cr2O7(aq), heat under reflux Step 4: PCl5(s) OR PCl3, heat OR SOCl2, heat OR PBr3, heat OR SOBr2, heat (iii) (iv) E(C–I) = + 240 kJ mol–1 E(C–Cl) = + 340 kJ mol–1 Since E(C-I) is smaller than E(C–Cl) and thus C– I bond is weaker than the C–Cl bond, it is easier to break C– I bond. Hence, F should contain the C l atom, rather than I atom. OR (Atomic) radius of I = 0.133 nm; (Atomic) radius of Cl = 0.099 nm Due to larger I atom, C– I bond is longer and weaker than the C–Cl bond, it is easier to break C– I bond. Hence, F should contain the Cl atom, rather than I atom. (v) Test : (1) Add NaOH(aq) to a sample of F and heat. (2) Acidify the mixture with HNO3(aq) to remove excess NaOH. (3) To the resulting mixture, add AgNO3(aq). Observation: If the student’s suggestion is not correct, white ppt of AgC l will be formed instead of yellow ppt of AgI. 632

4 © Jurong Junior College 9647/03/PRELIM/2013 2 (c) (vi) The electron–withdrawing C=O in H is directly bonded to benzene ring, making the benzene ring less electron–rich and thus less susceptible towards electrophilic attack. Since C+ from J is directly attached to an electronegative O atom, it is highly electron deficient and thus a stronger electrophile than C+ from H. Hence, H would require a harsher condition such as higher temperature or longer period of heating for Friedal–Crafts reaction to occur, as compared to J. 3 (a) (i) 1 (ii) CH3CH2OH(aq) + O2(g)  CH3COOH(aq) + H2O(l) (iii) When [CH 3CH2OH] increases, the equilibrium position of CH3COOH + 4e  + 4H +  CH 3CH2OH + H 2O shifts left so as to react away some CH3CH2OH, causing E(CH3COOH/CH3CH2OH) to become more negative. Hence, the voltage of the breathalyser will become more positive. (b) Silver nitrate catalyst provides an alte rnative reaction path of lower activation energy ( Ea’) than that of the uncatalysed reaction. Thus, more particles have the minimum energy required to react. Therefore, the frequency of effective collisions between particles with energy  Ea’ increases and hence, the reaction rate increases (i.e. test result is obtained within a short period of time). (ii) K: K3Cr(OH)6 OR Cr(OH)6 3 L: K2CrO4 OR CrO4 2 (iii) Cr2O7 2 + 14H+ + 6e  Cr3+ + 7H2O E= +1.33 V O2 + 2H+ + 2e  H2O2 E= +0.68 V E cell = (+1.33) – (+0.68) = +0.65 V >0  energetically feasible Cr2O7 2 will be reduced by H2O2 to form back Cr3+ if H2O2 is not removed. EaEa’ No. of particles 0 No. of particles with energy  Ea No. of particles with energy  Ea’ Energy/ kJ mol1 633

5 © Jurong Junior College 9647/03/PRELIM/2013 3 (b) (iv) When heated, H2O2 readily decomposes to O 2 and H2O. Heating can be stopped when there is no more oxygen gas evolved to relight a glowing splint. (c) (i) (ii) Extreme heat will disrupt the van der Waals’ forces in the tertiary and quaternary structures and the hydrogen bonds in the secondary, tertiary and quaternary structures of the protein. This alters the shape of the active site of the enzyme. The enzyme is denatured and loses its catalytic activity. Hence, the rate falls. (d) (i) S is positive as protein unfolding proceeds with an increase in disorder when the protein unfolds from its regular structure into random coils. (ii) G = H  TS 0 = (+200) – T(+0.600) T = 333 K 3 (d) (iii) Using G = H  TS, G = (+200) – 340(+0.600) = 4.00 kJ mol1 Using G = RT ln Kc, 4.00  103 =  8.31  340  ln Kc Kc = 4.12 Kc = eqm eqm [unfolded state] [folded state] = x x 4.12 Proportion of folded protein = eqm eqm eqm [folded state] [folded state] + [unfolded state] = x x x+ 4.12 = 0.195 (e) Since –CH(CH 3)CH2CH3 side chain of isoleucine residue is larger and thus has a greater number of electrons/ larger electron cloud as compared to –CH 3 side chain of alanine residue, its side chain forms stronger van der Waals’ forces with the side chain of another amino acid residue. Hence, magnitude of H for the unfolding of the mutant protein is most likely higher. 634

6 © Jurong Junior College 9647/03/PRELIM/2013 4 (a) (i) Crushing of solid into fine powder would increase the total surface area and increase the frequency of effective collisions between reactant particles. Hence, this increases the rate of dissolving solid. (ii) M(OH)2 + 2HCl  MCl2 + 2H2O  M l2Since OH 2HC , M  3 2 12 5 . 0amt of OH in 27.90 cm 0.0002502 1000  M x     6 6 2 3.13 10 mol 27.90solubility of OH , 3.13 10 1000 = 1.20 x 10 4 mol dm3 M(OH) 2(s)  M2+(aq) + 2OH –(aq) eqm conc /mol dm3 x 2 x where x is the solubili