SAJC H2 CHEM P1 Ans Prelim

1 2016 SAJC H2 CHEM PRELIM PAPER 1 (Worked solutions) 1 2 3 4 5 6 7 8 9 10 C B A D B D C C C C 11 12 13 14 15 16 17 18 19 20 A D B A A B A B D B 21 22 23 24 25 26 27 28 29 30 B B B B B C B A A C 31 32 33 34 35 36 37 38 39 40 D D D C D D B B A D 1. Ratio of methane: ethene = 40 : 60 = 2 : 3 Let the total volume of mixture be y cm3 i.e. fraction of methane in mixture = 5 2 ; fraction of ethene in mixture = 5 3 CH4 + 2O2  CO2 + 2H2O C2H4 + 3O2  2CO2 + 2H2O Volume/cm3 5 2 y 5 2 y 5 3 y 2 x ( 5 3 y) Hence, total volume of CO2 = 5 2 y + 2 ( 5 3 y) = 5 8y cm3 2. From 2MnO4- (aq) + 5C2O42- (aq) + 16H+ (aq)  2Mn2+ (aq) + 10CO2 (g) + 8H2O (l), Reacting mole ratio of MnO4- : C2O42- = 2 : 5 --- (*) 1 mol of NaHC2O4.H2C2O4 gives 2 mol of C2O42-. 1.0 x 10-3 mol of NaHC2O4.H2C2O4 gives 2.0 x 10-3 mol of C2O42-. From (*), no. of moles of MnO4- reacted with 2.0 x 10-3 moles of C2O42- = 5 2 x 2.0 x 10-3 = 0.008 mol Hence, volume of KMnO4 solution required = 2.0 008.0 = 0.400 dm3 = 40 cm3

2 3. pV = nRT pV = rM m RT Option A: Shape of graph is correct. rM m R is the gradient. Since X has a higher Mr, its gradient is less steep. (Correct) Option B: Shape of graph is correct. From above, since X has a higher Mr, its gradient should be less steep. Hence graph is wrong. Option C: Shape of graph is wrong. The graphs should be a horizontal line. Since T is constant, pV is constant for each compound. Option D: Shape of graph is wrong. The graphs should be a vertical line. Since T is constant, pV is constant for each compound. 4. Element A belongs to Group V. This is because the first big jump in ionisation energy occurs between the 5th and 6th I.E., indicating that the 6th electron is removed from the inner quantum shell. Valence configuration: ns2np3 Element B belongs to Group IV. This is because the first big jump in ionisation energy occurs between the 4th and 5th I.E., indicating that the 5th electron is removed from the inner quantum shell. Valence configuration: ns2np2 Element C may belongs to Group VIII. This is because there is no noticeable big jump between the ionisation energies. Element D belongs to Group II. This is because the first big jump in ionisation energy occurs between the 2 nd and 3rd I.E., indicating that the 3 th electron is removed from the inner quantum shell. Hence its valence electronic configuration is ns2 and has no p electrons in the valence shell. 5. Option A is correct as the 2 hydrogen bonds and 2 O-H covalent bonds would form a bond angle of 109.5. Option B is wrong as two lone pairs from each oxygen atom are involved in hydrogen bonding. Option C is correct as intermolecular forces of attraction are wea ker than strong covalent bonds. Option D is correct and this is the explanation why ice floats on water.

3 6. Option A is wrong as 1st ionisation energy is endothermic. Option B is wrong as lattice dissociation energy (or opposite of lattice energy) is endothermic. Option C is wrong as bond breaking (breaking the N≡ N bond) is endothermic. Option D is correct as 1 st electron affinity releases heat due to electrostatic forces of attraction between the nucleus of F and electron. 7. NH3 deviate the most from ideal gas because it has stronger hydrogen bonding between the molecules than induced dipole-induced dipole interactions which exist in C2H6 and Ne. Large and heavy gas particles with significant electron cloud size deviate more than smaller gas particles (Ne vs C2H6 ; C2H6 has a larger electron cloud size). The greater the electron cloud size, the stronger the van der Waals’ forces of attraction between the gas particles. Thus Ne has the least deviation. Hence, gas 1 is Ne, gas 2 is C2H6, gas 3 is NH3. 8. Option A has the wrong shape for a first order reaction. Option B is wrong as the constant half-life is not stated. Hence, it does not confirm whether it is a first order or second order reaction. Option C is correct. Rate is directly proportional to concentration of NH4Cl. Option D is wrong as rate should increase when concentration of NH4Cl is high. 9. Enzyme-substrate complex is an intermediate. Hence it cannot appear in the rate equation. The rate equation should be: rate = k [enzyme] [substrate]. 10. Option A is wrong as a catalyst will lower the activation energy of both forward and backward reactions by the same magnitude. Option B is wrong as equilibrium constant is only affected by a change in temperature. Option C is correct as when the temperature decreases, the rate constants of both forward and backward reactions will decrease. The position of equilibrium will shift to the right to release heat. Since forward reaction is favoured, the rate constant of the backward reaction will decrease more tha n that for the forward reaction.

4 Option D is wrong as the rate constants of both forward and backward reactions should increase when temperature increases. 11. Option A is correct as this is a weak acid -strong base titration, which will result in the for mation of carboxylate salt that undergoes basic hydrolysis to give a pH > 7 at equivalence point. 12. S = [Mg2+]3 [AsO43-]2 Let the solubility of Mg3(AsO4)2 be x mol dm-3 S = (3x)3 (2x)2  108 x5 = S x = (1/108 S)1/5 [AsO43-] = 2x = (32/108 S)1/5 = (8/27 S)1/5 13. Option A is wrong as when NaCN is added to the Fe3+/Fe2+ half cell, the half equation to look at is as given below. Therefore Eθcell is +0.36 – (+0.34) = + 0.02 V. Option B is correct. Since Cu2+/Cu has a more positive Eθ value, it will be the cathode where reduction takes place. Therefore, Cu will be the positive electrode. Option C is wrong. Eθcell before adding excess aq NH3 = (+0.77) – (+0.34) = + 0.43 V Eθcell after adding excess aq NH3 = (+0.77) – (–0.05) = + 0.82 V (more spontaneous)

5 Option D is wrong. Since Cu 2+/Cu is the anode, this means that Cu 2+ will be formed and to maintain electrical neutrality, anions in the salt bridge will move over instead. 14. period 3 elements Mg Al Si P S period 4 elements Ca Ga Ge As Se MgO dissolves in water to form alkaline solution. Al2O3 and SiO2 do not dissolve in water, hence the solution remains neutral. P4O6 P4O10 Both oxides of phosphorous react vigorously with water to form acidic solutions. P4O6(s) + 6H2O(l)  4H3PO3 *(aq) P4O10(s) + 6H2O(l)  4H3PO4(aq) * H3PO3 is a weak dibasic acid. Only 2 of the H atoms are acidic. SO2 SO3 SO2 dissolves readily in water to form an acidic solution of sulfurous acid. SO3 reacts exothermically with water to form sulfuric acid, which is a strongly acidic solution. SO2(g) + H2O(l)  H2SO3(aq) SO3(g) + H2O(l)  H2SO4(aq) 15.  W has the highest melting point among all the Period 3 elements. Hence W is Si.  The chloride of X is neutral. Hence X is Na.  The oxide of Z is amphoteric. Hence Z is Al.  The chloride of Y is acidic and the oxide of Y is basic. Hence Y is Mg. Option A is correct: The atomic radius increases in the following order: Si < Al < Mg < Na which is W < Z < Y < X. Option B is wrong: The first ionisation energy should decrease in the following order: Si > Mg > Al > Na which is W > Y > Z > X (From the data booklet). Option C is wrong: The pH of the chlorides should decrease in the following order: Na > Mg > Al > Si which is X > Y > Z > W. Option D is wrong: The electrical conductivity should increase in the following order: Si < Na < Mg < Al which is W < X < Y < Z; From Na to Al, number of delocalised valence electrons increases from 1 to 3, therefore there is an increase in electrical conductivity of these metals. Si is a metalloid (semi–metal), thus it is not a good conductor (it is a semi –conductor with low conducti