EJC 2020 JC2 Prelim H2 Chemistry Paper 1 Worked Solution

2020 JC2 Prelim Exam H2 Chemistry 9729 Paper 1 Worked Solution 1 angle of deflection q m  ion 7Li+ 9Be2+ 16O2– 14N3– q m . 1 7 0 143   . 2 9 0 222   . 2 16 0 125   . 3 14 0 215    B 2 1 The electronic configuration of Cr is [Ar] 3d5 4s1. The five d -electrons occupy each d-orbital singly. 2 The electronic configuration of Mn 2+ is [Ar] 3d5. The five d -electrons occupy each d-orbital singly. 3 The electronic configuration of both Fe2+ and Co3+ are [Ar] 3d6.  C 3  D 4 1 Al2O3 is amphoteric. 2 Charge density of A l3+ is higher than Na+, which gives it higher polarising power, hence imparting higher degree of covalent character to the Al–O bond. 3 latt qqH rr     Since 3A Naqq l and 3A Narr l     latt 2 3 latt 2A O Na OHH   l  Hlatt of Al2O3 is more exothermic.  A 5 mpV nRT pV RT M m RT RTp V M M        5 3 1 gradient 3 10 Pa 8.31 500 1.5 1000 g m 20.8 g mol RT M M M       Neon has a molar mass closest to this.  B 6  24 3Co SO 20.3 0.05 mol406n  A  3 24 3Co SOCo 2nn   and  2 2443 Co SOSO 3nn    32 2443ions Co SOCo SO 5 0.25 moln n n n     23 ionsno. of ions 1.5 10nL    B  3 24 3Co SOCo 2 2 0.05 0.1 molnn      C 2 58.9mass % of Co 100% 29.3%406    D 4 3 16.0mass % of O 100% 47.3%406     C 7 4BrO 20.0 reacted 0.02 0.0004 mol1000n     2NH OH 80.0 reacted 0.01 0.0008 mol1000n     42BrO 2NH OH 6 e  Every 1 mol of NH2OH lose 3 mol of e–  2O.N. of N in NH OH 3 1 2 1    O.N. of N in product 1 3 2    Product is NO (O.N. of N = +2)  A 8 1 Both processes refers to the enthalpy change to form one mole of gaseous metal atoms from one mole of the metal in its standard state. 2  sublimation fusion vapourisation ,m.p. b.p.H H H H      l The energy required to raise the temperature of the molten metal from its melting point to its boiling point is not taken into account. 3 Xenon can be a solid at temperatures below its melting point.  A 9 rS 0 since there is a decrease in the number of gaseous particles. rG rH rTS rH  rTS can be either positive or negative pending on T (which is always >0). rG r0, when HT  rS  C 10 Given that   rate k EF . A Gradient of graph = instantaneous rate, should decrease as reaction proceeds, since E and F are being consumed. B [G] which is the product concentration should increase as reaction proceeds. Since [E]0 = [F]0 and 1E ≡ 1F, therefore [E] = [F]. So the rate equation can be rewritten        22 rate k k k  E F E F . Plot of [ E][F] against t is the same as plot of [E]2 against t, for a second order reaction with rate equation   2 rate k E . Since plot of [E] against t is a curve. C Plot of [ E]2 against t will not be a straight line. D Plot of [E]2 against t is also a curve.  D 11 1 Step II is the rate-determining step. From Step II:           32 c 2 2 3 2 2 3 rate H O Br H O H O Br H O H O Br k kK k           where ck kK which corresponds to the experimentally- determined rate equation given. Summing up step I to III gives 2H2O2  O2 + 2H2O which is the equation for the decomposition of H2O2. H3O+ and Br– are not consumed in the reaction. 2 Since H 3O+ and Br –, arising from the dissociation of HBr(aq), appear in the rate equation (i.e. speed up reaction) , but are not consumed, HBr(aq) serves as a catalyst. 3  3H O and Br remains constant. Hence, reaction is pseudo -zero order w.r.t. H3O+ and Br–. The rate equation can be rewritten as  22rate H Ok , where  3H O Brkk    .  C 12 A Adding a catalyst lowers the activation energy of both the forward and backward reaction, hence increasing the rate for both. B Decreasing volume of vessel causes the concentration/partial pressure of all species to increase. The equilibrium shifts to the left and when equation is re-established, the [N 2], [H2] and [NH 3] are all still higher than before the change and hence the rate of both forward and backward reaction s are increased. C Increasing the temperature increases the rate of both the forward and backward reaction. D Removing gaseous NH 3 causes the equilibrium to shift to the left and hence the [N 2], [H 2] and [NH 3] when equilibrium is re -established will be lower than before the change and hence, the rate of both forward and backward reaction will decrease.  D 13 N2O4(g)  2NO2(g) initial partial pressure / atm 0.24 0 change in partial pressure / atm −x +2x equilibrium partial pressure / atm 0.24 − x 2x total 0.24 2 0.24 0.32 atm 0.08 atm p x x x x        24NOeqm 0.24 0.08 0.16 atmp   D 14 An acidic/alkaline buffer is made up of a weak acid/base with its conjugate base/conjugate acid. There is no conjugate acid-base pair in the mixture of potassium chloride and potassium ethanoate. Hence, the mixture is not a buffer.  C

15 Let the solubility be z mol dm–3.    3 3 3A mol dm ; Y 3 mol dmzz   l       1 4 3 33 3 4 sp 3 A Y 3 3 3 K z z z z      lK K  C 16 Since Zn is being oxidised to Zn 2+ oxidationE E  2Zn Zn 0.76 V  For the reduction by Zn to be spontaneous, reductionE 0.76 V , so that cellE 0 . E   2 2VO VO 1.00 V; E   23VO V 0.34 V E    32V V 0.26 V; E   2V V 1.20 V  Therefore V2+ is the final product formed.  A 17 A To form oxygen, water must be oxidised i.e. 2H2O  O2 + 4H+ + 4e– B Since Cr 2O72− is being reduced to chrome to chrome -plate the bumper, the bumper is the cathode. C Anode: 2H2O  O2 + 4H+ + 4e– Cathode: Cr2O72 + 14H + + 12 e  2Cr + 7H2O Cr2O72  12e  2Cr  3O2 Hence, for every 52g (1 mol) of Cr, 1.5 mol of O2 produced. D For every 52g (1 mol) of Cr, 6 mol of electrons are required. 6 96500 57900 s 16 h10 Q t nF nFt      I I  D 18 1 cellE E  32Mn Mn E      3 2MnO Mn 1.54 0.95 0.54 V 0       2 cellE E  32Mn Mn E      2Mn Mn 1.54 1.18 2.72 V 0       3 cellE E   2 2MnO Mn E      2Mn Mn 0.95 1.18 2.13 V 0        B 19 A As effective nuclear charge increases across Period 3, electronegativity also increases. B Electrical conductivity i ncreases across Groups 1 to 3 due to an increase in number of delocalised valence electrons contributed per atom. C The melting point decreases down Group 1 because metallic bonding weakens with decreasing charge density down the group. D Down the group, the first ioni sation energy of the elements generally decreases as the valence electron is further away from the nucleus and hence less electrostatic attraction.  D 20 1 The 2nd IE of Group 2 elements is only about twice that of the 1 st IE, while the 3rd IE is much higher than the 2 nd IE. Hence the only oxidation number exhibited is +2, involving loss of 2 e–. 2 As atomic radius increases down Group 2, tendency of the atoms to loss electrons (i.e. to be oxidised) increases, hence reducing strength increases. 3 22Mg Barr   charge density of Mg2+ is higher than that of Ba 2+. Stronger ion- dipole interactions with water and hence more exothermic Hhyd for Mg2+.  B 21 A H2SO4 is only reduced to either SO 2 or H2S, but not oxidised. B There is no transfer of H + in both equations shown. C Br− is a weaker reducing agent than I− as it can only reduce S to the +4 O.S. in SO2, while I− can reduce S to the −2 O.S. in H2S. D Reducing power increases down Group 17, so C l− should be an even weaker reducing agent than Br −.  D 22  B 23 A B (CH3)3CO+H2 (acid) loses a proton (H+)