2019 ASRJC Prelim H2 Chem P1 ANS

ASRJC JC2 PRELIMS 2019 9729/01/H2 1 ANDERSON SERANGOON JUNIOR COLLEGE H2 Chemistry 9729 2019 JC2 Prelim Exam Paper 1 Worked Solutions 1 Answer: C 68Ge has 32 protons and 36 neutrons. When ‘a proton changes into a neutron’, 68X contains 31 protons and 37 neutrons. X is gallium (Ga) which is in Group 13, i.e. has 3 valence electrons, only 1 of which is in the outer p orbitals. Ga: [Ar] 3d10 4s2 4p1 2 Answer: C X: The biggest jump in IE is between 4 th to 5 th electron hence, X has 4 valence electrons and belongs to Group 14. Y: The biggest jump in IE is between 6 th to 7 th electron hence , X has 6 valence electrons and belongs to Group 16. Possible covalent compounds formed are CO2, CS2 and SiO2. 3 Answer: C Notice that N1 and N3 only have two covalent bonds with carbon atoms. To fulfil octet, they need to gain one electron each from the central Mg atom to form N–. Hence, N1 and N3 are involved in ionic bonding. N2 and N 4 have three covalent bonds with carbon at oms. They have one lone pair each, which interacts with the central Mg2+ ion via co–ordinate bonding. 4 Answer: D Dipole moments cancel out. B: Dipole moment is smaller than that in D since S is less electronegative than O. C: Dipole moment is smaller than that in D as the net dipole moment of C–F bonds reduces the dipole moment of C=O bond

ASRJC JC2 PRELIMS 2019 9729/01/H2 2 D: 5 Answer: C V nRT mV RTM mM RTV M RT p p p p     p is directly proportional to . The graph should be a straight line that passes through the origin. Hence, A is wrong pV = nRT p is directly proportional to T / K. The graph should be a straight line that passes through the origin. Hence, B is wrong. V is also directly proportional to T / K. Hence, C is correct pV = nRT 1 V p nRT : 1 p  V at constant T. The graph should be a straight line that passes through the origin. Hence, D is wrong. 6 Answer: B The solution is one which is able to maintain the pH at about 5, hence it is an acidic buffer. Only the mixture in option B can form an acidic buffer.

ASRJC JC2 PRELIMS 2019 9729/01/H2 3 7 Answer: C H2O H+ + OH Kw = [H+][OH] Since the stoichiometric coefficients of H + and OH ions are equal in the above equation, [H +] = [OH–] at all temperatures. Hence, water is a neutral liquid at all temperatures. Option B is wrong. When temperature of water increases, Kw also increases. This shows that the position of equilibrium has shifted to the right , i.e. the forward reaction is favoured, and more H + and OH  ions are produced. Since the endothermic reaction is favoured when temperature increases, the dissociation of water (forward reaction) is an endothermic reaction. Option D is wrong. The value of Kw does not provide any information about hydrogen bonding between water molecules. Option A is wrong. 8 Answer: C A: Volatility of halogens (X2) decreases: F2 > Cl2 > Br2 > I2 X2 has simple molecular structure with weak id–id forces of attraction between molecules. Down the group, the number of electrons increases hence the electron cloud is bigger and more easily polarised resulting in stronger id–id forces of attraction. Thus, the boiling point increases down the group and the halogens become less volatile. B: Bond energy of X–X bond decreases down group 17 molecule bond energy of X–X (kJ mol–1) F2 + 158 Cl2 + 244 Br2 + 193 I2 + 151 C: Down group 17, Eo(X2/X–) becomes less positive.  increasing tendency for X– to be oxidised to X2 D: Thermal stability of hydrogen halides decreases down group 17 as the bond energy of H–X bond decreases down the group. molecules bond energy of H–X (kJ mol–1) H–F 562 H–Cl 431 H–Br 366 H–I 299 9 Answer: A There are 3 chlorine atoms in 1 molecule of trichloroisocyanuric acid. No. of moles of trichloroisocyanuric acid in the pool = 6 -32.50 x 10 x 1.50 x 10 232.5 = 16.1 No. of moles of chlorine atoms = 3 x 16.1 = 48.38 No. of chlorine atoms = 48.38 x 6.02 x 1023 = 2.91 x 1025

ASRJC JC2 PRELIMS 2019 9729/01/H2 4 10 Answer: B No. of moles of H2S = 0.03 24000 720  mol No. of moles of HNO3 = 0.02 0.500x 1000 40  mol Mole ratio of H2S: HNO3 = 3 : 2 From the given half–equation, 1 mole of H2S loses 2 moles of electrons  3 moles of H2S will lose 6 moles of electrons 2 moles of HNO3 gain 6 moles of electrons  1 mole of HNO3 will gain 3 moles of electrons original oxidation state of N in HNO3: (+1) + x + 3(–2) = 0 x = +5 Since the nitrogen atom in HNO3 gains 3 electrons, +5 – 3 = +2 new oxidation state of N = +2 (N in NO has an oxidation state of +2) 11 Answer: D Q’ (heat taken in by water) = 300 x 4.2 x T J Q (heat evolved by the reaction) = 1371 1000 J46.0 m  % efficiency = 300 4.2 300 4.2 46.0100% 100% 1371 10001371 100046.0 TT m m         12 Answer: A To calculate enthalpy change of formation of MgCO3(s), use the following: ∆Hrxn = ∆Hf(products) − ∆Hf(reactants) The enthalpy of combustion of C(s) is the same as the enthalpy change of formation of CO 2(g). Hence, option A is correct. C(s) + O2(g) → CO2(g) 13 Answer: A When a liquid boils, energy is needed to overcome the forces of attraction between liquid particles, hence ∆H is positive (i.e. endothermic). When a liquid boils, the state changes from liquid to gaseous state, which is more disordered, hence ∆S is positive.

ASRJC JC2 PRELIMS 2019 9729/01/H2 5 14 Answer: B rate = k[(CH3)3CCl] (first order reaction) When [(CH3)3CCl] is doubled, the rate is doubled. Given that 20% of 0.02 mol dm–3 (CH3)3CCl has reacted in 5 min, 0.004 mol dm–3 has reacted. Hence, for 0.04 mol dm–3 (CH3)3CCl, 0.008 mol dm–3 would have reacted as rate has doubled. % reacted = (0.008/0.04) x 100% = 20% 15 Answer: B . expt no. initial [O2] / mol dm–3 initial [NO] / mol dm–3 initial rate / mol dm–3 min–1 1 2.0 1.0 8.0 2 1.0 1.0 s 3 1.0 t 16.0 4 0.5 0.5 u Given: rate = k[O2][NO]2 Comparing expt 1 & 2, in which initial [NO] is constant, When [O2] is halved, rate will be halved  s = 8.0/2 = 4.0 Comparing expt 2 & 3, in which initial [O2] is constant, Since rate increases 4 times (from 4.0 to 16.0) and order of reaction w.r.t. NO is 2, [NO] must have doubled (from 1.0 to 2.0)  t = 2.0 Alternatively, 21.0 t 16.0( )( ) ( )1.0 1.0 4.0 t 2.0   Comparing expt 3 & 4, 20.5 0.5 u( )( ) ( )1.0 2.0 16.0 u 0.5   16 Answer: D 1: The rate equation can be derived as follows: Based on the slow step (rds): rate = k [OH+] [I–]2 [H+] ––––––– (1) For the first step (fast step), Kc = + + 22 [OH ] [H O ][H ] [OH+] = Kc [H2O2] [H+] ––––––– (2) Substituting (2) into (1): rate = k Kc [H2O2] [H+] [I–]2 [H+] rate = k1 [H2O2] [H+]2 [I–]2 which is not consistent with the experimentally obtained rate equation.

ASRJC JC2 PRELIMS 2019 9729/01/H2 6 2: The overall chemical equation obtained by summing up the 2 equations in the slow and fast steps is different from that given in the option. NO2 + NO2 → N2O3 + O (slow) N2O3 + CO → CO2 + 2NO (fast) CO + 2NO2 → CO2 + 2NO + O (given) CO + NO2 → CO2 + NO 3: Based on the slow step (rds), the rate equation should be: rate = k3 [H2] which is not consistent with the experimentally obtained rate equation. 17 Answer: C ∆Go < 0 at points 1, 2 and 3. When Kc is more than 1, it implies that the position of equilibrium lies more towards the right. This means that the forward reaction is spontaneous. Therefore, consider points where ∆Go < 0. 18 Answer: C At t1: When NO2 was removed, the position of equilibrium shifts right to produce more NO2, causing the amount and hence pressure of NO2Cl to decrease. At t2: when the