2020 TJC Answers Promo

©TJC 2020 TJC Promo H2 Chemistry MCQ Worked Solutions 1 Answer: B Since J has a larger angle of deflection than 16O2–, it would have a greater charge/mass ratio than 16O2–. 2 Answer: C P+ : 1s2 2s2 2p6 3s2 3p2 Cu3+: 1s2 2s2 2p6 3s2 3p6 3d8 Ca: 1s2 2s2 2p6 3s2 3p6 4s2 S: 1s2 2s2 2p6 3s2 3p4 3 Answer: A No. of moles of ethanol molecules = 23 24+16+6 = 0.500 mol No. of moles of atoms in g of ammonia = 2.125 17 x 4 = 0.500 mol No. of moles of molecules in 11.35 dm3 of sulfur dioxide = 11.35 22.7 = 0.500 mol No. of moles of sodium ions = 250 1000 x 2 x 2 = 1.00 mol 4 Answer: B V C H O % 19.21 45.3 5.29 30.2 Ar 50.9 12 1 16 Amt 0.377 3.77 5.29 1.88 rati o 1 10 14 5 Empirical formula = VC10H14O5 Z = (5 – 1)/2 = 2 5 Answer: A H2S → S + 2H+ + 2e SO2 + 4H+ + 4e → S + 2H2O Reacting ratio: 2H2S Ξ SO2 6 Answer: D Bond angle 1 is around a sp 2 carbon atom with 3 bp, 0 lp = 120 Bond angle 2 is around a sp 3 carbon atom with 4 bp, 0 lp = 109.5 Bond angle 3 is around a N -atom with 3 bp, 1 lp = 107 7 Answer: C There are no hydrogen bonds existing between methanal molecules. Methanal is a polar hence pd-pd interactions exist between methanol which is stronger than id -id interactions between non polar ethane molecules. 8 Answer: A Hr= 4(–241.8) – 2(50.6) – (9.2) = –1077.6 kJ mol–1 9 Answer: D  2: Combustion reactions are exothermic in nature.  4: The stoichiometric ratio of sodium oxide is defined as 1 and sodium oxide is formed from its elements. 10 Answer: B Heat absorbed by water = 80.2  100 80 = 64.16 kJ = 64 160 J Heat absorbed by water = mcΔT 64 160 = m (4.18)(80 – 25) Mass of water = 279 g 11 Answer: D Since Kc remains constant, there is no change in position of equilibrium when temperature changes and the enthalpy change, H is zero.

©TJC 12 Answer: B Kc = [NH3]2[H2SO4] = 2.57 x 10-5 (2x)2(x) = 2.57 x 10-5 hence x = 0.0186 13 Answer: B  1 Vol of O2 (product) increases from 0 to 50cm3 (½ of max vol) in 20 s Vol of O2 (product) increases from 50 to 75cm3 (½ of maxi vol to ¾ of max vol) in 20 s half-life is a constant at 20 s.  2 Since half-life is a constant at 20s, the reaction is first order w.r.t. [H2O2]. 14 Answer: C Both kf and kb will increase while Kc remains constant as Kc is only temperature dependent. 15 Answer: A When p increases 5 2 times, the V should decrease 5 2 times to 38.4 cm3 if the gas is ideal. But V of the gas only decreases to 46 cm 3  Vreal > Videal 16 Answer: B (0.68)VT = (0.7)(1.0) + (0.5)(2.5) + (1.0)(1.5) VT = 5.074 dm3 Vol of connecting tube = 5.074 – (1.0 + 2.5 + 1.5) = 0.074 dm3 17 Answer: C An electrophile is an electron pair acceptor. They are positively or partial positively charged. They are attracted to electron rich sites or regions with high electron density. 18 Answer: A H radicals are not formed in free radical substitution reactions. Hence H2 will not be formed. HCl is formed in the propagation step. CH2Cl2 is formed from further substitution of CH3Cl. CH3CH2Cl is formed from CH3 and CH2Cl in a termination step. 19 Answer: D 20 Answer: B Prismane undergoes free radical substitution with bromine in uv light. 2 hydrogen atoms are substituted with bromine atoms. There are 3 possible constitutional isomers:

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2 DO NOT WRITE IN THIS MARGIN ©TJC [Turn over DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN SECTION B (STRUCTURED) 1 (a) (i) Comparing expt 1 & 2, when [NO] is constant and [O2] doubles, rate of reaction doubles. The order of reaction wrt O2 is 1. (ii) Comparing expt 2 & 3, when [NO] doubles and [O2] constant, rate increases by 4 times. The order of reaction wrt NO is 2. (iii) Rate = k [O2][NO]2 (iv) [NO] = 3.90 x 10-2 mol dm-3 (b) (i) Kc = [𝐍𝐎]𝟐 [𝐍𝐎]𝟐[𝐎𝟐] = [𝟎.𝟎𝟐 𝟐 ]𝟐 [𝟎.𝟒𝟒 𝟐 ]𝟐[𝟎.𝟑𝟒 𝟐 ] = 0.0122 mol-1 dm3 (ii) By Le Chatelier’s Principle, the equilibrium position will shift left OR towards the backward reaction in o rder to produce greater no of moles of gas to increase pressure. Amount of NO2 decreases and the amount of NO and O2 increases. 2020 TJC H2 Promo Exam Section B Answers

3 DO NOT WRITE IN THIS MARGIN ©TJC [Turn over DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 2 (a) (i)  A catalyst provides an alternative pathway with lower activation energy or Ea . There are more molecules with energy greater than or equal to the lowered Ea, Frequency of effective collisions increases Rate constant increases and hence rate of reaction increases. (ii) Iron is a heterogeneous catalyst that provides a surface for H2 and N2 to be adsorbed. This increases the concentration of H2 and N2 at the catalyst surface. The intramolecular OR covalent bonds in H2 and N2 are weakened. Product NH3 desorbed from the iron surface and diffuses away. (iii) Porous solid provides a larger surface area for catalysis. (b) (i) ICl is polar ( has a permanent dipole) hence the electron rich C=C bond attacks the I+ in ICl more readily. (ii) (iii) Elimination Alcoholic KOH, heat No. of reactant molecules having energy  Ea No. of reactant molecules having energy  Ea’ Ea' Ea Energy Ea : activation energy for uncatalysed reaction Ea' : activation energy for catalysed reaction Number of molecules

4 DO NOT WRITE IN THIS MARGIN ©TJC [Turn over DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN 3 (a) (i) Element Y is in Group 16. Large increase in IE when the 7th electron is removed implies that 7th electron is removed from the inner shell (ii) I (iii) J has larger nuclear charge as it has more protons. Due to its additional electron shell, J has greater screening effect. The valence electrons of J are further from the nucleus and less strongly attracted by the nucleus. Less energy is required to remove the electron of J than that of B. (b) (i) NO3- + Sn2+ + 2H+  NO2- + Sn4+ + H2O (ii) No. of moles of S2O32- = 0.12 × 0.022 = 0.00264 mol No. of moles of Sn4+ = 0.00264/2 = 0.00132 mol No. of moles of Sn2+ in 50 cm3 solution = 0.00132 × 50/10 = 0.00660 mol (iii) Percentage purity of tin = (0.00660 x 118.7)/2.5 x 100% = 31.3% 4 (a) (i) Carbon 1: sp Carbon 2: sp3 (ii) Carbon 1: Linear, 180 Oxygen: Bent, 104.5 (b) (i) C4H6O(l) + 5O2(g)  4CO2(g) + 3H2O(l) (ii) Energy taken in for bond breaking = 6(410) + 1(350) + 1(840) + 2(360) + 5(496) = 6850 kJ mol-1

5 DO NOT WRITE IN THIS MARGIN ©TJC [Turn over DO NOT WRITE IN THIS MARGIN DO NOT WRITE IN THIS MARGIN Energy given out for bond forming = 8(805) + 6(460) = 9200 kJ mol-1 Hc = 6850 – 9200 = –2350 kJ mol-1 (iii) Compound X is a liquid whereas the bond energies in Data Booklet correspond to breaking covalent bonds in the gaseous state. C4H6O (l) + 5O2 (g)  4CO2 (g) + 3H2O (l) C4H6O (g) + 5O2 (g)  4CO2 (g) + 3H2O (g) Hvap = (–2444) + 3(+41) – (–2350) = +29.0 kJmol-1 (c) (i) Compound Y exhibits cis-trans isomerism as there are restricted (ii) rotation about the carbon-carbon double bonds and each carbon of the double bond is bonded to 2 different substituents. Total number of stereoisomers: 4 – 2444 – 2350 3(+41) Hvap

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