RI 2017 A-Level H2 Chem Solution

This document is copyrighted, please do not reproduce it without permission © Raffles Institution 2017 A-Level H2 Chemistry Paper 1 Suggested Solutions 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 C B D C D B B B C B A A D D A 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 B A D A C A A D C D A A B C A Q1 (C) Isotope Abundance / % 28Si 92.23 29Si x 30Si (100 – 92.23 – x) = (7.77 – x) 28(92.23) + 29(x) + 30(7.77 – x) = 28.10(100) x = 5.54% Q2 ( B) 1 Correct. E.g. 9 4Be has 5 neutrons and 4 protons. 2 Correct. E.g. 16 8 Ohas 8 neutrons and 8 protons. 3 Incorrect. There are no elements from Li to Mg which have more protons than neutrons. Q3 (D ) H C H O 3 regions of electron density around C atom of methanal ⇒ trigonal planar shape and bond angle of 120° Q4 (C) 1 Yes. The hydrogen bonds between water molecules are stronger than the id-id interactions between methane, requiring more energy to overcome, resulting in the higher boiling point of water. 2 No. O–H bonds are stronger than C–H bonds. However, boiling does not involve breaking the O–H and C–H bonds. It involves overcoming their intermolecular forces of attraction. 3 No. Water (10 e–) contains more 2 electrons than methane (8 e–). With a slightly larger electron cloud, water has slightly stronger id-id interactions. However, this does not account for the much higher boiling point of water, which can only be attributed to hydrogen bonding between water molecules. Q5 (D) A Incorrect. A change in pressure has no effect on temperature of a gas. B Incorrect. The kinetic energy of gas molecules depends on the temperature, not the pressure. C Incorrect. Pressure has no effect on the size of the molecules. D Correct. Increasing pressure moves the molecules closer, increasing the intermolecular forces of attraction so significantly that the gas becomes liquid. Q6 (B ) Recall: Lewis base is an electron pair donor. Bronsted-Lowry acid is a proton donor. A Incorrect. H+ does not have electron pairs and cannot be a Lewis base. B Correct. Electron pairs on O atom of H2O allow water to act as a Lewis base. H 2O can donate a H + (H2O → OH– + H+) ans acts as a Bronsted-Lowry acid. C Incorrect. O2– is unable to donate H+. D Incorrect. Electron pairs on O atom of OH – allow it to act as a Lewis base. OH – can donate a H + (OH– → O2– + H+) and acts as a Bronsted-Lowry acid. Q7 (B ) Mg, A l, Si and P are consecutive elements in Period 3. From the data booklet, the following information about their atomic radii can be obtained: • a tomic radius of Mg > Al > Si > P. • atomic radius of Si is closer to that of P than of Al. Since atomic radius is the x -axis, going from left to right, the elements should be in the order P, Si, Al, Mg, and Si is closer to P than Al (i.e. more like option B than D). Electronegativity increases across the period: Mg < Al < Si < P. Since electronegativity is the y -axis, going from bottom -up, the elements should be in the order Mg, Al, Si, P.

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Q8 (B ) From R to S, the difference in 5 th IE is extremely large. This is because the 5 th IE of R involves removing an electron from R 4+ which has a Group 18 electronic configuration (i.e. ns 2 np6). This also means that element R has 4 valence electrons and element Q would have 3 valence electrons i.e. Q belongs to group 13. Hence, the chloride of Q has the formula QCl3. Q9 (C ) The metal ions with the highest polarizing power will form the least stable peroxide. Polarizing power is proportional to charge density which is proportional to argch e radius . The charge of Mg2+ and Ba2+ is greater than that of Cs+ and Na+ (eliminate options B and D). The ionic radius of Mg 2+ is smaller than that of Ba 2+. Hence, Mg2+ has the greatest charge density and polarizing power, and forms the least stable peroxide. Q10 (B) Is statement correct? Does statement explain why diamond does not change into graphite? 1 Yes and yes. Despite the reaction being spontaneous (see option 2), diamond does not change into graphite due to high Ea which results in a small rate constant. 2 Yes. ∆G = ∆H – T∆S = –1900–298(3.4) = –2913 J mol–1 = –2.9 kJ mol–1 No. The negative ∆G implies that diamond should convert into graphite as it predicts the reaction to be thermodynamically feasible. 3 No. Since ∆G of forward reaction is –2.9 kJ mol–1, ∆G of reverse reaction is +2.9 kJ mol –1 > 0 i.e. reverse reaction is not spontaneous. No. This does not explain the phenomenon. Q11 (A ) X2 ⟶ 2 X Pressure Initial amt / mol 1 0 p Amt after 1 t1/2 / mol 0.5 1 1.5p Amt after 2 t1/2 / mol 0.25 1.5 1.75p When 75% of X2 has been converted to X, 25% (i.e. 0.25 mol) of X 2 remains. This happens after 2 half - lives. 1 Time elapsed = 2(30) = 60 min. 2 0.75 mol of X2 has reacted. Hence, the amt of X formed = 2(0.75) = 1.5 mol. 3 Total amt of gases after 2 half -lives = 0.25 + 1.5 = 1.75 mol At constant T and V, 1 mol of gas has a pressure of p, 1.75 mol of gas has a pressure of 1.75 p (i.e. 7p / 4) Q12 (A ) The overall rate equation can be determined from the slow step. For option A, the slow step involves 2 NO and 1 H2. The rate equation is rate = k[NO]2[H2] which agrees with the rate equation provided. Q13 (D) Density of water = 0.997 g cm–3 = (1000)(0.997) g dm–3 = 997 g dm–3 [H2O] = (997/18) mol dm–3 In pure water, [H3O+] = [OH–]. Hence, 2 3 2 2 2 32 3 3 [] [] 997[ ] [] 18 997no. of H O in 1.00dm (1 .00)( )18 997 18 c cc c c HOK HO HO K HO K KL LK + + + = = = = =

This document is copyrighted, please do not reproduce it without permission © Raffles Institution Q14 (D ) For [H 2]eqm > [H 2O]eqm, the position of equilibrium lies to the right when occurs when ∆G < 0 i.e. at points 3 and 4. Q15 (A) PbCrO4(s) ⇌ Pb2+(aq) + CrO42–(aq) Eqm conc / mol dm–3 – 1.3 x 10–7 1.3 x 10–7 Since solubility = 1.3 x 10 –7 mol dm –3, the solubility product = (1.3 x 10–7)2 = 1.7 x 10–14 mol2 dm–6. Q16 (B) 1 Same 2-aminopropane, CH3CH(NH2)CH3, has 9 H. 2-bromo-2-methylpropane, (CH3)3CBr, has 9 H. 2 Same Ethylpropanoate, CH3CH2CO2CH2CH3, has 10H. Butane-1,2-diol, HOCH2CH(OH)CH2CH3, has 10H. 3 Different Butanenitrile, CH3CH2CH2CN, has 7 H. 2-methylpropanal, CH3CH(CH3)CHO, has 8 H. Q17 (A) Reduction at cathode 2CH2=CHCN + 2H2O + 2e– ⟶ NC(CH2)4CN + 2OH– Oxidation at anode 4OH– ⟶ O2 + 2H2O + 4e– No of mol of acrylonitrile = 0.01 mol No of mol of e – taken in by acrylonitrile during reduction = 0.01 mol = no of mole of e – released from oxidation. No of mol of O2 = 0.01 / 4 = 0.0025 mol Volume of O2 released = 0.0025(24) = 0.06 dm3 = 60 cm3 Q18 (D) HBr Br Br Br Br Br Br Br Br tertiary bromoalkane tertiary bromoalkane tertiary bromoalkane When reacted with β -selinene, HBr adds across both C=C to give a mixture of products with molecular formula C 15H26Br2, three of which are tertiary bromoalkanes. Q19 (A) A Correct. The CN – nucleophile attacks the electrophilic carbonyl carbon from above and below the plane with equal probability, resulting in the formation of a mixture of 2 enantiomers. B Incorrect. R Cl CH2CH3 OH R OH CH2CH3 * * The S N2 reaction takes place at the – CH2Cl carbon which is not chiral. The reaction also does not affect the chiral centre indicated. Only one enantiomer is obtained as the product. C Incorrect. The reacting terti ary ch loroalkane has no chiral centre (note: the carbon bearing the C l atom has 2 ethyl groups). Since its reaction with OH– involves substitution of the – Cl for a –OH, the product d