2020 SAJC H2 Prelim Paper 1 (Answers)

1 [Turn over CANDIDATE NAME CLASS 19S ST ANDREW’S JUNIOR COLLEGE JC2 Preliminary Examination CHEMISTRY Paper 1 Multiple Choice 9729/01 September 2020 1 hour Additional Materials: Multiple Choice Answer Sheet Data Booklet READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write your name and class on the Answer Sheet in the spaces provided. There are thirty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the separate Answer sheet. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet. The use of an approved scientific calculator is expected, where appropriate. This document consists of 15 printed pages.

2 [Turn over 1 Carbon disulfide, CS2, burns in oxygen according to the following equation. CS2(g) + 3O2(g)  CO2(g) + 2SO2(g) 10 cm3 of carbon disulfide was completely burnt in 50 cm3 of oxygen. The resulting mixture was then treated with excess of aqueous sodium hydroxide. All measurements of volume were made at the same temperature and pressure in which all the reactant and products were in the gaseous state. What was the volume of the gas after burning and after adding NaOH (aq)? volume of gas after burning / cm3 volume of gas after adding NaOH(aq) / cm3 A 30 10 B 30 20 C 50 20 D 50 40 Ans: C Vol of excess O2 = 20 cm3 Volume of gas produced after burning = 20 + 10 + 20 = 50 cm3 CO2 and SO2 are both acidic gases that reacts with aq NaOH. vol of gas after adding aq NaOH = 50 – 30 = 20 cm3

3 [Turn over 2 1 mole of X2O32− oxidised 68.1 dm3 of sulfur dioxide gas to sulfate ions at standard temperature and pressure. What is the final oxidation state of X? A −2 B −1 C +1 D +2 Ans: B n(SO2) = 68.1/22.7= 3 SO2 + 2H2O  SO42- + 4H+ + 2e X2O32− : SO2 : e 1 : 3 : 6 1 mol of X2O32− accepts 6 mol of e 1 mol of X accepts 3 mol of e Initial OS of X is + 2, final OS of X is -1

4 [Turn over 3 Use of the Data Booklet is relevant to this question. Which species have the same number of unpaired electrons as chlorine atom? 1 Li+ 2 Mg+ 3 O− 4 F− A 2 and 3 B 3 and 4 C 1 only D 2 only Ans: A Chlorine has only one unpaired electron. Li+: 1s2 Mg+: [Ne]3s1 O−: [He] 1s2 2s22p5 F-: [Ne]

5 [Turn over 4 Hospitals require large amount of oxygen gas in the treatment of patients with respiratory symptoms. Oxygen is liquefied under pressure and stored in special cylinders. It vapourises when pressure is released. Which statement best describes why oxygen gas liquefies under pressure? A High pressure lowers the kinetic energy of oxygen molecules, creating less disordered liquid state. B High pressure pushes the molecules closer for more effective intermolecular interactions. C High pressure causes the temperature of oxygen gas to increase above its boiling point. D High pressure decreases the volume total occupied by the oxygen molecules. Ans: B The higher the pressure, the more particles there are per unit volume. Hence, the particles are closer in proximity. Hence, there are greater intermolecular forces between the gas particles. 5 Which species contain a dative (co-ordinate) bond? 1 CO 2 CO32− 3 N2H4 4 N2O4 A 2, 3 and 4 B 1 and 2 only C 1 and 4 only D 3 and 4 only Ans: C Option 1 has a dative bond from O to C while option 5 has a dative bond between N to O. The rest have no dative bond present.

6 [Turn over 6 One proposed strategy to reduce the amount of greenhouse gases in the atmosphere is to capture carbon dioxide emissions from fossil fuel power plants and pump them into the deep ocean bed. Under the extreme high pressure of the deep ocean bed, carbon dioxide would dissolve in water and stay out of the atmosphere. CO2(g) + nH2O(l) ⇌ CO2.nH2O(aq) What will be the signs of H and S for the forward reaction of this equilibrium at the deep ocean bed? H S A   B  + C +  D + + Ans: A S < 0 as there are no moles of gas in the products (more orderly) H < 0 since interactions are formed between carbon dioxide and water.

7 [Turn over 7 The table shows the charge and radius of each of six ions. Ion J+ L+ M2+ X Y Z2 Radius / nm 0.14 0.18 0.16 0.14 0.18 0.16 The ionic solids JX, LY and MZ are of the same lattice type. What is the correct order of their lattice energies, placing the least exothermic first? A JX, LY , MZ B JX, MZ, LY C LY, JX, MZ D LY, MZ, JX Ans: C LE is inversely proportional to sum of ionic radius and directly proportional to multiplication of charges. Hence LE of NP is the most exothermic as multiplication of charge outweighs the sum of the radii. Both JK and LM have the same ionic charge. However, JK have a more exothermic LE as J and K have a smaller ionic radii than L and M respectively.

8 [Turn over 8 Ammonium nitrate is commonly used in making of fertilisers. 2NH4NO3(s)  2N2(g) + 4H2O(g) + O2(g) The decomposition of ammonium nitrate at 230oC give gases. It is a first-order reaction with a half-life of 10 min. Explosion happens when the molar proportion of ammonium nitrate to gases is 1:14. How long would it take for ammonium nitrate to explode at 230oC? A 23 min B 35 min C 70 min D 80 min 2NH4NO3(s)  2N2(g) + 4H2O(g) + O2(g) initial 1 0 0 0 final 1- 2x 2x 4x x 2𝑥 + 4𝑥 + 𝑥 1 − 2𝑥 = 14 1 7x = 14 - 28x x = 0.4 % of NH4NO3 left = 1 – 2 x 0.4 = 20% (0.5)𝑛 = 0.2 No of half-lives = 2.32 Time taken = 2.32 x 10 = 23.2 min

9 [Turn over 9 The reaction of nitrogen monoxide and hydrogen gas, 2NO(g) + 2H2(g) N2(g) + 2H2O(g) involves the following steps: I NO + NO N2O2 (fast) II N2O2 + H2 H2O + N2O (slow) III N2O + H2 N2 + H2O (fast) Which statements about the reaction are correct? 1 N2O acts as the catalyst. 2 The reaction is second order with respect to [NO]. 3 The units of the rate constant is mol−3 dm9 s-1 4 When partial pressure of H2 doubles, rate of reaction quadruples. A 1, 2 and 3 B 1 and 3 only C 2 and 4 only D 2 only Ans: D  Option 1 is incorrect as N2O is an intermediate  Option 2 is correct. Based on the slow step: Rate = k [N2O2] [H2] However, N2O2 is not present in the final equation, this shows that it is an intermediate and should not be present in the rate equation. Based on step I, Rate = k[NO]2[H2]  Option 3 is incorrect. Based on the rate equation, overall order is 3. Hence units of rate constant = mol-2dm6s-1

10 [Turn over  Option 4 is incorrect. Since H2 is present in the rate equation, when [H2] is doubled, the rate is also doubled. 10 Given that the Kp for the following equilibrium is 125 at 65oC, X(s) + 3Y(g) ⇌ 3Z(g) What is the mole ratio of Y : Z at equilibrium at 65oC? A 1 : 1 B 1 : 5 C 25 : 1 D 125 : 1 Ans: B Kp= (PZ)3 (P𝑌)3 = (nZ)3 (n𝑌)3 = 125 nZ nY = 5 NZ : nY = 1 : 5