ACJC Prelim P1(Worked Solution)

1 © ACJC 2018 9729/01/Prelim/2018 [Turn over ANGLO-CHINESE JUNIOR COLLEGE DEPARTMENT OF CHEMISTRY Preliminary Examination C H E M I S T R Y 9729/01 Higher 2 Paper 1 Multiple Choice 29 August 2018 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 fluids. Write your name, index number and tutorial class on t he Answer Sheet in the spaces provided unless this has been done for you. 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. Read the instructions on the Answer Sheet very carefully. 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 18 printed pages. 9729/01/Prelim/18 ANGLO-CHINESE JUNIOR COLLEGE © ACJC 2018 Department of Chemistry [Turn over

2 © ACJC 2018 9729/01/Prelim/2018 [Turn over 1 D 11 C 21 D 2 B 12 B 22 C 3 B 13 D 23 B 4 A 14 D 24 A 5 B 15 D 25 A 6 C 16 B 26 D 7 C 17 A 27 A 8 C 18 D 28 B 9 A 19 A 29 C 10 B 20 D 30 B 1 The Basic Oxygen steel-making process is a method of preparing steel from carbon-rich molten pig iron. The process is basic because chemical bases are added to remove impurities. One such impurity is phosphorus pentoxide, P 4O10. Calcium oxide, CaO, is added to remove it and the only product of the reaction is the salt, calcium phosphate, Ca3(PO4)2. How many moles of CaO reacted with one mole of P4O10 in this reaction? A 1 B 1.5 C 3 D 6 P4Ol0 + 6CaO  2Ca3(PO4)2 Given that Ca3PO4 is the only product, write the balanced equation between P4O10 and CaO to form Ca 3PO4. Based on mole ratio, 6 moles of CaO is required to completely react with one mole of P4Ol0. Answer: D 2 Two moles of an oxidising agent, XO4− in the presence of excess acid oxidised 96 dm3 of nitrogen dioxide gas at room temperature and pressure to NO3−. What is the number of moles of electrons accepted by one mole of XO4−? A 1 B 2 C 3 D 4 n(NO 2) = 96/24= 4 NO2 + H2O NO3- + 2H+ + e XO4− ≡ NO2 2 : 4 1 : 2 2 mol of NO 2 donate 2 mol of e which is accepted by 1 mol of XO4−

2 © ACJC 2018 9729/01/Prelim/2018 [Turn over 3 Use of the Data Booklet is relevant to this question. The ion T+ contains 28 electrons and 35 neutrons. Which of the following statements about T+ or T is correct? A T and Ga3+ are isoelectronic species. B The elemental form of T can be oxidised by chlorine. C The electronic configuration of T+ ion is 1s22s22p63s23p63d94s1. D The angle of deflection of 27Al3+ is approximately three times that of T+ in an electric field. Since ion T+ contains 28 electrons, element T is copper which contains 29 electrons. Option A is wrong as Ga 3+ has 28 electrons, hence T and Ga3+ do not have the same number of electrons. Option B is correct as Cu can be oxidised to Cu2+ by chlorine. Cu2+ + 2e Cu E o = + 0.34 V Cl2 + 2e 2C l− E o = + 1.36 V Eocell = +1.36 – (+0.34) = +1.02 V Option C is wrong as the electronic configuration of T+ ion is 1s22s22p63s23p63d10. Option D is wrong as the angle of deflection of 27Al3+ is not approximately three times that of T+ in an electric field. The angle of deflection depends on the charge-to-mass ratio of the particles. Angle of deflection of Al3+ = k( 3 27) = 0.111 k Angle of deflection of T+ = k( 1 64) = 0.0156 k Answer: B 4 Which of the following species has a different bond angle from the rest? A ICl3 B SF3+ C ClO3– D N2H4 ~ 107o Cl O O O Answer: A

3 © ACJC 2018 9729/01/Prelim/2018 [Turn over 5 What will happen to the volume of a bubble of air submerged in water under a lake at 10.0 oC and 2.00 atm if it rises to the surface where the temperature is 20.0 oC and the pressure is 1.00 atm? A The volume will increase by a factor of 2.00. B The volume will increase by a factor of 2.07. C The volume will decrease by a factor of 2.00. D The volume will decrease by a factor of 1.93. Answer : B P1V1/T1 = P2V2/T2 2(V1)/ (273+10) = 1 (V2) / (273 + 20) V2 = 293 x (2) x V1 / 283 = 2.07 V1 6 The graph below shows the variation in the standard enthalpy change of fusion, ∆Hofus for 8 consecutive elements from period 2 to 3 in the periodic table. Standard enthalpy change of fusion is the heat absorbed when one mole of a substance changes its state from solid to liquid under standard conditions. Which of the following statements is true based on the information deduced from the above graph? A The chlorides become more acidic from A to C. B An oxide of E dissolves in water to form an alkaline solution. C Element G has a higher first ionisation energy than element F and H. D Element D has a lower electrical conductivity as compared to element F. Since F is an element from period 2 to 3, and it has the largest ∆Hofus (the rest of the elements have much lower ∆Hofus), F must be Si. Si has a giant covalent structure with high melting point and ∆Hofus. A G H ∆Hofus / kJ mol–1 Atomic number B C D E F

4 © ACJC 2018 9729/01/Prelim/2018 [Turn over So A is fluorine, B is neon, C is sodium, D is magnesium, E is aluminium, G is phosphorus and H is sulfur. Option A is wrong as sodium chloride is a neutral chloride and there is no chloride of fluorine (fluorine reacts with chlorine to form chlorine fluoride instead). There is no reaction between neon and chlorine. Option B is wrong as aluminium oxide is insoluble in water. Option C is correct as phosphorus has a hi gher first ionisation energy than silicon and sulfur. Option D is wrong as magnesium has a higher electrical conductivity as compared to silicon. Answer: C 7 Some enthalpy changes of combustion are given below. ΔHc / kJ mol−1 CO(g) −283 H2(g) −286 CH3OH(l) −715 What is the enthalpy change of the following reaction? CO(g) + 2H2(g) CH 3OH(l) A −146 B +146 C −140 D +140 Answer: C 2 3 O2(g) + CO(g) + 2H2(g) ⎯⎯→⎯ r∆Η CH3OH(l) + 2 3 O2(g) −283 2( −286) −715 C O 2(g) + 2H 2O(l) By Hess’ law: ΔHr =ΔHc (reactants) - ΔHc (products)

5 © ACJC 2018 9729/01/Prelim/2018 [Turn over = −283 + 2(−286) − (−715) = −140 kJ mol−1 8 The reaction of nitrogen monoxide and hydrogen gas 2NO(g) + 2H2(g) N 2(g) + 2H2O(g) is thought to involve the following steps: I NO + NO N 2O2 (fast) II N2O2 + H2 H 2O + N2O (slow) III N2O + H2 N 2 + H2O (fast) Which of the following about the reaction is true? A H2 acts as the catalyst. B The rate equation for the reaction is rate = k[N2O2][H2]. C The overall order of the reaction is 3. D Increasing the concentration of NO does not change the rate of reaction. Answer: C A is incorrect as there is insufficient information to deduce that H2 is the catalyst. B is incorrect. Based on the slow step: Rate = k [N 2O2] [H2] However, N2O2 is not present in the final equation, this show that it is an intermediate and should not be present in the rate equation. Based on step I, [N2O2] α k’[N