SRJC H1 CHEM P1 QNS ANS

1 SERANGOON JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 1 CHEMISTRY 8872/01 PRELIMINARY EXAMINATION 28 August 2009 Paper 1 50 min Additional Materials: Data Booklet Multiple Choice Answer Sheet READ T HESE INSTRUCTIONS FIRST Write your name, index number on the OMS Sheet in the spaces provided. Write in soft pencil. There are thirty questions in this paper. Answer all questions. For each question, there are four possible answers labelled A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the OMR answer sheet. Read very carefully the instructions on the OMR answer sheet. You are advised to fill in the OMR Answer Sheet as you go along; no additional time will be given for the transfer of answers once the examination has ended. 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 question paper. This document consists of 14 printed pages and 0 blank page. SRJC 2009 8872/01/PRELIM/2009 [Turn Over

2 Answer ALL Questions 1 0.5 g of zinc powder was found to reduc e an acidified solution of 25.50 cm 3 of 0.200 mol dm-3 VO2 +. Which one of the following is the reduced product of VO2 +? A VO3 - B VO2+ C V3+ D V2+ Solution: D Amount of Zn = 30.5 7.645 1065.4  mol Amount of VO2 + = 325.5 0.200 5.1 101000   mol 2 3 3 VO 7.65 10 3 5.1 10 2 Znn n      Reduction: Zinc is a reducing agent, henc e it will undergo oxidat ion. From the data booklet, we choose the following half-equation: Zn(s) → Zn2+(aq) + 2e- Given the 3 mol of Zn reacts with 2 mol of VO 2 +, we can deduce that the 2 mol of VO 2 + takes in (2e x 3 =) 6 mol of e- from Zn. Hence, 1 mol of VO2 + will take in 3 mol of e-. Oxidation state of V in VO2 + = + 5 Hence, the final oxidation state of V = +5 – 3 = +2 2 3.920 g of an oxide of formula MO was completely dissolved in 30.0 cm 3 of 2.00 mol dm -3 sulphuric acid. The resulting solution was made up to 100 cm 3. 25.0 cm 3 of this solution wa s neutralised by 27.5 cm 3 of 0.100 mol dm-3 sodium hydroxide. What is the relative molecular mass of M? A 48.6 B 54.9 C 55.9 D 101.0 Solution: C amount of NaOH = 327.5 0.1 2.75 101000   mol amount of H2SO4 in 25.0 cm3 of solution = ½(2.75 x 10─3) = 1.375 x 10─3 mol SRJC 2009 8872/01/PRELIM/2009 [Turn Over

3 amount of H2SO4 in 100 cm3 of solution = 3100 1.375 10 0.005525.0   mol amount of initial amount of H2SO4 = 30.0 2.0 0.061000 mol Therefore, the amount of H2SO4 that reacts with MO = 0.06 – 0.0055= 0.0545 mol Given that 1 mol of H 2SO4 reacts with 1 mol of MO (M has an oxidation state of +2), therefore, amount of MO in 3.518g of MO = 0.0545 mol Given amount of MO = mass of MO/ molar mass of MO 0.0545 = 3.920/(molar mass of M + 16.0) Molar mass of M = (3.920/0.0545) – 16.0 = 55.9 g/mol 3 One of the isotopes of carbon is carbon-14. Carbon-14 is radioactive and is used in carbon dating by archaeologists. Which one of the following species has the same number of neutrons and electrons as an atom of carbon-14? A 13C B 14N+ C 16O2+ D 17F+ Solution:C Number of protons in Carbon-14= 6p Number of neutrons in Carbon-14 = 14 – 6 = 8n Number of electrons = 6e A 13C 6p 7e 7n B 14N+ 7p 6e 7n C 16O2+ 8p 6e 8n D 17F+ 9p 8e 8n 4 The electronic configurations of two atoms, E and F , are 1s 22s22p3 and 1s 22s22p4 respectively. Compare the first and second ionisation energies of E and F. 1st I.E. 2nd I.E. A E < F E > F B E < F E < F C E > F E > F D E > F E < F Solution: D SRJC 2009 8872/01/PRELIM/2009 [Turn Over

4 E: Nitrogen; F: Oxygen 1st IE of N: N(1s22s22p3) → N+ (1s22s22p2) 1st IE of O: O(1s22s22p4) → O+ (1s22s22p3) Conclusion: The 1 st IE of O is lower than the 1 st IE of N as less energy is required to remove the 2p electron from the paired el ectrons in 2p orbital (inter-electronic repulsion). E > F. 2 nd IE of N: N+ (1s22s22p2) → N2+ (1s22s22p1) 2nd IE of O: O+(1s22s22p3) → O2+ (1s22s22p2) Conclusion: The 2nd IE of O is higher than the 2nd IE of N as both involves the removal of 2p electron from singly filled p-orbital. However, the removal of the 2p electron from O+ will experience a greater electr ostatic forces of attraction fr om the nucleus than that of N+ as O+ has greater nuclear charge (+8) than N+ (+7). E < F SRJC 2009 8872/01/PRELIM/2009 [Turn Over 5 Polyurethane is used in coatings, insulators and adhesiv es. Polyurethane What are the values of the bond angles marked x and y in polyurethane? x y A 90 90 B 120 120 C 107 120 D 109.5 90 Solution:C Around N: 3 bond pair and 1 lone pair  shape is trigonal pyramidal and bond angle is 107 Around C: 3 bond pair and 0 lone pair  shape is trigonal planar and bond angle is 120 C O N H CC H H H H N H C O OCC H H H H O y x n

5 6 Oxygen reacts with nitrogen monoxide in the equation shown. O2 (g) + 2 NO (g)  2 NO2 (g) In an experiment to investigate the effe cts of concentrations on the rate of the reaction, the following results were obtained. Expt [O 2] / mol dm−3 [NO] / mol dm−3 Rate / mol dm−3 s−1 1 1.0 1.0 0.0007 2 1.0 2.0 0.0028 3 2.0 1.0 0.0014 4 2.0 2.0 z The value of z is A 0.0007 B 0.0021 C 0.0056 D 0.0112 Solution:C From expt 1 & 2, when [NO] doubles, rate increase by 4 folds. Hence, 2nd order wrt. to [NO] From expt.1 & 3, when [O 2] doubles, rate also doubles. Hence, 1st order wrt. to [O2]. Rate = k [O2] [NO]2 Compare expt 3 & 4, when [NO] doubles, rate should increase by 4 folds. Rate= 0.0014 x 4 = 0.0056 moldm−3s−1 SRJC 2009 8872/01/PRELIM/2009 [Turn Over

6 7 What can be deduced from the following equilibrium? 2 SO2 (g) + O2 (g) 2 SO3 (g) H = 98 kJ mol-1 A Adding a catalyst increase the yield of SO3 (g). B Decreasing the pressure will cause the posi tion of equilibrium to shift to the right. C Decreasing the temperature will cause the position of equi librium to shift to the left. D The maximum mass of SO3(g) that can be made from 64 g of SO2(g) is 80 g Solution: D Adding catalyst only increase rate of forw ard and backward reaction equally. It does not affect the yield of SO3. Decreasing pressure will favour backward reaction which produces greater moles of gas. Decreasing temperature will favour forward reaction which produces heat. Amount of SO 2 in 64 g = 1 mol Mole ratio: SO2  SO3 Maximum mass of SO3 produced = 1 x (32 +16x3) = 80g 8 The syste m containing G, J and GJ3 is allowed to reach equilibrium in a 5 dm 3 vessel at a temperature of 1000K. G (g) + 3 J (g) GJ 3 (g) The diagram below shows the c hange in number of moles of G, J and GJ3 with time. Amount G J GJ3 Time SRJC 2009 8872/01/PRELIM/2009 [Turn Over

7 What is the equilibrium constant Kc for the reaction? A 3 0.5 0.1 (0.2) B 3 0.5 0.2 (0.2) C 3 3 0.5 5 0.1 (0.2)   D 3 3 0.5 5 0.2 (0.2)   Solution: C Kc = 3 3 [] [] [ ] GJ GJ = 3 (0.5 / 5) (0.1/ 5)(0.2 / 5) = 3 3 0.5 5 0.1 (0.2)   9 An enzyme was found to operate at maximum efficiency in an aqueous solution buffered at pH 5. Which of the following would give the necessary buffer solution when dissolved in 10 dm3 of water