NJC 2018 SH1 H1 Promo Paper 2 Answer

1 NJC Promotional Examination 887302/18 [Turn over SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper 2 Structured Questions Additional Materials: Data Booklet Writing Papers 8873/02 Mon 1 October 2018 1.5 hours READ THE INSTRUCTIONS FIRST Write your subject class, registration number and name on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use paper clips, highlighters, glue or correction fluid. Answer all questions in Section A (40 marks) in the space provided on the Question paper. Answer one question in Section B (20 marks) on the writing paper provided separately. The use of an approved calculator is expected, where appropriate. A Data Booklet is provided. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. Appropriate significant figures and units are expected for final numerical answers For Examiner’s Use 1 / 8 2 / 7 3 / 7 4 / 9 5 / 9 6 /20 Paper 2 / 60 Paper 1 / 20 Overall Total / 80 Overall percentage / 100 NATIONAL JUNIOR COLLEGE SH1 PROMOTIONAL EXAMINATION Higher 1 CANDIDATE NAME

2 NJC Promotional Examination 8873/02/18 Answer ALL questions in the spaces provided. 1 (a) The element nickel was discovered in 1751 by Swedish chemist A. F. Cronstedt. (i) Nickel is a transition metal. Describe the bonding in the element nickel. You should draw a labelled diagram to illustrate your answer. Metallic bonding is present in nickel, which is the electrostatic attraction between (positive) nickel ions and its delocalised (or a sea of) electrons. [1] [1] [2] (ii) State two physical properties that you would expect nickel metal to possess. Explain, in terms of the bonding present, why it possesses these properties. property Good electrical conductivity [1/2] explanation delocalised sea of electrons are mobile electrons that can act as charge carriers. [1] property high melting/boiling point [1/2] explanation strong electrostatic force of attraction between positive metal ions and sea of delocalised electrons requiring a large amount of energy to break. [1] [3] (b) A common battery is the nickel -cadmium cell. It has one electrode of cadmium and one electrode of nickel ( III) hydroxide, Ni(OH) 3. The two electrodes are connected using an electrolyte. The two half-equations for this cell are: Cd(OH)2 + 2e− Cd + 2OH− Ni(OH)3 + e− Ni(OH)2 + OH− (i) Combine these two half-equations to show the above overall reaction. Cd + 2 Ni(OH)3 → Cd(OH)2 + 2Ni(OH)2 ………………………………………………………………………………………...[1] Ni cation sea of delocalised electrons

3 NJC Promotional Examination 8873/02/18 [Turn over (ii) Using your equation in (b)(i), identify the species that have been reduced and oxidised respectively. Ni(OH)3 Species reduced: …………………………… Cd Species oxidised: …………………………… [2] [Total: 8] 2 Sulfur is a common element on Earth that forms many important chemical compounds. The table below shows the melting point of sulfur and some of its compounds. Compound Melting point / °C Sulfur, S8 115 Sodium thiosulfate, Na2S2O3 49 Sodium sulfide, Na2S 117 (a) Draw the dot-and-cross diagram of sodium sulfide, Na2S. [1] Na+ [1] S2− [2] (b) By considering the bonding and structure, explain why sulfur has a higher melting point than sodium thiosulfate. (i) Sodium thiosulfate has giant ionic structure [1/2] and sulfur has simple covalent structure [1/2]. The large electron cloud size in sulfur molecule leads to stronger [1/2] instantaneous dipole-induced dipole interactions between molecules [1/2] which require more energy [1/2] to overcome compared to ionic bonds in sodium thiosulfate [1/2]. …………………………………………………………………………………………... …………………………………………………………………………………………... …………………………………………………………………………………………... …………………………………………………………………………………………... ……………………………………………………………………………………..... [3]

4 NJC Promotional Examination 8873/02/18 (ii) How would you expect the magnitude of the lattice energy of Na 2S to compare with that of Na2S2O3? Explain your answer. Both compounds have the same product of charges but interionic distance for Na2S is smaller than that of Na2S2O3 [1]. Magnitude of LE of Na2S is larger [1]. Minus 0.5m for missing LE equation …………………………………………………………………………………………... …………………………………………………………………………………………... …………………………………………………………………………………………... …………………………………………………………………………………………... ……………………………………………………………………………………..... [2] [Total: 7]

5 NJC Promotional Examination 8873/02/18 [Turn over 3 The following reaction scheme shows how compound A, CH3CHClCH2CH3 could be formed and converted to other useful organic products. (a) Give the name for compound A. 2-chlorobutane ………………………………………………………. [1] (b) Suggest the reagents and conditions for step I and II. I: (limited) Cl2, uv ……………………………………………………………………………………….………. II: NaOH, ethanol, heat ……………………………………………………………………………………….……….[2] (c) Suggest the type of reaction for step II. Elimination ………………………………………………………………………………………………..[1] (d) Write a balanced equation for the reaction B with hydrogen. CH3CH=CHCH3 + H2 → CH3CH2CH2CH3 ………………………………………………………………………………………………..[1] (e) Compound B exhibits cis-trans isomerism. Draw and name the isomers. Draw and name “cis-isomer” [1] Draw and name “trans-isomer” [1] cis-isomer trans-isomer [2] [Total: 7] CH3CH2CH2CH3 CH3CHClCH2CH3 I CH3CH=CHCH3 A B II CH3 CH3

6 NJC Promotional Examination 8873/02/18 4 Nitrogen monoxide reacts with oxygen gas according to the equation: 2NO(g) + O2(g) → 2NO2(g) An experiment is performed to determine the order of reaction with respect to nitrogen monoxide and oxygen. The concentration of oxygen used for both experiments is 0.005 mol dm−3. The results are as follows. (a) Using the graphs above, determine the order of reaction with respect to O2 and NO, showing your workings clearly. When [NO] = 0.1 mol dm −3, t 1 = t 2  7 min . The half-lives of O 2 are approximately constant. Order of reaction wrt O2 is 1. [1] OR When [NO] = 0.05 mol dm −3, t 1 = t 2 28 min . The half-lives of O 2 are approximately constant. Order of reaction wrt O2 is 1. [1] When [NO] doubles, half-life decreases by 4 times. Rate increases by 4 times. Order of reaction wrt NO is 2. [1] OR Can solve by drawing tangent at t=0 to find initial rate for both graphs. [2]

7 NJC Promotional Examination 8873/02/18 [Turn over (b) Write a rate equation for the reaction. Rate=k[O2][NO]2 ……………………………………………………………………………………………..…[1] (b) Calculate the initial rate of reaction for