NYJC H1 Chemistry P2 Answers

[Turn over NANYANG JUNIOR COLLEGE JC 2 Preliminary Examinations Higher 1 CANDIDATE NAME Mark Scheme CLASS TUTOR’S NAME CHEMISTRY 8872/02 Paper 2 22 September 2014 2 hours Candidates answer on the Question Paper Additional Materials: Answer Paper Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and class 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 staples, paper clips, highlighters, glue or correction fluid. Section A Answer all questions. Section B Answer any two questions on separate answer paper. 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. For Examiner’s Use 1 / 20 2 / 8 3 / 12 Total / 40 This document consists of 14 printed pages and 0 blank page.

2 H1 Chemistry 8872/02 NYJC J2/14 PX For examiner’s use only Section A Answer all the questions in this section in the spaces provided. 1 A gas is said to exhibit ideal gas behavior if it obeys the ideal gas equation pV = nRT where p = pressure of gas in atm, V = volume of gas in dm3, n = amount of gas (mol), R = 0.0821 dm3 atm K1 mol1 (a constant), and T = temperature in K. (a) A 2.00 dm3 flask contains oxygen at 1.20 atm and 314 K. (i) Use the ideal gas equation to find the amount of oxygen in the flask. pVn RT 1.20 2.00 0.0821 314   [1] = 0.0931 mol [1] (ii) Hence calculate the mass of oxygen in the flask. 0.0931  32.0 = 2.98 g [1] [3] (b) The data below were derived using the ideal gas equation for 1 g of hydrogen. p / atm V / dm3 T / K 1.0 12.3 300 1.4 17.6 600 2.1 5.9 300 3.6 6.8 600 4.8 2.6 300 5.6 4.4 600

3 H1 Chemistry 8872/02 NYJC J2/14 PX For examiner’s use only (i) How does p vary with V for 1 g of hyd rogen at a constant temperature? p is inversely proportional to V [1] (ii) Sketch the graphs of p against V for 1 g of hydrogen at 300 K and at 600 K on the same axes below. Label each graph. shape of both graphs [1] relative positions of graphs [1]

4 H1 Chemistry 8872/02 NYJC J2/14 PX For examiner’s use only (iii) Sketch the graph of pV against p for 1 g of hydrogen at 450 K. [1] [4] (c) Oxygen gas behaves ideally because the intermolecular forces of attraction are very weak. On the other hand, the intermolecular forces in methanol vapour cause it to behave less ideally. (i) What are the intermolecular forces of attraction in each substance? Oxygen: van der Waals forces / dispersion forces [1] Methanol vapour: hydrogen bonds [1] (ii) Draw a labelled diagram to show the forces of attraction between two molecules of methanol in the vapour state. dipoles on four atoms [1] lone pair on O atom [1] correctly labelled hydrogen bond [1] [5]

5 H1 Chemistry 8872/02 NYJC J2/14 PX For examiner’s use only (d) The ideal gas equation can be used to derive a relationship for the relative molecular mass of a gas. r mRTM pV where m is the mass of the gas in grams. A 0.458 g sample containing a gaseous mixture of AlCl 3 and Al2Cl6 takes up a volume of 54 cm3 at a temperature of 98 °C and a pressure of 1 atm. (i) Draw a dot-and-cross diagram for the AlCl3 molecule. only three dot-cross pairs on Al and no lone pair on Al [1] lone pairs on Cl [1] (ii) Draw the structural formula of the Al2Cl6 molecule. [1] (iii) Calculate the average Mr of the mixture.   r 0.458 0.0821 98 273 541 1000 M     = 258 [1] Converts temperature 98+273 = 371 K [1] Converts volume 54/1000 dm3 [1] (iv) Hence calculate the percentage by mass of Al2Cl6 in the mixture. 267.0 133.5(100 ) 258.1100 xx  [1] x = 93.5% [1] [8] [Total: 20]

6 H1 Chemistry 8872/02 NYJC J2/14 PX For examiner’s use only 2 (a) Explain what is meant by the term order of reaction. For a general reaction aA + bB  products The rate equation is rate = k [A]x[B]y, order of reaction wrt A is x & order of reaction wrt B is y overall order of reaction is (x+y) OR The order of reaction with respect to a given reactant/catalyst is the power to which the concentration of that reactant/catalyst is raised in the experimentally determined rate equation. [1] (b) Hydrogen peroxide reacts with iodide ions in acidic solution as shown below. H2O2(aq) + 2H+(aq) + 2I(aq)  2H2O(l) + I2(aq) The results of some investigations of the rate of this reaction are shown below. Experiment number [H2O2] / mol dm3 [H+] / mol dm3 [I] / mol dm3 Relative initial rate / mol dm3 s1 1 0.030 0.050 0.060 1.80 2 0.020 0.060 0.050 1.00 3 0.0375 0.060 0.060 2.25 4 0.025 0.060 0.050 1.25 (i) Use the above data to determine the order of reaction with respect to  H2O2  H+  I Comparing expt 2 and 4, when [H2O2] increased by 1.25 times (0.025/0.020), rate increased by 1.25 times (1.25/1.00) Order wrt H2O2 is 1 [1] Comparing expt 1 and 3, when [H 2O2] increased by 1.25 times (0.0375/0.030), rate should increased by 1.25 times because order wrt H2O2 is 1. Hence rate increased from 1.80 in expt 1 to 2.25. When [H +] increased by 1.2 times, rate remains unchanged from the increased due to H2O2. Rate is independent of the [H+]. Order wrt H+ is 0 [1]

7 H1 Chemistry 8872/02 NYJC J2/14 PX For examiner’s use only Since order wrt H+ is 0, we can ignore [H+] as we deduce the order wrt [I]. Rate = k[H2O2][I]a Comparing experiment 1 and 2, 1.80 = k (0.03)(0.06)a ----------------------(1) 1.00 = k(0.02)(0.05)a ----------------------(2) (1) (2) 1.80 (0.06) (0.03) 1.00 (0.05) (0.02) a a a = 1; order with respect to I is 1[1] (ii) Hence, write a rate equation for the above reaction. Rate = k[H2O2][I] [1] [4] (c) With the aid of a sketch of the Boltzmann distribution, explain how an increase in temperature increases the rate of a chemical reaction. [2] An increase in temperature will lead to increase in the average kinetic energy of reactant molecules. More molecules will have energy larger than E a resulting in more collisions with energy greater than E a. Hence frequency of activated collisions will increase. Rate increases. [1] [3] [Total: 8]

8 H1 Chemistry 8872/02 NYJC J2/14 PX For examiner’s use only 3 Esters are compounds which provide the flavour of many fruits and the perfumes of many flowers. (a) The ester CH3(CH2)2CO2CH3 contributes to the aroma of apples. (i) State the reagents and conditions needed for the hyd rolysis of this ester. Aqueous HCl / H2SO4, heat under reflux OR aqueous NaOH, heat under reflux ............................................... [1] (ii) Write the equation for the hydrolysis of this ester. CH3(CH2)2CO2CH3 + H2O ⇌ CH3(CH2)2CO2H + CH3OH OR CH3(CH2)2CO2CH3 + NaOH ⇌ CH3(CH2)2CO2 Na+ + CH3OH ..... [1] (iii) Apart from their use as perfumes and food flavourings, state one major commercial use of esters. The most common uses of esters are (1) in artificial scents or perfumes, and (2) as artificial flavorings for sweets, ice -creams and soft drinks Industrial uses of esters include [1] any one (3) as solvents in the manufacture of fats, cellulose, varnishes and paints; (4) as solvents within pharmaceutical industries, and (5) as softeners or plasticizers in plastic and molding industries. [3]

9 H1 Chemistry 8872/02 NYJC J2/14 PX For examiner’s use only (b) Leaf alcohol is a stereoisomer that can form when insects such as caterpillars eat gree