DHS Prelim P2 SuggestedSolutions

 DHS 2014 9647/02 2014 DHS Year 6 Preliminary Examination H2 Chemistry 9647/02 Suggested Solutions 1 Planning (P) A student was provided with a spirit burner containing a ‘fuel mixture’ which was prepared by mixing equimolar amounts of pentane and ethanol. The enthalpy change of combustion of this ‘fuel mixture’ is 11.8 kJ per mole of ‘fuel mixture’. He was told to use the enthalpy change of combustion of this ‘fuel mixture’ to find the heat capacity of a metal calorimeter using the apparatus shown below. Heat capacity is defined as the number of joules of heat needed to raise the temperature of the calorimeter by one Kelvin or one degree Celsius. Additional information: 1. Specific capacity of water is 4.2 J g1 K1. 2. The maximum capacity of the metal calorimeter is between 100 to 150 cm3. 3. A temperature rise of 5 C is considered to be significant for this experiment. (a) Construct a balanced equation for the complete combustion of the ‘fuel mixture’ with state symbols. [1] C5H12(l) + C2H5OH(l) + 11O2(g)  7CO2(g) + 9H2O(l) (b) Identify one possible source of error and suggest an improvement to overcome this error in the experiment. [2] Error: Heat loss to the surroundings by the calorimeter and water. Improvement: Conduct the experiment in a draught-free room. Also accept: 1. Provide lagging on the metal calorimeter. 2. Cover the metal calorimeter with a cover/lid. (c) Calculate the minimum mass of the fuel mixture required to bring about a 5 C temperature rise. Assuming no heat loss, Q = mc∆T = (100)(4.2)(5) = 2100 J ∆Hc = – 2100 / nfuel mixture  – 11.8 x 103 = – 2100 / nfuel mixture  nfuel mixture = 2100 / (11.8 x 103) = 0.1779 mol spirit burner containing the fuel mixture metal calorimeter

Y6 H2 Chemistry Preliminary Exam 2 Suggested Solutions  DHS 2014 9647/02  minimum mass of fuel mixture = 0.1779 x 59.0 = 10.5 g [2] (d) Write a plan to determine the heat capacity of the metal calorimeter using the apparatus provided. In your plan you should give details of the procedure (number your steps) and provide a table to record the readings to be taken, including the units. Details about the appropriate mass of water used and temperature rise should also be included. [4] 1) Weigh the spirit burner containing the ‘fuel mixture’ 2) Using a measuring cylinder, measure 100 cm 3 of water into the metal calorimeter. 3) Measure the initial temperature of the water using a thermometer after the water is allowed to stand for a few minutes. 4) Light the burner and allow it to heat the water in the calorimeter. 5) Monitor the temperature of water using the thermometer, and extinguish the flame when the temperature of the water increases by about 5 oC. 6) Measure the highest temperature reached after the flame has been extinguished. 7) Cool and reweigh the spirit burner with the remaining ‘fuel mixture’. 8) Ensure that the mass of the fuel mixture is at least 10.5 g. Table: Initial temperature of water/ oC T1 Final temperature of water/ oC T2 Initial mass of spirit burner with ‘fuel mixture’ / g M Final mass of spirit burner with remaining ‘fuel mixture’ /g N (e) Show how you would calculate, from your above proposed plan and experimental results, the heat capacity of the metal calorimeter. [3] Since density of water is 1 g cm–3, mass of the water in the calorimeter = 100 g Let the heat capacity of the calorimeter be C J K–1 Temperature rise of the water in the calorimeter = 5C Heat gained by water and calorimeter = 100 x 4.2 x 5 + C x 5 = 2100 + 5C J Mass of ‘fuel mixture’ burned = (M-N) g = P g Average molar mass of ‘fuel mixture’ = (72 + 46)/2 = 59.0 gmol-1 Hence, amount of ‘fuel mixture’ = P/59 mol Heat lost by ‘fuel mixture’ = 11800 x P/59 = 200P J Heat lost by ‘fuel mixture’ = Heat gained by water and calorimeter 200P = 2100 + 5C C = 40P – 420 [Total: 12]

Y6 H2 Chemistry Preliminary Exam 3 Suggested Solutions  DHS 2014 9647/02 2 But–1–ene can be converted to compound E via the following series of reactions. CH2=CHCH2CH3 I A (C4H10O) II B C (C5H9NO) D IV LiAlH4,dry ether V ClCOCH2COCl E (C8H13NO3) HCN, trace NaCN 10 - 20 °C Both compounds A and B produce a yellow precipitate on warming separately with aqueous alkaline iodine. (a) Iodine undergoes a disproportionation reaction with hot dilute NaOH(aq). Describe what is observed and write a balanced equation for the reaction. Black solid dissolved (accept brown solution) to form a colourless solution. 3I 2 + 6NaOH → 5NaI + NaIO3 + 3H2O (b) State the reagents and conditions for steps I and II. Step I: cold conc H2SO4 followed by H2O and heat Step II: acidified K2Cr2O7(aq), reflux OR acidified KMnO4(aq), reflux (c) In the boxes below, draw the structural formulae of compounds A, B, D and E. A CH3 CH3 OH B CH3 CH3 O D CH3 CH3 OH NH2 E O N H CH3 CH3 O O (d) Explain why the reaction in Step III produces an equimolar mixture of two stereoisomers of compound C. Compound B is a carbonyl which is planar . The attacking nucleophile can approach the planar carbonyl from the top side or bottom side with equal probability hence resulting in a racemic mixture. [Total: 9]

Y6 H2 Chemistry Preliminary Exam 4 Suggested Solutions  DHS 2014 9647/02 3 2–chloroethyl methyl sulfide may be synthesised from ethane. CH2 CH2 C C H H Cl H H OH I Cl2(aq) II CH3SNa ethanol, heat C C H H S H H OH CH3 III C C H H S H H Cl CH3 2-chloroethyl methyl sulfide (a) Describe a simple chemical test to distinguish between CH 3SCH2CH2OH and CH3SCH2CH2Cl, of the reaction in Step III, stating the expected observation for each compound. Test: Add PCl5 to each of the sample. Observation: White fumes observed for CH3SCH2CH2OH. No white fumes observed for CH3SCH2CH2Cl. (b) The kinetics of the reaction in Step II was studied. The experimental results are given in the table below. Run [CH3SNa] / mol dm–3 [CH2ClCH2OH] / mol dm–3 Relative rate / min–1 1 0.100 0.150 6 2 0.150 0.150 9 3 0.200 0.200 16 Use the data to determine the order of reaction with respect to both CH 3SNa and CH2ClCH2OH. Hence, write a rate equation for the reaction and state the units for the rate constant. Comparing Run 1 & 2, when [CH 3SNa] increases by 1.5 times and the [CH2ClCH2OH] remains the same, the relative rate increases by 1.5 times. Hence Order with respect to CH3SNa is 1. = =  x = 1 Hence Order with respect to CH2ClCH2OH is 1. Rate equation: Rate = k[CH3SNa] [CH2ClCH2OH] Units for rate constant = mol–1 dm3 min–1 (c) In organic syntheses, the choice of solvent can affect the rate of a reaction. Ethanol is the solvent of choice in Step II above. The process of forming ion -dipole interactions between ions and solvent

Y6 H2 Chemistry Preliminary Exam 5 Suggested Solutions  DHS 2014 9647/02 molecules is called solvation. Polar protic solvents such as ethanol, contains at least one hydrogen atom directly bonded to an electronegative atom. These solvents solvate both cations and anions. It is known that the concentration of ions is inversely proportional to the degree of solvation of the ions involved. Polar aprotic solvents contain no hydrogen atom directly bonded to an electronegative atom. These solvents solvate