GEAR UP YOUR POTENTIAL – ANSWER SCHEME SK025 1 LABUAN MATRICULATION COLLEGE GEAR UP YOUR POTENTIAL – ANSWER SCHEME CHEMISTRY 2 hour DO NOT OPEN THIS QUESTION PAPER UNTIL YOU ARE TOLD TO DO SO. This question paper consists of 11 questions. Answer all questions.
GEAR UP YOUR POTENTIAL – ANSWER SCHEME SK025 2 RELATIVE ATOMIC MASSES OF SELECTED ELEMENTS Element Symbol Proton Number Relative Atomic Mass Aluminium Al 13 27.0 Silver Ag 47 107.9 Argon Ar 18 40.0 Arsenic As 33 74.9 Gold Au 79 197.0 Barium Ba 56 137.3 Beryllium Be 4 9.0 Bismuth Bi 83 209.0 Boron B 5 10.8 Bromine Br 35 79.9 Iron Fe 26 55.9 Fluorine F 9 19.0 Phosphorus P 15 31.0 Helium He 2 4.0 Mercury Hg 80 200.6 Hydrogen H 1 1.0 Iodine I 53 126.9 Cadmium Cd 48 112.4 Potassium K 19 39.1 Calcium Ca 20 40.1 Carbon C 6 12.0 Chlorine Cl 17 35.5 Cobalt Co 27 58.9 Krypton Kr 36 83.8 Chromium Cr 24 52.0 Copper Cu 29 63.6 Lithium Li 3 6.9 Magnesium Mg 12 24.3 Manganese Mn 25 54.9 Sodium Na 11 23.0 Neon Ne 10 20.2 Nickel Ni 28 58.7 Nitrogen N 7 14.0 Oxygen O 8 16.0 Platinum Pt 78 195.1 Lead Pb 82 207.2 Protactinium Pa 91 231.0 Radium Ra 88 226.0 Radon Rn 86 222.0 Rubidium Rb 37 85.5 Selenium Se 34 79.0 Cerium Ce 58 140.1 Cesium Cs 55 132.9 Silicon Si 14 28.1 Scandium Sc 21 45.0 Tin Sn 50 118.7 Antimony Sb 51 122.0 Strontium Sr 38 87.6 Sulphur S 16 32.1 Uranium U 92 238.0 Tungsten W 74 183.8 Zinc Zn 30 65.4
GEAR UP YOUR POTENTIAL – ANSWER SCHEME SK025 3 LIST OF SELECTED CONSTANT VALUES Ionisation constant for water at 25 ⁰C Kw = 1.00 × 10–4 mol2 dm–6 Molar volume of gases Vm = 22.4 dm3 mol–1 at STP = 24 dm3 mol–1 at room temperature Speed of light in a vacuum c = 3.0 × 108 m s–1 Specific heat of water = 4.18 kJ kg–1 K –1 = 4.18 J g–1 K –1 = 4.18 J g–1 ⁰C–1 Avogadro’s number NA = 6.02 × 1023 mol–1 Faraday constant F = 9.65 × 104 C mol–1 Planck’s constant h = 6.6256 ×10–34 J s Rydberg constant RH = 1.097 × 107 m–1 = 2.18 × 10–18 J Ideal gas constant R = 8.314 J mol-1 K –1 = 0.08206 L atm mol–1 K –1 Density of water at 25⁰C ρ = 1 g cm–3 Freezing point of water = 0.00 ⁰C Vapour pressure of water at 25⁰C Pwater = 23.8 torr UNIT AND CONVERSION FACTOR Volume 1 liter = 1 dm3 1 mL = 1 cm3 Energy 1 J = 1 kg m2 s –2 = 1 N m = 107 erg 1 calorie = 4.184 Joule 1 eV = 1.602 x 10-19 J Pressure 1 atm = 760 mmHg = 760 torr = 101.325 kPa = 101325 N m-2 Others 1 Faraday (F) = 96500 coulomb 1 newton (N) = 1 kg m s –2
GEAR UP YOUR POTENTIAL – ANSWER SCHEME SK025 4 Answer all questions. 1. The synthesis of mercurous chloride, Hg2Cl2 can be carried out by reacting mercury(II) chloride, HgCl2, with oxalic acid, H2C2O4. The chemical equation is given as follow. 2 HgCl2 (aq) + H2C2O4 (aq) → 2 HCl (aq) + 2 CO2 (g) + Hg2Cl2 (s) In investigation of this reaction, the following results were obtained: TABLE 1 Experiment Initial Concentration, M Initial rate, Ms-1 HgCl2 H2C2O4 1 0.573 0.252 0.0204 2 1.146 0.252 0.0817 3 0.573 0.504 0.0204 a) Determine the order of reaction with respect to HgCl2 and H2C2O4. b) Write the rate law for the reaction. c) Determine the time needed for 40% of a 0.06M HgCl2 to react, if the rate constant for the reaction is 0.0621 M-1 s -1 d) Give another factor that could affect the rate of production of mercurous chloride. [10 marks] NO. PART SCHEME MARKS 1. (a) Rate = k [HgCl2] x [H2C2O4] y To solve x, compare Experiment 1 and 2 Exp 2/ Exp 1 0.0817 / 0.0204 = ( 1.146 / 0.573 ) x 4 = 2x x = 2 The reaction is a 2nd order reaction w.r.t HgCl2 To solve Y, compare Experiment 1 and 3 Exp 3 / Exp 1 0.0204 / 0.0204 = ( 0.504 / 0.252 ) y 1 = 2y y = 0 The reaction is zero order w.r.t H2C2O4 1 1 1 1 1 M1 M2 M3 M4 M5 (b) Rate equation: Rate = k [HgCl2] 2 1 M6
GEAR UP YOUR POTENTIAL – ANSWER SCHEME SK025 5 NO. PART SCHEME MARKS 1. (c) 40% of a 0.06M HgCl2 to react means 60% HgCl2 left Hence, 60% of 0.06M = 0.036M 1 [2 ] − 1 [2 ] = @ 1 0.036 − 1 0.06 = (0.0621) t = 178.92 s (unit insist) 1 1 formula @ subs. 1 M7 M8 M9 (d) Temperature @ size of particle @ presence of catalyst 1 M10 TOTAL 10
GEAR UP YOUR POTENTIAL – ANSWER SCHEME SK025 6 2. (a) The enthalpy of combustion of fructose, C6H12O6 is -21.2 kJ mol-1 . An amount of 2.63g of C6H12O6 was completely burned in a bomb calorimeter at 25.0 oC. (i) Write the thermochemical equation for the combustion of fructose. (ii) Calculate the final temperature of the calorimeter if it contains 225.0 ml of water. [5 marks] (b) Use the data given below to calculate the first electron affinity of chlorine. Type of enthalpies Value (kJ mol-1 ) Enthalpy of atomisation of magnesium + 150 1 st Ionisation Energy of magnesium + 736 2 nd Ionisation Energy of magnesium +1450 Enthalpy of atomisation of chlorine +122 Enthalpy of formation of magnesium chloride -642 Lattice energy of magnesium chloride -2526 [5 marks] NO. PART SCHEME MARKS 2. (a)(i) C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O(l) ∆H = -21.2 kJ mol-1 (Correct species, phases, and ∆H value with negative sign are written) 1 M1 (a)(ii) Mol of fructose = 2.63 / 180 = 0.01461 mol ∆H = - (q / n) @ -21.2 = - (q / 0.01461) q = 0.3097 kJ @ 309.7 J @ 1 mol of fructose releases 21.2 kJ of heat 0.01461 mol of fructose releases 0.01461 x 21.2 = 0.3097 kJ @ 309.7 J of heat q = mw cw ∆T @ 309.7 = 225 x 4.18 x ( Tf - 25 ) Tf = 25.33 oC 1 1 1 1 1 1 1 M2 M3 M4 M3 M4 M5 M6 NO. PART SCHEME MARKS
GEAR UP YOUR POTENTIAL – ANSWER SCHEME SK025 7 2. (b) @ Correct species, phase and ∆H = 1 √ - 642 = (+150) + (+244) + (+736) + (+1450) + 2x + (- 2526) 2x = -696 x = -348 kJ mol-1 x = 1st electron affinity @ EA = -348 kJ mol-1 (unit insist!) 3 7 √ = 3m 5 – 6 √ = 2m 3 – 4 √ = 1m 1 1 M7 + M8 + M9 M10 M11 TOTAL MAX 11 10
GEAR UP YOUR POTENTIAL – ANSWER SCHEME SK025 8 3. (a) Based on the standard electrode potential given below: Fe3+ (aq) + e → Fe2+ (aq) Eo = +0.77V F2 (g) + 2e → 2F- (aq) Eo = +2.87V Predict the spontaneity of the voltaic cell above. Explain your answer. [3 marks] (b) In an electrolysis, a current of 2.50A is passed through an aqueous solution of tin salt for 3 hours. If. 8.32 g of tin are deposited at cathode, what is the formula of tin ion? [4 marks] NO. PART SCHEME MARKS 3. (a) E o cell = Eo cathode – E o anode = (+2.87) – (+0.77) = +2.10 V The reaction is a spontaneous reaction The value of Eo cell is positive 1 1 1 M1 M2 M3 (b) Q = It @ = (2.50) ( 3 x 60 x 60 ) = 27,000 C @ ≡ 0.28 F Mol of tin deposited = 8.32 / 118.7 = 0.07 mol 0.07 mol of Tin ≡ 0.28F 1 mol of Tin ≡ (0.28 / 0.07) = 4F Formula of Tin Ion = Sn4+ 1 1 1 1 (ratio) 1 M4 M5 M6 M7 M8 TOTAL MAX 8 7
GEAR UP YOUR POTENTIAL – ANSWER SCHEME SK025 9 4. (a) Write the equations for the reaction of 2-methyl-1-butene with the following reagents respectively. (i) HBr in the presence of CH3OOCH3 (ii) Acidified water (iii) Bromine water [3 marks] (b) Suggest a chemical test to differentiate between 2-methylbutane and 2-methyl-1-butene. Write the chemical equations and observation involved. [3 marks] NO. PART SCHEME MARKS 4. (a)(i) 1 M1 (a)(ii) 1 M2 (a)(iii) 1 M3 (b) Bromine test Observation: Reddish-brown of bromine decolourises. Observation: Reddish-brown of bromine remains unchanged. OR … 1 1 (Equation & observation) 1 (Equation & observation) M4 M5 M6
GEAR UP YOUR POTENTIAL – ANSWER SCHEME SK025 10 NO. PART SCHEME MARKS 4. (b) Baeyer’s test Observation: Purple colour of KMnO4 decolourises and brown precipitate is formed. Observation: Purple colour of KMnO4 remains unchanged and no brown precipitate is formed. TOTAL 6
GEAR UP YOUR POTENTIAL – ANSWER SCHEME SK025 11 5. Based on the reaction scheme below: (a) Suggest the structure of A. (b) Name the reaction for the formation of A. (c) Give reagents I and II. (d) Draw the structures of B and C. Label the major product. [7 marks] NO. PART SCHEME MARKS 5. (a) 1 M1 (b) Friedel-Crafts Alkylation reaction. 1 M2 (c) I : 2-bromopropane , FeCl3 @ 2-chloropropane , AlCl3 @ FeCl3 II : KMnO4 , H+ , heat @ Δ @ hot, acidified KMnO4 @ hot, acidified K2Cr2O7 1 1 M3 M4
GEAR UP YOUR POTENTIAL – ANSWER SCHEME SK025 12 NO. PART SCHEME MARKS 5. (d) Note: Structures of B & C are interchangeable. 1 1 1 M5 M6 M7 TOTAL 7
GEAR UP YOUR POTENTIAL – ANSWER SCHEME SK025 13 6. (a) Based on the reaction scheme below, draw structural formulae for D, E, and F. [3 marks] (b) The structural formula of 2-bromobutane which undergoes SN2 reaction with hydroxide ion, OH– , is given below: (i) Illustrate the SN2 mechanism for the reaction above. (ii) State the change of rate of reaction if the concentration of 2-bromobutane and concentration of hydroxide ion, OH– , are doubled respectively. [4 marks] NO. PART SCHEME MARKS 6. (a) M1 M2 M3 M1 M2 M3 (b)(i) 1 (Both curved arrows correct) 1 (Transition state) 1 (Inverted product) M4 M5 M6 (b)(ii) Quadrupled @ Increase 4 times 1 M7 TOTAL 7
GEAR UP YOUR POTENTIAL – ANSWER SCHEME SK025 14 7. (a) Compound G is the isomer of an alcohol with molecular formula C3H8O. Compound G gives cloudy solution with Lucas reagent within 5-10 minutes and also gives positive iodoform test. (i) Draw the structure of G. (ii) Write the chemical equations for all reactions involved. [3 marks] (b) Explain the following statement: “Butanol has a higher boiling point than pentane.” Compound Molecular mass Boiling point (⁰C) 1-Butanol 74 118 Pentane 72 36.1 [2 marks] NO. PART SCHEME MARKS 7. (a)(i) 1 M1 (a)(ii) 1 1 M2 M3 (b) - 1-butanol can form hydrogen bond between its molecules while pentane only has weak van der Waals forces. - Hydrogen bond is stronger than van der Waals forces. - Thus, 1-butanol requires more energy to overcome its intermolecular forces between its molecules. 1 1 1 M4 M5 M6 TOTAL MAX 6 5
GEAR UP YOUR POTENTIAL – ANSWER SCHEME SK025 15 8. (a) Draw the structure of compound K, L, M, N and O as shown in the following reaction scheme. [5 marks] (b) Suggest suitable reagents for the conversion of propanal to the following compounds: (i) 1-propanol (ii) Propanoic acid [2 marks] NO. PART SCHEME MARKS 8. (a) 1 1 1 1 1 M1 M2 M3 M4 M5
GEAR UP YOUR POTENTIAL – ANSWER SCHEME SK025 16 NO. PART SCHEME MARKS 8. (b) (i) NaBH4 in methanol @ H2, Pt/ Ni/ Pd @ LiAlH4 followed by hydrolysis, H3O + (ii) KMnO4, H+ , heat @ K2Cr2O7, H+ , heat @ Na2Cr2O7, H+ , heat 1 1 M6 M7 TOTAL 7
GEAR UP YOUR POTENTIAL – ANSWER SCHEME SK025 17 9. (a) When 2-phenylethanoic acid reacts with alcohol P, an ester Q with molecular formula C11H14O2 is formed. Oxidation of alcohol P with hot acidified potassium permanganate yield R. Treatment of R with alkaline iodine solution yields a yellow precipitate of compound S. (i) Draw the structures of P, Q and R. (ii) Write all chemical equations involved. (iii) Give the IUPAC name of S. [7 marks] (b) Arrange the following compounds in order of increasing acidity and explain your answer. 2-chloropropanoic acid, 3-chloropropanoic acid, 2,2-dichloropropanoic acid I II III [3 marks] NO. PART SCHEME MARKS 9. (a)(i) 1 1 1 M1 M2 M3 (a)(ii) yellow Precipitate 1 1 1 M4 M5 M6
GEAR UP YOUR POTENTIAL – ANSWER SCHEME SK025 18 NO. PART SCHEME MARKS 9. (a)(iii) Triiodomethane 1 M7 (b) II < I < III Increasing order of acidity 2,2-dichloropropanoic acid is the most acidic because there are a greater number of halogens @ high number of EWG. The inductive effect is much stronger compared to the acid containing one halogen atom. 2-chloropropanoic acid is more acidic than 3- chloropropanoic acid because the position halogen @ EWG is closer to the carboxyl group. The nearer the distance between halogen and carboxyl group, the greater the inductive effect and thus stronger the acid. 1 1 1 M8 M9 M10 TOTAL 10
GEAR UP YOUR POTENTIAL – ANSWER SCHEME SK025 19 10. (a) Based on the reaction scheme below: (i) Draw the structures of T and V. (ii) Identify reagent U. [3 marks] (b) Propanamine and N-ethylethanamine can be differentiated using nitrous acid test. Write the reaction equation and state the observation when both propanamine and N-ethylethanamine reacts with nitrous acid. [4 marks] NO. PART SCHEME MARKS 10. (a)(i) 1 1 M1 M2 (a)(ii) Zn, H+ @ Fe, H+ @ SnCl2, H+ 1 M3
GEAR UP YOUR POTENTIAL – ANSWER SCHEME SK025 20 NO. PART SCHEME MARKS 10. (b) 1 1 1 1 M4 M5 M6 M7 TOTAL 7
GEAR UP YOUR POTENTIAL – ANSWER SCHEME SK025 21 11. The structure of leucine is shown as below. (a) Name the compound according to IUPAC nomenclature. (b) Draw the structure of zwitterion for leucine. (c) Draw the structural formula of the product formed when leucine reacts with (i) ethanol in the presence of acid, (ii) HCl. [4 marks] NO. PART SCHEME MARKS 11. (a) 2-amino-4-methylpentanoic acid 1 M1 (b) 1 M2 (c)(i) 1 M3 (c)(ii) 1 M4 TOTAL 4