2019 TMJC H2 Chem Prelim P2 ANS

1 Tampines Meridian Junior College 2019 JC2 Prelim Exam H2 Chemistry 1 (a) Carbon forms compo unds with Group 16 elements such as oxygen, sulfur and selenium. The properties of some of these compounds are given below. compound structure net dipole moment boiling point / C CO2 O=C=O 0 sublimes CS2 S=C=S 0 46 COS S=C=O 0.71 –50 COSe Se=C=O 0.73 –22 (i) Explain, in terms of structure and bonding, the difference in the boiling points of CS2 and COS. CS2 has a higher boiling point. Both CS 2 and COS have simple molecular structures. CS2 has a larger electron cloud (or larger number of electrons ) than COS. More energy is required to overcome the stronger instantaneous dipole- induced dipole interactions (or IMF/Dispersion forces) between CS 2 molecules than the instantaneous dipole -induced dipole interactions between COS molecules. [2] (ii) Explain why  CO2 has no net dipole moment.  COSe has a greater net dipole moment than COS. CO2 has not net dipole moment because it is linear and the dipole moments cancel out. COSe has a greater net dipole moment than COS. There is a smaller difference between the dipole moment of C=O and C=S in COS than that between C=O and C=Se in COSe since S is more electronegative than Se. OR polarity of C=S is greater than that of C=Se. [2] TAMPINES MERIDIAN JUNIOR COLLEGE 2019 JC2 H2 Chemistry Prelim Exam Paper 2 (Suggested Answers)

2 Tampines Meridian Junior College 2019 JC2 Prelim Exam H2 Chemistry Carbon forms the backbone of organic compounds. Hydrocarbons are the simplest organic compounds that contain carbon and hydrogen. (b) Compound A is an isomer of the hydrocarbon octane, C8H18. A Controlled chlorination of compound A in the presence of UV light produces different mono-chlorinated products with a molecular formula of C8H17Cl. (i) Suggest the total number of constitutional isomers which can be formed from the possible mono-chlorination of compound A. Draw the structural formulae of any two of these products. Total number of possible mono-chlorinated products = 4 [3] (ii) Draw the skeletal formula of the isomer of octane which could produce only one possible mono-chlorinated product if it undergoes free radical substitution. [1]

3 Tampines Meridian Junior College 2019 JC2 Prelim Exam H2 Chemistry (c) Besides reaction of alkanes with halogens, the process of thermal cracking, in which large alkane molecules are broken down into smaller alkanes and alkenes, proceeds via free radical mechanism. The following are reactions involved when propane undergoes thermal cracking. CH3CH2CH3   ●CH3 + ●CH2CH3 -- (1) CH3CH2CH3   ●H + ●CH2CH2CH3 -- (2) ●CH3 + CH3CH2CH3   CH4 + ●CH2CH2CH3 -- (3) ●CH2CH3   ●H + CH2=CH2 -- (4) ●CH3 + CH2=CH2   ●CH2CH2CH3 -- (5) ●CH3 + ●CH3   CH3CH3 -- (6) 2 ●CH2CH3   CH3CH3 + CH2=CH2 -- (7) (i) Reactions (1) and (2) are termed initiation steps. By quoting relevant data from the Data Booklet, deduce which one is more likely to occur. Reaction (1) is more likely to occur as it is easier to break a C -C bond (350 kJ mol−1) compared to a C-H bond (410 kJ mol−1). [1] (ii) From reactions (3) to (7), identify those which may be termed propagation steps in the mechanism. Reactions (3), (4) and (5) [1] (iii) Which gas, if detected in the product mixture, would support the occurrence of both reactions (2) and (4)? Hydrogen [1] (iv) Suggest why reaction (7) may be termed a disproportionation reaction. The ●CH2CH3 radical loses a hydrogen (is oxidised) to form ethene and gains a hydrogen (is reduced) to form ethane. OR The oxidation number of carbon in ●CH2CH3 increases –3 from to –2 in ethane and decreases from to –3 to form ethane. [1]

4 Tampines Meridian Junior College 2019 JC2 Prelim Exam H2 Chemistry (d) Methanoic acid, H2CO2, is the simplest carboxylic acid. Draw a dot -and-cross diagram of methanoic acid. Suggest the shape around the carbon atom in methanoic acid. Shape around C atom: Trigonal planar [2] [Total: 14] 2 Hydrogen peroxide reacts with acidified iodide ions to liberate iodine, according to the following equation: H2O2(aq) + 2H+(aq) + 2I–(aq)  2H2O(l) + I2(aq) In investigations of this reaction, t he following results were obtained by varying the volumes of hydrogen peroxide and iodide ions. experiment volume of H2O2 / cm3 volume of I– / cm3 volume of H2O / cm3 initial rate / mol dm–3 s–1 1 20.0 20.0 20.0 1.2 x 10–2 2 20.0 30.0 10.0 1.8 x 10–2 3 50.0 10.0 0.0 1.5 x 10–2 (a) Explain why water was added to experiments 1 and 2. To ensure the total volume of the mixture remains constant , so that volume of reactant is directly proportional to [reactant] in the mixture . [1] (b) The reaction was determined to be zero order with respect to hydrogen ions. (i) Sketch the rate-concentration graph for H+ ions. [1] Rate [H+] 0

5 Tampines Meridian Junior College 2019 JC2 Prelim Exam H2 Chemistry (ii) Determine the order of reaction with respect to the other two reactants. Hence, write down the rate equation. Let rate= k[H2O2(aq)]m[I–(aq)]n Compare experiments 1 & 2, keeping [H2O2(aq)] and total volume constant 1.2𝑥10−2 1.8𝑥10−2 = 𝑘(20.0)𝑚(20.0)𝑛 𝑘(20.0)𝑚(30.0)𝑛 (or use inspection method) n = 1 Rate of reaction is 1st order with respect to I–(aq) . Or Compare experiments 1 & 2, keeping [H2O2(aq)] and total volume constant when [I–] x 3/2, rate of reaction x 3/2 Order of reaction w.r.t. I– = 1 Compare experiments 1 & 3, keeping total volume constant 1.2𝑥10−2 1.5𝑥10−2 = 𝑘(20.0)𝑚(20.0)1 𝑘(50.0)𝑚(10.0)1 m = 1 Rate of reaction is 1st order with respect to H2O2(aq) . Or Compare experiments 1 & 3, keeping total volume constant When [I–] x 1/2 and [H2O2] x 5/2, rate x 5/4 (1/2 x 5/2) Since the order wrt. I– is 1, order wrt to H2O2 is 1. rate= k[H2O2(aq)][ I–(aq)] [3] (c) In order to further investigate the kinetics of the reaction, experiments 4 and 5 were conducted. The following results were obtained by varying the concentrations of hydrogen peroxide and iodide ions. experiment initial [H2O2(aq)] / mol dm–3 initial [I–(aq)] / mol dm–3 4 0.020 0.500 5 0.050 1.000 The half-life of hydrogen peroxide was 9.6 min in ex periment 4. Explain and predict the half-life of hydrogen peroxide in experiment 5. For experiment 4 and 5, since [I–(aq)] >> [H2O2(aq)], [I–(aq)] is approximately constant. Thus, rate = k’[H2O2(aq)] (a pseudo first order reaction) where k’ = k[I–(aq)] t1/2 = = 𝑙𝑛 2 𝑘′ = 𝑙𝑛 2 𝑘[𝐼−] t1/2 of H2O2 in experiment 4 = 9.6 min (for [I–(aq)] = 0.500 mol dm–3) t1/2 of H2O2 in experiment 5 = 4.8 min (for [I–(aq)] = 1.00 mol dm–3) [2]

6 Tampines Meridian Junior College 2019 JC2 Prelim Exam H2 Chemistry (d) An alternative method of investigating the rate of the above reaction is by withdrawing aliquots at specified time intervals and titrating the iodine formed in each aliquot with sodium thiosulfate solution. Suggest how the reaction can be quench