TJC Prelim P1 P2 ANS

Answ ers to H2 Chemistry Preliminary Examinations 2009 Paper 1 and 2 Paper 1 1 B 6 D 11 D 16 A 21 B 26 A 31 B 36 C 2 A 7 C 12 B 17 C 22 C 27 B 32 B 37 D 3 A 8 D 13 C 18 C 23 C 28 D 33 B 38 A 4 A 9 D 14 A 19 C 24 B 29 C 34 A 39 C 5 B 10 C 15 D 20 C 25 C 30 A 35 A 40 C Paper 2 1 This question is about chlorine and its compounds. (a) A teacher instructed a st udent to react some chlorine gas with 250 cm 3 of 0.100 mol dm-3 aqueous sodium hydroxide, and left the lab. After he returned, he realized that he had not specified the reaction temperature. The teacher found that the pH of the solution had become 12.0. After acidifying the solution and adding excess silver nitrate, 2.69 g of silver chloride was obtained. He then concluded that the student had heated the sodium hydroxide. For this question, you may assume that all of the chloride ions are precipitated by the addition of silver ions. (i) Determine how the teacher arrived at his conclusion. ● Number of moles of NaOH used = 0.250 x 0.100 – 0.250 x 10 (14 – 12.0) = 0.0225 mol ● Number of moles of chloride = 2.69 / (107.8 + 35.5) = 0.0188 mol Number of moles of chloride : number of moles of NaOH = 0.0188 : 0.0225 ● = 5 : 6 Which is in line with the equation: 3Cl 2 + 6OH- → 5Cl- + ClO3 - + 3H2O 1

(ii) If the student had not heated the aqueous sodium hydroxide and the same amount of chlorine gas was used, what would be the resulting mass of silver chloride obtained when excess silver ni trate is added after acidifying the solution? [5] Number of moles of chlorine gas used = 0.0188 x 3 / 5 = 0.0113 mol Cl2 + 2OH  C l + ClO + H2O ● Number of moles of chloride formed (cold) = 0.0113 mol ● Mass of silver chloride formed = 0.0113 x (107.8 + 35.5) = 1.62 g (b) The melting point of POCl 3 is 1.25 oC whereas the melting point of PC l5 is 166.8 oC. ) ame and draw diagrams to illustrate the shapes of POCl3 and PCl5.   (ii) , in terms of structure and bonding, why PC l5 has a higher melting point. [4] ● ed-dipole-induced- ● resulting in stronger van der Waals’ forces. (i N Explain POCl 3 and PC l5 both are simple discrete molecules with weak van der Waals’ forces of attraction between the molecules (or POC l3 has permanent-dipole-permanent- dipole interactions while PC l5 has induc dipole interactions between molecules). PCl5 has a much larger M r than POC l3, and therefore a larger electron cloud that is mo re easily distorted, giving rise to larger partial charges, Cl P Cl O C Cl C l POCl3 Tetrahedral Trigonal bipyramidal PCl5 PCl C Cl l l 2

(c) hosp uch as the following: ) ith reference to th B ooklet a nthalpies, alculate the enthalpy of the abo y change of reacti on = (460 + 360 + 2 x 331) Bond Bond Ene kJ mol -1 P horous pentachloride is commonly used in organic reactions s CH3OH(l) + PCl5(s) CH 3Cl(g) + POCl3(l) + HCl(g) (i W e Data change nd the following bond e ve reaction. c Bonds broken: C-O, 2 P-Cl, O-H Bonds formed: C-Cl, P=O, HCl  Enthalp – (340 + 460 + 431) = 1482 – 1231 ● = + 251 kJ mol –1 of CH3OH rgy / P=O 460 P-Cl 331 i) educe whether the entropy change is positive or negative and hence curs at room temperature. ● 3 l), this gives rise to a greater degree of disorder in the products. (i D explain why the above reaction oc ● Entropy change is positive. As the reactants are in liquid and solid states but the products contain two gases (CH Cl and HC Go = Ho – TSo Since H● o is positive and So is positive, room temperature must be sufficiently high that the magnitude of T So exceeds the magnitude of Ho. 3

ii) Given that the actual standard enthalp 1 ● the data booklet are average values, each taking into account the bond energies of the same bond in different compounds. ● Bond energies are based on the compounds being in gaseous state (The discrepancy in value is due to the enthalpy change of fusion of ). (d) stud nd etha ) De to first i dentify ethanoyl chloride from among the test tubes. ● tube with ethanoyl chloride, while no pr ecipitate will be formed in the (i y change of reaction is 04.3 kJ mol1, suggest two reasons for the discrepancy. [7] The values in but in the above reaction, CH 3OH, PC l5 and POC l3 are non-gaseous. PCl5, and the enthalpy change of vapourisation of all three compounds A ent mixed up three test tubes cont aining chlorobenzene, trichloromethane noyl chloride separately. scribe a chemical test a (i three test tubes. ● Add acidified aqueous silver nitrate at room temperature to all three A white precipitate (of silver chlori de) will be formed in the test other two test tubes. 4

(ii) Describe a second chemical test to then identify trichloromet hane from among the remaining two test tubes. de to both test tubes and reflux. Then hile no precipitate will be formed in the other test tube. [Total: 20 marks] 2 The use of the Data Booklet is required for this question. (a) romine is the only liquid non-metallic element at room temperature . It reacts igorously with metals , especially in the presence of water, as well as most organic ompounds, especially upon illumination. min attery hich mine from reaching the zinc electrode where it would react with inc, c using the battery to self-discharge. During the charging of the battery, romid is converted to bromine. [4] ● Add aqueous sodium hydroxi add excess dilute nitric acid followed by aqueous silver nitrate. ● A white precipitate (of silver chloride) will be formed in the test tube with trichloromethane, w B v c Bro e is also used in the zinc-bromi ne flow battery which is a rechargeable . The two electrode chambers are separated by a micro-porous membrane prevents bro b w z a b e carbon zinc mixtu f Z Br 2 (aq) and Br2 (aq) ZnBr2 (aq) re o n micro-porous membrane 5

(i) In the table below, indi cate the polarity (+/ ) of the electrodes and write the half-equations for the electrode proc ess es that occur when the battery discharges. Electrode Polarity Half-equation Zinc   Zn(s)  Zn 2+(aq) + 2e Carbon +  Br (aq) + 2e  2Br(aq) 2 (ii) 4. th g of bromine.  2.0 538 s  2 g of bromine is formed in the ce ll when a current of 2.0 A is passed rough it during charging. Determine the length of time required to form 4.2  Number of moles of electrons passed = 2  (4.2 / 159.8) = 0.0526 mol Quantity of charge required = 0.0526  96500 = 5076 C = I  t  time required = 5076 = 2  6

(iii) In a research laboratory, a research worker first charged a zinc-bromine battery fully before adding 10 c m3 of 0.2 mol dm 3 sodium bromide solution to each of the two electrode chambers. What is the effect on the electrical output of the battery when it is discharged? [8]  There is no effect to the positi on of equilibrium in the chamber containing Zn electrode and Zn 2+(aq). Hence EZn2+/ Zn remains as 0.76 V.  Br 2 (aq) + 2e 2B