DHS H2 Chem 2012 Prelim P1 Soln

2012 DHS Preliminary Examination (H2 Chemistry 9647/01 Solutions) 2 6 C Calcium and chlorine form CaC l2 rather than CaCl because DHf(CaCl2) is more exothermic than DHf(CaCl). DHf is the sum of enthalpy changes involved in a series of processes leading from the elements to the compounds: 1. Forming separate atoms from elements (DHat) 2. Forming positive (from the metal atom) or negative (from the non -metal atom) ions. (Sum of I.E. or sum of E.A.) 3. Combining the ions together in an ionic lattice held together by the attraction between oppositely charged ions (LE). For CaCl, w less energy is required to ionise the Ca atom only once, to Ca+ w only one C l atom needs to be formed from Cl2 and converted into a Cl- ion w lattice energy released by forming a 1:1 lattice of singly charged ions is less exothermic than that for CaCl2, which involves Ca2+ ions. Option C is the best answer. 7 C -100 -75 -50 -25 0 25 50 75 100 reactants products progress of reaction Ea = +50 kJ mol-1 formation of intermediate DH = -100 kJ mol-1 8 A In 1.0 mol dm –3 sulfuric acid, [H +] = 2.0 mol dm–3. To make the electrode a standard hydrogen electrode, either change the acid to a 1.0 mol dm –3 monoprotic acid (option A), or halve the concentration of sulfuric acid used. 9 B Since forward reaction is exothermic, higher temperatures will favour the backward reaction. This increases [NH 3] and [O 2] and decreases [N2] and [H 2O] at higher temperatures. Thus the two downward– sloping graphs apply to either N2 or H2O. At a higher pressure, backward reaction is favoured, and [N2] and [H2O] decreases. Thus, z > y. 10 A H2O(l) Ý H+(aq) + OH–(aq) Kw = [H+][OH–] [H+] = wK A P At 25 oC, [H+] = 1.0 × 10–7 mol dm–3 At 10 oC, [H+] = 0.5 × 10–7 mol dm–3 At 0 oC, [H+] = 0.3 × 10–7 mol dm–3 B O Ionic dissociation of water increases by a factor of 3.3 between 0 oC and 25 oC. C O Extent of hydrogen bonding cannot be deduced from the given information. D O [H+] = [OH –] at 0 oC, 10 oC and 25 oC. Thus water remains a neutral liquid at these temperatures. However pH of neutral water is no longer 7.0, but increases with decreasing temperature.

2012 DHS Preliminary Examination (H2 Chemistry 9647/01 Solutions) 3 11 A Rate law is determined by slow step in proposed mechanism, i.e. Rate = k’[O3][O]. This rate law cannot be compared directly with the experimental rate equation because it contains the concentration of an intermediate, O. Thus we need to express rate law in a way that removes the intermediate O. From Step 1, ] O [ ] O ][ O [ 3 2=K Þ [O] = ] O [ ] O [ 2 3K Ass uming that Step 1 equilibrium is established quickly before O is reacted with O3 in Step 2, Rate = k’[O3][O] = ] O [ ] O [ ' 2 23K k = ] O [ ] O [ 2 23k where k = k’K 12 C [S(C2O4)2(NH3)2]– Let the oxidation number of S be x. x + 2(–2) + 2(0) = –1 Þ x = +3 S3+ : ­ ­ ­ (d3) S0 : ­ ­ ­ ­ ­ ­¯ 13 D A (Wrong) NH3 acts as a weak base and a ligand in Reaction I. B (Wrong) G is Cu(OH) 2 and H is [Cu(NH3)4]2+ Oxidation state of Cu remains at +2. \ It is not a redox reaction. C (Wrong) H is a deep blue solution containing [Cu(NH3)4(H2O)2]SO4. D (Correct) When edta 4– binds to the Cu2+during ligand exchange, 4 moles of NH3 and 2 moles of H2O is released. \ The entropy of the system increase s when reaction III occurs. 14 D Z has a greater atomic radius than W Þ Z is earlier on in the period (Z…..W …..) W has greater electrical conductivity than Y and Y has higher boiling point than W Þ Y i s in Group IV and hence W is a metal (not Si) 15 B First trace of precipitate appears when ionic product = Ksp. ZnCO3(s) Ý Zn2+(aq) + CO32–(aq) Ksp = [Zn2+][CO32–] = 1.4 ´ 10–11 \ [CO32–] at which first trace of ZnCO3 appears = 2 . 0 104 . 1 11-´ = 7 ´ 10–11 mol dm–3 Ag2CO3(s) Ý 2Ag2(aq) + CO32–(aq) Ksp = [Ag+]2[CO32–] = 8.1 ´ 10–12 \ [CO32–] at which first trace of Ag 2CO3 appears = 2 12 1 0 101 . 8 . -´ = 8 ´ 10–10 mol dm–3 Thus ZnCO 3 will precipitate first when [CO 32–] reaches 7 ´ 10–11 mol dm–3. When [CO 32–] reaches 8 ´ 10–10 mol dm –3, Ag2CO3 will precipitate next. 16 B A (Wrong) Down the group, oxidising power decreases. B (Correct) Down the group, K sp value decreases. C (Wrong) Down the group, lattice energy of AgX becomes less exothermic due to the increasing anionic radius of halides. D (Wrong) Down the group, the hydration energy decreases due to the increasing anionic radius of halides.

2012 DHS Preliminary Examination (H2 Chemistry 9647/01 Solutions) 4 17 C n formula distribution of Cl atoms no. of chloroethanes C1 C2 1 C2H5Cl 1 0 1 2 C 2H4Cl2 2 0 2 1 1 3 C 2H3Cl3 3 0 2 2 1 4 C 2H2Cl4 3 1 2 2 2 5 C2HCl5 3 2 1 6 C2Cl6 3 3 1 Total no. of different chloroethanes = 9 18 A The stronger an acid, the lower the pKa value. The following lists the four acids used in the options: w CH3CO2H w CCl3CO2H w C 2H5OH w C 6H5OH Increasing acid strength: C 2H5OH < C6H5OH < CH3CO2H < CCl3CO2H Decreasing pKa value: C2H5OH > C6H5OH > CH3CO2H > CCl3CO2H 19 D product(s) of oxidation A O OH O H B O OH O C O O + CO2 D P O O 20 B Alkyl side -chains on a benzene ring are susceptible to oxidative degradation if they possess at least one benzylic hydrogen atom. C H HH benzylic hydrogen Thus, side-chains on Z which can be oxidised are: I II I : It is only possible to oxidise the carbon atom in – CH2 to a lower oxidation state of +2 in – C=O, rather than +3 in –COOH. There is only 1 C atom connecting both benzene rings. If I is oxidised to –COOH, one of the benzene rings has to be reduced by replacing the oxidised carbon with a hydrogen atom. II : II may be regarded as 2 separate alkyl groups, with each being oxidised to –COOH.

2012 DHS Preliminary Examination (H2 Chemistry 9647/01 Solutions) 5 21 C A (Account is correct) Electrophilic substitution of phenol by bromine in aqueous medium yields a tri -substituted product, rather than a mono- substituted one. B (Account is co rrect) Ppt is 2,4,6 - tribromophenol, which is white. C (Account is wrong) Resultant solution contains 2,4,6- tribromophenol, which is acidic. D (Account is correct) Resultant solution is yellow, due to the presence of excess dissolved bromine. 22 D A O 1 mol –COOH reacts with 1 mol PCl5, and 1 mol – OH group reacts with 1 mol PCl5. \ 1 mol citric acid reacts with 4 mol PCl5. B O 3 mol tribasic acid reacts with 1 mol Na2CO3. (Alcohols are neutral, and do not react with Na2CO3.) \ 1 mol citric acid rea cts with 31 mol Na2CO3. C O 1 mol –COOH reacts with 1 mol NaOH. (Alcohols are neutral, and do not react with NaOH.) \ 1 mol citric reacts with 3 mol NaOH. D P 1 mol –COOH reacts with 1 mol Na, and 1 mol –OH group reacts with 1 mol Na. \ 1 mol citric acid reacts with 4 mol Na. 23 B Nitrile groups on CS are hydrolysed to either salts of carboxylic acids (alkaline hydrolysis) or carboxylic acids (acid hydrolysis). 24 A Product of Claisen condensation of ethyl esters must be an ethyl b-keto ester: A O O CH3 Þ Options B and D are not possible. 4th member of series has 2 isomers: I : C H3 O CH 3 O II : C H3 O CH 3 O CH3 Þ Option C is not possible, since it is derived from an isomer of the 5th member: O CH 3 O C H3 Explanation for option A: Option A is one of four p ossible products resulting from the condensation of two different esters, i.e. ethyl ethanoate and isomer II: R O O O R CH3 L R source segment L segment R ethyl ethanoate ethyl ethanoate ethyl ethanoate II II II ® A II ethyl ethanoate 25 A A substitue nt group (f luorine atom), and not a hydrogen, is substituted without disruption to the aromaticity of the benzene ring. This is therefore not an electrophilic substitution reaction. F is highly electronegative. The C ato