AJC H2 CHEM P3 Solutions

©2014AndersonJC/CHEM 1 H2 Chemistry 9647 2014 JC2 Prelim P3 Suggested Solutions 1 (a) (i) C: Mg(OH)2 D: [Al(OH)4]– E: Al(OH)3 F: Al2O3 [1] [1] [1] [1] (ii) acid–base reaction [1] (iii) No. of moles of Mg(OH) 2 = 2(17.0)24.3 0.18  = 3.087 x 10–3 mol No. of moles of Mg = 3.087 x 10–3 mol Mass of Mg in the alloy = 3.087 x 10–3 x 24.3 = 0.07501 g % composition of Mg in the alloy = 1001.75 0.07501 = 4.29% [1] (b) (i) BeCl2 + H2O BeO + 2HCl accept BeCl2 + 2H2O Be(OH)2 + 2HCl [1] (ii) Al atom in A lCl3 molecule is electron–deficient as it has only 6 valence electrons around it. Hence, the A l atom can accept another 2 electrons from the lone pair of N atom in one molecule of NH 3 to achieve the (stable) octet configuration. [1] (iii) Al Cl Cl Cl N H H H [1] (iv) NH3 NH3 BeCl Cl accept condensed formula: BeCl2(NH3)2 [1]

©2014AndersonJC/CHEM 2 (c) (i) Electrophilic substitution Generation of electrophile: CH3Cl + AlCl3 CH3 + + AlCl4 – CH3 + HC H 3 CH3 + H++ slow fast Regeneration of catalyst: AlCl4 – + H+ AlCl3 + HCl [1m] for correct equations for generation of electrophile & regeneration of catalyst [1m] for curly arrows & charges + slow & fast [1m] for correct structure of intermediate [3] (ii) H : ClCH 2CH2Cl OH CH2CH2Cl J: O CH2CH2OH K: OH CH2COOH L: OH CH2COCl M: Step 2: NaOH(aq), heat under reflux Step 3: K 2Cr2O7(aq), dil. H2SO4, heat under reflux Step 4: PCl5 or SOCl2 [1m] for correct structure of M and reagents/conditions for step 4 [1] [1] [1] [1] [1] [1] [1]

©2014AndersonJC/CHEM 3 2 (a) (i) When the temperature is increased, the forward endothermic reaction is favoured. The position of equilibrium will shift to the right in order to reduce temperature by absorbing extra heat. Partial pressure of hydrogen will increase. When the pressure is increased, the reverse reaction is favoured. The position of equilibrium shifts to the left in order to decrease pressure by producing less gaseous particles. Partial pressure of hydrogen will decrease. [1m] for each change [2] (ii) The two assumptions are: There are negligible intermolecular forces between the gaseous particles. The volume of the gaseous particles is negligible compared to the volume it occupies. H 2O(g) has stronger hydrogen bonding between molecules compared to the weaker van der Waals’ forces between H 2 molecules, and the volume of H 2O molecules is greater than that of H 2, hence H2O(g) deviates more from ideality than H2. [1m] for each assumption [1m] for correct explanation [3] (iii) Kp = 2 2 O 2 CO 2 CO pp p 2CO(g) + O 2(g) 2CO2(g) Initial amt / mol 5.00 2.50 0 change in amt / mol – 4.95 – 2 95.4 + 4.95 amt at eqm / mol 0.05 0.025 4.95 total amt at eqm = 4.95 + 0.05 + 0.025 = 5.025 mol p(CO2) at eqm = 025.5 95.4 (101) = 99.5 kPa p(CO) at eqm = 025.5 05.0 (101) = 1.00 kPa p(O2) at eqm = 025.5 025.0 (101) = 0.502 kPa Kp = )502.0()00.1( )5.99( 2 2 = 19700 kPa–1 [1m] for correct Kp expression and calculated value (ignore units) [1m] for correct amt at eqm / partial pressures at eqm [2]

©2014AndersonJC/CHEM 4 (iv) Use of an alkaline material / sorbent e.g. amine / hydroxide to react with the acidic CO2. Removing the CO2 formed will shift the position of equilibrium for reaction (2) to the right and hence eliminating the CO present and possibility of the catalyst being ‘poisoned’. This improves the efficiency of the purification of H2. [1] [1] (v) Gold nanoparticles are in the solid phase, different from the gaseous reactants. Due to the availability of vacant (or partially filled) d–orbitals in the nanoparticles, the CO and O 2 molecules can form temporary bonds on the catalyst surface. This weakens the bonds within the molecules, hence lowering the activation energy. By concentrating the CO and O 2 molecules on the catalyst surface, the number of these molecules with energy greater or equal to Ea increases and hence the frequency of effective collision increases. [1m] for differentiating solid catalyst from gaseous reactants [1m] for either factor that brings about increase in rate [2] (b) (i) anode: H2 2H+ + 2e– cathode: O2 + 4H+ + 4e– 2H2O [1] [1] (ii) 2H2 + O2 2H2O Eo = EO2H2O – EH+H2 = +1.23 – 0.00 = +1.23 V [1] (iii) shape: square planar bond angle: 90 o accept tetrahedral; 109 o [1] (iv) [Pt(NH3)4]2+ + 2e– Pt + 4NH3 accept Pt2+ + 2e– Pt [1] (c) (i) CH3CH2OH + NaH CH3CH2O–Na+ + H2 accept ionic form w/o Na+ but not H+ + H– H2 [1] (ii) Nucleophilic substitution (SN2) C Cl H CH3 CH3 CH3CH2O C ClCH3CH2O CH3 H CH3 C H CH3 CH3 CH3CH2O Cl+  - [1m] for correct movement of electrons and lone pair on CH3CH2O– [1m] for correct charges, and transition state [2] (iii) T is CH2=CHCH3 [1]

©2014AndersonJC/CHEM 5 3 (a) (i) Since Ka1 >>> K a2, only the first dissociation makes an appreciable contribution to the pH of the solution.   A][H ]O][H[HA 2 3- a1  K i2 23 A][H ]O[H     1.30 x 10–2 0.10 O[H 23 ]    [H3O+] = 0.03606 mol dm–3 pH = –log[H3O+] = 1.44 [1] (ii) 2NaOH + H2A A2– + 2Na+ + 2H2O No. of moles of A 2– formed = (25/1000) x 0.10 = 0.0025 mol No. of moles of NaOH required for complete neutralisation = 0.0025 x 2 = 0.00500 mol Volume of NaOH required = (0.0050 / 0.10) x 1000 = 50 cm 3 [A2–] =  100050 25 0.0025 0.0333 mol dm–3 [1] (iii) A2– + H2O HA– + OH– ][A ]][OH[HA -2b1  K i-2 2 ][A ][OH    [1] Kb1 = Kw / Ka2 = 10–14 / (5.90 x 10–7) = 1.695 x 10–8 mol dm–3 1.695 x 10–8 0.03333 [OH 2]         (allow ecf marking)  [OH–] = 2.377 x 10–5 mol dm–3 pOH = –log[OH–] = 4.62 pH = 14 – 4.62 = 9.38 [1]

©2014AndersonJC/CHEM 6 (iv) [1m] for correct shape of curve until 50 cm3. [1m] for correct indication of 2 equivalence points at 25.0 cm 3 and 50.0 cm3 with corresponding pH at 9.38. [1m] for correct indication of 2 pK a values (1.89 and 6.23) and corresponding volumes (12.5 cm3 and 37.5 cm3) [3] (v) Indicator: thymolphthalein The working pH range of thymolphthalein lies within the range of rapid pH change at the end point. or The pH at the second equivalence point is within the working pH range of thymolphthalein. [1] [1] (b) (i) CC H H C C O O O O H H CC C H H C O O H O OH trans cis [1m] for correct displayed formulae [1m] for correct labelling [2] volume of NaOH / cm3 25.0 50.0 pH 12.0 1.44 9.38 12.5 37.5 1.89 6.23 60.0

©2014AndersonJC/CHEM 7 (ii) There is intramolecular hydrogen bonding involving H of –OH group with the neighbouring C=O group. Hence, there are fewer sites available for intermolecular hydrogen bonds to be formed between the molecules. Less energy is needed to overcome the less extensive intermolecular H–bonding during melting. or Cis–butenedioic acid results in kinks which cause the molecules to be less closely packed in the structure. Less energy is required to overcome the intermolecular forces. [1] (c) O O O [1] (d) geraniol C CH3 CH3 C H CH2CH2 C CH3 CC H 2 OH H accept CC H CH2CH2 C CH3 C CH3 CH3 H CH2HO or H CC H CC CH3 CH3 CH3 CH2CH2CH2HO N C O CH3 CH3 P C O HO CH 2CH2 C O CH3 Q C O -O CH2CH2 C O O- R C O HO CH 2CH2 C O OH [1m] each [7]

©2014AndersonJC/CHEM 8 Reaction Type of reaction Deduction 1. 1 mole of geraniol decolourises 2 moles of Br 2(aq) electrophilic addition Presence of two C=C bonds in geraniol 2. Heating geraniol with excess conc. acidified KMnO 4(aq) produces N, P and a colourless gas. oxidative cleavage Presence of more than one C=C bond in geraniol The colourless gas is CO 2. 3.