RVHS H2 CHEM P3 ANS Prelim

1 River Valley High School 9647/03/PRELIM/11 Preliminary Examination 2011 RIVER VALLEY HIGH SCHOOL YEAR 6 PRELIMINARY EXAMINATION H2 CHEMISTRY 9647/03 Paper 3 Free Response Suggested Answers 1ai ∆Hr = BE(bonds broken) – BE(bonds form) = [12BE(N-H) + 3BE(O=O)] – [2(NΞN) + 12(O-H)] = [12(+390) + 3(+496)]  [2(+994)+12(+460)] = 1340 kJ mol-1 1aii The calculated value differs from the theoretical value because the bond energies given in the Data Booklet are only average energies. 1aiii ∆S is expected to be positive as the reaction proceeds with an increase in disorder due to increase in the number of gaseous molecules. ∆G = ∆HT∆S, since ∆S is positive and ∆H is negative, ∆G will always be a negative value. Hence we can conclude that the given reaction is spontaneous. 1aiv Increasing deviation: N2 < O2 < NH3 < H2O Deviation from ideal behaviour depends on the strength of intermolecular forces and number of electrons in the molecule. Weak instantaneous dipole-induced dipole/van der Waals’ forces exist between nitrogen molecules and between oxygen molecules. Strong hydrogen bonds exist between ammonia molecules and water molecules. Water shows more extensive hydrogen bonding than ammonia while oxygen has a larger electron cloud, and therefore stronger instantaneous dipole-induced dipole/van der Waals’ forces than nitrogen. 1bi Haemoglobin consists of four protein chain subunits: two -chain subunits and two - chain subunits, held by non-covalent interactions between R groups: ionic interactions, van der Waals’ forces and hydrogen bonds. 1bii Assuming oxygen is an ideal gas, pV = nRT (98.7 x 10 3)(3.06 x 10-6) = n(8.314)(37+273) n = 1.172 x 10-4 mol

2 River Valley High School 9647/03/PRELIM/11 Preliminary Examination 2011 Amount of haemoglobin = n/4 = 2.930 x 10-5 mol Molar mass of haemoglobin = 2.00/2.930 x 10-5 = 6.83 x 104 g mol-1 1biii Oxygen is not an ideal gas. Intermolecular forces of attraction are present between oxygen molecules. 1ci Valine exhibits optical isomerism. 1cii Van der Waals’ forces between ile, leu, met, phe, val (any two). Hydrogen bonds between lys, thr, trp (any two). C H O C H H H H C H H C H H C H H C H H N H H + + + - - hydrogen bond CH(CH3)(CH2CH3) CH(CH3)2 C H HOOC NH2 CH(CH3)2 C H H2N CH(CH3)2 COOH van der Waals’ forces

3 River Valley High School 9647/03/PRELIM/11 Preliminary Examination 2011 1ciii

4 River Valley High School 9647/03/PRELIM/11 Preliminary Examination 2011 2a A is not soluble in acid : A is not basic. [] Acid hydrolysis [] of A took place to give B and C. A is likely to be an ester [] while B is likely to be an alcohol. [] B rotates plane-polarized light. B has at least 1 chiral C. [] B is oxidized [] by hot acidified KMnO 4 to give D, with the removal of 2 H in D. D is a ketone [] while B is a secondary alcohol. [] D gives a positive iodoform test [] with alkaline I 2: D is a methyl ketone [] while B is a methyl alcohol. [] B undergoes dehydration [] with Al2O3 to give alkene E. [] E undergoes oxidative cleavage to give butanoic acid. E contains a fragment with structure. C CH2CH2CH3 H [] CC H 2 II CC H 2 CN CN H2 + 2NaCN ethanol H2 + 2 NaI heat [] CC H 2 CN CN CH2 CH2 COOHHOOC H2 + 4H2O + 2H+ dil H2SO4 heat R [] A B C D C

5 River Valley High School 9647/03/PRELIM/11 Preliminary Examination 2011 H3C C C H2 C C H2 CH3 H H E 2bi Concentration of (R 3NH)+ = 0.20 mol dm-3 Let concentration of H+ at eqm be x mol dm-3 x2 / (0.20 – x) = Kw / 1.62 × 10-6 Solving, x = 3.514 × 10-5 mol dm-3 pH = -lg x = 4.45 2bii At equivalence point, only R 3N is present. Let concentration of OH- at eqm be y mol dm-3 y2 / (0.1 – y) = 1.62 × 10-6 Solving, y = 4.025 x 10-4 mol dm-3 pOH = -lg y = 3.40, pH = 14.0 – 3.40 = 10.60 2biii 1 mark for correct shape with correctly labeled axes. 1 mark for initial pH and pH at equivalence point. ECF. 1 mark for showing pH at maximum buffer capacity at 250 cm 3 of NaOH added. 500250 Volume of NaOH/ cm3 pH 0 X 10.60 8.21 4.45

6 River Valley High School 9647/03/PRELIM/11 Preliminary Examination 2011 2c In the presence of a catalyst, presence of an alternate reaction pathway with lower Ea. [] Greater proportion of particles will possess the minimum amount of energy, [ ] Ea, for a successful collision to occur. Frequency of effective collisions increases[] , rate of reaction increases. No of particles with Energy E Energy E Ea Ea’ (catalysed)

7 River Valley High School 9647/03/PRELIM/11 Preliminary Examination 2011 3ai 3aii Comparing Expt 1 and 3: in itial rate of formation of B remained unchanged when only [H+] was changed. The reaction is zero order with respect to H+. Comparing Expt 2 and 4: When [ A] increased to 1.25 times while [H +] and [CN] were unchanged, initial rate of formation of B increased to 1.25 times. The reaction is first order with respect to A. Comparing Expt 2 and 3: When [CN ] increased to 1.2 times while [H +] and [ A] were unchanged, initial rate of formation of B increased to 1.2 times. The reaction is first order with respect to CN. 3aiii Rate = k[A][CN] 3aiv Since H+ does not appear in rate equation, it is not in volved in slow/r.d. step. / Only 1 CN - and 1 A is involved in the r.d.s/slow step as indi cated by their order of reaction in the rate equation. [1] Or Step I involve breaking of π bond of C=O , which involved higher Ea compared to Step II. δ+

8 River Valley High School 9647/03/PRELIM/11 Preliminary Examination 2011 3b Step 1: HX(g) (X= Cl, Br, X) [1] Step 2: alcoholic KCN/ NaCN, heat (under reflux) [1] Step 3: dilute H2SO4, heat (under reflux) [1] Intermediate from step 1 [1] Br Intermediate from step 2 [1] 3ci Ibuprofen + propanol  ester + water Initial amt 0.00500 0.500 0 0 Change - 0.00230 - 0.00230 + 0.00230 + 0.00230 Eqm amt 0.00270 0.498 0.00230 0.00230 c 0.00230 0.00230() () VVK 0.00270 0.498() ( )VV 0.00393   . 3cii When temperature is increased, the proportion of ibuprofen reacted decreased. The equilibrium position shifted to the left to absorb some of the energy. Since increased in temperature favours endothermic reaction , the forward reaction is exothermic. 3ciii Propanol is very volatile/ evaporates rapidly at higher temperature. (The equilibrium position is shift to the left to replenish some of the propanol lost. Therefore, a smaller proportion of ibuprofen reacted to form ester.) 3civ In the original experiment, t he equilibrium position is shift to the left to consume some of the water added. Therefore, a smaller proportion ibuprofen reacted in the original experiment. KOH (aq), an alkali, neutralized some of the sulfuric acid which is the catalyst for the reaction. Equilibrium is established at a slow rate/less negative gradient than the original.

9 River Valley High School 9647/03/PRELIM/11 Preliminary Examination 2011 4ai Disproportionation (Reject wrong spelling) 3Cl2 + 6NaOH  5NaCl + NaClO3 + 3H2O 4aii Trigonal pyramidal 4bi Element K Br O Mass ratio 23.4 47.8 28.8 Mole ratio 23.4 39.1 0.598 47.8 79.9  0.598 28.8 16 1.8 Simplest ratio 1 1 1.8 0.598  3.01 3 Formula of solid is KBrO3 4bii nKBrO3 = 0.835 167 = 0.00500 mol nPbCl2 = 4.17 278 = 0.0150 mol nKBrO3 nPbCl2  0.00500 0.0150 = 1 3 4biii 3Cl2 + KBr + 6KOH  6KCl + KBrO3 + 3H2O 4ci pKa of 3-chloropropanoic acid will be lower than 4.87/that of propanoic acid. 3-chloropropanoic acid is a stronger acid [ ] than propanoic acid because the electron- withdrawing Cl atom [] disper