JJC H2 Chem 2013 Prelim P2 Soln

© Jurong Junior College Page 1 of 5 2013 Prelim Paper 2 Answers Suggested Mark Scheme for 2013 JJC Prelim Exam Paper 2 (9647/02) 1. (a) KNO3(s)  K+(aq) + NO3 –(aq) H > 0 ---(*) When the temperature is increased, the equilibrium position of (*) shifts right to favour endothermic reaction so as to absorb some heat. Hence, the solubility of KNO3 increases with increasing temperature. (b) The method does not apply to solid that decomposes on heating as it will result in greater mass loss. (c) 1. Using a 50 cm3 burette/measuring cylinder, add 50 cm3 of water into a small beaker. 2. Place the small beaker containing water into water–bath with thermostat set at 30 °C. 3. Using a spatula, add KNO3(s) into the water. Stir to dissolve all solid. 4. Repeat step 3 until some solid remains undissolved. 5. Stir the mixture until temperature of solution reaches 30 °C. Let the mixture stand in the water–bath at 30 °C for some time. 6. Using an electronic weighing balance, measure and record the mass of an empty , dry crucible. 7. Using a dry filter funnel and filter paper , filter the mixture and collect the filtrate in the crucible. 8. Using a Bunsen Burner, heat the filtrate to dryness. 9. Using an electronic weighing balance, measure and record the mass of crucible with solid residue. 10. Repeat step 1 to 9 at 40°C, 50°C, 60°C and 70°C. (d)   3 3mass of KNO dissolved in 50 cm of water gyx   3solubility of KNO diss olved in 100 g of water g /100g    100 250 yx yx (e) Use oven/heat resistant gloves or tongs to handle the hot beaker/crucible. OR Cool hot crucible before handling. 2. (a) (i) It is to quench the reaction by removing H 2SO4/H+ in the reaction mixture via acid– carbonate reaction. (ii) I2(aq) + 2S2O3 2–(aq)  2I–(aq) + S4O6 2–(aq) (iii) Since H2SO4 is a catalyst , it will be regenerated such that [H2SO4] will remain constant throughout the reaction. Hence, it is not necessary to use H2SO4 in large excess in order to make [H2SO4] constant. solubility / (g/100g) temperature / °C (or K) 614

© Jurong Junior College Page 2 of 5 2013 Prelim Paper 2 Answers 2. (b) (i) (ii) Order with respect to [iodine] = 0 (iii) Order with respect to [propanone] = 1 (iv) Rate = k [H+] [propanone] (c) (i)   r bonds broken bonds formedHE E             410 151 240 299  122.0 kJ mol (ii) The bond energy values quoted from the Data Booklet are only average value derived form the full range of molecules that contains the particular bonds. OR The reactants are in aqueous states while the bond energies from the Data Booklet are for gaseous species. (iii) (d) (i)      pVn2 .03 molRT     33101 10 50 10 28.31 300 time /min volume of Na2S2O3 /cm3 Reaction coordinate Energy/ kJ mol1 (CH3)2CO + I2 + H+ CH3COCH2I + HI + H+ Ea [1m] 615

© Jurong Junior College Page 3 of 5 2013 Prelim Paper 2 Answers 2. (d) (ii) H3C CH2I 3. (a) HO O HO O cyclic compound B non–cyclic compound C (b) HO O HO O (c) CH 3OH (d) C C C C C CC O O C O O C C H H H H H H H H H H H H HH HH (f) Both have simple molecular/covalent structures. Smaller amount of energy is required to overcome the less extensive hydrogen bonds between 2–hydroxyphenylamine molecules than that between 4–hydroxyphenylamine since 2–hydroxyphenylamine is able to form intramolecular hydrogen bonds due to close proximity of the –OH and –NH2 groups. Hence, 2–hydroxyphenylamine has a lower melting point than 4–hydroxyphenylamine. 4. (a) (i) (ii) (b) In neopentane, there is only 1 type of replaceable/substitutable H atoms . Hence, only 1 type of monochlorinated product will be formed, giving a better yield of neopentylchloride. In pentane, there is 3 types of replaceable/substitutable H atoms and hence, a mixture of 3 types of monochlorinated product will be formed, giving a low yield of 1–chloropentane. 616

© Jurong Junior College Page 4 of 5 2013 Prelim Paper 2 Answers 5. (a) (b) Element X: chlorine/Cl Element Y: fluorine/F Element Z: phosphorus/P (c) (i) The hydrogen bonds between N 2H4 molecules is stronger than the permanent dipole– permanent dipole interaction between ZY3 molecules. Hence, N2H4 deviates more from ideality than ZY3. (ii) 6. (a) (i) It is more difficult to remove H+ from negatively charged anion than from molecule. Hence, it is less likely to form –OOC–R–COO– than HOOC–R–COO– and pKa,2 is higher. (ii) p–p orbital overlap results in the delocalisation of lone pair of electrons on O atom over the two O and into benzene ring of (COOH)C 6H4COO–. This disperses the negative charge and stabilises (COOH)C6H4COO– more. Hence, (COOH)C6H4COOH is a stronger acid and has a lower pKa,1. (b) (i) To obtain the maximum buffering capcity ( i.e. pH = pKa) of the acidic buffer of HOOC–R–COO – /–OOC–R–COO–, the volume of NaOH required is 22.5 cm3. (ii) Since HOOC–R–COOH  2NaOH,  amount of HOOC R COOH used 0.00300 mol 25HOOC R COOH used 0.00300 1000           3 13 00.2002 1000 0.120 mol dm (iii) System: weak acid At initial pH of 2.7, a,11 Hc K    22 2.7 a 10H c0 . 1 2 0K       533.32 10 mol dm (iv)   5 a,1 10p log 3.32 10 4.48K    Identity of unknown acid = suberic acid (c) (i) Compound : M Reagent and conditions : acidified KMnO4(aq), heat under reflux 1 ideal gas ZY3 at 500 K N2H4 at 500 K N2H4 at 800 K pV RT p 617

© Jurong Junior College Page 5 of 5 2013 Prelim Paper 2 Answers 6. (c) (ii) (A) L and N (B) L, M and N (C) M and N (iii) Type of mechanism: electrophilic addition 7. (a) Cu: [Ar] 3d10 4s1 (b) (i) [CuC l4]2– (ii) Since Cl– has a larger size/radius than F–, there will be steric hindrance around Cu2+. Hence, Cu2+ cannot accommodate more than four Cl– ions. (c) Observation in step I : pale blue ppt formed . Equation : Cu2+(aq) + 2OH–(aq)  Cu(OH)2(s) Observation in step II : Pale blue ppt dissolves to give a dark blue solution. Equation : Cu(OH)2(s) + 4NH3(aq) + 2H2O(l)  [Cu(NH3)4(H2O)2]2+(aq) + 2OH–(aq) (d) (i) Since Kstab, 2 is the largest among the three, ion S is [Cu(H2O)2(en)2]2+ (ii) Cu H2O H2O CH2 CH2N N HH HH H2C H2C N N HH HH 2+ (d) (iii) If N2H4 is used, an unstable 3–membered ring complex will be formd due to ring strain, resulting in the bond angle in the complex to be too small. (e) Stronger ligand displaces weaker ligand to give a more stable complex by forming stronger dative bond. Since H2NCH2CH2NH2 displaces NH3 and NH3 displaces H2O, the ligand strength of H2O < NH3 < H2NCH2CH2NH2. 618