JJC H2 CHEM P2 Ans

[Turn over JURONG JUNIOR COLLEGE 2015 JC 2 PRELIMINARY EXAMINATION Higher 2 Paper 2 (solutions) 1 Preparation of 5 diluted solutions from 0.0500 mol dm 3 Ca2+ (aq) Dilution and measurement: 1. Using a burette, run Vcm 3 of Ca 2+ stock solution into a 25.0 cm 3 graduated flask. 2. Make up to the graduated mark with deionised water. 3. Stopper and shake the flask to obtain a homogeneous solution. 4. Repeat steps 1-3 using the volumes specified in the table above in the preparation of solutions 2 to 5. 5. Place each of the prepared solutions into the atomic absorption spectrophotometer and record the absorbance value for each solution. 6. Plot absorbance value for each solution against concentration of Ca 2+ Solution [Ca2 (aq)] / mol dm3 Volume of Ca2+ stock solution, V / cm3 1 0.0250 12.50 2 0.0200 10.00 3 0.0150 7.50 4 0.0100 5.00 5 0.00500 2.50

2 © Jurong Junior College 9647/02/ J2 PRELIMINARY EXAMINATION/2015 1 Preparation of saturated solution of Ca(OH) 2(aq) 1. Using a measuring cylinder, place 100 cm 3 of deionised water into a 250 cm3 beaker. 2. Using a spatula, add a few tips of solid Ca(OH) 2 into the flask. Stir to dissolve all the solids using a glass rod. Keep adding more solids, with stirring after each addition, until some solids are left undissolved. 3. To ensure that the solution is saturated, stir the solution for a while and leave the conical flask containing the solution to stand in a water bath at 25 oC for some time. There must be some solids left undissolved. 4. To remove undissolved solids, filter the saturated solution into a clean, dry conical flask using a dry filter funnel and a piece of dry filter paper. 5. Place the saturated Ca(OH) 2 solution into the atomic absorption spectrophotometer and record the absorbance value for the solution. Using the calibration line, read the corresponding concentration of Ca2+, for the absorbance value for the saturated Ca(OH)2 solution. (b) Solubility product of Ca(OH)2 = 4 x3 mol3 dm9

3 © Jurong Junior College 9647/02/ J2 PRELIMINARY EXAMINATION/2015 [Turn over 2 (a) (b) Nucleophile (c) (d) Nucleophilic addition (e) (f) (i) Since half life are constant (t½  76 min), the reaction is first order wrt [HCN]. (ii) When [CH2=NH] is halved, rate is halved. The reaction is first order wrt [CH2=NH]. (iii) Rate = k[HCN][CH2=NH] (iv) Rate = k’[HCN] where k’=k[CH2=NH] k’ = ln2 /t1/2 = ln2 / 76 = 0.00912 min–1 k = 0.00912 / 0.2 = 0.0456 mol-1 dm3 min-1 Or Initial rate = gradient = 0.01/ 124 = 0.0000807 mol dm–3 min–1 0.0000807 = k(0.01)(0.2) k = 0.0403 mol-1 dm3 min-1 (v) 78 min

4 © Jurong Junior College 9647/02/ J2 PRELIMINARY EXAMINATION/2015 3 (a)   22 2 c 32 [Pb ][Cr ]K [Cr ] units: mol dm3 (b) (i) [Pb2+(aq)] = ½ (2.96  104) = 1.48  104 mol dm–3 (ii)  44 2 c 2 (1.48 10 )(2.96 10 )K (0.200) = 3.24  1010 (iii) Kc value is significantly smaller than 1. Equilibrium position lies to the left, reaction hardly proceeds. (c) (i) PbSO4 (ii) [Pb2+] decreases as PbSO 4 is precipitated. Equilibrium position will shift right to form more Pb2+. No change in Kc as Kc is only dependent on temperature.

5 © Jurong Junior College 9647/02/ J2 PRELIMINARY EXAMINATION/2015 [Turn over 4 (a) (i) Cis isomer Trans isomer Structure (ii) In the cis isomer, there will be intramolecular molecular hydrogen bonding between the -COOH and –OH as they are close to each other. Hence less energy is required to overcome the less extensive hydrogen bonding between the cis isomer molecules as compared to the trans isomer. (b) (c) (i) Stage I: NaBr + H2SO4  NaHSO4 + HBr Stage II: CH(OH)=CHCOOH + HBr  CH2(OH)CHBrCOOH or CH(OH)=CHCOOH + HBr  CHBr(OH)CH2COOH (ii) Inorganic by-product: Br2 Equation: 2HBr + H2SO4  Br2 + SO2 + 2H2O

6 © Jurong Junior College 9647/02/ J2 PRELIMINARY EXAMINATION/2015 5 (a) Functional groups present: alkene and alcohol (b) Compound P Compound Q CH3CH2C CH3 CHCH2OH P (C6H12O) OH Q (C5H8O) (c) (i) Functional groups present in T: carboxylic acid Functional groups present in U: alcohol (ii) Compound R Compound S R (C5H10O) S (C7H14O) Compound T Compound U CH3CCOOH O T (C3H4O3) U (C6H12O2) (d) (i) Reagents and condition: NaOH(aq), Heat under reflux. Type of reaction: Nucleophilic Substitution (ii)

7 © Jurong Junior College 9647/02/ J2 PRELIMINARY EXAMINATION/2015 [Turn over 6 (a) (i) Cu+ ion: 1s22s22p63s23p63d10 Cu2+ ion: 1s22s22p63s23p63d9 (ii) [Cu(H2O)6]2+ : pale blue [Cu(NH3)4(H2O)2]2+ : deep blue (b) (i) White precipitate J: CuI Equation: Cu2+(aq) + 4I (aq)  CuI (s) + I2(aq) (ii) Formula of K: [Cu(NH3)2]+ (iii) Cu+ is oxidised to Cu2+ by oxygen in air. (iv) Solid L: Cu Type of reaction: Disproportionation (c) (i) Cu(NO3)2  CuO + 2NO2 + ½O2 (ii) Ionic radius of Cu2+ = 0.069nm Ionic radius of Ca2+ = 0.099 nm Since Cu2+ has a smaller ionic radius, Cu 2+ has a larger charge density and is able to polarise the large NO3 – to a larger extent than Ca2+ ion. Hence, Cu(NO3)2 is less thermally stable than Ca(NO3)2. (iii) Decomposition temperature for Cu(NO3)2 = any temp between 330°C and 561°C.