SAJC Prelim P1 Worked Solution

1 2017 H2 Chemistry Paper 1 Worked Solution 1. C 2. A 3. B 4. C 5. C 6. A 7. A 8. B 9. C 10. A 11. B 12. A 13. D 14. D 15. C 16. C 17. D 18. A 19. D 20. B 21. D 22. A 23. C 24. B 25. C 26. B 27. B 28. A 29. C 30. D 1 Amt of KMnO4 used for first experiment = 0.1 x 15/1000 = 1.5 x 10–3 mol 5Fe2+ + MnO4– + 8H+  5Fe3+ + Mn2+ + 4H2O Amt of Fe2+ present = 1.5 x 10–3 x 5 = 7.5 x 10–3 mol [Fe2+] = 7.5 x 10–3 /(25/1000) = 0.3 mol dm–3 Amt of KMnO 4 used for second experiment =0.1 x 34.2/1000 = 3.42 x 10 –3 mol Amt of Fe2+ present = 0.0171 mol Amt of Fe3+ originally in 25 cm3 = 0.0171 – (7.5 x 10–3) = 9.6 x 10–3 mol [Fe3+] = 9.6 x 10–3 / (25/1000) = 0.384 mol dm–3 Ans: C 2 9 x 4 = 36 nuclides were lost, hence the mass number of element A is now 241 – 36 = 205. 9 x 2 = 18 protons were lost, hence the atomic number of element A before accounting for the electrons is 94 – 76. However, the implication of e–1 0 is that each electron lost increases the atomic number by 1, so 5 electrons means that the atomic number increases by 5. Hence 76 + 5 = 81. Ans: A 3 A is not correct. From NH3 to H2S, the shape changes from trigonal pyramidal to bent, hence the bond angle actually decreases. B is correct. PH 3 has a smaller bond angle than NH 3 as P is less electronegative than N, hence the bond pairs are further away from the central atom and experience less bond–pair–bond pair repulsion, hence leading to a

2 smaller bond angle in PH 3. AlCl3 has a larger bond angle than NH 3 as it is trigonal planar with a bond angle of 120 º. C is not correct. From PH3 to PF3, H is less electronegative than F, hence the bonding electron pairs are closer to the central atom in PH3, hence experience greater bond–pair–bond pair repulsion, hence leading to a greater bond angle in PH3. From PH3 to PF3, there is a decrease in bond angle. D is not correct. XeF 4 is square planar with bond angle of 90 º. SCl6 is also square planar with bond angle of 90 º. Hence there is no increase in bond angle. Ans: B 4 Glycine forms zwitterions and has an ionic lattice structure with strong electrostatic forces of attraction between the zwitterions, hence has a higher melting point than 2–hydroxyethanoic acid. Ans: C 5 pV = nRT. For fixed mass of gas at constant T, pV α k. Hence graph should be horizontal line with gradient k. At lower T, pV should be be lower than original. Ans: C 6 Ans: A -110 -393 + 1 2O2 (g) CO(g) C(s) + O2 (g) CO2(g) Energy

3 7 ∆G = ∆H – T∆S ∆H will be positive, because interactions in the protein are broken going to the unfolded state. ∆S will also be positive, because the unfolded state is more disordered with coil being more random. ∆G is negative as denaturation is spontaneous when egg is cooked. Ans: A 8 At a higher temperature, the graph is skewed towards the right. Area under the graph represents the number of molecules with energy greater than or equal to EA. Ans: B 9 Since total vol is constant, [reactants] is α to its vol Rate α 1/time Comparing expt 1 & 2, when [ I–] increases 1.5 times, rate increases 170/113 = 1.5 times. Hence 1st order wrt [I–] Comparing expt 2 and 3, [S2O82–] increases 2 times, rate increases 113/56.5 = 2 times. Hence 1st order wrt [S2O82–]  Rate = k [S2O82–][I –] Comparing Expt 1 & 4, when [I–] increases 2 times and [S2O82–] increases 4 times, rate should increase 8 times. Hence time in Expt 4 = 170/8 = 21.3 s. Since Rate = k [S2O82–][I –], slow step should involve 1 mole of (NH4)2S2O8 and 1 mole of KI. Since overall eqn involves 2 moles of I–, the reaction must have more than 1 step. Hence reaction should involve intermediates. Fe2+ can act as homogenous catalyst in the reaction which is slow due to the reaction between two anions: Step 1: 2Fe2+ + S2O82‒  2Fe3+ + 2SO42‒ Ecell = +2.01 – (+0.77) = + 1.24 V (> 0, reaction is feasible) Step 2: 2Fe3+ + 2I‒  2Fe2+ + I2 Ecell = +0.77 – (+0.54) = + 0.23 V (> 0, reaction is feasible)

4 Ans: C 10 A: When acid is added, NH3 is removed. Hence POE will shift left and white precipitate of AgCl is formed. B: Kc expression does not include solids. C: Kc expression is not affected by concentration D: Removing solid does not shift POE. Ans: A 11 2SO3 (g) 2SO2 (g) + O2(g) Initial/atm x 0 0 Eqm/atm 0.5x 0.5x 0.25x 0.5x+0.5x+0.25x = 1.2 x = 0.96 Mole fraction of O2 = 0.25(0.96) / 1.2 = 0.2 Alternatively, 2SO3 (g) 2SO2 (g) + O2(g) Initial/mol x 0 0 Eqm/mol 0.5x 0.5x 0.25x Mole Fraction of O2 = (0.25x)/ (0.5x+0.5x+0.25x) = 0.2 Ans: B 12 1) HI is a strong acid. CH3CH2NH2 will be protonated. No more weak base present. Not a buffer. 2) 2 mol of HCl and 2 mol of CH3COOH will be formed from the hydrolysis of CH3COCl . 2 mole of HCl reacts with 2 moles of NaOH, leaving 1 mole of NaOH. 1 mol of CH3COONa will be formed from the acid base reaction of 2 moles of CH3COOH with 1 mole of NaOH, leaving 1 mol of CH3COOH. Since there is 1 mol of

5 CH3COONa and 1 mol of CH3COOH, a buffer is formed (Think about the species and the possibility of reacting) 3) H2SO4 is a stong dibasic acid which will cause CH3CH2NH2 to be protonated. Not a buffer Ans: A 13 MX2 (s) M2+ (aq) + 2X- (aq) LY (s) L+ (aq) + Y- (aq) A: Ksp increase only when the fwd reaction is endothermic while Ksp decrease when the reaction is exothermic. B: For solute containing different number of ions, solubility should be calculated to determine which is more soluble. Ksp can only be used for comparison of the no. of ions that made up the two solutes are similar. C: To calculate for any concentration of ion, it should be IONIC PRODUCT D: If the solublity of LH is exothermic, a higher temperature will lower the Ksp since the kb increase a greater extent than kf. Ans: D 14 Graph 1: Electronegativity increases across the period. Graph 2: Si has the highest melting point as it requires the most energy to overcome the strong network of covalent bonding in its giant covalent lattice. P4 are held by weak id-id interaction, hence low mpt. Graph 3: NaCl MgCl2 AlCl3 SiCl4 PCl5 pH 7 6.5 3 1-2 1-2 15 Ans: D 1. From the more positive EƟ, it can be seen the X2 has the greatest tendency to be reduced. Hence, it has the greatest oxidising power.

6 2. The EƟcell will be +1.36 – (+0.54) = +1.90 > 0. Hence, the reaction will occur. 3. From the data booklet, Z2 is I2 which is the largest atomic radius, so it has the least tendency to attract electrons to form negatively charge species. Hence, lowest E.A. Ans: C 16 Which of the following shows the correct type of reaction(s) occurring for each step of the synthesis? 17 The reaction does not have an activation barrier (zero E A ) which means it must be a very energetically favourable reaction which does not involve any bond breaking. Ans: D 18 A: HCl (white fumes) is formed when the alcohol reacts with ethanoyl chloride. B: is wrong as alkene does not undergo reduction with LiAlH4. In Step I, –COOH undergoes reduction to form an alcohol. Step 2 involves an increase in carbon chain. The alcohol group probably underwent a substitution reaction to form an alkyl halide before undergoing another substitution reaction with CN– followed by acid hydrolysis. Step 3 involves the formation of an acid chloride followed by the condensation of the amine with acid chloride. Ans: C

7 C: is wrong since CO2 would not be produced when the compound undergoes strong oxidation. No presence of ethanedioic acid that can break down into CO2. D: 1 mol of Vitamin A produces 0.5 mol of H2 and that is 22.7/2 = 11.35 dm3 at stp. Ans: A 19 Option 1 is wrong since CH 3CH2CHClCH3 undergoes S N2 nucleophilic to g