SAJC Prelim P1 ANS

JC2 Prelims 2009 H2 Chemistry (Paper 1) Worked Solutions 1. Ans: A MO + H2SO4  MSO4 + H2O H2SO4 + 2NaOH  Na2SO4 + 2H2O No. of moles of H2SO4 = 11000 100  = 0.1 mol No. of moles of excess H2SO4 = ½ × No. of moles of NaOH = ½ 11000 45 21 . = 0.0107 mol No of moles of H 2SO4 reacted with MO = no. of moles of MO = 0.1- 0.0107 =0.0892 mol M r of MO = 0892 0 0 5 . . = 56.0 Ar of M = 40.0 2. Ans: C Let the volume of CH4 be x. Then the volume of C2H6 = 70- x Volume of carbon dioxide gas formed in total = x + 2(70-x) = 140 – x Change in volume of residual gases after passing through NaOH = volume of carbon dioxide formed = 130 -35 = 95 cm 3 Hecne, 140- x = 95 x = 45 cm 3 Volume of methane = 45 cm3 ; volume of ethane = 25 cm3

3. Ans: C A: Si: [Ne] 3s2 3p2 2 unpaired electrons in the 3p orbital B: S [Ne] 3s2 3p4 3 unpaired electrons in the 3p orbital C: Fe 2+ : [Ar]3d6 4 unpaired electrons in the 3d orbital D: Cr 3+: [Ar]3d3 3 unpaired electrons in the 3d orbital 4. Ans: B Element X is in Group III while element Y is in Group VI. Hence the formula formed between X and Y is X 2Y3. 5. Ans: C NO2, SO2 are simple covalent molecules with weak van der Waals forces, while water, also a simple molecule, has hydrogen bonding. SiO2 is a giant covalent molecule with strong covalent bonds between Si and O atoms, hence it requires the largest amount of energy to overcome the strong covalent bonds. Hence, it has the highest energy. 6. Ans: A In ice, each water molecule is hydrogen-bonded to four other water molecules, in a tetrahedral arrangement, hence having a bond angle of 109.5°. 7. Ans: B no. of moles of gas in small spacecraft = RT RT RT pV 800) 20 )( 40 ( no. of moles of gas in large spacecraft = RTRTRT pV 7500) 50 )( 150 ( total no. of moles of gas = RTRT 83007500 800 Pressure in the combined arrangement =    atm RTRT V nRT 11950 20 8300  8. Ans: D pV=nRT  p nRTV  when the mass is fixed, n is constant. when pressure is constant, V is proportional to T in Kelvin. Since the x-axis shown is T in °C, the straight line should cut the x-axis at -273 9. Ans: C Q=mcT = (25+25)(4.2)(20) no. of mol of H 2O formed = 0.04 numerical value of H = 04 . 0 20 2 . 4 50 04 . 0  Q

10. Ans: C H O O H  2H (g) + 2O(g) Hatm = energy required to break 2 O-H bonds and 1 O-O bond = + (2)(460) + 150 = +1070 kJ mol-1 11. Ans: A Students need to balance their own half equation as follows: MnO4 - + e  MnO4 2- H2O + HCOO-  CO3 2- + 3H+ + 2e 3 4 4 4 4 505 . 0 10 5 . 2 moles of number , 10 5 . 2 10 5 1 2 cmx ionconcentratvolumeSince x x nHCOO nMnO         12. Ans: B Ni2+ + 2e Ni E= -0.25 V Since Nickel can only be oxidized, species A – D can only be reduced. Hence, Options C and D is eliminated as Cu and Cl- cannot be reduced. Option A: Cr3+ + e  Cr2+ E= -0.41 V is chosen over Cr3+ + e  Cr E= -0.74 V as a more positive reduction potential indicates that preferred reduction. Cr3+ + e  Cr2+ E= -0.41 V Ni2+ + 2e Ni E= -0.25 V E overall = -0.41 – (-0.25) = -0.16 V (not feasible) Option B: Pb2+ + 2e  Pb E= -0.13 V Ni2+ + 2e Ni E= -0.25 V E overall = -0.13 – (-0.25) = +0.12 V (feasible) 13. Ans: A 372 . 0 )2 35 . 0( )2 45 . 0( 2 3  Kc P (g) Q(g) R (g) Initial 0.50 0.50 0 Change -0.15 -0.15 +0.45 Equilibrium 0.35 0.35 +0.45

14. Ans: C Condition Position of equilibrium Kp Rate of forward reaction A Increase in temperature Right (False) Position of eqm will shift to favour the endothermic side which is the left hand side. Increase (False) Kp = kf / kb Hence, backward rate constant will increase more than the rate forward constant. Hence Kp should decrease. Increase (True) Rate of forward and rate of backward both increases. However, backward rate will increase more than the rate forward. B Addition of catalyst No change (True) Rate of forward and backward inceases by the same amount. Hence eqm position does not change. No change (True) Kp is independent of catalyst and dependant only on Temperature. No change (False) Rate of forward and backward increases by the same amount. C Decrease in pressure Left (True) Position of eqm will shift to favour the side with lesser no. of moles of gaseous molecules which is the left hand side. No change (True) Kp is independent of pressure and dependant only on Temperature. Decrease (True) Rate of forward and rate of backward both decreases by the same amount. Decrease in pressure infers lesser no. of effective collisions. D Addition of H2 (g) Right (True) Position of eqm will shift to favour the side with no hydrogen which is the right hand side. Increase (False) Kp is independent of concentration and dependant only on Temperature. Increase (True) Rate of forward increases as more hydrogen infers more no. of effective collisions. 15. Ans: C Ksp = [Ag+][OH-] = 1.52 x 10-8 mol2dm-6 Hence at equilibrium, [OH-]2 = 1.52 x 10-8 mol2dm-6 [OH-] = (1.52 x 10-8 )1/2 = 1.23 x10-4 pOH = 3.90 pH = 14 – 3.90 = 10.1 16. Ans: B 8.2 g of sodium ethanoate = 0.1 mol [sodium ethanoate] = [salt] = 0.2 mol dm -3 Ka = Kw / Kb = 1.75 x 10-5 pH = pKa + lg ([salt] / [acid]) = -lg(1.75 x 10-5) + lg ((0.2) / (0.1)) = 5.06

17. Ans: B No. of counts per minute is proportionate to the concentration of the radioactive substance. n o f C C          2 1 where n is the number of half-lives and Cf is the no. of counts per minute for the final product, Co is the no. of counts per minute for the initial substance. n = 0.58496 therefore the age of the ship is n x t½ = 3276 years 18. Ans: D Base on experiments 2 and 3, 2nd order with respect to [R] Base on experiments 1 and 2, and that it is 2nd order with respect to [R], 0th order with respect to [Q] Rate’s unit is mol dm-3 s-1 Therefore units of k, rate constant, is mol-1 dm3 s-1 19. Ans: B  electrical conductivity would increase first and then decrease  bonding with oxides does change from ionic to covalent  melting point of the oxides would increase first and then decrease  electronegativity of the element would increase across the period 20. Ans: A  Na would combust to form Na 2O, which would produce an alkali, NaOH, when dissolved in water  A l would combust to form Al2O3, which is insoluble in water, resulting in a neutral solution  S would combust to form SO 2, which would produce an acid, H2SO3, when dissolved in water  P would combust to form P 4O10, which would produce an acid, H3PO4, when dissolved in water 21. Ans: A Lattice energy is always negative. Down the group, the magnitude of the lattice energy decreases due to greater cationic radius. Lattice energy becomes less exothermic down the group. The enthalpy change of hydration is the sum of hydration energy of the cation and anion. The difference lies in the cations. Down the group, the charge density decreases as size increases. Therefore, the hydration energy becomes less exothermic. 22. Ans: C Heterogeneous catalyst works by the availability of the 3d and 4s electrons to allow ready exchange of electrons between the transition metal catalyst and the reactant molecules to form weak bond.

23. Ans: A CH3CH2Cl can undergo nucleophilic substitution to give CH3CH2CN while CH3COCH2CH3 can undergo nucleophilic addition to give CH3