SAJC H2 Chem 2013 Prelim P1 Soln

20     [Turn Over   1 B 11 A 21 C 31 D 2 C 12 A 22 C 32 D 3 B 13 C 23 A 33 D 4 C 14 C 24 A 34 D 5 B 15 C 25 B 35 B 6 C 16 A 26 D 36 C 7 A 17 A 27 B 37 A 8 D 18 C 28 A 38 D 9 D 19 C 29 A 39 C 10 B 20 D 30 C 40 B 1130

1 [Turn Over Worked Solutions for SAJC 2013 Prelim Exam Paper 1 1 The metal, M (Ar = 52.8), combines with 1.2 g of oxygen to form the oxide M2O. The metal, M, also forms a second oxide in which the metal and oxygen are present in the ratio 14:1 by mass. What is the molecular formula of the second oxide? A M2O3 B M4O C M11O3 D M7O In M2O For 2nd oxide, assuming 15g of oxide, 1 g of O = 0.0625 mol 14 g of M = 0.265 mol Ratio of M:O is 0.265:0.0625 /barb2right/barb2right /barb2right/barb2right 4.24:1 Ratio is 4.24:1. Ans: B 2 0.0100 mol of an oxide of an element, L, represented by L2On is found to react with exactly 8. 00 x 10I3 mol of acidified KMnO 4 solution. In the reaction, aqueous H LO4 2I ion is formed. Wh at is the initial oxidation state of L? A + 1 B + 2 C + 3 D + 4 L is oxidised to HLO4 2R, MnO4 R is reduced. L loses electrons. Amt of KMnO4 used = 8.00 x 10R3 mol Am t of electrons donated by L = 0.04 mol Amt of L atoms = 0.01 x 2 = 0.02 mol Ratio of electrons donated : L is 0.04:0.02 /barb2right/barb2right /barb2right/barb2right 2:1 Each L has its oxidation state increased by +2. L has final oxidation state of 5+ in HLO4 2R, so initial oxidation state should be +3. An s: C 1131

2 [Turn Over V V PV P P Density 3 Which of the following diagrams correctly describes the behavior of a fixed mass of an ideal gas at constant T? (T is measured in K.) A C B D Ans: B. PV = nRT, so for a fixed mass of gas, the value of nRT is constant and does not change on the graph regardless of the value of V. 4 With reference to its electronic configuration, which of the following statements is true about the electrons in the atoms of the element with atomic number 29? A The electrons in the outermost principle quantum shell experience interIelectronic repulsion. B The d electrons are in the same principle quantum shell as the outermost s electrons. C There are p electrons in 2 different principle quantum shells. D There are s electrons in 3 different principle quantum shells. Element with atomic number 29 is copper Cu. A is not true because the only electrons in the outermost principal quantum shell are the 4s electrons – there is no interRelectronic repulsion in 4s orbital. B is not true as the electrons are in 3d and 4s orbitals. C is true as there are electrons in the 2p and 3p subshells. D is not true as there are electrons in the 1s, 2s, 3s, and 4s subshells. Ans: C 1132

3 [Turn Over 5 The graph shows the logarithm, lg, of the first eight ionisation energies of two elements in Periods 2 and 3. What is the most likely compound that will be formed between the two elements? A CO2 B SiO2 C CS2 D SiCl4 The graph shows two lines. One line (triangle markers) shows a jump from 4 to 5, which means the 5 th electron removed needs high IE, so it is from an inner shell. He nce, it is a group IV element. The other line (square markers) shows a jump from 6 to 7, which means the 7th electron re moved needs high IE, so it is from an inner shell. Hence, it is a group VI element. The group 4 element shows a lower IE for the first 4 electrons removed than the group 6 element, so this indicates that the group IV element lies lower on the Periodic table, so it should be from Period 3 – it is silicon. The group VI element is from period 2 – it is oxygen. The compound is therefore SiO2. Ans: B 6 Covalent bonds are formed by orbital overlap. Which of the following is not present in the following molecule? A sp3Isp2 overlap between C1 and C2 B sp2Isp2 overlap between C2 and C3 1 2 3 4 5 6 7 8 lg (ionisationenergy) no. of electrons removed 1133

4 [Turn Over C sp3Isp overlap between C3 and C4 D spIsp overlap between C4 and C5 Hybridisation state of each carbon: C1 – sp3 C2 – sp2 C3 – sp2 C4 – sp C5 – sp Option C states a sp3Rsp overlap between C3 and C4, and C3 is definitely not sp3. An s: C 7 Which of the following molecules is planar? A Benzene – planar B SCl6 – octahedral C Butanone – has carbons in tetrahedral arrangement D Cyclohexene – has carbons in tetrahedral arrangement Ans: A 8 Given the following enthalpy changes: Enthalpy change of atomisation of carbon = +715 kJ mol I1 Enthalpy change of atomisation of oxygen = +248 kJ mol I1 Enthalpy change of combustion of carbon = I394 kJ mol I1 What is the bond energy of the C=O bond? A 617 kJ molI1 B 679 kJ molI1 C 740 kJ molI1 D 803 kJ molI1 Using Hess’ Law, clockwise = antiRclockwise R394 = (+248 x 2) + (+715) + [B.E. of C=O x 2] [B.E. of C=O] = 802.5 kJmolR1 An s: D C (g) + 2O(g) C (s) + O2(g) C (s) + 2O(g) ∆Hatom of C ∆Hatom of O x 2 ∆Hcombustion of C 2 C=O bond formation CO2(g) 1134

5 [Turn Over 9 The enthalpy changes of formation of gaseous oxides of nitrogen are positive. Which of the following statements accounts for the positive enthalpy changes? A Oxygen has a high tendency to form O2I ions. B Oxygen and nitrogen have similar electronegativity. C Nitrogen has high ionisation energies. D The nitrogen molecule has a high bond energy. A is not correct though it is true, because a high tendency actually means the process is less exothermic (negative enthalpy change). Also, the oxides are covalent, not ionic. B is not correct though it is true, because electronegativity has no effect on enthalpy change here. C is not correct though it is true, because the oxides of nitrogen are covalent and do not involve ionisation. D is correct and true because the N≡N bond requires more energy to break, hence the overall reaction is endothermic (positive energy change) Ans: D 10 The equation below represents a monomerIdimer system: kf kb 2NO2 (g) N2O4 (g) brown colourless Which statement about the equilibrium is correct? A kf increases and kb decreases when the equilibrium mixture is heated. B Increasing the temperature decreases the equilibrium constant Kc. C Addition of argon at constant volume shifts the equilibrium position to the right. D Addition of a catalyst will increase the colour intensity of the mixture. A is not correct because rate constants always increase when temperature increases. B is correct as the enthalpy change is negative – when temperature increases, the system will counter by absorbing heat, favouring the endothermic reaction. In this case, the endothermic reaction is backwards. Hence equilibrium shifts left, and the concentration of reactants increases while the concentration of products decreases /barb2right/barb2right /barb2right/barb2right the equilibrium constant value decreases. C and D are not correct as the addition of catalyst has no effect on the equilibrium and do not result in more products or reactants formed. Ans: B ∆H < 0 1135

6 [Turn Over 11 A 1:3 molar mixture of X2 and Y2 is heated at 673 K and 100 atm of constant pressure so that it comes to equilibrium. The equilibrium mixture contains 20 % of XY3. X2 (g) + 3Y2 (g) 2XY 3 (g) Usi ng the equation above, what is the numerical value of Kp for the reaction at 673 K? A 9.26 x 10I5 B 2.93 x 10I4 C 2.50 x 10I4 D 5.68 x 10I4 As this is a chemical equilibria question, an ICE table is required. But as pressure is constant throughout, the units for the ICE table cannot be pressure, particularly so in this case when the number of gases on each side of the equilibrium are not the same. The units for the table therefore has to be in mol. Let amount of XY3 formed be 2a. X2(g) + 3Y2(g) 2XY3(g) Total mol Pressure Initial/mol 1 3 0 4 100 Change/mol Ra R3a +2a 100 Eq/mo