MI H2 Chem 2013 Prelim P2 Soln

Class Adm No Candidate Name: This document consists of XX printed pages and X blank page. [Turn over 2013 Preliminary Examination II Pre-university 3 H2 CHEMISTRY 9647/02 Paper 2 Structured Questions Wednesday, 18 Sep 2013 2 hours Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name, class and index number in the spaces provided at the top of this page. Write in dark blue or black pen in the spaces provided. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. A Data Booklet is provided. You are reminded of the need for good English and clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 /12 2 /10 3 /13 4 /12 5 /13 6 /12 Total /72 822

2 1 Planning (P) For Examiner’s Use You are to design an experiment to determine the enthalpy change of reaction between solid Mg(OH)2 and hydrochloric acid. You are provided with the following list of apparatus and reagents. [You may assume that all other common laboratory apparatus not listed below are also available.] Reagents:  FA1: 2.50 mol dm-3 of HCl  FA2: Mg(OH)2 solid Apparatus:  A thermometer with 0.2 °C interval  A 100 cm 3 styrofoam cup with a lid  50 m l measuring cylinder (a) Write a balanced equation (including state symbols) for the reaction between solid Mg(OH)2 and hydrochloric acid. [1] Mg(OH)2(s) + 2HCl(aq) → MgCl2(aq) + 2H2O(l) 1m for balanced equation with state symbols (b) Given that the enthalpy change of reaction between solid Mg(OH) 2 and hydrochloric acid is approximately – 30.0 kJ mol1, prove with calculations whether 50.0 cm3 of 2.5 mol dm–3 HCl and 1.50 g of Mg(OH)2 are suitable quantities to use at the start of the experiment. Justify your answer. [Assuming the specific heat capacity of the solution is 4.2 J g-1 K-1] [4] Amount of HCl = 50/1000 x 2.5 = 0.125 mol Amount of Mg(OH)2 = 1.50 ÷ (24.3 + 17.0 x 2) = 0.02573 mol (L.R.) Heat released by reaction = heat absorbed by solution 30000 x 0.02573 = 50 x 4.2 x ∆T ∆T = 3.68 °C Amount of solid Mg(OH)2 is not suitable as ∆T is not within 5-10 °C. A very small ∆T will give rise to a large experimental error. 1m for calculating amount of both reagents 1m for determining L.R. 1m for determining ∆T 1m for justification 823

[Turn over 3 (c) Hence give a detailed procedure to determine the enthalpy change of reaction between solid Mg(OH)2 and HCl. [4] For Examiner’s Use Note: Use double the mass of Mg(OH) 2 solid (i.e. 3.00 g) so that ∆T is doubled to 7.36 °C and dilute HCl is still present in excess. 1. Weigh accurately about 3.00 g of Mg(OH) 2 using a weighing bottle. 2. Using a 50 m l measuring cylinder, measure 50 cm 3 of HC l into a styrofoam cup supported in a beaker. 3. Measure the initial temperature of HCl solution. 4. Add solid Mg(OH) 2 into the styrofoam cup and replace the lid immediately. 5. Stir with the thermometer and measure the highest temperature reached. 6. Weigh the mass of weighing bottle and residual solid and calculate the actual mass of Mg(OH)2 used. 7. Wash the styrofoam cup and dry it thoroughly before the next experiment. 8. Repeat experiment for consistent results. 1m for M1 (accept any mass of Mg(OH) 2: 2.05 g < m < 3.60 g in order for it to still be the L.R.) 1m for M2 – M4 1m for M5 – M6 1m for M7 – M8 (d) Both Group I and Group II hydroxides are white solids. A student accidentally mixes up two unlabelled bottles of white solids, known to contain either KOH or Ba(OH)2. Briefly describe an experiment in which the student can perform in order to determine the identity of the white solids. You are provided with two unknown colourless solutions both at the same concentration of 1.00 mol dm -3, either containing KOH (aq) or Ba(OH) 2 (aq) and FA1: 2.50 mol dm-3 of HCl. [3] 1. Using a measuring cylinder, measure 25 cm 3 of each unknown solution into separate Styrofoam cups and measure its initial temperature. 2. Add 50 cm3 of FA1 into each cup. Stir the mixture with a thermometer and record the highest temperature reached. 3. Determine the temperature change for both solutions. The two temperature changes should be mathematically related by: ∆T1=2∆T2. 4. The mixture which give rises to the higher temperature change (twice) contains Ba(OH)2, due to two moles of water produced during the neutralisation reaction. 824

4 Note: Volume of FA1 chosen must allow FA1 used to be in excess. 1m for M1&M2 1m for M3 1m for M4 [Total: 12] 825

[Turn over 5 2 The graph shows the logarithm, lg, of the ionization energies for the outermost electrons of element X in Period 4. For Examiner’s Use (a) From the graph above, deduce the group in the Periodic Table to which X is likely to belong to and explain your reasoning. Hence, suggest an identity for element X. [2] Group I. Largest increase in energy to remove the second outermost electron. The second electron was removed from a lower principal quantum shell ; X is potassium. ; (b) (i) Sketch a graph for elements of the third period (sodium to chlorine) to show how each property changes across the period. I. Atomic and Ionic Radius ( on the same graph) II. Melting Point Explain your sketches. 0 5 10 15 Number of electrons removed 0 1 2 3 4 5 lg I.E. 826

6 1m for each correctly labelled sketch (total 2m) Across the period, nuclear charge ↑ as proton number ↑change in the screening effect is negligible (Same no. of inner shells of e-s across the period), the outer e-s are more strongly attracted by the nucleus  atomic radii ↓ cations have one shell less than neutral atoms, the outer e-s are more strongly attracted by the nucleus, thus smaller in radius than the parent atom. anions have more e -s than protons and so, the effective attractive force on the outer e-s is less than that in neutral atoms, the outer e-s are less strongly attracted by the nucleus, thus bigger in radius as compared to the parent atom. 1m for explanation of atomic radius 1m for explanation of both cationic and anionic radius Na to Al giant metallic structure strong electrostatic forces of attraction between metal cations and delocalised electrons, large amount of energy required to overcome those forces  high m.p. delocalised electrons ↑, size of cations ↓, (charge on cation ↑) strength of metallic bond ↑m.p. ↑ Si giant covalent structure  large amount of energy required to break strong covalent bonds between Si atoms. high m.p. P to Cl 827

[Turn over 7 Simple molecular structure weak temporary dipole – induced dipole forces of attraction between molecules  relatively smaller amount of energy required to overcome those forces  low m.p. size of electron cloud of molecules varies, S 8>P4>Cl2 size of electron cloud of molecules ↑, strength of temporary dipole – induced dipole forces of attraction ↑.  m.p.: S8>P4>Cl2 1m for exp