MJC H2 CHEM P3 answers

Name _________________________ Class: 13S________ Reg Number: _____ MERIDIAN JUNIOR COLLEGE JC 2 Preliminary Examination Higher 2 ___________________________________________________________________________ Chemistry 9647/03 Paper 3 Free Response 17 September 2014 2 hours Additional Materials: Data Booklet Writing Paper _____________________________________________________________________ INSTRUCTIONS TO CANDIDATES Write your name, class and register number in the spaces provided at the top of this page. Answer 4 out of 5 questions in this paper. Begin each question on a fresh page of writing paper. Fasten the writing papers behind the given Cover Page for Questions 1, 2 & 3 and Cover Page for Questions 4 & 5 respectively. Hand in Questions 1, 2 & 3 and 4 & 5 separately. You are advised to spend about 30 minutes per question only. INFORMATION FOR CANDIDATES The number of marks is given in brackets [ ] at the end of each question or part question. You are reminded of the need for good English and clear presentation in your answers. This document consists of 12 printed pages (excluding the cover pages for answer scripts).

2 [Turn Over Answer any 4 out of 5 questions in this paper. Begin each question on a fresh sheet of writing paper. 1 The following tables show some standard enthalpies change of solution, ∆H o sol, regarding some Group I chlorides and Group II carbonates, as well as their corresponding solubility in water under standard conditions. Group I chlorides NaCl KCl RbCl ∆Ho sol / kJ mol–1 +3.9 +17.2 +16.7 Solubility / mol per 100 cm3 of water 0.615 0.481 0.781 Group II carbonates MgCO3 CaCO3 SrCO3 ∆Ho sol / kJ mol–1 –25.3 –12.3 –3.4 Solubility / mol per 100 cm3 of water 1.26 x 10–4 1.30 x 10–5 7.45 x 10–6 Use relevant data from the tables above, if applicable, to answer the following questions. (a) It is a well -known fact that the solubility of Group II carbonates decreases down the group. (i) Using the ∆Ho sol data from both tables, comment on whether ∆H o sol is a good predictor for the solubilities of salts in water. Group II carbonates have negative / exothermic ∆H o soln even though they are insoluble, On the other hand, ∆H o soln of soluble Group I chlorides are positive / endothermic. Hence, ∆Ho soln is not a good predictor of solubility. (ii) Calculate the solubility product of calcium carbonate, stating its units. [3] Solubility of CaCO3 = 1.30 x 10–5 x 10 = 1.30 x 10–4 mol dm–3 Ksp of CaCO3 = [Ca2+] [CO3 2–] = (1.30 x 10–4)2 = 1.69 x 10–8 mol2 dm–6 (b) (i) Assuming a 100% efficient system, calculate the maximum temperature change that could take place when potassium chloride is dissolved in 100 cm 3 of water. State clearly whether the temperature increases or decreases. 0.481 mol of KCl dissolves in 100 cm3 water to give a saturated solution. Quantity of heat evolved by solution = 0.481 x 17.2 = 8.273 kJ = mc∆T

3 [Turn Over Maximum temperature decrease, ∆T = 38.273 x 10 100 x 4.18 = 19.8 oC / K (ii) The enthalpy change of solution assumes that an infinitely dilute solution is formed when one mole of ionic compound dissolves in water . Based on this assumption, suggest a reason to account for the maximum temperature change n ot being achieved in practice other than t he system is not 100% efficient. [3] A very concentrated solution results in ions being in close proximity in the solution allowing repulsive and attractive forces to exist. (c) The standard entropy change of solution, ∆S o sol, of sodium chloride is +42.6 J mol–1 K–1. (i) Calculate the standard Gibbs free energy change of solution, ∆G o sol, of sodium chloride, explaining why sodium chloride is soluble. ∆Go sol = ∆Ho sol – T∆So sol = +3.9 – 298 +42.6 1000   = –8.79 kJ mol–1 < 0, hence reaction is (energetically) feasible and salt dissolves. (ii) The standard Gibbs free energy change of solution, ∆G o sol, of magnesium chloride is –126.3 kJ mol –1. Some standard enthalpy change of formation, ∆Ho f, related to magnesium chloride are provided below. Substances MgCl2(s) Mg2+(aq) Cl –(aq) ∆Ho f / kJ mol–1 –641.6 –462.0 –167.5 Calculate the ∆S o sol of magnesium chloride. Hence, explain with reasoning, whether the sign of ∆So sol obtained is as expected. [5] ∆Ho soln(MgCl2) = ∆Ho f(Mg2+(aq)) + 2∆Ho f(Cl – (aq)) – ∆Ho f(MgCl2(s)) = –462.0 + 2(–167.5) – (–641.6) = –155.4 kJ mol–1 ∆Go soln = ∆Ho soln – T∆So soln ∆So soln = 155.4 ( 126.3) 298    = –0.0977 kJ mol–1 K–1 = –97.7 J mol–1 K–1

4 [Turn Over The negative sign is unexpected. When solid MgC l2 is d issolved, aqueous ions are formed which caused entropy to increase and system is supposed to be more disordered. (d) The melting point of CaCO3 is 825 oC and its thermal decomposition temperature is 837 oC. (i) The melting point of BeCO 3 is 54 oC. Briefly explain why the melting point of BeCO3 is much lower than that of CaCO3.  BeCO3 is a simple molecular compound with weak van der Waals’ forces between molecules. CaCO3 is a giant ionic lattice compound with strong electrostatic forces of attraction be tween oppositely charged ions of Ca2+ and CO3 2–. (ii) By quoting data from the Data Booklet , predict whether the thermal decomposition temperature of SrCO 3 would be higher or lower than that of CaCO3. [5]  ionic radius of Ca2+ = 0.099 nm, ionic radius of Sr2+ = 0.113 nm.  ionic radius: Sr2+ > Ca2+  charge density and polarising power: Sr2+ < Ca2+  ease of cation to polarise CO3 2– anion and break C–O bond: Sr2+ < Ca2+  energy required to break C–O bond and decompose compound: SrCO3 > CaCO3  thermal stability: SrCO3 > CaCO3  thermal decomposition temperature: SrCO3 > CaCO3 (e) Using relevant data from the Data Booklet and information given below, construct an energy cycle to calculate the lattice energy of MgCO3. enthalpy change of formation of magnesium carbonate = –1096 kJ mol–1 enthalpy change of formation of CO2 (g) = –393 kJ mol–1 enthalpy change of atomisation of magnesium = +148 kJ mol–1 sum of first and second electron affinities of oxygen = +657 kJ mol–1 CO2 (g) + O2– (g) CO3 2– (g) ; ∆H1 = –778 kJ mol–1 [4] [Total: 20]

5 [Turn Over By Hess’ Law, ∆Hlatt o(MgCO3) = –1096 – (–393) – 148 – ½(496) – 736 – 1450 – 657 – (–778) = –3164 kJ mol–1 2 This question is about phosphoric acid. (a) Phosphoric acid, H3PO4, is a triprotic acid and its structure is given below. phosphoric acid The p Ka values of the two successive dissociations of H 3PO4 at 25 oC are as shown. pKa1 pKa2 Phosphoric acid 2.1 7.2 Small quantities of H 3PO4 are used to impart the sour taste to many soft drinks. A typical can of soft drink with density of 1.00 g cm –3 contains 0.05 % by mass of H3PO4. (i) Briefly explain why the pKa2 of phosphoric acid is higher than its pKa1. H2PO4 – is a weaker acid than H3PO4. It is more difficult to dissociate H+ from a negatively charge species / ion.

6 [Turn Over (ii) Calculate the concentration, in mol dm–3, of phosphoric acid in the soft drink. No. of moles of H3PO4 in 1 cm3 of soft drink = . x . = 5.10 x 10–6 [H3PO4] = 5.10 x 10–3 mol dm–3 (iii) Hence, by considering only the first dissociation of H 3PO4, determine the pH of this soft drink. Assume that the acidity of the soft drink arises only from H3PO4. H3PO4 H+ + H2PO4 – Ka = 10–2 .1 mol dm–3 [H+] = √ -2. x . x - = 6.36 x 10–3 mol dm–3 pH = –log [H+] = –log (6.36 x 10–3) = 2.20