SRJC H2 CHEM P3 ANS

SRJC 2009 9746/03/Prelim/2009 [Turn over SERANGOON JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 2 CHEMISTRY 9746/03 Preliminary Examination Paper 3 Free Response 20th August 2009 MARK SCHEME 2 hours CONFIDENTIAL DOCUMENT Teachers to keep this document under lock and key until 20 August 2009 12pm. This document consists of 19 printed pages and 1 blank page

2 SRJC 2009 9746/03/Prelim/2009 [Turn over 1 (a) A student dissolved 8.4 g of sodium fluoride in 250 g of water. Given the following data, calculate the initial temperature of water if the final temperature of the solution is 20.0 °C. Assume that the specific heat capacity of sodium fluoride solution is 4.2 J g–1 K–1. Lattice energy of NaF –918 kJ mol–1 Enthalpy change of hydration of F- –457 kJ mol–1 Enthalpy change of hydration of Na+ –390 kJ mol–1 [3] ½ m ½ m ½m ½m ½m ½m ∆H soln NaF (s) Na+ (aq) + F- (aq) ∆H latt ∆Hhyd ∆Hhyd N a + (g) + F- (g) ∆Hsoln (NaF) = -∆Hlatt (NaF) + ∆Hhyd (Na + ) + ∆Hhyd (F - ) = -(-918) + (-457) + (-390) {give method marks, if answer is wrong} = +71.0 kJ mol-1 (since ∆Hsoln > 0, temp of surroundings decreases) Amt of NaF dissolved = 8.4 / (23+19) = 0.2000 mol Q = ∆H soln x Amt of NaF = 71 x 0.2000 = 14.2 kJ {give method marks, if answer is wrong} Q = mc ∆T 14.2 x 1000 = 250 x 4.2 x (T i - 20) Note: since temp surr decreases, Ti > Tf {give method marks, if answer is wrong} 14200 = 1050 (T i – 20) 14200 = 1050T i - 21000 1050T i = 35200 Ti = 33.5°C (b) In the manufacture of fuel for industries, liquid MTBE is added as a solvent to petroleum to reduce pollution. C CH3 CH3 CH3 O CH3 Methyl t-butyl ether (MTBE) (i) What is meant by the term standard enthalpy change of formation of MTBE? Support your answer with the aid of an appropriate equation with state symbols. 1m 1m Standard enthalpy change of formation of MTBE is the energy evolve/change when 1 mole of MTBE is formed from its elements under standard conditions. 5C(s) + 6H 2 (g) + ½O2 (g) → C5H12O (l) {½ for state symbol, ½ for equations} M T B E

3 SRJC 2009 9746/03/Prelim/2009 [Turn over (ii) Use the following data and by means of an appropriate energy cycle, prove that the bond energy for the C–H bond in MTBE is +418 kJ mol-1. Enthalpy change of atomisation of C(s) +715 kJ mol–1 Enthalpy change of formation of MTBE –383 kJ mol–1 Enthalpy change of vapourisation of MTBE +30.4 kJ mol–1 Bond Energy of H-H +436 kJ mol-1 Bond Energy of O=O +496 kJ mol-1 Bond Energy of C-O +360 kJ mol-1 Bond Energy of C-C +350 kJ mol-1 2m 1m 1m ∆H f (MTBE) 5C(s) + 6H 2 (g) + ½O2 (g) C 5H12O (l) 5 ∆Hat (C) 12∆Hat(H) ∆Hat (O) ∆Hvap (MTBE) 5C (g) + 12H(g) + O(g) C 5H12O (g) ∆Hat (C5H12O) correct energy cycle {one mistake minus ½m} By Hess’ Law, ∆Hf (MTBE) = 5∆Hat (C) + 12∆Hat (H) + ∆Hat (O) – ∆Hat (C5H12O)- ∆Hvap (MTBE) -383 = 5(715) + 12 [½ BE(H-H)] + ½BE(O=O) - ∆Hat (C5H12O) – 30.4 -383 = 3575 + 12(½ x 436) + ½ (496) -30.4 - ∆Hat (C5H12O) -383 = 3575 + 2616 + 248 – 30.4 - ∆Hat (C5H12O) ∆Hat (C5H12O) = + 6791.6 kJ mol-1 C OC C HHH H H H C H H H C HH H Displayed formula of MTBE Bond energy of C5H12O involves breaking a total of: 2(C-O), 3(C-C) and 12(C-H) bonds. 6791.6 = 2BE(C-O) + 3BE(C-C) + 12BE(C-H) 6791.6 = (2 x 360) + (3 x 350) + 12BE(C-H) 6791.6 = 1770 + 12BE(C-H) 12 BE(C-H) = 5021.6 BE(C-H) = +418 kJ mol -1 {Italics give full method marks}

4 SRJC 2009 9746/03/Prelim/2009 [Turn over (iii) Suggest a reason for the difference in the C–H bond energy in (b)(ii) from the value given in the Data Booklet. [7] 1m The bond energy values from the Data Booklet are average values and would differ from experimental values. (c) Magnesium chloride, MgCl 2, is an important coagulant used in the preparation of soy products. The lattice energy of magnesium chloride is given to be –2490 kJ mol–1. (i) How would you expect the numerical magnitude of the lattice energy of barium chloride, BaCl2(s) to compare with that of MgC l2(s)? Hence predict a likely value for the lattice energy of barium chloride. ½m ½m 1m ½m ∆H latt is proportional to qq rr   , Ionic radius: Mg2+ < Ba2+ |∆Hlatt | : MgCl2 > BaCl2 or Numerical value of lattice energy of BaCl2 will be smaller. or Lattice energy of BaCl2 will be less exothermic. Accept any value smaller than 2490 kJ mol-1. The enthalpy change of reaction, for the reaction below is given to be -391 kJ mol-1. 2MgCl (s) → MgCl2(s) + Mg (s) (ii) Comment on the stability of MgCl (s) relative to that of MgCl2(s). [4] ½m 1m Since ∆H reaction is exothermic, Energy of products is lower than that of reactants. MgCl2(s) is energetically more stable than MgCl (s). (accept proper energy profile diagram) (d) Part of the research on Supramolecular Chem istry focuses on Host-Guest Chemistry, in which cationic transition metal complexes are involved in second-sphere coordination which is provided by ligands with an organised set of donors. For example, a transition metal rhodium complex that contains hydrogen bond donor groups (ammonia or water ligands) in its primary coordination sphere can be considered as a guest capable of binding to a hydrog en bond acceptor host via second sphere coordination. (i) State the bonding formed between rhodium and ammonia. 1m dative / coordinate/covalent bond (ii) Describe the chemical structure and bonding in solid rhodium. ½m ½m Rh has a giant metallic structure with strong electrostatic forces of attraction between cations and sea of delocalised electrons. Deleted: ¶

5 SRJC 2009 9746/03/Prelim/2009 [Turn over (iii) Show with aid of a diagram the intermolecular bonding in ammonia. ½m ½m Partial charge Others (iv) Predict and account for the boiling point of ammonia relative to water. ½m ½m ½m ½m Both NH 3 and H2O have simple molecular structures with intermolecular hydrogen bonds. For the identification of hydrogen bonds, either mentioned here or in part (iv). Number / Extent of hydrogen bonds: NH3 < H2O Energy required: NH3 < H2O Boiling point: NH3 < H2O (v) Predict and account for the solubility of ammonia in water. [6] ½m ½m NH3 is soluble in water as favourable hydrogen bonds between NH3 and water / solute-solvent interaction can be formed. {Do not accept the NH3 + H2O ⇌ NH4 + + OH- as the extent of dissociation is too small} [Total: 20]

6 SRJC 2009 9746/03/Prelim/2009 [Turn over 2 (a) The table below gives some data on some oxides of elements in Period 3 of the Periodic Table. oxide Na2O MgO Al2O3 SiO2 P4O6 SO2 Boiling point / K 1548 3873 3253 2503 448 263 (i) Explain, in terms of chemical bonding and structure, why the boiling point of MgO is higher than P4O6. ½m ½m 1m ½m ½m 1m MgO  has a giant ionic lattice structure  large amount of energy is needed to  overcome the strong ionic bonds OR electrostatic forces of attraction between oppositely charged ions.  has a high boiling point P 4O6  has a simple molecular structure  small amount of energy is needed to  overcome the weak intermolecular Van d