NYJC Prelim CHEM P2 ANS

NANYANG JUNIOR COLLEGE JC 2 PRELIMINARY EXAM Higher 1 CANDIDATE NAME CLASS TUTOR’S NAME CHEMISTRY 8872/02 Paper 2 23 Sep 2009 Candidates answer Section A on the Question Paper. 2 hours Candidates answer Section A on the Question Paper Additional Materials: Answer Paper Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. For Examiner’s Use A1 A2 A3 B1 B2 B3 Section A Answer all questions. Section B Answer two questions on separate answer paper. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. Total This document consists of 14 printed pages. [Turn over

2 H1 Chemistry 8872/02 NYJC J2/09 Prelim For examiner’s use only Section A Answer all questions in this section in the spaces provided. 1(a) A sample of carbon containing different isotopes was artificially enriched. The relative isotopic masses and its relative abundances determined in a mass spectrometer are given in the table below. Relative isotopic mass Relative abundance 12C 85.0 % 13C 10.0 % 14C 5.0 % Calculate the relative atomic mass of carbon in the sample to 1 decimal place. [1] A r = 12 x 0.850 + 13 x 0.100 + 14 x 0.050 = 12.2 [1] (b) A 0.100 mol of a hydrocarbon W was burnt completely to produce 26.4 g of carbon dioxide gas and 10.8 g of water. The same mass of W when burnt under a container with 300 g of water at 30 oC was found to bring the water to boil. The process was known to be only 70% efficient. The specific heat capacity of water is 4.2 J g –1 K–1. (i) Determine the molecular formula of the hydrocarbon W. [2] CxHy + (x + ¼ y)O 2  xCO 2 + ½ y (H 2O) Amt of CO2 formed = 26.4 / 44.0 = 0.600 mol Amt of H2O formed = 10.8 / 18.0 = 0.600 mol C xHy CO2 H2O n 0.100 0.600 0.600 Lowest ratio 1 6 6 Mol ratio 1 x ½ y Comparing coefficient x = 6 6 = ½ y  y = 12 Molecular formula T is C6H12 [1] [1] (ii) Calculate the enthalpy change of combustion of W. [2] Heat lost from combustion = heat gained by water [1] [1]

3 H1 Chemistry 8872/02 NYJC J2/09 Prelim For examiner’s use only c -1 -1 m c T 300 x 4.2 x (100-30) 70H = = - n 0.100 100 = -1260 000 J mol = -1260 kJ mol  (c) Arranging the following in order of increasing boiling points. Explain your choice in terms of structure and bonding. H 2O, NH 3, NaCl, CH 4 [5]  H 2O, NH3, CH4 and C6H14 have simple molecular structure consisting of molecules held together by weak forces.  CH 4 consists of simple non–polar molecules held together by weak van der waals forces.  NH 3 and H2O molecules are held together by stronger hydrogen bonds thus have higher boiling points than C6H14.  H 2O has higher boiling points than NH3 because it has 2 hydrogen–lone pair units per molecule thus has more hydrogen bonds per molecule.  NaCl has giant ionic structure consisting of Na + and Cl– ions held together by strong electrostatic forces of attraction thus has highest boiling point. [1] [1] [1] [1] [1] [Total: 10 marks]

4 H1 Chemistry 8872/02 NYJC J2/09 Prelim For examiner’s use only C C C C C OH H O H OH C O H H H HO H OH H H C OO H C C C C H O H C O H H H HO H OH H H OH C O CO C C OHO H C CH2 OH OH H H HH Step I controlled oxidation Step II cyclic formation Step III C O CO C C OHO H C CH2 OH OH H H 2 (a) (i) Complete the graph below for the melting points of elements in period 3. [1] [1] (ii) Explain the shape of your graph in terms of bonding and structure. [5]  Na, Mg and Al have giant metallic structures with strong electrostatic attraction between the cations and the sea of delocalized electrons.  The metallic bonds increases as more electrons are added to the sea of delocalized electrons.  Si has a giant molecular stru cture consisting of Si atoms held together by very strong covalent bonds.  P, S, Cl and Ar have simple molecular structure consisting of molecules or atoms held by weak van der Waals’ forces.  The melting points ar e in the order of S 8> P4 > Cl 2 > Ar as van der Waals’ forces are proportional to the number of electrons in the molecules / atoms. [1] [1] [1] [1] [1] (b) The building block for ascorbic acid is the glucose molecule. The following synthetic pathway was proposed: Ar Cl S P Si Mg Al Na 11 12 13 14 15 16 17 18 Melting points A scorbic Acid Intermediate product 2 Intermediate product 1 Glucose

5 H1 Chemistry 8872/02 NYJC J2/09 Prelim For examiner’s use only (i) State the type of reaction found in step II and hence circle the functional group(s) present in the intermediate product 1 that is/are involved in the reaction. [3] Condensation [1]; both circles correct [2] (ii) State the type of reaction present in Step III. [1] Elimination (c) Aspirin, also known as acetylsalicylic acid is often used as an analgesic (pain- remover) to relieve minor aches and pai ns. It is readily absorbed from the intestines since it diffuses rapidly into the tissues. CO2H OCOCH3 The molecule is hydrolysed by acids in the stomach. (i) Draw the structural formulae of the hydrolysis products. [2] C H3 C O OH CO2H OH+ [2] Aspirin can be synthesised from 2-methylphenol via the following steps: CO2H OCOCH3 CH3 OH A acidified KMnO4(aq) heat 2-methylphenol Aspirin (ii) Draw the structural formula of the intermediate A. [1] CO2H OH [1] (d) The enthalpy of combustion of Compound A, CH 2=CHCO2H, can be determined either by direct measurem ent of the heat evolved usin g a bomb calorimeter or by indirect method using Hess Law. The energy cycle involving Compound A is given below.

6 H1 Chemistry 8872/02 NYJC J2/09 Prelim For examiner’s use only CH2 CHCO2H 3CO2 + 2H2O CH3CH2 CO2H + H2 3O2 + 1/2 O2 H3= -286 kJ mol-1 H2 = -1450 kJ mol-1 H1 = -380 kJ mol-1 +7/2 O2 Hc + H2 3CO2 + 3H2O + (l) (aq) (g) (g)(l) (l) (g) (l) (i) Name the enthalpy change represented by H2 [1] H2 : Enthalpy of combustion of propanoic acid [1] (ii) State the type of reaction for H1. [1] Reduction reaction [1] (iii) Using Hess’ Law, calculate the enthalpy change of combustion of compound A. [1] c -1 c H + (-286) = -380 + (-1450) H = -1544 kJ mol   [1] [Total: 16 marks]

7 H1 Chemistry 8872/02 NYJC J2/09 Prelim For examiner’s use only 3(a) Hydrogen peroxide, H2O2, is a strong oxidizing agent and is used as an antiseptic. However, it is not stable at room temperature and will undergo decomposition reaction. (i) Draw the dot and cross diagram of H2O2. [1] OO H H x x xx x xx [1] (ii) State, with reason, the shape about the oxygen atom. [2] 2 lone pair and 2 bond pair  shape is bent [1+1] (iii) State the oxidation number of O in the reactants and products. [1] O in H2O2 : –1 O in H2O = –2 O in O2 = 0 [1] (iv) Hence write two balanced half and overall equations for the decomposition of H2O2. [2] H2O2 + 2H+ + 2e  2H 2O H2O2  2H + + 2e + O2 2H2O2  2H 2O + O2 [1] [1] (b) The rate of H 2O2 decomposition can be catalysed by adding small amounts of MnO2. Using an appropriate diagram, expl ain in molecular terms, how the presence of MnO2 catalyst increases the rate of decomposition of H2O2. [4]  The presence of a catalyst, lowers the activation energy compared to the uncatalysed reaction.  More molecules will possess energy gr eater than or eq