AJC H1 CHEM P1 P2 ANS

Anderson Junior College 2009 H1 Chemistry Preliminary Examinations Paper 2 Mark Scheme 1 (a) Empirical formula is the simplest formula that shows the relative number of atoms of each element present in a compound. (b) (i) CH2 (ii) CnH2n + 2 3 n O2  n CO2 + n H2O Volume of H2O = 40 cm3 Mole ratio of H 2O : CnH2n = n : 1 = 40 : 10 Hence n = 4 Molecular formula of A is C 4H8. (c) CC CH3 H H3C H CC H CH3 H3C H A B CC CH3 CH3 H H CC H CH2CH3 H H C D A (or B), C a nd D are structural isomers. A and B are cis-trans isomers. 2 (a) (i) Let the rate equation be rate = k [CH3COCH3]x [I2]y [H+]z As observed from the graph, rate of reaction remains constant when [I2] changes => rate of reaction is independent of [I2] => zero order wrt [I2], y = 0 Using the 1 st graph, when [H+] doubled while [CH3COCH3] is constant,  z 0.006 0.10240 ()0.006 0.20 120 z = 1

Using the 2 nd graph, when [CH 3COCH3] increased by 1.5 times while [H +] is constant,  x 0.006 0.10240 ()0.006 0.15 160 x = 1 (ii) rate = k [CH3COCH3] [H+] 0.006 120 = k (0.10)(0.20)  k = 2.50 X 10 –3 mol–1 dm3 min–1 (iii) Hydrogen ions speed up the rate of the reaction by providing a different reaction path which has lower activation energy. As shown on the diagram, the number of reactant molecules with energy greater than or equal to the activation energy (E a’) will increase. This results in an increase in the frequency of effective collisions. Hence, the rate of reaction increases. (b) : I2 molecule Simple molecular lattice structure with weak van der Waals forces of attraction between I2 molecules. (c) NaOH(aq), warm Yellow precipitate observed for propanone but not propanal Represents no. of molecules with energy greater or equal to Ea (uncatalysed) Number of molecules with energy < Represents no. of molecules with energy greater or equal to Ea’ (catalysed) Energ y Ea Ea’

3 (a)  Silicon has the highest melting point, followed by magnesium, then sulfur.  Magnesium has a giant metallic structure with strong electrostatic attraction between magnesium ions and delocalized electrons.  Silicon has a giant macromolecular structure with strong and extensive covalent bonds between silicon atoms.  Sulfur has a simple molecular structure with weak van der waals’ forces of attraction between the molecules. (b) (i) Maximum temperature reached = 54.5 0C [1] Maximum Temperature reached = 54.5 oC [1] Maximum temp reached = 54.5oC (ii) Mg + 2HCl  MgCl2 + H2 (iii) nHCl = 0.04 mol nMg = 3 . 24 25 . 0 = 0.0103 mol Mg is the limiting agent. Heat evolved = Heat absorbed by solution= 40 X 4.2 X (54.5 – 28.0) = 4452 J H = – 0103 . 0 452 . 4 = – 432 kJ mol–1 (no mark if no –ve sign) (iv) Using Hess’ Law, –432 + (–286) = –600 + H1 H1 = –118 kJ mol–1

4 (a) sp2 (b) (c) (i) Brown solution turns colourless. CH 2= CH–CH=CH2 +2Br2  CH2BrCH(Br)CH(Br)CH2Br (ii) Purple s olution turns colourless with black solid formed. CH2=CH–CH=CH2 + 2[O] + 2H2O  CH2(OH) CH(OH)CH(OH)CH2OH (d) (i) H 2C2O4  2CO2 + 2H+ + 2e 5H2C2O4 + 2MnO4 – + 6H+  10CO2 + 2Mn2+ + 8H2O (ii) No. of moles of MnO 4 – = 0.0200 x (20.0 / 1000) = 4 x10–4 mol No. of moles of H2C2O4 = 4 x 10–4 x 5/2 = 1.00 x 10–3 mol Concentration of H2C2O4= 1.00 x 10–3/ (25.0/1000) = 0.0400 mol dm–3 5. (a) (i) (ii) S +: 1s2 2s2 2p6 3s2 3p3 Cl+: 1s2 2s2 2p6 3s2 3p4 (iii) The removal of the 2 nd electron from C l+ is easier due to the inter–electronic repulsion between the paired electrons in the 3p orbital, hence 2 nd IE of C l is less than that of S. sharp increase ionisation energy / kJ mol-1 no1 2 3 4 5 6 7 8 . of electrons removed

(b) (i) As the [H2O2] increases by 3 times ( 0.30 0.10 ), the time taken for 100 cm3 of O2 to be produced decreases by about 1 3 ( 13 40 ) => rate of decomposition of H2O2 is directly proportional to [H2O2] => order of reaction wrt [H2O2] is 1 (ii) When equal volumes of H 2O2 and water are mixed, the [H 2O2] will be halved as volume doubled. hence rate of reaction will be halved . (time taken will be doubled). (iii) (c) (i) Na(g) Na+(g) + e– (Or 2Na(g) 2Na+(g) + 2e–) H4 = 2(494) = + 988 kJ mol–1 (ii) H = 1 2 BEO=O = 1 2 (496) = + 248 kJ mol–1 (iii) H6 = H1 – (H2 + H3 + H4 + H5) O HH H O O H    HB = – 414 – (214 + 248 + 988 + 657) = – 2521 kJ mol–1 (iv) Na + has a smaller charge and larger ionic radius , hence the electrostatic forces of attraction between Na + and O 2– will be weaker. Therefore melting point of Na2O is lower than that of Al2O3. (v) Al + NaOH + 3H2O NaAl(OH)4 + 3 2 H2

(vi) nH2 evolved = 75 24000 = 3.125 x 10–3 mol nAl present = 33.125x10 3/2 = 2.083 x 10–3 mol mass of Al present = 2.083 x 10–3 x 27.0 = 0.05625 g % purity of Al sample = 0.05625 x100%0.10 = 56.3 % 6. (a) (i) Bond Angle: 109.5o (ii) Aluminium is electron deficient which can accept a lone pair of electrons from chlorine to attain an octet structure. (iii) AlCl3 + 6H2O  [Al(H2O)5(OH)]2+ + H3O+ + 3Cl– (iv) Effervescence is observed and the zinc sheet dissolves. 2H+ + Zn  H2 + Zn2+ (b) (i) The salt form is soluble and is more easily absorbed in the body. (ii) The Cl+ is electron deficient and it attacks the electron rich benzene ring. (iii) Substitution (iv) CO2H Cl CO2HCl (v) Add 2,4–DNPH to the two drugs separately and warm. Naprofen: No orange precipitate Ketoprofen: Orange precipitate

C CHCO2HCH3 N N H NO2 O2N (c) (i) The forward reaction is endothermic, heat is absorbed. For the equilibrium position to be shifted to the right hand side, the reaction must be heated to absorb the external heat supplied. (ii) Kc = ] [ ] ][ [ 5 2 3 PCl Cl PCl [PCl3]eqm = [Cl2]eqm = 2 30 . 0 = 0.15 mol dm–3 [PCl5]eqm = 2 ) 30 . 0 40 . 0 ( = 0.05 mol dm–3 Kc = 05 . 0 15 . 0 15 . 0X = 0.45 mol dm–3 7 (a) (i) [H+] = 10–1.93 = 0.0117 mol dm–3 < 0.20 mol dm–3 Citric acid undergoes partial dissociation in water to give H+ , hence it is a weak acid. (ii) Ka = ][ ] ][[ 4 7 5 4 7 5 COOH O H C H COO O H C  (iii) K a = [H+][C5H7COO–]/ [C5H7O4COOH] = 7.27 x 10–4 mol dm–3 C 5H7O4COOH <–> C5H7O4COO– + H+ Initial 0.20 0 0 Change – 0.0117 + 0.0117 + 0.0117 Equilibrium 0.1833 0.0117 0.0117 (b) (i) [OH –] = 0.100 mol dm–3 pOH = 1 pH = 14 – pOH = 13

(ii) No. of moles of sodium hydroxide = 0.100 x (20.00 /1000) = 0.002 mol No. of moles of sodium hydroxide = No. of moles of citric acid = 0.002 mol Volume of citric acid needed for complete neu