RVHS H2 Chem 2013 Prelim P3 Soln

Mark Scheme 1 H2 Chemistry (9647) Prelims 2013 Paper 3 1 (a) Orthocaine A B C [9] 1 mark for each correct structure. C:H ratio in orthocaine  1:1 Presence of benzene ring. Orthocaine gives violet complex with neutral FeCl3 (aq)  Presence of phenol. Orthocaine undergoes electrophilic substitution with aq Br2.  Presence of phenylamine/ phenol. Orthocaine undergoes acid – base rxn with NaOH(aq).  Presence of phenol/ carboxylic acid. Orthocaine undergoes acid – base rxn with HCl(aq).  Presence of amine. Orthocaine undergoes acid hydrolysis in hot acid to give CH3OH. CH3OH is oxidized by hot acidified KMnO4 to give CO2 , colourless gas. Allow alternative answers (b) (i) Using Expt 2 & 3, [NO] x 1.5 times while keeping [O 2] constant, rate x 2.25 times (i.e. (1.5) 2 times). Therefore second order with respect to NO. Using Expt 1 & 2, let rate = k[NO]2[O2]n n2 n2 5 5 ] 003 . 0 [ ] 002 . 0 [ k ] 002 . 0 [ ] 001 . 0 [ k 10 4 . 8 10 4 . 1 ) 2 ( rate ) 1 ( rate      1089

Mark Scheme 2 3 2 3 2 n      so n = 1 First order with respect to O2. Using Expt 1, rate = k[NO]2[O2] 1.40  105 = k(0.001)2(0.002) k = 7000 mol2dm6s1 (ii) rate of formation of NO2 = k[NO]2[O2] = 7000(0.002)2(0.002) = 5.60 x 105 moldm3s1 2NO(g) + O2(g)  2NO2(g) rate of depletion of O2 = ½ x rate of formation of NO2 = ½ x 5.60 x 105 = 2.80 x 105 moldm3s1 [5] (c) (1): CH3COOAg (s) + aq CH3COO + Ag+ (2): CH3COOH CH3COO + H+ CH3COO is a strong conjugate base of a weak acid. Cl and Br  are weak conjugate bases of strong acids. When nitric acid is added, position of eqm (POE) in (2) shifts to the left , concentration of CH3COO decreases. As a result, POE in (1) shifts right to counter the decrease in concentration of CH 3COO. This continues until all of CH 3COOAg dissolves. [3] (d) K sp = (1.30 × 10-5)2 = 1.69 × 10 10 mol dm3 [Ag+] = (0.025/ 170) ÷ 2 = 7.353× 105 mol dm3 (7.353 × 10 5 + s)(s) = 1.69 × 1010 Since Ksp is small, assume s is small, 7.353× 105 + s  7.353× 105 Solving, s = 2.298 × 106 mol dm3 Mass of AgCl ppt [3] 1090

Mark Scheme 3 = (1.30 × 105  2.298 × 106) (2) (108 + 35.5) = 3.07 × 103 g 2 (a) (i) The enthalpy change of solution of a substance is the enthalpy change when one mole of the substance is completely dissolved in enough (or excess) solvent (or water) so that no further enthalpy change takes place on adding more solvent. (ii) Using Hess’ Law, ∆Hsoln =  H(LE) +  Hhyd = 365 + (-132-207) = +26.0 kJmol-1 (iii) Ammonium nitrate is a giant ionic lattice structure and water molecules are simple covalent molecules . When ammonium nitrate dissolves in water, the energy released in the formation of ion- dipole interactions between the ions and water molecules is less that the energy absorbed in breaking up the ionic bonds between ammonium and nit