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 nitrate ions and hydrogen bonds between water molecules. (iv) ∆S = 186 + 256 – 151 = +291 Jmol-1K-1 ∆G = ∆H – T∆S For the instant cold pack to work, ∆G < 0 i.e. ∆H – T∆S < 0 T > ∆S ∆H T > 291 10 26.03    )( i.e. T > 89.3 K [7] (b) (i) Method 1: NH4NO3(s) + aq  NH4 +(aq) + NO3 -(aq) ∆Hsoln NH4 +(g) + NO3 -(g) -365 -132 + (-207) 1091

Mark Scheme 4 ∆Hr = [2(-285.8) + 81.6] – [-174.1 - 80.8 - 57.0 - 26.0] = -152.1 kJmol-1 Method 2: ∆H f(NH4NO3(s)) = -174.1 – 80.8 – 57.0 - 26.0 = -337.9 ∆Hr = 2∆Hf(H2O(l)) + ∆Hf(N2O(g)) – ∆Hf(NH4NO3(s)) = -152.1 kJmol-1 (ii) or [4] (c) (i) Denaturation is the process by which a protein loses its tertiary or secondary / quarternary structure by application of external stress and loses its function (ii) pH changes: -Disrupts hydrogen bonds in the tertiary structure. At low pH, NH2 becomes NH3 +.At high pH, phenol becomes phenoxide ion. Or - Disrupts ionic interactions in the tertiary structure. At low pH, 0 N2(g) + 2 3 O2(g) + 2H2(g) N2O(g) + 2H2O(l) 2(-285.8) + 81.6 NH3(aq) + HNO3(aq) (-174.1) + (-80.8) NH4NO3(aq) -57.0 +26.0 NH4NO3(s) ∆Hr 2H2(g) + N2(g) + 3/2O2(g)  NH4NO3(s) ∆Hf HNO3(aq) + NH3(aq) → NH4NO3(aq) -174.1 + (-80.8) +26.0 -57.0 1092

Mark Scheme 5 phenoxide becomes phenol while at high pH, NH3 + becomes NH2 . (iii) 7 [4] (d) (i) or Linear [2] (i) Volume of gas = 3.0 x 24.0 = 72.0 dm3 (ii) 11 1 pV T = 22 2 pV T i.e. 10) (273 ) (0.8)(V 25 273 (1)(72.0) 2  Therefore, V2 = 79.4 dm3 It is assumed that N2 was behaving as an ideal gas. [3] 3 (a) (i) nNaOH = nHCl in 5.00 cm3 sample At equilibrium, VNaOH = 30.00 cm3 nNaOH = nHCl = 0.0300  0.100 = 0.00300 mol [HCl] = 0.00300  0.005 = 0.600 mol dm3 C2F4  2HCl [C2F4] = ½ [HCl] = 0.300 mol dm3 CHC lF2  HCl [CHClF2] = 1.00 – [CHClF2]reacted = 1.00 – 0.600 = 0.400 mol dm3 (ii) 2 2 2 4 2 ] [ ] ][ [ F CHC HC F C c l lK  675 . 0] 400 . 0 [ ] 600 . 0 ][ 300 . 0 [ 2 2 cK mol dm3 (iii) Comparing Experiment 1 and 2, when the temperature is increased, the equilibrium [HCl] increased/ the volume of NaOH required to neutralise the HCl increases. The equilibrium position shifted to the right to absorb energy. Therefore, forward reaction is endothermic. [6] 1093

Mark Scheme 6 OR H = 2BE(C-H) + 2BE(C-Cl) + 4BE(C-F) – BE(C=C) – 4BE(C-F) – 2BE (H-Cl) = +28.0 kJ mol-1 Since H is positive, the forward reaction is endothermic. (b) (i) At the start of reaction, there is only forward reaction since there is no product for backward reaction to take place. Therefore, to find the rate of forward reaction, draw a tangent to the graph at t = 0 min . The gradient of the tangent is the initial rate of forward reaction. (ii) From the respective tangent of the graphs for experiment 2 and 3, 1 : 4 30 9 :30 35 Expt3 : Expt2 When [CHClF2] is halved, the rate of forward reaction decreased by ¼ . Therefore, the reaction is second order with respect to CHClF2(g). (iii) If conclude first order in (b)(ii): CHClF2  CF2 + HCl (slow) 2 CF2  C2F4 OR CHClF2  CF2 + HCl (slow) CF2 + CHClF2  C2F4 + HCl If conclude 2nd order in (b)(ii): 2nd step is the slow step [6] 1094

Mark Scheme 7 (c) (i) (ii) Electrophilic Substitution Step 1: Generation of Cl+ electrophile. If FeCl 3 is used, FeCl3+ Cl2 FeCl4  + Cl+ If Fe is used, Fe + 3/2Cl2 FeCl3 FeCl3+ Cl2 FeCl4  + Cl+ If AlCl3 is used, A lCl3+ Cl2AlCl4  + Cl+ Step 2: Electrophilic attack of benzene by Cl+. [8] 4 (a) (i) Fe(OH)2(s) ⇌ Fe2+(aq) + 2OH(aq) [1] 1095

Mark Scheme 8 (ii) Ni(s) + Fe2+(aq) → Ni2+(aq) + Fe(s) E⦵= 0.44  (0.25) = 0.19V, < 0V Since E is less than 0V, the reaction is not spontaneous and will not be feasible. [2] (iii) When excess NH3(aq) is added: Oxidation: [Ni(NH3)6]2+(aq) + 2e → Ni(s) + 6NH3(aq) 0.51V Reduction: Fe2+(aq) + 2e → Fe(s) 0.44V Overall: Ni(s) + 6NH3(aq) + Fe2+(aq) → Fe(s) + [Ni(NH3)6]2+(aq) Overall E ⦵ = 0.44  (0.51) = +0.07 V Hence, since E > 0V, the reaction can take place. Fe(OH)2(s) ⇌ Fe2+(aq) + 2OH(aq) Fe2+ is consumed/ [Fe 2+] drops. (By le chatelier's principle) more Fe(OH)2(s) is able to dissolved/ The equilibrium position shift right to replenish the loss of Fe2+ and removing Fe(OH)2(s). [3] (iv) Grey ppt: Iron/ Fe(s) Pale blue solution: [Ni(NH3)6]2+ OR [Ni(NH3)6]Cl2 [2] (b) (i) SӨ > 0 as there is an increase in disorder due to the increase of number of moles of aqueous/liquid produts from 4 to 7. This increase in disorder will result in ΔG value to be more likely negagtive, and hence the reaction is expected to be favourable. [2] (ii) Fe(OH)2(s) ⇌ Fe2+(aq) + 2OH