RVHS H2 CHEM P3 ANS RV Prelim

1 Answer key to Paper 3 1 (a) (i) A buffer solution is one that can resist a change in pH when a small amount of acid or base is added to it. (ii) 1. pKb = 5.93 Let [OH] be x x2  (0.10  x) = 105.93 Since x is small, x2  0.10 = 105.93 x = 3.43 × 104 mol dm3 pOH = 3.47 pH = 10.5 2. At equivalence point, 12.5 cm3 of HCl solution is required. pKa of conjugate acid = 8.07 Amt of RNH3 + = (25/1000) × 0.1 = 2.50 × 103 mol dm3 [RNH3 +] = 2.50 × 103  (37.5/1000) = 6.67 × 102 mol dm3 Let the concentration of [H+] be y y2  (6.67 × 102 –y) = 108.07 Since y is small, y = 2.38 × 105 mol dm3 pH = 4.62 3. Amt of HCl remaining = (30/1000) × 0.20 - 2.50 × 103 = 3.50 × 103 mol [HCl] = 3.50 × 103  (55/1000) = 6.36 × 102 mol dm3 pH = 1.20

2 (iii) Maximum buffering capacity occurs at 8.07, which is close to physiological pH of 7.4/ The buffering action occurs near physiological pH of 7.4/ Non-toxic [6] (b) (i) Rate = k[HONO]2[C6H5NH2]0 (ii) Units for k  mol1 dm3 s1 or mol1 dm3 min1 [2] (c) Observations/Reactions Deduction G does not undergo (nucleophilic) substitution with PCl5 to give white fumes G does not contain carboxylic nor alcohol functional group. G undergoes alkaline hydrolysis to give H and salt of I. G is an amide I is a carboxylic acid G has a fishy smell and does not react with HONO. G is an alkyl amine I rotates plane-polarised light I is chiral I undergoes positive iodoform test I contains –COCH3 or –C(OH)CH3 groups I undergoes condensation reaction with 2,4-DNPH I does not have -C(OH)CH3 groups / I must contain –COCH3 functional group I undergoes electrophilic substitution with Br2(aq) I undergoes 1 electrophilic substitution on benzene ring containing phenol functional group /other ortho & para positions are blocked from ES G

3 H I J [9] (d) N2 + 3H2 ⇌ 2NH3 Heterogeneous catalyst: Fe S2O8 2 + 2I   2SO4 2 + I2 Homogeneous catalyst: Fe2+ or Fe3+ Or C6H6 + Br2  C6H5Br + HBr (or C6H6 + Cl2  C6H5Cl + HCl) Homogeneous catalyst: FeBr3 (or FeCl3) [3] [Total: 20]

4 2 (a) Nitrogen is inert and nitrogen compounds are hard to form. Ammonium salts and nitrates are soluble in water and thus woul d only accumulate in dry areas. [2] (b) Low temperature and high pressure. Ammonia gas has hydrogen bonding between its molecules and thus significant intermolecular forces are present at low temperatures. At high pressure, the ammonia molecules would be very close to each other and the volume of the gas molecules would be significant when compared to the volume of the container / gas. [3] (c) pV = nRT pV = mRT/Mr density = m/V = pMr/RT = (1.01  105  17.0) / (8.31  303) = 682 g m3 = 682 mg dm3 Concentration is below lethal value. [3] (d) (i) Gradient = [(5.44  (8.72)] / (3.33  103  2.50  103) = 3.95  103 K H = 3.95  103  8.31 = 32825 J mol1 = 32.8 kJ mol1 (ii) 5.44 = (3.95  103)(3.33  103) + S/8.31 S = 155 J mol1 K1 (iii) G = H – TS T = 32825/155 = 212 K [6]

5 (e) (i) Transport, structure, catalysts, defence, signalling / control (any 3) (ii) β-pleated sheet consists of adjacent polypeptide strands stabilised by hydrogen bonds between the backbone C=O group of one strand and the backbone N-H group of the adjacent strand. N CH C N CH C N CH C N CH O H O H O H C O H N CH C N CH C N CH C N CH O HO HO H C O H       hydrogen bond R R R R R R R R (iii) Charged and polar R groups / side chains of amino acid residues will form ion-dipole and hydrogen bonds with water molecules. Addition of ammonium ions and sulfate ions can disrupt these interactions which will cause the proteins to precipitate out. [6] [Total: 20] +  + + +   

6 3 (a) (i) (ii) From the graph, t 1/2 is (approximately) constant at 5.0 min. (Since the reaction occurs in aqueous medium, the concentration of water (the other reactant) can be assumed to be constant .) Therefore the reaction is first order with respect to ethyl diazoethanoate. (iii) Rate constant = 2/1t 2ln = 0.139 (min1) [5] (b) (i) Nucleophilic Addition (ii) Formation of gaseous molecule results in an increase in entropy. [2] (c) (i) Reduction Oxidation (ii) Dilute HCl/H2SO4(aq), heat under reflux 0.0 10.0 20.0 30.0 40.0 50.0 60.0 0.0 10.0 20.0 30.0 40.0 50.0 Volume of nitrogen / cm3 time / min

7 (iii) C D E F G H [9] (d) (i) B + Methanol ⇌ Ester + Water Initial amount / mol 0.25 0.34 0 0 Change in amount / mol  0.13  0.13 + 0.13 + 0.13 Eqm amount / mol 0.12 0.21 0.13 0.13 Kc = ]OHCH][HCOCOCHCH[ ]OH][CHCOCOCHCH[ 3223 23223 = )V/21.0)(V/12.0( )V/13.0)(V/13.0( = 0.671

8 (ii) Concentrated sulfuric acid removes the water formed / is a dehydrating agent , causing the equilibrium position to shift to the right to produce more water. Or concentrated sul furic acid acts as a catalyst and does not change the equilibrium position. Increasing the temperature favours the endothermic reaction to absorb some of the added heat . Since the forward reaction is exothermic, the equilibrium position shifts to the left to favour the backward endothermic reaction. [4] [Total: 20]

9 4 (a) (i) (ii) E⦵ / V Fe3+ + e– ⇌ Fe2+ + 0.77 [Fe(CN)6]3– + e– ⇌ [Fe(CN)6]4– + 0.36 Since E⦵(Fe3+/Fe2+) is more positive than E⦵([Fe(CN)6]3–/ [Fe(CN)6]4–), Fe 3+/[Fe(H2O)6]3+ is more easily reduced than [Fe(CN)6]3–. Hence, [Fe(H2O)6]3+ is a stronger oxidising agent. (iii) There will be repulsion between the negatively charged [Fe(CN)6]3– and the electron to be added . / CN  is a stronger ligand than H 2O, hence it stabilises the complex ion to a greater extent. Hence, [Fe(CN)6]3– is less easily reduced and a weaker oxidising agent. [6] (b) (i) 1s2 2s2 2p6 3s2 3p6 3d6 (ii) The energy gap between the two sets of d -orbitals is greater than the repulsion energy, favouring the pairing of electrons. (iii) [Fe(CN)6]4– will adopt a low spin state. since CN– ligands will result in a large ∆E so electrons will pair up in the lower energy orbital. Voltmeter

10 (iv) electron arrangement in [Fe(CN)6]4– The 6 d -electrons are all paired up in the lower energy orbitals/There are no unpaired electrons in [Fe(CN)6]4–. Hence [Fe(CN)6]4– will not be paramagnetic. [7] (c) Amount of Fe2+ reacted = (30/1000)  0.200 = 6.00  10–3 mol Amount of FeOx 2– reacted = (10/1000)  0.200 = 2.00  10–3 mol Since Fe2+  e– Amount of e– gained by 1 mol of FeOx 2– = 6.00  10–3 / 2.00  10–3 = 3 mol Oxidation state of Fe in FeOx 2– = +3 + 3 = +6 +6 + (–2)x = –2 x = 4 [3] (d) (i) NH4SCN gives a blood/deep red colouration with iron(III) chloride but NH4NO3 does not give blood/deep red colouration. (ii) A violet colouration/solution will be formed. (iii) Positive test: Negative test: [4] [Total: 20] OH CH3 CH2OH