2008 PJC H2 Chemistry Prelim Paper 3 (Suggested Answers)

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1 2008 PJC H2 Chemistry Prelim Paper 3 (Suggested Answers) 1 (a) (i) Using pv = nrt (38.9 x )(2.0 x 10-3 ) = n total x 8.31 x ( ) Total no. of moles at equilibrium, n total = 3.12 mol SO 2 Cl 2 (g) SO 2 (g) + Cl 2 (g) Initial no of mol Change in no of mol -x +x +x Eqm no of mol 2 x x x At equilibrium, n total = 2 x + x + x = (2 + x) mol x = 1.12 mol [1] for table no. of mol of SO 2 Cl 2 = = 0.88 mol [1] no. of mol of SO 2 = no. of mol of Cl 2 = 1.12 mol [1] (ii) [SO 2 Cl 2 ] = ½(0.88) = 0.44 mol dm -3 [SO 2 ] = [Cl 2 ] = ½(1.12) = 0.56 mol dm -3 K c = [ SO ][ ] 2 Cl2 [ SO Cl ] (0.56) = = mol dm -3 (0.44) Similar approach to MJC MYE 09 Paper 3 Q2. (iii) If a system at dynamic equilibrium is subjected to a change in which disturbs the equilibrium, the system will respond in such a way as to counteract the effect of the change. (iv) SO 2 Cl 2 (g) SO 2 (g) + Cl 2 (g) H negative By Le Chatelier s principle, the equilibrium position shifts to the left towards the endothermic reaction where heat is absorbed. Thus, the value of K c will decrease. (b) NaCl has a high melting point whereas SiCl 4 has a low melting point. NaCl is a giant ionic compound with strong electrostatic forces of attraction between Na + and Cl - ions, hence large amount of energy is required to break the bonds during melting. SiCl 4 has simple molecular structure with weak intermolecular Van der waals forces. Lesser energy is required to overcome these intermolecular forces of attractions during melting. Another possible physical property: electrical conductivity.

2 2 (c) (i) NaCl gives Na + and Cl - on dissolution in water and both do not undergo hydrolysis thus a solution of sodium chloride is neutral. However, SiCl 4 undergoes hydrolysis to give an acidic solution. NaCl(s) + aq Na + (aq) + Cl - (aq) SiCl 4 (l) + 2H 2 O(l) SiO 2 (s) + 4HCl(aq) I II III CH 3 CH 2 OH CH 3 CH 2 Cl CH 3 CH 2 CN CH 3 CH 2 CH 2 NH 2 Step I: PCl 5 Step II: KCN in ethanol, heat Step III: LiAlH 4, dry ether (ii) I II III CH 3 CH=CH 2 CH 3 CH(OH)CH 2 Br CH 3 COCH 2 Br CH 3 COCH 2 OH Step I: Br 2 (aq) Step II: KMnO 4 (aq), H 2 SO 4 (aq), heat Step III: NaOH(aq), heat

3 3 2 (a) (i) Standard enthalpy change of neutralisation ( H n θ ) is the energy released when an acid and a base react to form one mole of water at 298K and 1 atm. (ii) Ethanoic acid is weak acid which undergoes partial dissociation. Some of the energy liberated from the neutralisation is used for the dissociation of the weak acid unlike HCl and HNO 3 which dissociate completely. (iii) CH 3 CH 2 NH 2 + HCl CH 3 CH 2 NH 3 + Cl - Heat evolved, Q = mc T = ( ) x 4.2 x 5.2 = 1747 J = kj No of moles of salt produced = 35.0/1000 x 1.0 = H reaction = Q / n = /0.035 = kj mol -1 (b) (i) CH 3 CH 2 NH 2 + H 2 O CH 3 CH 2 NH OH - Initial [ ] Change in [ ] -x x x [ ] eqm / mol dm x x x for table + [CH3CH2NH3 ][OH ] K b = [CH CH NH ] x 10-4 = x x Since CH 3 CH 2 NH 2 is a weak base, it undergoes partial dissociation. Hence, 0.10 >> x. Thus, 0.10 x 0.10 x = [OH - ] = 6.71 x 10-3 mol dm -3 ph = 14 poh = 14 - [-lg(6.71 x 10-3 )] = 11.8 (ii) A solution that is equally effective in resisting changes in ph is a buffer at its maximum buffer capacity, which occurs at ½ V e in this case. Volume of HCl required for complete neutralisation = cm 3 Volume of HCl required to neutralise half the amount of CH 3 CH 2 NH 2 = cm 3. At the maximum buffer capacity, poh = pk b poh = -lg(4.5 x 10-4 ) = 3.35 ph = = 10.7

4 4 (iii) ph Max. Buffer capacity (10, 10.7) <7 Equivalence point Vol of HCl / cm 3 [4] for graph, - for each missing/wrong labels (c) phenylamine < 4-methylphenylamine < ethylamine [1] Both the phenylamines are weaker bases than ethylamine due to the delocalisation of the lone pair of electrons on their N atoms into their benzene rings. Hence, the lone pair on their N atoms are less available to accept a proton. 4-methylphenylamine is a stronger base than phenylamine due to its electron-donating CH 3 group which increases the electron density on the lone pair of N atom. Hence, the lone pair on N atom is more available to accept a proton. (d) (i) CH 3 COCl, r.t.p. (ii) Phenol is not a good nucleophile [1] to react with ethanoyl chloride to form the ester. [Note: To form the ester, phenol is usually treated with NaOH(aq) to form sodium phenoxide. Phenoxide ion is a better nucleophile.] (iii) NaOH(aq) at r.t.p. Product: CH 3 COHN O - Na + NaOH(aq), heat Products: H 2 N O - Na + [1] CH 3 CO 2 - Na +

5 5 3 (a) (i) The order of reaction with respect to a reactant is the power to which the concentration of that reactant is raised in the experimentally determined rate equation. [1] Half-life, t ½, is the time taken for the reactant concentration to decrease to half of its original value. [1] (ii) Let rate= k[h 2 O 2 (aq)] m [ I - (aq)] n Compare experiments 1 & 2, keeping [H 2 O 2 (aq)] constant 4 m n 1.2x10 k(0.020) (0.040) = (or use inspection method) 4 m n 1.5x10 k(0.020) (0.050) n = 1 Rate of reaction is 1st order [1] with respect to I - (aq) (or rate α[i - (aq)]). Compare experiments 1 & 3, keeping [I - (aq)] constant 4 m n 1.2x10 k(0.020) (0.040) = (or use comparing method) 4 m n 3.0x10 k(0.050) (0.040) m = 1 Rate of reaction is 1st order [1] with respect to H 2 O 2 (aq) (or rate α [H 2 O 2 ]) Rate= k[h 2 O 2 (aq)][ I - (aq)] [1] (iii) Using experiment 1, Rate= k[h 2 O 2 (aq)][i - (aq)] 1.2 x 10-4 = k(0.020)(0.040) k = mol -1 dm 3 min -1 [1] (iv) For experiments 4 and 5, since [I - (aq)] >> [H 2 O 2 (aq)], [I - (aq)] is approximately constant. Thus, rate = k [H 2 O 2 (aq)] (a pseudo first order reaction) where k = k[i - (aq)] ln2 ln2 t 1/2 = = = k' k[ I ] t 1/2 of H 2 O 2 in experiment 4 = 9.2 min (for [I - (aq)] = mol dm -3 ) t 1/2 of H 2 O 2 in experiment 5 = 4.6 min [1] (for [I - (aq)] = 1.00 mol dm -3 ) (b) (i) The reaction can be quenched by: adding NaOH/NaHCO 3 /Na 2 CO 3 to remove the H + (aq) sudden cooling of the reaction mixture sudden dilution through the addition of large volume of water any of the above method [1] (ii) I 2 (aq) + 2S 2 O 3 2- (aq) 2I - (aq) + S 4 O 6 2- (aq) [1]

6 6 (c) (i) Reason: Down the group, Size of electron cloud of HX increases (HI > HBr > HCl) Extent of distortion of the electron cloud increases Intermolecular Van Der Waals forces of attraction becomes more extensive Larger amount of energy required (ii) Down the group, Covalent bond length of H-X increases (HI > HBr > HCl) Covalent bond strength decreases Bond dissociation energy decreases Lesser amount of energy required (d) Information given P, C 10 H 11 O 2 Br, has comparable no. of C and H atoms, no. of C > 6. P, C 10 H 11 O 2 Br, does not undergo neutralization with aq Na 2 CO 3 P undergoes alkaline hydrolysis and nucleophilic substitution to form Q and R(C 7 H 8 O). R undergoes oxidation to give benzoic acid. Deduction Presence of benzene ring P does not contain a carboxylic acid group. P is an ester. Q is salt of carboxylic acid and contains OH group. R is an alcohol with aromatic ring. R Is phenylmethanol. Q is optically active. Q has a chiral carbon with 4 different groups attached. Q undergoes nucleophilic substitution with PCl 5 to form steamy fumes of HCl. Q contains a carboxylic acid and an alcohol group.

7 7 Structures: C H 3 H O C C OH OH CH 2 OH Q R, C 7 H 8 O O CH 2 O C CH Br CH 3 P (C 10 H 11 O 2 Br) Equations: CH 2 OH+ 2[O] COOH + H 2 O C H 3 H O C C OH OH H O + 2 PCl 5 H 3 C C C + 2POCl 3 + 2HCl Cl Cl O CH 2 O C CH Br CH 3 + 2NaOH(aq) C H 3 H O C C OH O-Na + CH 2 OH + + NaBr(aq)

8 8 4 (a) (i) 3Mg(NH 2 ) 2 (s) Mg 3 N 2 (s) + 4NH 3 (g) [1] (ii) Ionic size of Mg 2+ < Ba 2+ Charge density of Mg 2+ > Ba 2+ Polarising power of Mg 2+ > Ba 2+ Extent of polarising effect on the anion charge cloud: Mg 2+ > Ba 2+ Hence, magnesium amide is less thermally stable than barium amide (iii) (Group II Tutorial Question) Mg 3 N 2 (s) + 6H 2 O(l) 3Mg(OH) 2 (s) + 2NH 3 (aq) NH 3 (aq) + HCl NH Cl - Amount of acid = 12/1000 x 0.50 = mol = amount of ammonia Amount of Mg 3 N 2 formed = ½ x = mol M r of Mg 3 N 2 = Mass of Mg 3 N 2 = x = g Percentage of magnesium nitride = 0.303/1.00 x 100% = 30.3% (b) (i) At 298K, G = H - TS = (174/1000) = kj mol -1 [1] The reaction is not energetically feasible as the G is positive [1]. (ii) Yes [1] justified by reason; since S > 0, -T S < 0 hence at higher temperatures, the negative -T S term will outweigh the positive H term to make G < 0 [1]

9 9 (c) Enthalpy Zn 2+ (g) + O 2- (g) 1st + 2nd E.A. (O) Zn 2+ (g) + 2e + O(g) 1st + 2nd I.E. (Zn) H lattice (ZnO) Zn(g) + O(g) 0 Zn(g) + 1/2O 2 (g) Zn(s) + 1/2O 2 (g) H o f (ZnO) 1/2 B.E. (O=O) H o at (Zn) ZnO(s) By Hess Law, H lattice (ZnO) = ½(496) (-348) = kjmol -1 [1] [3] correct Born Haber cycle

10 10 (d) Information given in Qn L, C 9 H 8 O, has a high C:H ratio (comparable no. of C and H atoms) and number of C > 6. Compound L is insoluble in water and aqueous sodium hydroxide. L undergoes electrophilic addition with liquid bromine to form M, C 9 H 8 OBr 2. Deduction A benzene/aromatic ring present L does not contain a phenol group L contains an alkene group L undergoes oxidation with Tollens reagent to give a silver mirror but not with Fehling s solution. L undergoes oxidation with acidified potassium manganate(vii) to give N, C 8 H 6 O 4 with the loss of 1 C as CO 2. Compound N undergoes acid-base reaction with aqueous sodium hydroxide to form a soluble sodium salt. N undergoes nucleophilic substitution with SOCl 2 to form two acid chloride groups which undergoes further nucleophilic substitution with ethylamine. L is an aromatic aldehyde / benzaldehyde L has the C=C is at the terminal position. N is a diacid due to the presence of 4 oxygen atoms. N contains carboxylic acid groups. The final product O is a diamide. L M N O

11 11 5 (a) surface area of artefact = 800 cm 2 volume of Al 2 O 3 deposited = surface area x thickness = 800 x (0.1/10) = 8.00 cm 3 mass of Al 2 O 3 deposited = 4.0 x 8.00 = 32.0 g amount of Al 2 O 3 deposited = 32.0 / [2(27.0) + 3(16.0)] = mol During anodising, 2H 2 O O 2 + 4H + + 4e 4Al + 3O 2 2Al 2 O 3 Hence, 2Al 2 O 3 3O 2 12 e - 1Al 2 O 3 3/2 O 2 6 e - amount of e - needed to deposit mole of Al 2 O 3 = x 6 = mol Q = nf = x = C 1.82 x 10 5 C [1] (b) (i) (ii) A complex ion contains a central metal ion bonded to ligands [1] by coordinate / dative bonds [1]. An example of a complex ion is [Fe(H 2 O) 6 ] 3+. The d orbitals are split into two groups due to the ability of the ligands to split them into the energy levels as shown. This effect is known as d orbital splitting. [1] d orbitals in free ion d orbitals in octahedral complex E The d electrons undergoes d-d transition and is promoted to the higher d orbital [1] During the transition, the d electron absorbs a certain wavelength of light from the visible region of the electromagnetic spectrum and emits the remaining wavelength which appears as the colour of the complex observed. [1]

12 12 (c) Fe 3+ (aq) + e Fe 2+ (aq) I 2 (aq) + 2e 2I - (aq) E = V E = V [Fe(CN) 6 ] 3- + e [Fe(CN) 6 ] 4- E = V 2Fe I - 2Fe 2+ + I 2 E cell = (-0.54) = V > 0 Hence, reaction is energetically feasible. Iodide is oxidised to iodine which accounts for the reddish brown solution. 2[Fe(CN) 6 ] I - 2[Fe(CN) 6 ] 4- + I 2 E cell = (-0.54) = V < 0 The reaction is not energetically feasible. [Fe(CN) 6 ] 3- cannot be reduced by I -, this is because CN - ligands are able to stabilise the +3 oxidation state of iron (relative to the +2 oxidation). (d) Haemoglobin consists of 4 sub-units: 2 -sub-units and 2 -sub-units, each being non-covalently linked to a haem residue Iron in haem group binds reversibly to oxygen : Haemoglobin + O 2 Oxyhaemoglobin Both the and -sub-units have 70 % -helical regions There is little contact within the pairs of - and -sub-units There is much contact between an -sub-unit and its neighbouring - -sub-unit (e) (i) Dative bonding in which the lone pair of electrons on the oxygen atom is donated to the vacant orbitals of iron. (ii) The oxidation state of iron remains as +2 [1] as the reaction to form oxyhaemoglobin is a ligand exchange not a redox reaction [1]. (f) Glycine, Gly Aspartic acid, Asp Lysine, Lys O O O O H 3 N + CH C H N CH H 3 N C CH C H 3 N + CH 2 C OH CH O- 2 (CH 2 ) 4 O- O- C + NH 3 O O- A Aspartic acid ; B Glycine ; C Lysine

voltmeter salt bridge

voltmeter salt bridge 2012 H2 Chemistry Preliminary Examination Paper 3 Solutions 1 1 (a) (i) 4FeCr 2 O 4 + 8Na 2 CO 3 + 7O 2 2Fe 2 O 3 + 8Na 2 CrO 4 + 8CO 2 a = 8, b = 7, c = 2, d = 8, e = 8 Any dilute acid e.g. dilute H 2