ACJC Chemistry Paper 3 Answer Prelims

Mark Scheme 2008 H2 Chemistry Preliminary Examination Paper 3 Mark Scheme 1 (a)(i) Order of rxn wrt to Br2 Comparing expt 1 & 3, when [Br 2] triples from 0.01 to 0.03 moldm -3, the rate of rxn also triples. Hence, [Br 2] is directly proportional to the rate of rxn. Order of rxn wrt Br 2 = 1 Order of rxn wrt to HCOOH Comparing expt 1 & 2, 8.0 x 10 -6 (0.02)1(0.04)x 2.0 x 10-6 (0.01)1(0.02)x x = 1 Order of rxn wrt HCOOH = 1 * can used any 2 sets of expts to compare too. Rate = k[Br 2][HCOOH] (a)(ii) 6.0 x 10-6 = k (0.03) (0.02) k = 0.01 mol-1 dm3 s-1 (a)(iii) The reaction involves 1 mole of Br 2 and 1 mole of HCOOH in the rate determining step. Overall, this is a one-step reaction. (a)(iv) After mixing, [Br 2] = 0.005 mol dm-3 [HCOOH] = 0.01 mol dm-3 Rate = (0.01)(0.005)(0.01) = 5.0 x 10 -7 mol dm-3 s-1 1 (b)(i) Methanal, methanol and methanoic acid are all polar molecules and exist as simple molecular structure. Methanal has permanent dipole-permanent dipole interaction between molecules. Methanol and methanoic acid have hydrogen bonds between molecules. Since permanent dipole-permanent dipole is weaker than hydrogen bonds, less energy is needed to over the attractions between methanal. Hence, methanal has the lowest boiling point. Methanoic acid forms more extensive hydrogen bonds than methanol. = © ACJC 2008 9746/03/Prelim/08 1

Mark Scheme Thus, more energy is needed to overcome the attractions between methanoic acid. Hence, boiling point of methanoic acid is higher than that of methanol. [Generally, compare: *difference between methanal and ‘methanol and methanoic acid’ *difference between methanol and methanoic acid] (b)(ii) In benzene solution, methanoic acid dimerises through hydrogen bonds. CH3 C O OH CH3C O HO hydrogen bond 1 (c)(i) (c)(i) con’t (1) 2,4-DNPH, at room temp - methanal ⇒ orange ppt - methanoic acid ⇒ no orange ppt HCHO + NO2 NHNH2 NO2 NO2 NHN NO2 CH2 +H 2O (2) Na2CO3(aq), room temp - methanal ⇒ no CO2(g) - methanoic acid ⇒ C O 2(g) evolved which forms white ppt with Ca(OH)2 (aq) 2 HCOOH + Na2CO3 (aq) 2 HCOONa + CO2 + H2O (3) Na(s), room temp - methanal ⇒ no H2(g) - methanoic acid ⇒ H 2(g) which gives a ‘pop’ sound with lighted splinter HCOOH + Na (s) HCOO-Na+ + ½ H2 (g) (c)(ii) (1) KMnO4/H+, heat - methanoic acid ⇒ decolourise purple KMnO 4 and give CO 2(g) which forms white ppt with limewater. - ethanoic acid ⇒ no decolourisation of purple KMnO4 HCOOH + [O] CO2 + H2O (2) aq. silver nitrate in excess ammonia solution, warm © ACJC 2008 9746/03/Prelim/08 2

Mark Scheme - methanoic acid ⇒ silver mirror - ethanoic acid ⇒ no silver mirror HCOOH + 2[Ag(NH3)2]+ + 2 OH- (NH4)2CO3 + 2 Ag + 2 NH3 + H2O *Reagent and conditions [1] *Observations(+ve and –ve) [1] *Balanced equations [1] 2 (a)(i) Standard electrode potential is defined as the potential difference between a standard hydrogen electrode and an electrode immersed in a solution containing ions at 1 mol dm -3 concentration at 25oC and 1 atmospheric pressure (a)(ii) Eθ cell = +0.77 – 0.34 = +0.43 V (a)(iii) Cu → Cu2+ + 2e Fe3+ + e → Fe2+ Cu + 2Fe3+ → Cu2+ + 2Fe2+ (reversible arrow-no marks) (a)(iv) [Fe3+] decreases [Fe2+] increases (a)(v) Cu → Cu2+ + 2e Q = I x t = 30 X 1 = 30 C Q =nFe 30 x 1 = mass/Mr x 96500 x2 mass of Cu dissolves = 9.87 x 10-3 g (a)(vi) Either :The direction of electron flow will be reversed or the direction of current flow will be reversed or voltage increase (b) Equilibrium considerations for mention of shift of POE (Using Le Chatelier’s principle to predict) suggest the use of a low temperature as the forward exothermic reaction liberates energy for the synthesis of ammonia Whereas a high temperature is favoured for the endothermic reaction in the synthesis of nitrogen monoxide. A catalyst is used to speed up the reaction in the synthesis of ammonia as too high temperature is not favoured 2 (c)(i) Step II : KCN in ethanol, reflux Step III : dilute acid or H+(aq), boil under reflux (c)(ii) Fe in the presence of excess Cl 2 gives FeCl3 and Cl2 (not necessary) FeCl3 + Cl2 → Cl+(FeCl4)- (not necessary) Electrophilic substitution of methyl benzene with Cl+ © ACJC 2008 9746/03/Prelim/08 3

Mark Scheme C H3 H Cl + C H3 Cl C H3 + H +slow fast Intermediate Cl + Arrows [1] Intermediate [1] (c)(iii) P = C6H5CH2COOH (c)(iv) Step III : acidic hydrolysis Step IV : reduction 3 (a) Hydrogen Halides HCl HBr HI Observations No observable reactions. White fumes of HCl remains. Reddish brown fumes of bromine observed. OR Reddish brown liquid of bromine observed. Violet or purple fumes of iodine observed. OR Black deposits of iodine observed. ** [½] for 1-2 correct observations of 3 Equations No applicable See below See below HBr 2HBr(g) + H2SO4(l) → 2H2O(l) + SO2(g) + Br2(g) HI 2HI(g) + H2SO4(l) → 2H2O(l) + SO2(g) + I2(g) 8HI(g) + H2SO4(l) → 4I2(g) + H2S(g) + 4H2O(l) 6HI(g) + H2SO4(l) → 3I2(g) + S(s) + 4H2O(l) **[1] for 1 correct equations of 2 **[½] for missing state symbols **[0] for equations that are not balanced 3 (b)(i) & (ii) The breakdown of marks are as follows: • Max of 1m for description • Max of 1m for balanced equation Any 1 equation is acceptable. © ACJC 2008 9746/03/Prelim/08 4

Mark Scheme If state symbols are missing or wrong, [ 21 ] for each balanced equation. 3 (b)(i) Cl2 (g) + 2I- (aq) → 2Cl- (aq) + I2 (s) I2 (s) + I- (aq) ⇌ I3 - (aq) ---- if not given, it is still acceptable. The iodine that is precipitated gives dark blue colouration with starch solution. If equation is not provided, observations like chlorine displaces iodide are acceptable. 3 (b)(ii) If descriptions are missing, award [2m] for the balanced disproportionation equations. If equations are not provided, award max of 1 m for description like • Disproportionation reactions occur at different temperatures • Disappearance of chlorine smell (b)(ii) With cold aqueous alkali, Cl2(g) + 2OH-(aq) → Cl-(aq) + ClO-(aq) + H2O(l) With hot aqueous alkali, 3Cl2(g) + 6OH-(aq) → 5Cl-(aq) + ClO3 -(aq) + 3H2O(l) OR 3ClO-(aq) → 2Cl-(aq) + ClO3 -(aq) (c)(i) Identification+deductions=Max 3 m Element X : Magnesium Element Y: Sulphur Formula of the oxide of Element Y: SO3 6.99g of white ppt correspond to the mass of 0.03 mol of BaSO4 Upon addition of BaCl2, white ppt formed is BaSO4 ƒ MgCl 2 hydrolyses slightly to form a weakly acidic solution of pH ~ 6.5 Balanced chemical equations MAX: 4 × 1m = 4 m) [1] for each of any 4 correct equations from equations (i)-(vi) MgO(s) + H2O(l) → Mg(OH)2(aq)-----------(i) MgCl 2(s) + 6H2O → [Mg(H20)6]2+ (aq) + 2Cl– (aq)--------(ii) [Mg(H2O)6]2+ (aq) ⇌ [Mg(H2O)5(OH)]+ (aq) + H+(aq)--------(iii) OR [Mg(H2O)6]2+ (aq) + H2O(l) ⇌[Mg(H2O)5(OH)]+ (aq)+ H3O+(aq)-------(iii) © ACJC 2008 9746/03/Prelim/08 5

Mark Scheme SO3(g) + H2O(l) → H2SO4(aq)-------(iv) H2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2HCl(aq)--------(v) H 2SO4(aq) + MgO(aq) → MgSO4(s) + H2O(l)-------(vi) 3 (c)(ii) Mg (Element X) has a higher melting point than S8 (Element Y) During melting, more energy is required to overcome the strong metallic bonds