MI H2 Chem 2012 Prelim P3 Soln

- 2 - Answer ANY FOUR quest ions 1 (a) When studying Hydrocarbons, we recognised the environmental consequences of carbon monoxide, oxides of nitrogen and unburnt hydrocarbons arising from the internal combustion engine. Because of these environmental concerns, alongside with high oil prices and oil being a limited natural resource, development of cleaner alternative fuels and advanced power systems for vehicles has become a high priority for many governments and vehicle manufacturers around the world. An alternative fuel vehicle is a vehicle that runs on a fuel othe r than "traditional" petroleum fuels (petrol or diesel); and also refers to any technology of powering an engine that does not involve solely petroleum (e.g. electric car, hybrid electric vehicles, solar powered). One such source of alternative fuel is the hydrogen / oxygen fuel cell, which is increasingly used in space crafts. Describe, providing details, the hydrogen / oxygen fuel cell. Include the following in your answer: · Draw a well-labeled diagram for the cell. · Outline the reactions taking place at each electrode, assuming an alkaline medium. · State one advantage of using this fuel cell. [6] ( 1 mark for correct electrolyte, 1 mark for labelled anode, 1 mark for labelled cathode) In alkaline electrolyte Reaction at anode: H 2(g) + 2OH–(aq) ® 2H2O(l) + 2e Reaction at cathode: ½O2(g) + H2O(l) + 2e ® 2OH–(aq) O verall cell reaction: H 2(g) + ½O2(g) ® H2O(l) ;;; ; ;

- 3 - Advantage: Clean products are produced. OR Water produced in hydrogen-oxygen fuel cell used in spacecraft can be used for drinking and washing. ; (b) Cyanogen is a highly toxic gas, composing of 46.2% carbon and 53.8% nitrogen by mass. At 25 ºC and 1 atm, 1.05 g of cyanogen occupies 0.500 dm3. (i) Determine the molecular formula of cyanogen. C N % mass 46.2 53.8 Mole ratio 3.85 3.84 Simple ratio 1 1 E mpirical formula of cyanogen is CN. Using PV = nRT 1.01 x 105 x 0.5 x 10-3 = (1.05 / Mr) x 8.31 x 298 Mr = 51.5 (1 d.p) Thus, molecular formula of Cyanogen = C2N2 ; ; ; (ii) Draw the dot-and-cross diagram for cyanogen. Hence, state its shape. N≡C-C≡N , linear ; ; (iii) ‘Cyanogen is soluble in ethanol.’ Explain this statement with the aid of a suitable diagram. [7] Cyanogen can form hydrogen bonds with ethanol as it has a lone pair of electrons on nitrogen, thus soluble in ethanol. (lone pairs on N and O, +, - must all be clearly shown) ; ; (c) Sketch and explain the trend observed for the atomic and ionic radii of the elements (from Na to Cl) in Period 3. [4] ;

- 4 - Atomic Radius: Across the period, nuclear charge ↑ as proton number ↑ change in the screening effect is negligible (Same no. of inner shells of e-s across the period). Thus the outer e-s are more strongly attracted by the nucleus \ atomic radii ↓ Cations (Na+ to Si4+) cations have one shell less than neutral atoms. Thus, the outer e-s are more strongly attracted by the nucleus therefore having a smaller radius than its parent atoms. Anions (P 3- to Cl-) · anions have more e-s than protons and so, the effective attractive force on the outer e-s is less than that in neutral atoms. Þ the outer e-s are less strongly attracted by the nucleus, thus having a radius that is larger than its parent atoms. ; ; ; (d) Phosphine, PH3, a gas at room temperature can be prepared by action of sodium hydroxide on phosphonium iodide, PH 4I. When 1.00 g of phosphonium iodide reacted with solid sodium hydroxide, 0.925 g of white solid was formed, together with steam and 150 cm3 of PH3 gas. All measurements were taken at room temperature and pressure. Identify the white solid and use the information given to write a balanced equation with state symbols, for the preparation of PH3. [3] The white solid is NaI. Molar mass of PH4I = 162.0 Amount of PH4I = 1 / 162 = 0.00617 mol Amount of Na I = 0.925 / (23.0 + 127.0) = 0.00617 mol Amount of PH3 = 150 / 24000 = 0.00625 mol Mole ratio of PH4 I : PH3 : Na I ≈ 1 : 1 : 1 Thus, PH4 I (s) + NaOH (s) → Na I (s) + PH3 (g) + H2O (g) (state symbols must be correct to earn this mark) ; ; ; [Total: 20 marks]

- 5 - 2 (a) Halogen derivatives can be used to synthesize alcohols. The overall reaction is shown below: CH3CH2Br + NaOH → CH3CH2OH + NaBr State the type of reaction mechanism for the above reaction and illustrate how the reaction proceeds via the mechanism you stated. [3] Nucleophilic Substitution, SN2 1 mark for name of mechanism SN2 1 mark for clearly shown arrow pushing 1 mark for correct intermediate and the final product must have the structure of an ‘inverted umbrella’ (b) Describe the reactions of chloride, bromide and iodide ions with the following reagents: I. Aqueous silver nitrate, followed by aqueous ammonia II. Concentrated sulfuric acid You are required to write equations where appropriate and give explanations for the differences in their reactions. [7] Ppt Colour Reaction with NH3 (aq) AgCl White AgCl(s) + 2NH3 (aq) → [Ag (NH3)2]+ (aq) +Cl- (aq) diamine silver (I) ion White precipitate readily dissolves in NH 3 (aq) to give a colourless solution, diamine silver (I) ion. AgBr Pale yellow/ Cream AgBr (s) + 2NH3 (aq) → [Ag (NH3)2]+ (aq) + Br- (aq) diamine silver (I) ion Cream precipitate ONLY dissolves in Conc. NH3 solution. AgI Deep yellow Precipitate insoluble in NH3 (aq) ; ; ; · · C Br H H3C H d+ d- OH slow w C H H CH 3 BrHO d+ d-d- C HO H CH3 H + Br- -

- 6 - NaCl + H2SO4 → HCl + NaHSO4 NaBr + H2SO4 → HBr + NaHSO4 2HBr + H2SO4 → Br2 + SO2 + 2H2O NaI + H2SO4 → HI + NaHSO4 6HI + H2SO4 → 3I2 + S + 4H2O OR 8HI + H2SO4 → 4I2 + H2S + 4H2O The ease of oxidation of halide ions, X - → X 2, increases from C l- to I- (as the reducing power of the halides increases from C l- to I-). Hence, I - is readily oxidised by conc. sulphuric acid to I 2, Br- is oxidised to Br 2 to a lesser extent and C l- is not oxidised at all. (Students can also quote E values to substantiate their answers) ; ; ; ; (c) When a primary aromatic amine is treated with nitrous acid in a cool solution, the product is unstable compound, known as a diazonium salt. NH2 N2 + Cl - One r eaction the diazonium cation undergoes is the substitution of halides. The reaction is shown below, where X r epresents the halogen. C6H5N2+ + KX → C6H5X + K+ + N2 diazonium ion Compound P can be synthesised from benzoic acid in the reaction shown below. Suggest a synthetic route for the conversion of benzoic acid to compound P. In each case, identify all the intermediate compounds and state clearly the reagents and conditions used for each transformation. [*Note: In your proposed synthesis route, two of the stages of the synthesis requires the formation of the diazonium ion and the substitution of the halide.] [5 ] + HNO2 + HCl → diazonium salt 5 ºC + H2O

- 7 - (1 mark for every underlined answer) (d) Arrange the following halogen containing compounds according to increasing pKa values. Explain your answer. CH3CHClCOOH , CH3CCl2COOH , CH3CHBrCOOH [3] Increasing pKa value: CH3CCl2COOH , CH3CHClCOOH , CH3CHBrCOOH CH3CCl2COOH has the smallest pKa (thus most acidic) as there is the presence of two electron withdrawing Cl atoms. This, the negative charge on the O atom in (CH3CCl2COO-) is more dispersed than in (CH3CHClCOO-). Thus the CH3CCl2COO- anion is more stable, and the acid is more willing to donate a proton, increasing the acidity of the solution. CH 3CHBrCOOH has the largest pKa value (least acidic) as Br is less electronegative than Cl. Thus, the electron withdrawing ability of Br is less than Cl. Thus the CH3CHBrCOO- anion is less stable, and the acid is less willing to donate a proton, decreasing the acidity of the solution. (no marks for correct arrangement without explanation) ; ; ; (e)