ACJC Chemistry Answerto P2 Prelims

2008 H2 Chemistry Preliminary Examinations Paper 2 Mark Scheme 1 (a) On heating, Group I metal nitrates such as sodium nitrate(V) decompose giving the metal nitrate(III) and oxygen, while Group II metal nitrates, for example magnesium nitrate(V), decompose giving different products. (i) Write balanced equations for the decomposition of sodium nitrate(V) and magnesium nitrate(V) respectively. NaNO 3 NaNO2 + ½O2 Mg(NO3)2 MgO + 2NO2 + ½O2 15.35 g of a mixture of sodium nitrate(V) and magnesium nitrate(V) was heated in a fume cupboard until no more gases were evolved. The water soluble part of the residue was dissolved in water to prepare 1.00 dm 3 of solution. 10.00 cm3 of this solution was reacted with 20.00 cm3 (in excess) of 0.0200 mol dm-3 potassium manganate(VII) solution, acidified with dilute sulphuric acid. (ii) The nitrate(III) half equation is NO 2 - + H2O NO3 - + 2H+ + 2e-. Write a balanced equation for the reaction between nitrate(III) ions and manganate(VII) ions. 2MnO 4 - + 5NO2 - + 6H+ 2Mn2+ + 5NO3 - + 3H2O The excess potassium manganate(VII) required 12.00 cm3 of 0.0500 mol dm-3 ethanedioic acid solution for complete reaction. [ 2MnO4 - + 5C2O4 2- + 16H+ 2Mn2+ + 10CO2 + 8H2O] (iii) Calculate the amount in moles of the nitrate(III) ions in the 10.00 cm3 solution. no of mols of C2O4 2- ions in 12.00 cm3 = 0.012 x 0.05 = 6.00 x10-4 no of mols of excess MnO4 - = (2/5) x 0.0006 = 2.40 x 10-4 no of mols of MnO4 - in 20.00 cm3 = 0.02 x 0.02 = 4.00 x 10-4 no of mols of MnO4 - reacted with NO2 - in 10.00 cm3 = 4 x 10-4-2.40 x 10-4 = 1.60 x 10-4 l no of mols of NO2 - in 10.00 cm3 = (5/2) x 1.60 x 10-4 = 4.00x 10-4 1

(iv) Hence, calculate the mass of each nitrate in the mixture. NaNO3≡ NaNO2 mass of NaNO3 in 1 dm3 = (1000/10) x 4.00 10-4 x (23.0+14.0+16.0x3) = 0.0400 x 85 = 3.40g mass of Mg(NO3)2 = 15.35-3.40 =12.0g ( 3s.f.) (or 11.95 g) (b) Magnesium nitrate(V) and strontium nitrate(V) decompose similarly on heating. Magnesium nitrate decomposes at a lower temperature than strontium nitrate. Explain why these two nitrates decompose at different temperatures. • Mg 2+ has a smaller ionic radius than Sr2+ • higher charge density and greater polarising power • hence Mg2+ distorts electron cloud around NO3 - to a greater extent, thus decomposing at a lower temperature. (c) Ammonium nitrate(V) decompose to produce nitrous oxide, N2O. Nitrous oxide is relatively inert at room temperature but at 500oC, it decomposes to oxygen, nitrogen and nitric oxide, NO. In the spaces provided, draw the dot and cross diagrams of these two oxides of nitrogen. Formula Nitrogen oxidation state Dot and Cross Diagram N2O +1 N NO x x x x x x x OR N N O x x x x x NO +2 N O x x x x x x N O x x x x OR 2

2 (a) The first ionisation energies of nine el ements from sodium to potassium are shown in the sketch below. Give reasons for: (i) the general trend across the period from Na to Ar. First IE increases across a period. Across a period, number of protons increases. Shielding effect remains the same as number of inner shell electrons is the same. Hence effective nuclear charge on valence electrons is stronger across a period. Atomic size is also smaller. More energy is required to remove an outermost electron due to increasing electrostatic attraction. (ii) the discontinuity between Mg and Al. Mg 1s 2 2s2 2p6 3s2 Al 1s 2 2s2 2p6 3s2 3p1 Al has a lower first IE. Less energy is required to remove a 3p electron in Al than a 3s electron in Mg since 3p electron is further away from the nucleus. (iii) the discontinuity between P and S. P 1s 2 2s2 2p6 3s2 3px 1 3py 1 3pz 1 OR 1s2 2s2 2p6 3s2 3p3 S 1s 2 2s2 2p6 3s2 3px 2 3py 1 3pz 1 1s2 2s2 2p6 3s2 3p4 S has a lower first IE. Less energy is required to remove an electron from paired 3p electrons in S since inter-electron repulsion is experienced between the paired electrons. (iv) the difference between the first ionisation energies of Na and K. Na 1s 2 2s2 2p6 3s1 K 1s 2 2s2 2p6 3s2 3p6 4s1 K has a lower first IE than Na. Less energy is required to remove a 4s electron in K than a 3s electron in Na since 4s electron is further away from the nucleus. 11 12 13 14 15 16 17 18 19 number of protons First IE / kJ mol- • 1 •• ••• •• •Na Mg Al Si P S Cl Ar K 3

(b) (i) Sketch the melting point trend of elements in period 3. (ii) Explain the difference in melting points for sodium and silicon in terms of their structures and bondings. Silicon has higher melting point than sodium. Silicon has covalent bond and giant molecular structure. Sodium has metallic bond and giant lattice structure . More energy is required to break the relatively stronger bonds in silicon than in sodium. atomic radius of silicon is smaller than atomic radius of sodium; hence silicon atoms are more closely packed together. (c) Sodium was first produced commercially in 1855 by thermal reduction of sodium carbonate with carbon in what is known as the Deville process. Na 2CO3(l) + 2C(s) → 2Na(g) + 3CO(g) The standard entropy change of reaction, ∆S r θ, is +549 J K-1 mol-1 (i) Explain why the entropy change of above reaction is positive. Change in phase: S gas >> Sliquid >> Ssolid There is a change in phase from solid or liquid reactants to gaseous products. Entropy of a gas is much greater than that of a liquid or solid as its particles are free to move and system becomes more disorderly. Entropy increases as the number of gaseous particles increases and system becomes more disorderly. (ii) Determine the range of temperatures for the above reaction to be feasible. Na 2CO3(l) CO(g) Na(g) ΔHf Ө / kJ mol-1 -1103 -111 +107 ΔHr Ө = ∑nΔHf Ө(products) − ∑mΔHf Ө(reactants) = 2 ΔHf Ө(Na) + 3 ΔHf Ө(CO) - ΔHf Ө(Na2CO3) - 2 ΔHf Ө(C) = (2)(107) + (3)(-111) – (-1103) – 0 = +984 kJ mol -1 11 12 13 14 15 16 17 18 number of protons Melting point / ºC • •• •• • • • Na Mg Al Si P S Cl Ar 4

Assume that ΔH and ΔS remain constant. For reaction to be feasible, ΔGr < 0 ΔHr - TΔSr < 0 +984,000 - (T)(549) < 0 T > 1792 K 3 (a) (i) Calcium is a fairly soft, silvery-grey metal which quickly tarnishes in air; hence metallic calcium has no commercial uses. However titanium is a commercially important engineering metal. State two physical properties which make titanium a very useful material in the aircraft industry and suggest another property that allows titanium to be used in artificial hip joints. (Low density/ light) ; (strong/high tensile strength) Does not corrode (ii) Calcium can only exist as Ca 2+ ions in its compounds but titanium forms ions with different charges (+2,+3 and +4) in its solid compounds. TiCl 3 is coloured while TiF4 ( an ionic compound) is a white powder. Unlike calcium, expla