HCI H2 Chem 2013 Prelim P2 Soln

2013 C2 Chemistry Prelim Paper 2 – Answers 1 (a) Diagram: [1] Example of procedure: [5] 1. Use a burette to transfer _(volume)_ cm3 of __(H2SO4)__ into a styrofoam cup. (Place the cup in a 250 cm3 beaker to prevent it tipping over.) Record the initial temperature of H 2SO4 solution using a thermometer of precision (± 0.1 oC). 2. From a second burette, measure _(volume)_ cm3 of __(NaOH)__ into another styrofoam cup. Record the initial temperature of NaOH solution. 3. Add H 2SO4 solution to NaOH solution (or vice versa). Use the thermometer to stir the mixture gently. 4. Record the highest temperature reached. 5. Repeat steps 1 to 4 using the volumes in the table below. 6. The temperature change, ΔT = Tfinal – Tinitial, where Tinitial is the weighted average by volume. Experiment Volume of H2SO4 / cm3 Volume of NaOH / cm3 Initial temperature of H2SO4 / oC Initial temperature of NaOH / oC Highest temperature / oC 1 5.00 45.00 (given data) 10.00 40.00 11.0 2 15.00 35.00 (given data) 20.00 30.00 18.4 3 30.00 20.00 4 35.00 15.00 5 40.00 10.00 Sketch : [1] thermometer lid styrofoam cup calorimeter (250 cm3) beaker H2SO4 (or NaOH or mixture) 639

Calculations: [3] (i) Let volume of NaOH at ΔTmax (from the graph) = a cm3 Then OH– = 1000 a x 2.00 mol and H2SO4 = ½ x 1000 a x 2.00 = 1000 a mol Therefore, [H2SO4] = v n = 1000 )50( 1000 a a  = )50( a a  mol dm–3 (ii) Heat change for the reaction Q = m x c x ΔTmax J where m = 50 g, c = specific heat capacity of the final solution in J g–1 K–1, ΔTmax is read from graph (in K or oC) ΔHneut = − n Q J mol–1 where n = no of mol of H2O formed = 2 x 1000 a Therefore, ΔHneut = − max 1000 2 1000 m c T a   = − a T c m 2 max kJ mol–1 (b) The results will be the same for the concentration of sulfuric acid (because the neutralization reaction occurs according to stoichiometry / the same no of moles of OH– and H+ is present. However, the molar enthalpy change of neutralization will be different because the precipitation of barium sulfate occurs with an enthalpy change / is exothermic. [2] 2 (a) (i) no. of moles of AgNO3 = 1.12 x 10–2 no. of moles of Cl– = 1.12 x 10–2 no. of moles of A = 1.00/268.4 = 3.73 x 10–3 no. of moles of free Cl– ions per mole of A = (1.12 x 10–2)/( 3.73 x 10–3) = 3 [2] 0 5 10 15 20 0 10 20 30 40 50 ΔTmax ΔT / oC vol of NaOH / cm3 640

(ii) [1] (iii) Complex ion in B: [Co(NH3)5Cl]2+ [1] (iv) The different ligands in A and B cause the splitting of the d–orbitals to be different /energy gap between the splitted d –orbitals is different/ both absorb wavelength of different frequency, thus showing different complement colours. [1] (b) (i) Pink Co(II) (aq) is oxidised to green Co( III) (aq) by aqueous hydrogen peroxide , while hydrogen peroxide is reduced to water. After a while, the green Co(III) (aq) is reduced back to pink Co(II) (aq) by hydrogen peroxide while hydrogen peroxide is oxidised to oxygen gas. [2] (ii) Role: Co(II) (aq) is acting as a homogeneous catalyst. Evidence:  There is an intermediate Co(III) formed  The rate of reaction speeds up  Co(II) (aq) is regenerated [3] (iii) 2 H2O2  O2 + 2H2O [1] (iv) Co3+ + e Co2+ +1.82V O2 + 4H+ + 4e 2H2O +1.23V Eo cell = +1.82 – (+1.23) = + 0.59 V Co3+ reduced easily in the presence of water to Co2+. Tartaric acid stabilises Co3+/ act as a complexing agent / act as a ligand. [3] (c) (i) [1] Leaf coated with graphite Co(II) sulphate (aq) Pt electrode 641

(ii) Cathode: Co2+(aq) + 2e  Co(s) Anode: 2H2O(l)  O2(g) + 4H+(aq) + 4e [2] (iii) No. of moles of Co coated = (8.90 x 0.2 x 10–1 x 10) / 58.9 = 0.0302 No. of moles of O2 = (0.0302) / 2 = 0.0151 Volume of gas = 0.0151 x 24 = 0.363 dm3 (or 363 cm3) [2] (iv) No of moles of Co to be coated on the leaf = 0.0302 mol No of moles of electrons to be transferred = 0.0302  2 = 0.0604 mol Q = It = nF 0.5  t = 0.0604  96500 t = 11657.2s = 3.24 hr [1] 3 (a) (i) Sodium. At 600oC, sodium is still a liquid. [1] (ii)  Charge density of Mg2+ is greater than Na+  Number of delocalized electrons in magnesium is greater than sodium  More energy is required to overcome the stronger metallic bonding (attraction between Mg2+ cations with its sea of delocalized electrons), hence melting point of magnesium is higher than sodium. [2] (ii) [1] – correct shape of curve (first gap is larger than the second gap) No. of electrons removed × × × × × × × × × × Energy/ kJ mol–1 642

(b) (i)  The activation energy of a catalysed (Ea cat) reaction is lower than that of an uncatalysed (Eauncat) reaction.  The catalyst provides an alternate pathway with lower activation energy.  This increases the proportion of molecules with KE greater than or equal to Ea and increases frequency of effective collision and hence increases rate of reaction. [3] (ii)  CC consists of 1 bond and 2  bonds  The bond is formed from a head on overlap of the sp hybridised orbitals  The  bonds are formed from a side on overlap of the p orbitals. If a diagram is drawn, diagram must be labeled to show the orbitals involved in each overlap. [2] (c) (i) The boron atom is electron deficient and is able to accept an electron pair from the CC bond. [1] (ii) [2] Kinetic Energy Number of particles EA cat. EA 643

(iii) [1] 4 (a) (i) The peptide bond is not formed using the –CO2H on the -carbon. OR The peptide bond is formed using the side chain acid group of the N–terminal amino acid. [1] (ii) [1] for each correct structure (b) Ile-His-Cys-Pro-Gly-Val-Leu-Pro-Val-Lys-Val [1] (c) (i) At low pH, N / NH becomes NH+ / NH2 +. This disrupts the hydrogen bonds in the tertiary structure, and hence leads to denaturation. [2] (ii) In aqueous media, histidine exists as a zwitterion. Thus it is able to form ion-dipole interactions with the water molecules, hence allowing it to dissolve well in water. [2] (iii) The sample of histidine extracted from cartilage protein contains only one enantiomer. The sample obtained from laboratory synthesis is a racemic mixture OR contains both enantiomers in equal amounts. [2] (iv) [1] 5 (a) (i) Bond angle strain as the P-P-P bond angle (60o) is more acute that the ideal 107o. [1] (ii) P4 (s) + 5O2 (g)  P4O10 (s) [1] (b) (i) P4 (l) + 6H2O (g) + 5O2 (g)  4H3PO4 (l) / (g) [1] (ii) Mass of H3PO4 in 500 cm3 of Coke = 0.05 500 1100  = 0.25 g 644

H3PO4 = 0.25 98.0 = 2.55 x 10-3 mol [H3PO4] = 32.55 10 5001000  = 5.10 x 10-3 mol dm-3 Ka1 = 2+ 24 34 H PO H H PO         7.50 x 10-3 = 2+ -3 + H 5.10 10 H    [H+] = 3.48 x 10-3 mol dm-3 pH = 2.46 [3] (c) PCl5 has a large electron cloud/ has a large number of electrons and hence dispersion forces exist between molecules. PCl5 is large molecule and hence its volume is not negligible compared to the volume of the container. Collisions between PCl5 molecules are inelastic as energy is expended to overcome the dispersion forces of attraction. [3] (d) (i) [3] (ii) Either C or D may undergo nucleophilic substi