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 differen