NYJC H2 Chem 2013 Prelim P2 Soln

1 H2 Chemistry 9647/02 NYJC J2/2013 Prelim [Turn Over For Examiner's Use Answer all questions in the spaces provided. 1 Planning (P) Acids can be classified in terms of t heir basicity depending on the number of protons that can be donated to a base. A st udent is investigating the basicity of a solution of 0.80 mol dm-3 acid, A, using just the acid and sodium hydroxide of the same concentration. He conducted two ex periments by mixing different volumes of A and sodium hydroxide, and taking the temperature rise for each experiment. (a) Based on the definition of the enthalpy change of neutralisation (∆Hn) and his results, deduce the basicity of acid A. Experiment Volume of acid Volume of NaOH ∆T /oC 1 40 20 3.6 2 20 40 7.2 If it is a dibasic acid, n(H 2O) formed in experiment 2 is twice that of experiment 1. Since total volume is constant, ∆T for experiment 2 is doubled. Hence it is dibasic. If it is a m onobasic acid, n(H 2O) formed in expt 1 = n(H 2O) formed in expt 2 Since total volume is constant, ∆T must be the same for both experiments. (b) (i) The temperature rise of the reaction can be det ermined graphically by measuring how the temperature changes with time.   You are to plan an experiment to determine the ∆Hn between a solid sample of A and sodium hydroxide. You are provided with a solid sample of A (Mr = 174) and sodium hydroxide solution of 0.80 mol dm-3. In your plan, you should give:  details including calculations to determine the quantities of the reactants to use;  choice of apparatus (You may use apparatus normally found in the school laboratory.);  the essential details for obtaining t he graph of temperature against time in order to determine the temperature rise;  table(s) of readings to taken. 51

Calculations: Volume of NaOH to use = 40 cm3 Amount of acid used in experiment 2 = 0.016 mol Mass of acid used in experiment 2 = 0.016 x 174 = 2.784 g To ensure acid is limiting here, we use 2.5 g. Procedure: 1. Using a weighing balance, we igh accurately about 2.5 g of A. 2. Using a 50 cm 3 measuring cylinder, measure 50 cm 3 of NaOH and add into a Styrofoam cup and cover with a lid. 3. Using a 0.2 oC division thermometer, note the tem perature of the solution at 1 min interval for the first 3 minutes. 4. At 3.5 min, rapidly add sample A into the cup, close the lid. 5. Stir the mixture. 6. Measure the temperature of the mixture at the 4 th minute, and at every 1 minute interval until the 10th minute. Apparatus: weighing balance, 50 cm 3 measuring cylinder, Styrofoam cup, 0.2 oC division thermometer. (2 items missing, 0) Timing: take initial temp for at least 3 min, time intervals (at most 1 min), read minimum of 5 readings after mixing. (2 items missing, 0) General procedure: rapidly pour, stir, close lid. (2 missing, 0) Quantities: vol. of NaOH, mass of X (no penalty since calculation was done) Table of readings: Mass of X and weighing bottle /g Mass of weighing bottle and residue /g Mass of empty weighing bottle /g Mass of X used /g Time /min Temperature / oC 52

3 H2 Chemistry 9647/02 NYJC J2/2013 Prelim [Turn Over For Examiner's Use (ii) Sketch the graph that you would ex pect, showing clearly how you can determine the temperature rise. ∆T 3.5 (or point of mixing) [8] (c) Explain why temperature rise in the experiment should not be too low. Too low, result in high percentage error. [1] (d) An alternative method to determine the temp erature rise is to measure the highest temperature reached. Explain if this is a better method. Poorer method as it does not take into account heat loss to the surroundings. [1] [Total: 12] T/ °C Time / s 53

2 One of the most impo rtant features of the transition elements is that they exhibit variable oxidation states. This question illustrates the various oxidation states shown by iron in its compounds. (a) The reaction between iodide ions, I -, and peroxidisulfate ions, S2O8 2-, is slow. The reaction can be catalysed by adding a small amount of Fe2+ ions. The initial rate of the slow reaction between iodide ions and peroxidisulfate ions can be studied by using thiosulfate ions. The equations for the reactions are as follows. 2I - + S2O8 2- I2 + 2SO4 2- (slow) reaction I I2 + 2S2O3 2- 2I- + S4O6 2- (fast) reaction II In the presence of a cons tant amount of th iosulfate ions, the iodine being slowly produced by reaction l will immediately react in reaction ll until all the thiosulfate ions has been used up. At that point, free iodine will be present in the solution, which will cause a sudden appearance of a dee p blue colour if starch is present. A series of experiments was carried out using different volumes of the five reagents. The following results were obtained. Expt Volume of S 2O8 2- /cm3 Volume of I- /cm3 Volume of S2O3 2- /cm3 Volume of distilled water /cm3 Volume of Starch /cm3 Time for the appearance of deep blue colour/s 1 20 20 10 5 5 30 2 20 15 10 10 5 40 3 5 25 10 15 5 t 3 4 10 15 10 20 5 80 (i) If the orders of reaction with respect to peroxidisulfate ions and iodide ions are both one respectively, deduce an expre ssion relating the volume of these two reactants and time taken for the appearance of deep blue colour. Explain your reasoning. Rate = k [S 2O8 2-][ I-]  Since total volume of solution is constant, concentration of a reactant  volume of reactant  rate  1/time 1/time = k (volume of S 2O8 2-)(volume of I-) (volume of S 2O8 2-)(volume of I-) x time = constant 54

5 H2 Chemistry 9647/02 NYJC J2/2013 Prelim [Turn Over For Examiner's Use (ii) Hence, predict the time, t3, required for the appearance of deep blue colour in experiment 3. time for expt 3, t 3 = 96 s [3] (b) (i) In the Haber process, the rate of forma tion of ammonia is increased by using an iron catalyst. State the type of catalysis occurring here and describe using the Maxwell-Boltzmann Distribution curve, how the iron catalyst increases the rate of reaction. Heterogeneous catalysis Boltzmann Distribution Diagram: In the presence of a catalyst,  a reaction has a different mechanism with a lower activation energy compared to the uncatalysed reaction.  More molecules will possess energy greater than this lowered activation energy,  hence frequency of effective collisions will increase. Hence, rate of reaction increases. (ii) Ammonia shows significant deviations from the ideal behaviour that is predicted by the kinetic theory of gases. State two assumptions of the kinetic theory of gases. 2 assumptions for ideal gas behaviour are: - Particles in gaseous stat e do not exert any force or negligible forces of attraction; - volume of particles is negligibly sma ll compared with that of the container. Key: No of molecules with energy ≥ Ea No of molecules with energy ≥ Ea c (catalysed) 0 No of molecules Eac Catalysed rxn Ea uncatalysed rxn Energy 55

(iii) A sample of ammonia can be liquefied at room temperature just by increasing the pressure of the sample. Why does the application of pressure cause the gas to liquefy? At high pressure, volume of gas decreases, and the particles are much closer to one another. They are able to form more significant intermolecular forces (hydrogen bond