DHS Prelim_H2_CHEM_P2_ans

© DHS 2016 9647/02 [Turn over Answer all questions in the spaces provided. 1 (a) Repulsion between two negatively charged ions causes the activation energy to be high. [1] (b) Preparation of reaction mixture 1. Using a 50.0 cm 3 measuring cylinder, add 25.0 cm 3 of peroxodisulfate ions into a 250 cm3 conical flask. 2. Using a 10.0 cm 3 measuring cylinder, add 2.0 cm 3 of aqueous copper ( II) ions into the same conical flask. 3. Using a 50.0 cm3 measuring cylinder, add 45.5 cm 3 of deionised water into the conical flask. 4. Using a 10.0 cm 3 measuring cylinder, measure out 2.5 cm 3 of ethanedioate ions. 5. Transfer the ethanedioate ions into the conical flask and insert a rubber bung into the conical f lask. The rubber bung has a delivery tube connected to rubber tubing with the other end of the tubing inserted into an inverted burette filled with water. 6. Start the stopwatch immediately and gently swirl the conical flask continuously. 7. Monitor the water level in the burette and stop the stopwatch when 40.00 cm3 of CO2 is produced. Record the time taken. 8. Repeat step 1 to 7 using the following volumes of reactants, copper ( II) ions and deionised water shown in experiment 2 to 4, at the same temperature. Experiment Volume of S2O82 / cm3 Volume of C2O42 / cm3 Volume of Cu2+ / cm3 Volume of water / cm3 1 25.0 2.5 2.0 45.5 2 25.0 5.0 2.0 43.0 3 2.5 25.0 2.0 45.5 4 5.0 25.0 2.0 43.0 Treatment of results 1. Relative rates (= 1/t) of expt 1 to 4 are calculated. Since total volume of mixture is constant in expt 1 to 4 , concentration of reactant is proportional to its volume used. 2. Compare the relative rates of expt 1 to 2 and 3 to 4, to find the order of reaction with respect to (w.r.t) C2O42 and S 2O82 respectively. If volume of C2O42 doubles and rate remains the same, it is zero order w.r.t C 2O42. If volume of C 2O42 doubles, and rate remains doubles, it is first order w.r.t C2O42. If volume of C 2O42 doubles and rate quadruples, it is second order w.r.t C2O42. The same applies for S2O82. 3. The rate law of reaction can then be found, rate = k[S2O82]n[ C2O42]m, where n and m are the orders of reaction w.r.t to S2O82 and C2O42 respectively. [8]

2 © DHS 2016 9647/02 [Turn over (c) 1. Repeat one of the experiments (e.g. expt 2) at two (at least) different temperatures, other than that for experiment 1 to 4 above. This can be done by immersing conical flasks in water bath maintained at different constant temperatures e.g. 40 °C and 60 °C. 2. Using the results in experiment 2 and that of the two further experiments, relative rate, followed by ln (relative rate) for each expt is calculated. 3. A graph of ln (relative rate) against T 1 is plotted. The gradient of the best fit line is then determined, where gradient = – R Ea . E a is given by gradient x R. [3] [Total: 12] 2 (a) (i) H