CJC Prelim H1 Chem 2008 P2 (worked soln) (updated)

CJC H1 Prelim Exam 2008 Paper 2 Solutions CATHOLIC JUNIOR COLLEGE H1 CHEMISTRY PRELIM EXAM PAPER 2 SOLUTIONS SECTION A 1 The equation for the decomposition of gaseous X at 373 K is shown below. 2X (g) Y (g) + Z (g) The values for the initial rates of decrease in concentration of X at various initial concentrations have been determined. These are shown in the table below. Initial concentration / mol dm-3 1.67 3.34 5.01 6.68 Initial rate / mol dm-3 s-1 0.41 1.64 3.69 6.56 (a) (i) Determine order of reaction with respect to X and write down the rate equation for this reaction. By inspection, when initial concentration of X increases 2 times, from 1.67 to 3.34 mol dm-3, the initial rate of reaction increases 4 times from 0.41 to 1.64 mol dm-3 s-1. ∴ Order of reaction wrt X is 2nd order. Rate equation: rate = k [X]2 (ii) Calculate the rate constant for this reaction, stating the units clearly. Using data from 1 st expt: 0.41 = k (1.67)2 ∴k = 0.41 / (1.67)2 = 0.147 dm3 mol-1 s-1 [4] (b) (i) At 373 K, the activation energy for the forward reaction is 130 kJ mol -1 and that for the reverse reaction is 72 kJ mol -1. Sketch and label the energy profile diagram for this reaction. (ii) Hence, determine the value of the enthalpy change of the decomposition reaction. Energy Ea of forward rxn = 130 kJ mol -1 Ea of catalysed reverse rxn = 43 kJ mol-1 ∆H Ea of reverse rxn = 72 kJ mol-1 1(d) ∆Hr = 130 – 72 = +58 kJ mol-1 [3]

CJC H1 Prelim Exam 2008 Paper 2 Solutions (c) What will be the effects of increasing pressure on the decomposition of X? By Le Chatelier’s Principle, increasing pressure will favour will not shift the position of equilibrium in either direction as both forward and reverse reactions produce the same number of gaseous molecules. Hence the decomposition of X will not be favoured. However , the rate of decomposition will still increase since pressure is increased. [2] (d) In the presence of a catalyst, the activation energy of the reverse reaction is 43 kJ mol -1. On the same axes in (b)(i), draw the energy profile of the catalysed reaction and label it clearly. [1] [Total: 10] 2 2-chloroethylbenzene can be formed stepwise from benzene via the following reaction scheme. CH2CH3 CH2CH3 Cl Step I Step II (a) Step I shows how ethylbenzene is synthesised from benzene via a reaction known as Friedel-craft alkylation. The equation below represents the chemical reaction that occurs in Step I. C 6H6 (l) + CH3CH2Cl (g) → C6H5CH2CH3 (l) + HCl (g) (i) Use the following data to calculate the enthalpy change of the above Friedel-craft alkylation reaction. ∆Ho f (C6H6) = +49.0 kJ mol-1 ∆Ho f (CH3CH2Cl) = -109 kJ mol-1 ∆Ho f (C6H5CH2CH3) = -12.5 kJ mol-1