RI H2 Chem 2013 Prelim P3 Soln

-1- Raffles Institution 2 Oct 2013 H2 Chemistry Prelim Exam 2013 – Paper 3 (Suggested Answers) Question 1 (a) The reaction involves autocatalysis where CH 3CH2COOH (or H + from the dissociation of CH3CH2COOH) is the autocatalyst. The rate of reaction gradually increases at first as the autocatalyst pr oduced catalyses the reaction. However, the rate decreases subsequently as the [CH 3CH2COOCH2CH3] decreases. (b) (i) Plotted the graph for Experiment 2.  smooth curve with 6 correctly plotted points (ii) To find the order of reaction with respect to the ester. Consider the graph for experiment 1. 1st t1/2 = 31.0 min (with construction lines shown on the graph) 2nd t1/2 = 32.0 min (with construction lines shown on the graph) Since t1/2 is constant, order of reaction with respect to ethyl propanoate = 1 Alternative answer Can also use the graph for experiment 2 to find two t1/2 values. 1st t1/2 = 63.0 min (with construction lines shown on the graph) 2nd t1/2 = 63.0 min (with construction lines shown on the graph) Since t1/2 is constant, order of reaction with respect to ethyl propanoate = 1 To find the order of reaction with respect to NaOH. For experiment 1, initial rate = 0.010/24.0 = 4.17 x 10–4 mol dm–3 min–1 For experiment 2, initial rate = 0.010/48.0 = 2.08 x 10–4 mol dm–3 min–1 When [NaOH] is doubled, initial rate is also doubled (4.17x10-4/2.08x10-4 = 2). Hence order of reaction with respect to NaOH = 1 Alternative answer Consider the graph for experiment 2. t1/2 = 63.0 min  t1/2 for experiment 2 = (2)(t1/2 for experiment 1)  reaction rate is halved when [NaOH] is halved. Hence order of reaction with respect to NaOH = 1 (iii) Rate = k [CH3CH2CO2CH2CH3] [NaOH] (iv) For experiment 1, initial rate = 0.010/24.0 = 4.17 x 10–4 mol dm–3 min–1 initial [NaOH] = 2.0 mol dm–3 initial [CH3CH2CO2CH2CH3] = 0.0200 mol dm–3 Hence k = 4.17 x 10–4/[(2.0)(0.0200)] = 0.0104 mol–1 dm3 min–1 (v) Refer to the graph. Sketch with correct co-ordinates and 2 half-lives. Half-life = 31 to 32 min 226

-2- Question 1 (b) (vi) (c) (i) The conjugate base of the ester is stabilised as the negative charge can be dispersed by delocalisation into the adjacent C=O group. Note: Accept all sensible answers (ii) (1) The conjugate base of ethane is unstable as the electron-donating methyl group will intensify the negative charge. Note: Accept all other sensible answers. (ii) (2) The –COOH group of ethanoic acid is more acidic than the -hydrogen atom. Note: Accept all other sensible answers. (iii) O OCH2CH3 O O OCH2CH3 O O OCH2CH3 O O OCH2CH3 O (iv) OCH3CH3O O O 227

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