NYJC 2015 H1 CHEM P2 Answer Prelims

[Turn over NANYANG JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 1 CANDIDATE NAME ANSWERS CLASS TUTOR’S NAME CHEMISTRY 8872/02 Paper 2 16 September 2015 2 hours Candidates answer Section A on the Question Paper. Additional Materials: Answer Paper Data Booklet Graph Paper READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. The use of an approved scientific calculator is expected, where appropriate. Section A Answer all the questions. Section B Answer two questions on separate answer paper. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use Section A B5 B6 B7 Total This document consists of 15 printed pages.

2 H1 Chemistry 8872/02 NYJC J2/2015 Prelim Section A Answer all the questions in this section in the spaces provided. 1 (a) Predict and explain which of the following isomers has a higher solubility in water. C C H CH2OHNH2 CH3 C C H CH2OHCH3 NH2 Isomer A Isomer B [3] Isomer B is more soluble in water. Isomers A and B both have simple molecular structures with hydrogen bonding between molecules. There is intramolecular hydrogen bonding in isomer A as the –OH and –NH2 groups are close to each other (and hence, less intermolecular hydrogen bonding). More energy is required to overcome the more extensive intermolecular hydrogen bonding between isomer B. (b) (i) Draw the ‘dot-and-cross’ diagram of NH2OH. (ii) Using VSEPR theory, e xplain why the bond angle about the N atom is greater than the bond angle about the O atom. About N atom – 3 bond pairs, 1 lone pair About O atom – 2 bond pairs, 2 lone pairs Lone pair-lone pair repulsion is greater than lone pair-bond pair / bond pair-bond pair repulsion. [3] [Total: 6]

3 H1 Chemistry 8872/02 NYJC J2/2015 Prelim [Turn over 2 (a) Describe the bonding in ethene in terms of orbital overlap. You may draw a diagram to illustrate your answer. Each C atom forms 2 sigma bonds with 2 neighbouring H atoms and 1 sigma bond with the other C atom when their orbitals overlap head -on. The two C atoms forms a pi bond when their p orbitals overlap side-on above and below the plane of the molecule. C H H bond C H H bond bond bond bond (b) (i) Ethene can be converted into ethane -1,2-diol. State the reagents, conditions and type of reaction for this reaction. reagents: dilute KMnO4, H2SO4(aq) or dilute KMnO4, NaOH(aq) conditions: cold type of reaction: mild oxidation (ii) Ethene and ethane bo th react with chlorine. Suggest with particular reference to bonding, why ethane needs the presence of UV light but ethene does not. Ans: Ethene has pi electron cloud along the C=C double bond that is electron rich enough to polarise Cl2, thus increasing the reactivity of Cl2 as an electrophile to react with ethene without the presence of UV light. Ethane contains C-C and C -H bonds which are strong and non -polar. Hence, ethane is unreactive. UV light is required to cause homolytic fission of C l2 to generate a more reactive Cl● radical to react with ethane. [6] [Total: 9]

4 H1 Chemistry 8872/02 NYJC J2/2015 Prelim 3 When compound C is placed in a 2 dm3 closed container, the following equilibrium is established. The backward reaction is exothermic. 2C ⇌ D (a) (i) Write an expression for Kc for this equilibrium. c 2 [D]K [C] (ii) Calculate the value of the equilibrium constant for the reaction at the 1st minute, stating its units. At the 1st min: 10 ]2 0.4[ ]2 0.8[ K 2 c  mol1 dm3 (iii) A change was made to the system at the 2 nd and 4 th minute respectively. State the change and explain your reasoning. Change at 2nd minute: Temperature is increased. Reason: Since backward reaction is exothermic, forward reaction is endothermic. By Le Chatelier’s P rinciple, the system will react to remove the heat. The endothermic forward reaction is favoured and position of equilibrium shifts to the right. Therefore, the amount of C decreases and D increases. C D 6 0.4 Amount / mol Time / min 1 2 3 4 5 7 8 0 0.2 0.6 0.8 1.0 1.2 1.4

5 H1 Chemistry 8872/02 NYJC J2/2015 Prelim [Turn over Change at 4th minute: Compound C was added. Reason: By Le Chatelier’s Principle, when C is added, the system will react to reduce some of the added C. The forward reaction is favoured and position of equilibrium shifts to the right. Therefore, the amount of C decreases and C increases. [7] (b) 10.0 cm 3 of 0.100 mol dm 3 of ethan oic acid is titrated against 0.0300 mol dm 3 of Ba(OH) 2. The graph of pH against volume of Ba(OH) 2 added is shown: (i) Write the equation for the reaction between ethanoic acid and Ba(OH)2. 2CH3COOH + Ba(OH)2  (CH3COO)2Ba2+ + 2H2O (ii) Calculate the volume of Ba(OH)2 required at equivalence point E. n(ethanoic acid) = 10/1000 × 0.100 = 1 × 103 mol n(Ba(OH)2) = (1 × 103) / 2 = 5 × 104 mol v(Ba(OH)2) = (5 × 104) / 0.03 = 0.0167 dm3 = 16.7 cm3 (iii) Hence, calculate the pH value at point E. pH at point Y is due to excess Ba(OH)2 V(excess Ba(OH)2) = 25  16.7 = 8.33 cm3 n(OH) = 2 × n(Ba(OH)2) = 2 × 8.33/1000 × 0.03 = 5 × 104 mol [OH] = (5 × 104) / (35/1000) = 0.0143 mol dm3 pH = 14  pOH = 14  (lg 0.0143) = 12.2 [4] [Total: 11] Volume of Ba(OH)2/cm3 25 E pH

6 H1 Chemistry 8872/02 NYJC J2/2015 Prelim 4 (a) Define order of reaction. [1] The order of reaction with respect to a given re actant is the power to which the concentration of that reactant is raised in the experimentally determined rate equation. (b) The reaction kinetics of a reaction using CH 3CH2Br and excess NH 3 to synthesise an amine is determined by monitoring the change in the concentration of CH3CH2Br with time. The results are shown below. (i) Write a balanced equation for the above reaction. CH3CH2Br + NH3  CH3CH2NH2 + HBr (ii) Use the graphs to deduce the rate equation. Since t 1/2 is c onstant at 30 min (using [NH 3] = 3 .0 mol dm 3) graph, reaction is 1st order with respect to CH3CH2Br. OR t1/2 is constant at 60 min (using [NH3] = 1.5 mol dm3) graph, reaction is 1st order with respect to CH3CH2Br. Using initial rate method, Initial rate for graph where [NH3] = 1.5 mol dm3 = 0.2/9