2017 RVHS H2 Chemistry P3 Soln

River Valley High School 9729/03/PRELIM II/17 [Turn over 2017 Preliminary Examination II RIVER VALLEY HIGH SCHOOL YEAR 6 PRELIMINARY EXAMINATION II CANDIDATE NAME CLASS 6 CENTRE NUMBER S INDEX NUMBER H2 CHEMISTRY 9729/03 Paper 3 Free Response 19 September 2017 2 hours Candidates answer on separate paper. Additional Materials: Answer Paper Cover Page Data Booklet READ THESE INSTRUCTIONS FIRST Write your name, class, centre number and index number on all the work you hand in. Write in dark blue or black pen on both sides of paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Section A Answer all questions. Section B Answer one question. Begin each question on a fresh sheet of paper. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. Do not write anything on it. You are reminded of the need for good English and clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together, with the cover page on top. This document consists of XX printed pages.

2 River Valley High School 9729/03/PRELIM II/17 2017 Preliminary Examination II Section A Answer all the questions in this section. 1 (a) Under suitable conditions, SCl2 reacts with water to produce a yellow solid and an acidic solution A. Solution A contains a mixture of SO 2(aq) and another compound. (i) State the oxidation number of S in SCl2 . [1] +2 (ii) Construct an equation for the reaction between SCl2 and water. [1] 2SCl2 + 2H2O → S + SO2 + 4HCl (iii) In the Contact Process, one important step is the conversion of SO2 to SO3 as shown below. 2SO2(g) + O2(g) ⇌ 2SO3(g) A 2.00 L flask was filled with 0.0400 mol SO 2 and 0.0200 mol O 2. At equilibrium, the flask contained 0.0296 mol of SO3. Determine the value of Kc, stating its unit. [3] 2SO2 + O2 ⇌ 2SO3 I / mol 0.04 0.02 0 C / mol −0.0296 0.0148 +0.0296 E / mol 0.0104 0.0052 0.0296 Kc = [0.0296 2⁄ ]2 [0.0104 2⁄ ]2[0.0052 2⁄ ] = 3116 = 3120 mol–1 dm3

3 River Valley High School 9729/03/PRELIM II/17 [Turn over 2017 Preliminary Examination II (b) During the electrolysis of dilute sulfuric acid using a current of 0.75 A for 90 min and platinum electrodes, the volume of oxygen gas collected wa s recorded and is shown in the Table 1.1 below. Table 1.1 Time / min Volume of O2 gas / cm3 20 55 40 110 60 165 80 220 (i) Plot a graph of volume of O2 gas over time. Use x axis: 2 cm for 10 min ; y axis: 2 cm for 50 cm3 [2] See graph paper (ii) Give equations for the reactions that occur at each electrode in the electrolysis of sulfuric acid. [2] Cathode: 2H2O(l) + 2e–  H2(g) + 2OH–(aq) Anode: 2H2O(l)  4H+(aq) + O2(g) + 4e– (iii) On the same graph, draw and label a line (H2) to predict the volume of hydrogen that would be given off during the same experiment. [1] See graph paper. Each point is twice the value of graph in (ii). (iv) On the same graph, draw and label a line (O2) to predict the volume of oxygen that would be given off if a current of 0.3 A was used instead in the original experiment. [1] See graph paper. Each point is (3/7.5) the value of graph in (ii). (v) In a 2 nd experiment, the platinum electrodes were replaced with graphite electrodes. The volume of gas collected at the anode was 150 cm 3 while the volume of hydrogen gas collected was 220 cm3. The difference in volume of gas collected at the anode between the two experiments was due to production of CO gas at the anode. Calculate the volume of CO gas produced at the anode. [2] C + ½ O2 (g) CO (g) If no reaction with anode, volume of gas is 110 cm3 Let volume of O2 reacted to form CO be x 110  x + 2x = 150 x = 40 cm3

4 River Valley High School 9729/03/PRELIM II/17 2017 Preliminary Examination II VCO = 2(40) = 80 cm3 (c) About 100 years ago, in a reaction discovered by German chemist Karl Fries, compound B was converted into compound C when heated with AlCl3. Compound C is a structural isomer of B. It is insoluble in water but dissolves in aqueous sodium hydroxide. It give s a yellow ppt with alkaline aqueous iodine and a white ppt with aqueous bromine. (i) Suggest the structure for compound C. [1] Accept 1,2 and 1,3 isomers. The various reactions of compound C can be represented as follows:

5 River Valley High School 9729/03/PRELIM II/17 [Turn over 2017 Preliminary Examination II (ii) Suggest the structures for D to G. [4] CHI3 D and E are interchangeable. D E F G Accept 1,2- and 1,3 isomers for all. Compound H, as shown below, is another structural isomer of B. It has a ether functional group whose general formula is RO−R’. Compound H can be formed via a reaction between a substituted phenoxide ion and an alkyl halide molecule.

6 River Valley High School 9729/03/PRELIM II/17 2017 Preliminary Examination II (iii) Describe the mechanism when compound H is formed as described above. [3] Let be RO− [Total: 21] 2 In the late 1940s, Willard Libby developed the radiocarbon dating method for determining the age of an object containing organic material by using the properties of radiocarbon (14C), a radioactive isotope of carbon. The principle of carbon dating is as such: During its life, a plant or animal is exchanging carbon with its surroundings, so the carbon it contains will have the same proportion of 14C as the atmosphere. Once it dies, it ceases to acquire 14C, but the 14C within its biological material at that time will continue to decay, and so the ratio of 14C to 12C in its remains will gradually decrease. Because 14C decays with first order kinetics, the proportion of radiocarbon can be used to determine how long it has been since a given sample stopped exchanging carbon – the older the sample, the less 14C will be left.

7 River Valley High School 9729/03/PRELIM II/17 [Turn over 2017 Preliminary Examination II (a) A sample of carbon dioxide gas (that contained both 12CO2 and 14CO2) was analysed to determine the proportion of 14CO2 found within. Analysis results showed that there is one 14CO2 molecule for every 1012 CO2 molecules. (i) Calculate the number of 14CO2 molecules in a 10.0 dm 3 carbon dioxide gas sample, measured under s.t.p. [2] Number of moles of CO2 = 10 22.7 = 0.441 mol Number of 14CO2 molecules = 0.441 × 6.02×1023 1012 = 2.65 × 1011 molecules (ii) Calculate the mass of 14CO2 in the 10.0 dm3 sample. [1] Mass of 14CO2 = 2.65×1011 6.02×1023 × (14.0 + 16.0 × 2) = 2.03 × 10−11 g (iii) Hence, explain why it would be difficult to determine the proportion of 14CO2 by means of mass measurement. [1] The amount/mass of 14CO2 is too small to be accurately measured. (b) To more accurately determine the proportion of 14C in a sample of graphite, the graphite is vaporised and ionised to C +(g) ions. These ions were then passed through 2 electric plates. Given that H + is deflected with an angle of 8.4°, what is the angle of deflection for 14C+ ions under the same experimental set-up? [1] 14C+ deflected by ( 1 14) (8.4) = 0.60° (c) The half-life of 14C is 5730 years. Determine the time that has elapsed for a piece of wood from a dead tree to contain 30.0% of its original 14C. [2] Let the number of half-life be n 30.0 100 = (1 2) 𝑛 𝑛 = 𝑙𝑔 (30.0 100) 𝑙𝑔 (1 2) 𝑛 = 1.74 [1] Time taken = 5730  1.74 = 9970 years (d) The age of crude oil is far older than what could be determined fro