IJC H2 CHEM P1 worked solution

INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION 2 in preparation for General Certificate of Education Advanced Level Higher 2 CANDIDATE NAME CLASS INDEX NUMBER CHEMISTRY Paper 1 Multiple Choice Additional Materials: Data Booklet Multiple Choice Answer Sheet 9647/01 30 August 2016 1 hour READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in soft pencil. Do not use staples, paper clips, highlighters, glue or correction fluid. There are forty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the separate Answer Sheet. Read the instructions on the Answer Sheet very carefully. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet. This document consists of 16 printed pages and 0 blank page. Innova Junior College [Turn over

2  IJC 2016 Prelim 2/9647/01 [Turn over Section A For each question there are four possible answers, A, B, C, and D. Choose the one you consider to be correct. 1 Carbon disulfide, CS2, is a volatile liquid used in the production of cellophane which is used for food packaging. On combustion, CS2 is oxidised as follows. CS2(g) + 3O2(g) CO2(g) + 2SO2(g) A 20 cm3 sample of carbon disulfide vapour is ignited with 100 cm3 of oxygen. The final volume of gas after burning is treated with an excess of aqueous alkali. Which percentage of this final volume dissolves in alkali? (All volumes are measured at room temperature of pressure.) A 20% B 40% C 60% D 80% Answer: C 1 mole of CS2 reacts with 60cm3 of O2, remaining volume of O2 = 40 cm3 produces 1 mole of CO2 20 cm3 of CS2 produces 20 cm3 of CO2 and 40 cm3 of SO2 Total volume of gases at the end of reaction = 40 + 20 + 40 = 100 cm3 Both CO2 and SO2 are acidic gases which will react with NaOH Hence % of final volume dissolved in alkali 60/100 x 100% = 60% 2 Consider the following half-equations C2O42– 2CO2 + 2e– Fe2+ Fe3+ + e– MnO4– + 8H+ + 5e– Mn2+ + 4H2O What volume of 0.01 mol dm–3 potassium manganate(VII) is needed to completely oxidise 25.0 cm3 of an acidified solution of 0.01 mol dm–3 FeC2O4? A 5 cm3 B 7.5 cm3 C 10 cm3 D 15 cm3 Answer: D Combining eqns: FeC2O4 Fe3++ 2CO2 + 3e– x5

3  IJC 2016 Prelim 2/9647/01 [Turn over MnO4– + 8H+ + 5e– Mn2+ + 4H2O x3 Overall equation: 3MnO4– + 5FeC2O4 10CO2 + 5Fe3+ + 3Mn2+ + 12H2O Amount of FeC2O4= 25/100 x 0.01 = 0.00025 mol Mole ratio of FeC2O4 : MnO4– = 5 : 3 Amount of MnO4– = 3/5 x 0.00025 = 0.00015 mol Volume of MnO4– = 0.00015 / 0.01 = 0.015 dm3 = 15 cm3 3 Which electronic configuration represents an element that forms a simple ion with a charge of – 3? A 1s22s22p63s23p1 B 1s22s22p63s23p3 C 1s22s22p63s23p63d14s2 D 1s22s22p63s23p63d34s2 Answer: B A : Group III element : + 3 ion B: Group V : – 3 ion C and D are transition elements – forms positively charged ions 4 Carbon dioxide is a gas at room temperature while silicon dioxide is a solid because A carbon dioxide contains double covalent bonds while silicon dioxide contains single covalent bonds. B instantaneous dipole – induced dipole attractions are weaker than permanent dipole – permanent dipole attractions. C carbon-oxygen bonds are less polar than silicon-oxygen bonds. D van der Waals’ forces are much weaker than covalent bonds. Answer D Carbon dioxide is a simple molecules with weak can der waal’s forces of attraction between molecules while silicon dioxide is a giant covalent structure with strong covalent bonds. Thus, less amt of energy is required to overcome the wea ker VDW forces of attraction between CO2 molecules, resulting in lower boiling point.

4  IJC 2016 Prelim 2/9647/01 [Turn over 5 Under which conditions will the behaviour of a gas be most ideal? pressure temperature A high high B high low C Low high D Low low Answer: C Gas behave most ideally at high temperatures and low pressures. At high temperatures, particles have sufficient energy to overcome the intermolecular forces of attraction. At low pressures, particles are very far apart. The volume of the particles is neg ligible to the volume of container which they moves in; intermolecular forces of attraction is negligible. 6 The enthalpy changes involving some oxides of nitrogen are given below: N2(g) + O2(g) → 2NO(g) ∆H = +180 kJ mol–1 2NO2(g) + 1 2O2(g) → N2O5(g) ∆H = –55 kJ mol–1 N2(g) + 5 2O2(g) → N2O5(g) ∆H = +11 kJ mol–1 What is the enthalpy change, in kJ mol–1, of the following reaction? 2NO(g) + O2(g) → 2NO2(g) A –114 B +114 C –136 D +136 Answer: A Reverse Equation 1: 2NO(g) → N2(g) + O2(g) ∆H = –180 kJ mol–1 Reverse Equation 2: N2O5(g) → 2NO2(g) + 1 2O2(g) ∆H = +55 kJ mol–1 Equation 3 remains: N2(g) + 5 2O2(g) → N2O5(g) ∆H = +11 kJ mol–1 Sum up these 3 equations: 2NO(g) + O2(g) → 2NO2(g) ∆H = –180+55+11 = –114 kJ mol–1

5  IJC 2016 Prelim 2/9647/01 [Turn over 7 A typical protein forms hundreds of hydrogen bonds and thousands of van der Waals’ force s in folding from primary to tertiary structures. Which of the following thermodynamic state functions of the protein best represents the folding process? G / kJ mol–1 H / kJ mol–1 S / J K–1 mol–1 A    B +  + C  +  D + +  Answer: A When hydrogen bonds and van der Waals’ forces are formed, heat will be released. Thus H is negative. When the protein is folded from primary to tertiary structures, the system becomes less disordered. Thus S is negative. The process of protein folding is spontaneous. Thus, G is negative. 8 When 1 mole of carbon dioxide gas solidifies as dry ice, 25.2 kJ of heat energy is evolved. The sublimation temperature of carbon dioxide is 78.5 C. What is the entropy change when 132 g of carbon dioxide gas solidifies at this temperature? A +130 J K1 B 130 J K1 C +389 J K1 D 389 J K1 Answer: D This is a phase change reaction. G = 0 G = H  TS S = 𝐻− 𝐺 𝑇 S = −25200 − 0 −78.5+273 S = −25200 194.5 S =  129.56 J K-1 S when 132 g of carbon dioxide gas solidifies =  129.56 x 132 44 =  389 J K-1

6  IJC 2016 Prelim 2/9647/01 [Turn over 9 A student set up the hydrogen electrode shown in the diagram below. What would have to be changed to make this a standard hydrogen electrode? A the acid solution used B the temperature of the gas and of the acid solution C the pressure of the gas D the metal comprising the electrode Answer:A In 1.0 mol dm –3 sulfuric acid, [H +] = 2.0 mol dm –3. To make the electrode a standard hydrogen electrode, either change the acid to a 1.0 mol dm–3 monoprotic acid (option A), or halve the concentration of sulfuric acid used. 10 Use of the Data Bo