NJC Prelims Paper 2 Answers Final

NJC/H2 Chem Preliminary Examination/02/2021 1 [Turn over NATIONAL JUNIOR COLLEGE SH2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper 2 Structured Questions Candidates answer on Question Paper. Additional Materials: Data Booklet 9729/02 24 August 2021 2 hours READ THE INSTRUCTIONS FIRST Write your subject class, registration number and name on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use paper clips, highlighters, glue or correction fluid. Answers all questions. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 /8 2 /12 3 /8 4 /11 5 /17 6 /19 Paper 2 Total /75 Marks Weightings Paper 1 /30 15% Paper 2 /75 30% Overall Percentage Paper 3 /80 35% Paper 4 /55 20% Grade This document consists of 23 printed pages and 1 blank page.

NJC/H2 Chem Preliminary Examination/02/2021 2 Answer all the questions in the spaces provided. 1 (a) Iodine can undergo a reaction with aqueous potassium hydroxide to form iodate and iodide ions as shown in the following equation. 3I2(s) + 6KOH(aq) o KIO3(aq) + 5KI(aq) + 3H2O(l) This reaction has a potential application in times of nuclear catastrophe when radioactive and volatile iodine-131 is produced. Spraying alkalis into the nuclear reactors can convert the volatile iodine into non-volatile iodate and iodide ions, thus minimizing the damaging radioactive effects. (i) Name the type of reaction for the above equation. Disproportionation Note : I is simultaneously oxidized from O.S. = 0 in I2 to O.S. = +5 in IO3− and reduced from O.S. = 0 in I2 to O.S. = −1 in I− [1] (ii) Write the two balanced half-equations for the above reaction. [O] I2 + 12OH– o 2IO3– + 6 H2O + 10e– (construct by following the steps in balancing half equation) [R] I2 + 2e– o 2I– (obtained from Data Booklet) State symbols not required [2]

NJC/H2 Chem Preliminary Examination/02/2021 3 [Turn over (b) The reaction of iodide and peroxodisulfate ions is very slow and can be catalysed by using a homogeneous catalyst. S2O82– (aq) + 2I– (aq) o 2SO42– (aq) + I2 (aq) (i) Explain why the rate of the above reaction is slow. The rate of the reaction is very slow because the activation energy is very high due to the collision of two ions of the same charges. [1] (ii) With the aid of the Boltzmann distribution, explain how addition of a homogenous catalyst helps to increase the rate of a reaction. A catalyst provides an alternative reaction pathway with lower activation energy (Ea2). The fraction of particles with K.E. ≥ Ea increases as shown in the Boltzmann distribution. The frequency of effective collisions increases hence rate of reaction increases. 1m diagram 1m explanation [2]

NJC/H2 Chem Preliminary Examination/02/2021 4 (iii) By considering relevant Eꝋ values from the Data Booklet, explain how Fe3+(aq) can act as a homogenous catalyst in this reaction. Step 1: Formation of intermediate (Fe3+ colliding with I−) Eꝋ/V Fe3+(aq) + e− Fe2+(aq) +0.77 I2 (aq) + 2e− 2 I− (aq) +0.54 Overall for Step 1: 2I− (aq) + 2Fe3+(aq) o I2 (aq) + 2 Fe2+(aq) [1/2m] Eꝋcell = +0.77 −0.54 = +0.23 V [1/2m] Step 2: Regeneration of catalyst Eꝋ/V Fe3+(aq) + e− Fe2+(aq) +0.77 S2O82− (aq) + 2e− 2SO42− (aq) +2.01 Eꝋcell = +2.01 − 0.77 = +1.24 V[1/2m] Overall for Step 2: S2O82− (aq) + 2Fe2+(aq) o 2SO42− (aq) + 2Fe3+(aq) [1/2m] −1/2 m if missing/wrong state symbols. It is important to show Fe3+(aq) as a homogeneous catalyst. [2] [Total: 8]

NJC/H2 Chem Preliminary Examination/02/2021 5 [Turn over 2 (a) The Period 3 elements vary in their physical properties. (i) On the axes below, sketch the melting point and electrical conductivity trends for the stated elements. Note: Melting point trend For metals, Na to Al, mp increases because of stronger metallic bond strength between cations and sea of delocalized electrons as the charge of cations increases and number of delocalised electrons increases. Si has giant covalent lattice structure. Much energy is required to break the strong covalent bonds between Si atoms. P4, S8, Cl2 and Ar have simple molecular structure. Little energy is required to overcome the weaker id-id between molecules. As strength of id-id ∝ no. of electrons in a molecule/atom, mp decrease in order: S8 > P4 > Cl2 > Ar Electrical conductivity trend For metals, electrical conductivity increases as no. of delocalized electrons (mobile charge carriers) increases from Na to Al. Si is a metalloid which behaves as a semi-conductor and have some electrical conductivity. P4, S8, Cl2 and Ar are non-conductor of electricity as they do not have mobile charge carriers. [2]

NJC/H2 Chem Preliminary Examination/02/2021 6 (ii) Sulfur is an element in Period 3 of the Periodic Table. The graph below shows the second ionisation energies of eight elements with consecutive proton number. Which of the elements A to H represents sulfur? Explain your answer. The sharp drop in 2nd I.E. from G to H indicates that H is in Group 2 where its 2nd I.E. involves the removal of the most loosely held electron from an outer principal quantum shell as compared to that of G. OR The sharp drop of 2nd I.E. from G to H indicates that G is in Group 1 where its 2nd I.E. involves the removal of an electron from the inner quantum shell, which is much closer to the nucleus as compared to that of H. 1m explanation Hence D is a Group 16 element and it is sulfur. [1] [2]

NJC/H2 Chem Preliminary Examination/02/2021 7 [Turn over (b) A 0.400 g solid sample of a mineral, XY(CO3)2 (where X and Y are Group 2 elements) was heated strongly to give a mixture of oxides of X and Y and carbon dioxide. The solid mixture has a total mass of 0.275 g. The solid mixture was added to excess water and stirred. The suspension was filtered, and the oxide of X was obtained as a residue. The dried residue weighed 0.057 g. (i) Write a balanced equation for the decomposition of XY(CO3)2 XY(CO3)2 o XO + YO + 2 CO2 [1] (ii) Calculate the mass of carbon dioxide produced. Mass of CO2 = 0.400 – 0.275 = 0.125 g [1] (iii) Hence, or otherwise, identify the metals, X and Y, showing your working clearly. Mass of YO = 0.275 – 0.057 = 0.218 g XY(CO3)2 o XO (insoluble) + YO (soluble) + 2 CO2 Mass / g : 0.400 0.057 0.218 0.125 Let Ar of X be a and Ar of Y be b Amt of CO2 = ଴.ଵଶହ ସସ.଴ = 0.00284 mol [1/2] Amt of XO = Amt of YO = ଴.ଶ଼ସ ଶ = 0.00142 mol [1/2] For XO, ଴.଴ହ଻ ௔ ା ଵ଺.଴ = 0.00142 a = 24.1 , X is Mg [1] For YO, ଴.ଶଵ଼ ௕ ା ଵ଺.଴ = 0.00142 b = 137.5 , Y is Ba [1] [3]

NJC/H2 Chem Preliminary Examination/02/2021 8 (c) Aluminium is commonly extracted from its oxide, Al2O3. (i) Al2O3 dissolves in hot aqueous solution of sodium hydroxide. Write an ionic equation to explain the reaction. Al2O3(s) + 2OH−(aq) + 3H2O(l) o 2[Al(OH)4]− (aq) Must include state symbol for ionic equation and exclude Na+(aq) spectator ion [1] (ii) Al2O3 is dissolved in molten cryolite. The mixture is electrolysed using graphite electrodes. The cell operates at a very high current of 50 000 A. Calculate the time needed to obtain 1 kg of pure aluminium. [R] Al3+ + 3e− o Al Amount of Al to be produced = ଵ଴଴଴ ଶ଻.଴ = 37.04 mol [1/2] Amount of electrons required = 3 × 37.04 =