2021 MI Prelim H2 Chemistry P2 (9729 02) Answers

Class Adm No Candidate Name: This question paper consists of 17 printed pages and 1 blank page. 2021 Preliminary Exams Pre-University 3 H2 CHEMISTRY 9729/02 Paper 2 Structured Questions 16 Sep 2021 2 hours Candidates answer on the Question paper. Additional materials: Data Booklet READ THESE INSTRUCTIONS FIRST Do not turn over this question paper until you are told to do so Write your name, class and admission number on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. 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. Question 1 2 3 4 5 6 Total Marks 13 7 18 13 9 15 75

2 Answer all questions in the spaces provided. 1 (a) Explain why the ions of Group 2 elements increase in size down the Group. [2] Although both nuclear charge and shielding effect increases, the distance between the valence electron and the nucleus increases. ; The electrostatic forces of attraction between the positive nucleus and the outermost electron decreases and valence electrons are pulled less closely. ; (b) Magnesium is a Group 2 element. Magnesium nitrate decomposes to form an oxide and two gases when heated strongly. (i) Write a balanced chemical equation, with state symbols, for the decomposition of magnesium nitrate. [1] 2 Mg(NO3)2 (s) ⟶ 2 MgO (s) + 4 NO2 (g) + O2 (g) (ii) Describe and explain the relative thermal stability of magnesium nitrate and barium nitrate. [2] Ba2+ ion is larger than Mg2+ ion while charge is the same, thus Ba2+ has a lower charge density and polarises NO3– less. ; N–O bond is weakened less and BaNO3 is more thermally stable. ; (iii) 2.0 g of magnesium nitrate was heated strongly with a non-luminous Bunsen flame until it was completely decomposed. On the same axes, sketch a graph showing the progress of reaction as 2.0 g of barium nitrate was strongly heated with the same Bunsen flame until it was completely decomposed. [1] Mg(NO3)2 mass of solid remaining / g 2.0 time Ba(NO3)2 less steep everywhere flatten and higher mass remaining

3 [Turn over (c) One of the products of the decomposition of Group 2 nitrates is the metal oxide. Metal oxides are ionic compounds with attractive forces existing within them that can be quantified by the lattice energy. (i) Using the following data and relevant data from the Data Booklet, construct a Born-Haber cycle and use it to calculate the lattice energy of calcium oxide. standard enthalpy change of formation of calcium oxide = –636 kJ mol–1 standard enthalpy change of atomisation of calcium = +178 kJ mol–1 first electron affinity of oxygen = –141 kJ mol–1 second electron affinity of oxygen = +744 kJ mol–1 [4] – 636 = + 178 + 12 (496) + 590 + 1150 – 141 + 744 + LE LE(CaO) = –3405 kJ mol–1 [1] accept –3410 Ca (s) + !" O2 (g) Ca (g) + O (g) Ca2+ (g) + 2 e– + O (g) Ca2+ (g) + e– + O– (g) CaO (s) +178 +!"(496) +590 +1150 -141 –636 +744 Ca2+ (g) + O2– (g) LE [1] [1] [1]

4 (ii) The melting point of calcium oxide is 2886 K. Using appropriate values from the Data Booklet, estimate the melting point of magnesium oxide. Show all your working clearly. [3] LE(CaO) = k q+ × q-r+ + r- = k (+2)×(-2)0.099 + 0.140 = –3405 k = 203.44875 ; LE(MgO) = 203.44875 (+2)×(-2)0.065 + 0.140 = –3969.73 kJ mol–1 ; estimated melting point of MgO = 2886 × 3969.733405 = 3360 K ; [Total: 13]

5 [Turn over 2 Car batteries generate electricity to power the appliances in cars. The electrical flow is generated with a spontaneous redox reaction across the two electrodes. As the battery operates, it becomes discharged, and the two electrodes are converted into by-products. A common arrangement for a car battery would be 6 cells each generating 2 volts, connected in series to form a 12-volt battery. (a) Using suitable electrode reactions, prove that the Ecellɵ of one battery cell discharging is +1.60 V. [3] cathode: PbO2 + 4 H+ + 2 e– ⟶ Pb2+ + 2 H2O +1.47 ; anode: Pb ⟶ Pb2+ + 2 e– +0.13 ; Ecellɵ = +1.47 + 0.13 = +1.60 V ; (b) Aqueous sulfuric acid is used as the electrolyte. Suggest one possible reason why the voltage of one car battery cell is larger than 1.60 V. [1] any 1: lead(II) sulfate precipitate is formed shifting equilibrium position right hence increasing Ecell electrolyte / acid concentration is higher than 1.00 mol dm–3 engine temperature is higher than 25 °C protective casing positive terminal negative terminal cell divider positive electrode (lead(IV) oxide) negative electrode (lead) acidic electrolyte

6 (c) A fully-charged 12-volt battery contains 18.6 kg of electrodes with equal masses of positive and negative electrodes. Calculate the quantity of charge provided by the 12-volt battery after it is fully discharged. [3] since PbO2 has a higher Mr than Pb and ##$%"##$=$$ limiting reagent is PbO2 ; mass of PbO2 = $%&'' )* = 9300 g 𝜂+,-" = ./'' )*/..* )/234 = 38.8796 mol since #&'##$%"=*$, 𝜂5' = 38.8796 mol × 2 = 77.759 mol ; Q = nF = 77.759 mol × 96500 C/mol = 7.50 × 106 C ; [Total: 7]

7 [Turn over 3 Hydrogen peroxide decomposes on its own due to the weak O–O bond. 2H2O2(aq) ⟶ 2H2O(l) + O2(g) However, at room temperature and pressure, the decomposition is very slow. The decomposition of hydrogen peroxide is a first-order reaction. (a) By measuring the volume of oxygen formed over time, the initial rate of the decomposition of a certain concentration of hydrogen peroxide solution can be found. initial concentration of H2O2 / mol dm–3 initial rate of decomposition / mol dm–3 s–1 0.100 1.44 × 10-6 Calculate the time taken for the concentration of hydrogen peroxide to drop to half its original value. [2] rate = k[H2O2] k = 6785[:"-"] = $.<< × $''( 234 >2') ?'!'.$'' 234 >2') = 1.44 × 10-5 s–1 ; 𝑡$*⁄ = AB2C = AB2$.<< × $''* ?'! = 48100 s ; (b) When aqueous potassium iodide is added into hydrogen peroxide, the decomposition occurs rapidly. Potassium iodide speeds up the decomposition of hydrogen peroxide in two reaction steps with the following reversible reaction: IO– + 2H+ + 2e– ⇌ I– + H2O (i) Hydrogen peroxide reacts with I– in the first step, and I– is regenerated in the second step. Write two equations to show the mechanism of iodide ions speeding up the decomposition of hydrogen peroxide. [2] step 1: H2O2 + I– ⟶ H2O + IO– step 2: H2O2 + IO– ⟶ O2 + H2O + I–

8 (ii) Use bond energies to calculate the enthalpy change for the first step, ΔH1. [Bond energy of I–O = 201 kJ mol–1] [2] ΔH1 = 150 – 201 = –51 kJ mol–1 [1] correct Eabsorbed or Ereleased [1] correct final answer (iii) Explain why the enthalpy change calculated using bond energy values in (ii) is only an estimation of the actual value. [1] any 1: bond energy values assume gaseous state, but chemicals are aqueous / liquid state bond energy values are only averages the IO– ion is charged, affecting bond energy (iv) The enthalpy change for the second step, ΔH2, is –145 kJ mol–1. Draw an energy profile diagram to show the energy changes and chemical species involved in each of the two reaction steps. [3] intermediate products and