VJC 2023 H2 Chem Yearend Consolidation Practice Paper Ans

1 Victoria Junior College 2023 JC1 H2 Chemistry December Self Practice Paper (Answers) Recommended Questions For Consolidation of Concepts and Skills 1 The chalcogens, or the oxygen family, are the elements in group 16 of the Periodic Table. These elements are common in both organic and inorganic compounds. (a) The graph below shows the trend in the first ionisation energies of oxygen, sulfur and selenium. (i) Explain the trend in the first ionisation energies of oxygen, sulfur and selenium. The first ionisation energy decreases from oxygen to selenium. This is because down the group while the nuclear charge increases, number of electronic shells increases and valence electrons are further away from the nucleus. Hence, the valence electrons experience weaker electrostatic force s of attraction from the nucleus and so a smaller amount of energy is required to remove this electron from the atom. [2] (ii) On the same grid above, sketch the trend in the first ionisation energies of nitrogen, phosphorus and arsenic. [1] Note: Fewer electronic shells lead to higher ionisation energies. 1st I.E. / kJ mol–1 Oxygen Sulfur Selenium Nitrogen Phosphorus Arsenic

2 (b) A common chalcogen-containing reagent used in both organic and inorganic synthesis is hydrogen peroxide, H2O2. Hydrogen peroxide readily decomposes at room temperature. Iodide ions, I–, catalyse this decomposition, as shown below: Step I: H2O2 + I – → H2O + IO– (slow) Step II: H2O2 + IO– → H2O + O2 + I – The overall equation for the decomposition of hydrogen peroxide is shown below: 2H2O2 → 2H2O + O2 The enthalpy and entropy changes for the reaction above are shown in the table below: Enthalpy change / kJ mol–1 –98 Entropy change / J K–1 mol–1 +71 (i) Using the data above, complete the diagram below to show the energy profile diagram for the decomposition of hydrogen peroxide in the presence of iodide ions labelling clearly the activation energies and enthalpy change of the reaction. Note: Activation energy of slow step must be greater than that of fast step. [2] (ii) An unknown amount of hydrogen peroxide was allowed to decompose in a 5 dm 3 closed vessel at 120 ºC. When all the hydrogen peroxide was decomposed, a pressure of 177 kPa was measured in the vessel. Determine the amount of hydrogen peroxide that decomposed in the vessel. (Assume that H2O and O2 are ideal gases under the above reaction conditions) pV = nRT Total amount f gases (O2 & H2O) in vessel, n = 177000 × 5 ×10−3 8.31 ×(273+120) = 0.27098 mol Hence, amount of hydrogen peroxide that decomposed = 0.27098 ÷ 3 × 2 = 0.181 mol [2] Energy Progress of reaction Ea2 2H2O2 + I – 2H2O + O2 + I – –98 kJ mol–1 Ea1 H2O + H2O2 + IO –

3 2 (a) Vitamin C is an essential nutrient also known as ascorbic acid. A deficiency of vitamin C leads to a disease known as scurvy. Ascorbic acid is known to have a Mr of 176.0 and contains 40.9% of carbon and 54.5% of oxygen by mass. (i) Determine the molecular formula of ascorbic acid. C O H Mass of atoms in 100 g 40.9 54.5 4.6 Divide by Ar 3.41 3.41 4.6 Simplest mole ratio 3 3 4 The empirical formula of ascorbic acid is C3O3H4. n x (empirical formula) = molecular formula n[(3 x 12.0) + (3 x 16.0) + (4 x 1.0)] = 176.0 n = 2 • The molecular formula of ascorbic acid is C6O6H8. [2] (b) Ascorbic acid is a monobasic acid, HA, and has a pKa of 4.10. The amount of ascorbic acid contained in dietary supplement tablets can be verified by titration. A tablet containing 500 mg of ascorbic acid was dissolved in 25.0 cm3 of deionised water and titrated against 0.100 mol dm-3 sodium hydroxide. (i) Calculate the volume of 0.100 mol dm -3 sodium hydroxide required for complete neutralisation. Amount of moles of HA = Amount of NaOH = 0.500  176.0 = 2.841x10−3 mol Volume of NaOH needed = 2.841x10−3  0.100 = 0.0284 dm3 [1] (ii) Calculate the initial pH of the ascorbic acid solution. [𝐻𝐴] = 2.841×10−3 (25.0)÷1000 = 0.1136 mol dm-3 𝐾𝑎 = 10−4.10 [𝐻+] = √𝐾𝑎 × [𝐻𝐴] = 3.00 × 10−3 mol dm-3 𝑝𝐻 = − lg[3.00 × 10−3] = 2.52 [2] (iii) Suggest a suitable indicator for the titration and describe the colour change at end- point. Phenolphthalein, solution changes from colourless to pink at end point. or Thymolphthalein, solution changes from colourless to blue at end point. Note: These indicators are sui table for strong base titration . Methyl orange is suitable for strong acid titration. [1] (iv) With the aid of a suitable equation, explain your choice of indicator in (iii). A− + H2O HA + OH− At end-point, the conjugate base or salt of ascorbic acid undergoes hydrolysis to form hydroxide ions causing the [OH−] to be greater than [H+] and so pH > 7. The rapid pH change at equivalence point coincides with the working range of the indicator. [2]

4 (c) When a vitamin C tablet is swallowed, it dissolves in the stomach. The pH of the stomach is 2. (i) Determine the percentage of ascorbic acid that is ionised in the stomach. 𝐾𝑎 = [𝐻+][𝐴−] [𝐻𝐴] 10−4.10 = [10−2][𝐴−] [𝐻𝐴] [𝐴−] [𝐻𝐴] = 10−2.1 = 7.93 × 10−3 Let α represent the percentage ionisation. 𝛼 100 − 𝛼 = 7.93 × 10−3 𝛼 = 0.788% Alternative method: Since [HA]>>[A-] or since Ka is small, % ionisation = 0.793% [2] The pH of blood is maintained at 7.35 by a H2CO3/HCO3- buffer. (ii) Using appropriate equations, explain how the buffer minimises changes in pH. When a little acid is present in blood, it will be removed by hydrogen carbonate. HCO3− + H+ → H2CO3 When a little base is present in blood, it will be removed by carbonic acid. H2CO3 + OH− → HCO3− + H2O [2] 3 (a) Singapore has been affected by severe smoke haze due to forest fires in the region periodically. The National Environment Agency (NEA) is taking action to ensure that its population is better equipped to deal with haze. During one of the sample analysis of air, the air sample was found to contain elevated amounts of NO2 and gas S. By careful measurements and extrapolation, the value of 𝑝 ρ for gas S has been found to be approximately 110.3, at 100 °C and at very low pressure. [𝑝 is the pressure of the gas in Pa and ρ is the density of the gas in g m–3] 𝑝 110.3 𝑝 ρ

5 (i) State two main assumptions of the kinetic theory as applied to an ideal gas , and use these to explain why you might expect the behaviour of nitrogen dioxide to be less ideal compared to that of hydrogen. [3] • The gas particles have zero or negligible volume compared to the volume of the container. • There are no intermolecular forces of attraction between gas particles. NO2 is less ideal than H2 as it has a larger size since it has a larger electron cloud. Both NO 2 and H 2 have simple molecular structure but NO 2 is a polar molecule while H 2 is a non-polar molecule . Hence there is stronger permanent dipole -permanent dipole interaction s in NO 2 and weaker instantaneous dipole-induced dipole interactions in H2. (ii) Calculate an approximate value for the relative molecular mass of S. [1] 𝑝V = nRT = (m/M)RT 𝑝 = (m/V)(RT/M) 𝑝 ρ = RT/M At 100 °C and very low pressure, RT/M = 110.3 M = (8.31 x 373) / 110.3 = 28.1 g mol–1 Mr = 28.1 (iii) Hence, identify gas S, where S is a diatomic neutral pollutant. [1] Gas S is CO (b) Atmospheric sulfur dioxide is a major air pollu