DHS Prelim P2 Ans

1 H2 Chemistry Prelims P2 Answers Scheme © DHS 2017 9729/02 Answer all questions in the spaces provided. 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 quantum shells increases and valence electrons are further away from the nucleus. Hence, down the group, the valence electrons experience weaker attraction to the nucleus and 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] 1st Ionisation Energy / kJ mol-1 1st I.E. / kJ mol–1 Oxygen Sulfur Selenium Nitrogen Phosphorus Arsenic Oxygen Sulfur Selenium

2 H2 Chemistry Prelims P2 Answers Scheme © DHS 2017 9729/02 (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. [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 moles of gas in vessel, n = 177000 × 5 ×10−3 8.31 ×(273+120) = 0.27098 mol Hence, moles 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 H2 Chemistry Prelims P2 Answers Scheme © DHS 2017 9729/02 (c) Chalcogens are also very commonly found in organic compounds. Table 1 below shows some common oxygen or sulfur containing organic functional groups. Oxygen- containing functional groups Alcohols R–OH Carboxylic Acids O OH R Acyl chlorides O Cl R Sulfur- containing functional groups Thiols R–SH Sulfonic Acids S O OHR O Sulfonyl chlorides S O ClR O Table 1 (i) Explain why carboxylic acids are generally more acidic than alcohols. This is because the lone pair of electrons on the negatively charged oxygen of the carboxylate anion is able to delocalise over two oxygen atoms. This results in the dispersal of the negative charge and hence, the conjugate base of carboxylic acids are more stable than that of alcohols. [1] (ii) Hence, suggest a reason why carboxylic acids are generally less acidic than their corresponding sulfonic acids. The conjugate base of the sulfonic acids are more stable due to the presence of an additional (electronegative) oxygen atom, which allows the delocalisation of electrons over more atoms.This reduces the intenisty of the negative charge on a single atom. [1] (iii) A reaction scheme for the synthesis of dimethyl sulfide (CH3SCH3) from methylsulfonyl chloride (CH3SO2Cl) is shown below: S O ClCH3 O S O OHCH3 O SHCH3 Step I KI Step II SCH3 CH3 1. Na 2. Reagent X Step III Given that sulfur -containing functional groups undergo similar reactions as their corresponding oxygen-containing functional groups (Table 1), suggest: I. The reagent(s) and condition(s) required for Step I. H2O [1] II. The role of KI in Step II. Reducing agent

4 H2 Chemistry Prelims P2 Answers Scheme © DHS 2017 9729/02 [1] III. The identity of reagent X in Step III. CH3Cl or CH3Br or CH3I [1] IV. The structure of the product(s) formed when methylsulfonic acid (CH3SO3H) is reacted with ethylamine (CH3CH2NH2) at room temperature. CH3 S O O O - NH3 + CH2CH3 [1] [Total: 13] 2 The electrolysis of dilute sulfuric acid was carried out using two different currents at room temperature and pressure. Current / A Duration / min Experiment 1 0.75 90 Experiment 2 0.45 90 Oxygen gas is collected at one of the electrodes. (a) (i) Calculate the final volume of oxygen produced in experiment 2. Q = I t = 0.45 x (90 x 60) = 2430 C At the anode, 2H2O  O2 + 4H+ + 4e 4F  1O2 Moles of oxygen produced = nF Q = 96500x4 2430 = 6.30 x 103 Final volume of O2 produced = (6.30 x 103) x 24 000 cm3 = 151 cm3 [2] (ii) The volume of oxygen collect at one of the electrodes for experiment 1 is shown below.

5 H2 Chemistry Prelims P2 Answers Scheme © DHS 2017 9729/02 On the graph above, draw a line to show each of the following: I. the volume of H 2 gas that would be given off in experiment 1 (Label this line 1) II. the volume of oxygen that would be produced in experiment 2 (Label this line 2) [2] I. A straight line from the origin which has double the oxygen volume at a given time. II. A straight line from the origin which has 0.45/0.75 of the volume of oxygen at a given time. (b) In another experiment, electrolysis of aqueous potassium butanedioate, (OOCCH2CH2COO)K2 as the electrolyte was carried out. It was found that two gases, Y and Z, were liberated at the anode in a 2:1 ratio by volume. Gas Y formed is absorbed by soda lime while gas Z is able to decolourise bromine water. (i) Suggest the identities of gases Y and Z. [1] Gas Y is CO2 Gas Z is C2H4 (ii) Construct the half-equation for the reaction that occurs at the anode and the cathode respectively. [2] Anode OOCCH2CH2COO  2CO2 + CH2CH2 + 2e Cathode 2H2O + 2e  H2 + 2OH (iii) Predict the main organic product that would be obtained at the anode when a solution of potassium pentanedioate is electrolysed. [1] Propene

6 H2 Chemistry Prelims P2 Answers Scheme © DHS 2017 9729/02 (c) Nicotinamide adenine dinucleotide (NAD +) is involved in redox chemistry throughout the respiratory system. Aerobic respiration is the process of producing ce llular energy involving oxygen. The electrode potential for the reduction of NAD+ in a biological system, E(pH 7), at 1 mol dm–3, 25 oC and pH 7, is as shown. Its oxidised and reduced forms are represented as NAD+ and NADH respectively. NAD+ + H+ + 2e– NADH E(pH 7) = –0.320 V The reduction electrode potential of oxygen at different pH is given below. (i) With reference to the Data Booklet and the graph given above, calculate the value