2022 TMJC H2 Chem Prelim P3 (Ans)

2022 H2 Chem Prelim P3 (Ans) Section A Answer all the questions from this section. 1(a) Samples of three different oxides were added to water separately. (i) State the Arrhenius theory of acids and bases. [1] Acids are hydrogen-containing substances that produce H+ in aqueous solution while bases are hydroxide-containing substances that produce OH− in aqueous solution. [1] (ii) The pH value of the solution formed when sodium oxide is shaken with water is greater than the pH value of the solution formed when magnesium oxide is shaken with water. The pH of the solution formed when sulfur trioxide is shaken with water is less than b oth of these solutions. Explain these observations using the Arrhenius theory. Write equations for all the reactions described. [3] Both Na2O and MgO produced OH– when dissolved in water. Na2O + H2O → 2NaOH OR Na2O + H2O → 2Na+ + 2OH– MgO + H2O Mg(OH)2 OR MgO + H2O Mg2+ + 2OH− [1] for both equations However, NaOH is much more soluble in water than Mg(OH)2 so the [OH−] is higher and thus pH value is higher. [1] OR NaOH dissociates fully while Mg(OH)2 dissociates partially so the [OH−] is higher and thus pH value is higher. SO3 dissolves in water to produce H+, an acidic solution. Hence, its pH value is lower than the other two oxides. SO3 + H2O → HSO4− + H+ Or SO3 + H2O → SO42− + 2H+ [1] for equation and explanation Markers’ Comments • Some candidates confuse the Arrhenius theory with the Bronsted Lowry and Lewis Theory of acids and bases. Candidates are reminded to do their bookwork and learn the definitions well. Markers’ Comments • Candidates need to make clear reference to the Arrhenius theory in answering this part to link to the H+ and OH− produced in the solution. • Candidates need to take note that the higher [OH−] and hence higher pH of the solution formed from dissolving Na2O as compared to that of MgO is due to full dissociation of the NaOH and partial dissociation of MgOH)2 respectively.

2 2022 H2 Chem Prelim P3 (Ans) Tampines Meridian Junior College 2022 JC2 Preliminary Examination H2 Chemistry (b) Diazomethane, CH2N2, reacts with water to give methanol and nitrogen gas. CH2N2 + H2O → CH3OH + N2 When 2.50 × 10–3 mol of CH2N2 was added into water , the volume of nitrogen gas evolved at various time intervals after the start of the reaction were measured. At the end of the reaction, 60 cm3 of nitrogen gas was collected. The experiment results are plotted in Fig. 1.1 below. Fig. 1.1 (i) Use Fig. 1.1 and the information given to show that the order of reaction with respect to [CH2N2] is 1. [2] Max volume of N2 gas collected = 60 cm3 (100%) The two half-lives are determined from graph at volume = 30 cm3 (50%) & 45 cm3 (75%) 1st t1/2 is at a volume of ½ × 60 = 30 cm3 2nd t1/2 is at a volume of ¾ × 60 = 45 cm3 [1] for clear construction lines on graph with labelling of t½ From the graph, t1/2 is constant at about 14.5 min [1] hence order of reaction with respect to [CH2N2] is 1. 0.0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 0 10 20 30 40 50 60 70 Volume of N2 (g) / cm3 time / min t½ = 14.5 min t½ = 14.5 min 45.0 (75%) 30.0 (50%) 60.0 (100%) Markers’ Comments • Candidates need to take note that this is a product−time graph. To determine a set of 2 half-lives, the first one is at half the maximum volume of 60 cm3. 0 12 t ⎯⎯ → x 2 12 t ⎯⎯ → 3x 4 12 t ⎯⎯ → 7x 8 (where x is 60 cm3 in this case) • t½ is constant

3 2022 H2 Chem Prelim P3 (Ans) Tampines Meridian Junior College 2022 JC2 Preliminary Examination H2 Chemistry The reaction of CH2N2 with water takes place in the presence of an acid. Two experiments were conducted at different pH values to investigate the kinetics of this reaction. T he results are shown in Table 1.1. Table 1.1 Experiment [CH2N2] / mol dm−3 pH Relative rate 1 1.00 × 10−3 1.00 1 2 4.00 × 10−3 1.30 2 (ii) Calculate the concentration of H+(aq) in experiment 1 and 2. [1] Experiment pH [H+] / mol dm−3 1 1.00 10−1.00 = 0.100 2 1.30 10−1.30 = 0.0501 [1] for both (iii) Use the data provided to determine the order of reaction with respect to [H+], and hence write the rate equation for the reaction. [2] Experiment [CH2N2] / mol dm−3 pH [H+] / mol dm−3 Relative rate 1 1.00 × 10−3 1.00 0.100 1 2 4.00 × 10−3 1.30 0.0501 2 Let rate = k [CH2N2] [H+]x ( ) 1 11 22 11 212 2 4 2 1+ 22 1+ 22 -3 1 -3 1 [CH N ] [H ] [CH N ] [H ] (1.0 10 ) (0.100) 1 (4.0 10 ) (0.0501) x x x x x krate rate k x = =  =  = OR From Experiments 1 and 2, when [CH2N2] is increased 4 times and [H+] is halved, overall rate is doubled. Since order of reaction w.r.t. [CH 2N2] is 1, when [CH 2N2] is increased 4 times, rate should have increased 4 times. Since (4)1 (½)1 = 2, this means that when [H+] is halved, rate is halved. Hence order of reaction w.r.t. [H+] is 1. [1] Rate = k [CH2N2] [H+] [1] (allow for ECF) Markers’ Comments • pH = 1.00  −log10 [H+] = 1.00  [H+] = 10−1.00 = 0.100 Markers’ Comments • Candidates are reminded to read the questions carefully and not miss out on answering either of the two parts asked in the question to determine both the order of reaction w.r.t H+ as well as to give the rate equation.

4 2022 H2 Chem Prelim P3 (Ans) Tampines Meridian Junior College 2022 JC2 Preliminary Examination H2 Chemistry (c) Pure magnesium needed for making alloys can be obtained by the electrolysis of molten magnesium chloride as shown in Fig. 1.2. Fig. 1.2 (i) Write the half -equations, including state symbols , for the reactions occurring at the graphite and iron electrodes. Label your equations clearly to indicate the reaction occurring at the graphite and iron electrode respectively. [2] Graphite electrode / anode: 2Cl− (l) → Cl2 (g) + 2e− [1] Iron electrode / cathode: Mg2+(l) + 2e− → Mg (l) [1] (ii) Draw and label the direction of electron flow in the cell on Fig. 1.2. [1] (iii) Calculate the mass of magnesium obtained if a current of 3.00 A is supplied for 10.0 h. [2] Q = I × t = 3.00 × 10.0 × 3600 [1] = 108 000C Amount of Mg = It nF = 108 000 2 96500 = 0.5596 mol Mass of Mg = 0.5596 × 24.3 = 13.6 g [1] iron electrode inert gas out Gas produced molten magnesium chloride inert gas in molten magnesium graphite electrode to external circuit (c)(ii) Correct direction of electron flow [1] Markers’ Comments • Since Cl2 gas is produced at the graphite electrode, oxidation of Cl− took place and hence the graphite electrode is the anode. Flow of electrons is from the graphite anode to the iron electrode (cathode). Markers’ Comments • Candidates need to take note that the states of Mg2+ , Cl− and Mg are liquid since it was given in the question that magnesium and magnesium chloride are in the molten state.

5 2022 H2 Chem Prelim P3 (Ans) Tampines Meridian Junior College 2022 JC2 Preliminary Examination H2 Chemistry (iv) A gas is continuously passed over the molten magnesium in the electrolytic cell to provide an inert environment. Suggest a gas that can be used for this. [1] Helium, argon or any suitable noble gas. [1] Do not accept nitrogen gas. (v) Molten magnesium chloride in the cell is being replaced with aqueous magnesium chloride. Using relevant data from the Data Booklet, state and explain the reactions taking place at both electrodes when this change is made. [2] At the cathode, (reduction) Mg2+ + 2e Mg Eo = −2.38 V 2H2O + 2e– H2 + 2OH− Eo = −0.83 V At the anode, (oxidation) Cl2 + 2e– 2Cl– Eo = + 1.36 V O2 + 4H+ + 4e– 2H2O