2021 NYJC Prelim 9729 P4 (Ans)

Nanyang Junior College 2021 J2 H2 Chemistry Prelims Answers. 1 NYJC J2 H2 Chemistry Prelim Answers Paper 4 Answers 1 (a) (i) Dilution of FA 1 Final burette reading for FA 1 / cm3 Initial burette reading for FA 1 / cm3 Volume of FA 1 used / cm3 24.20 2 d.p. Titration Results Final burette reading for FA 4 / cm3 Initial burette reading for FA 4 / cm3 Volume of FA 4 used / cm3 19.15 19.05 2 d.p. 2 d.p. Correct header and units for both tables [1] Correct evaluation and precision for both tables [1] 24.25scaled mean titre = x mean titre volume diluted Scaled mean titre diff d r0.5 cm3 [2] or d r1.0 cm3 [1] (ii) Ave titre value = 19.10 (2 d.p.) Choice of titre values are consistent and average correctly evaluated [1] (b) (i) 2- 23 22 2 -3 SO 3 HO -3 -3 -3 -2 -3 22 19.10n reacted = 0.200 x = 3.82 x 10 mol1000 n in 25.0 cm reacted = n formed 1= x 3.82 x 10 mol2 = 1.91 x 10 mol [1] 1.91 x 10[H O ] = = 7.64 x 10 mol dm [1] 25.0 1000 FA 5 FA 5 I (ii) -2 22 22 -2 -3 22 250 24.207.64 x 10 x = [H O ] x ecf for [H O ] from 2(b)(i)1000 1000 2507.64 x 10 x 1000[H O ] = = 0.789 mol dm [1] 24.20 1000 FA 1 FA 5 FA 1

Nanyang Junior College 2021 J2 H2 Chemistry Prelims Answers. 2 (c) 2H2O2 J 2H2O + O2 2 2 2 3 O 3- 4 O 3- 4 O 1n evolved from 1 dm of = x 0.789 2 = 0.394 mol 0.3946n evolved from 1 cm of at r.t.p. = = 3 .946 x 10 mol [1] 1000 V evolved from 1 cm of at r.t.p. = 3.94 6 x 10 mol x 24000 c FA 1 FA 1 FA 1 3- 1 3 33 2 m mol = 9.47 cm [1] Hence 1 cm of gives 9.47 cm of O Vol st rength of is 9.47ŸFA 1 FA 1 (d) H2O2 in FA 5 is the limiting reagent while FA 3 is measured in excess. Using a measuring cylinder to measure FA 3 will not affect the amount of I2 produced hence will not affect the volume / accuracy of FA 4 used. (e) Impact: The oxidation of I– by H2O2 is incomplete when titration is carried out as the oxidation reaction is slow hence the concentration of H2O2 calculated is smaller than expected.[1] Evidence: The reaction mixture needs to stand for at least 7 minutes before titration. OR slow return of the blue black colour at end point suggest oxidation of I– still continues.[1] 2 (a) Temperature readings Time / min Temperature / oC Time / min Temperature / oC 0 29.3 7 ½ 36.2 1 29.3 8 36.4 2 29.2 8 ½ 36.4 3 29.2 9 36.4 4 –– 9 ½ 36.4 5 32.0 10 36.3 5 ½ 33.8 10 ½ 36.3 6 34.6 11 36.2 6 ½ 35.4 11½ 36.1 7 36.0 12 36.0 Correct heading and units [1] Record time to nearest half minute and temp to 1 d.p except at 4th min.[1]

Nanyang Junior College 2021 J2 H2 Chemistry Prelims Answers. 3 (b)

Nanyang Junior College 2021 J2 H2 Chemistry Prelims Answers. 4 Temperature change, 'T = oC [4] Correctly labelled axes and units, no odd scale used [1] Correctly plotted all points within ½ sq except at 4th min [1] Drew and extrapolated 2 best–fit lines to the 4th min, allowing 1 anomalous pt after mixing [1] Correctly determined and labelled 'T clearly [1] (c) = mc = 100 x 4.18 x Ans fr J [1] including units ' 2(b) qT 22 decomp HO -1 mc= -n 100 x 4.18 x Ans fr = - 50 x 0.950 [1]1000 = - 8800 x Ans fr J mol [1] with sign a nd units '' 2(b) 2(b) TH (d) (i) 2H2O2(aq) J 2H2O(l) + O2(g) (ii) Iodide is acting as a catalyst. It is reacted in step 1 but regenerated in step 3 hence is not involved in the overall decomposition reaction.[1] (iii) The enthalpy changes for both experiments will be the same / very similar.[1] Energy evolved in forming bonds between H2O2 and the ions will require the same amount of energy to eventually break it up hence both catalysts are not involved in the overall reaction.[1] (e) (i) The results would be equally reliable. Enthalpy change of reaction, 'Hrxn, and specific heat capacity, c, remain the same, while the mass of reactants, m, and the amount of reactants, n, used are half the original quantity hence 'T would remain the same.[1] 1 2 rxn 1 2 (100) x 4.18 x H = n '' T (ii) The 'T value would still be reliable as cooling happened throughout the experiment at a constant rate hence extrapolating the cooling portion of the graph to the point of mixing would compensate for the heat loss due to cooling.[1]

Nanyang Junior College 2021 J2 H2 Chemistry Prelims Answers. 5 3 (a) Planning 1. Fill a burette with aq KMnO4 solution. 2. Using a 100 cm3 measuring cylinder, transfer 100 cm3 of H2O2 into a 250 cm3 conical flask (allow beaker). [1] 3. Add a small spatula of MnO2 into the conical flask, swirl the mixture and start th e stopwatch at the same time. Allow the mixture to stand. [1] 4. Before 2 minutes, use a 10.0 cm3 pipette to transfer a sample into a clean 250 cm3 conical flask. step 1 + step 4 [1] 5. At 2 min, use a clean beaker to transfer 100 (to 150) cm3 of deionised water into the conical flask (containing the 10 cm3 sample) and record the timing. [1] 6. Using a 25.0 cm3 measuring cylinder, add 20.0 (to 25) cm3 of H2SO4 into the conical flask. [1] 7. Titrate the resulting solution against KMnO4 from the burette until 1 drop gives a permanent pink colour. [1] 8. Repeat steps 4 to 7 for another 4 (or 5) times but at time intervals of approximately 4 (to 5) min. [1] 9. Calculate the amount of H2O2 remaining at each timing using the volume of KMnO4. (b) Sketch labelled graph of VKMnO4 against time with units, downward sloping curve [1] If the decomposition is first order with respect to H2O2, the graph would show constant half–life when volume of KMnO4 decreases by half. Note: VKMnO4 reacted D nKMnO4 reacted D nH2O2 reacted D [H2O2] Explains constant half–life and marked out on graph [1] time / min t1 t2 V ½ V ¼ V t1/2 [H2O2] remaining / mol dm-3 OR Vol of KMnO4 / cm3 t1/2

Nanyang Junior College 2021 J2 H2 Chemistry Prelims Answers. 6 4 (a) (i) test 1 To 1 cm3 of FA 7 in a test–tube, add about 2–4 cm3 of H2SO4 / HCl / HNO3. If effervescence observed and white ppt is formed when gas evolved is bubbled into limewater, carbonate ions are present. test 2 To 1 cm3 of FA 7 in a test–tube, (add 1 cm3 of HNO3 and) add 1 cm3 of aq AgNO3 followed by aq NH3 dropwise till in excess. If white ppt is formed and is soluble in excess aq NH3, Cl– ions are present. If cream ppt is formed and is partially soluble in excess aq NH3, Br– ions are present. If yellow ppt is formed and is insoluble in excess aq NH3, I– ions are present. test 3 To 1 cm3 of FA 7 in a test–tube, add 1 cm3 of aq BaCl2 followed by 2 cm3 of aq HNO3 / HCl. If white ppt is formed and is insoluble in acid, SO42– ions are present. (ii) test 1: No effervescence observed test 2: No ppt observed. Note: There are no halides present, but FA 7 contains Fe2+. When aq NH3 is added, green ppt is formed and is insoluble in excess aq NH3. (Ppt turns brown upon standing in air.) The presence of AgNO3 and aq NH3 may also cause a black ppt to form. test 3: White ppt formed is insoluble in excess acid. identity of anion: SO42– (iii) Table 4.1 Test Observation Add 2 cm3 of FA 7 into a boiling tube followed by aq NaOH dropwise. Warm the mixture. Green ppt formed and is insoluble in excess aq NaOH. Ppt turns brown upon standing in air. [1] Gas evolved turns damp red litmus paper blue.[1] Cations present in aqueous FA 7: NH4+ and Fe2+ [1] Identity of solid FA 7: (NH4)2Fe(SO4)2 [1] (b) (i) Table 4.2 test observation FA 8 FA 9 1. Add 2 cm3 of aqueous sulfuric acid and FA solution into a test–tube followed by 1 cm3 of potassium manganate(VII) solution. Purple KMnO4 decolourised. Effervescence. Gas evolved relights glowing splinter. No observable change

Nanyang Junior College 2021 J2 H2 Chemistry Prelims Answers. 7 Warm the test–tube. No further / observable change Pu