DEPARTEMENT CHEMIE DEPARTMENT OF CHEMISTRY CMY 117 SEMESTERTOETS 2 / SEMESTER TEST 2 DATUM / DATE: 13 May / Mei 2013 PUNTE / MARKS: 100 TYD / TIME: 3 ure / hours Afdeling A / Section A: 40 Afdeling B / Section B: 60 ************************************ AFDELING A / SECTION A Memorandum VAN EN VOORLETTERS: ----------------------------------------------------------------------------- SURNAME AND INITIALS: REGISTRASIENOMMER: ---------------------------- GRAADKURSUS: ------------------------ REGISTRATION NUMBER: DEGREE COURSE: HANDTEKENING / SIGNATURE : ----------------------------------------------------- VRAAG QUESTION PUNTE MARKS 1 10 EKSAMINATOR EXAMINER 2 10 3 10 4 10 TOTAAL: AFDELING A TOTAL: SECTION A 40 TOTAAL: AFDELING B TOTAL: SECTION B 60 TOTAAL / TOTAL 100 INSTRUCTIONS All answers (calculations, sketches, and diagrams) must be given in ink. All calculations must be shown in full. Answers must be given to the correct number of significant figures. An information page is attached to Section B. INS TRUKS IES Alle berekeninge, sketse en diagramme moet in ink gegee word. Alle berekeninge moet volledig getoon word. Antwoorde moet tot die korrekte aantal betekenisvolle syfers gegee word. n Datablad is aangeheg aan Afdeling B.
Question 1 Dilutions and Titrations [10] Vraag 1 Verdunnings en Titrasies [10] The concentration of a given 5.000 dm 3 solution of phosphoric acid (H 3 PO 4 ) must be determined. The following procedure was carried out in the laboratory: 15.00 cm 3 of the given solution was transferred with a pipette into an empty 250.00 cm 3 volumetric flask. Distilled water was added to the graduation mark of this volumetric flask. The content of the flask was well mixed. 20.00 cm 3 of the solution of the volumetric flask was transferred with a pipette into an empty conical flask. Indicator was added to this solution. The content of the conical flask was titrated with a 0.4051 mol.dm 3 solution of sodium hydroxide. At the endpoint of the titration the reading on the burette was 22.22 cm 3. From this information, calculate the concentration of the given solution of phosphoric acid in mol.dm 3. [10] Die konsentrasie van n gegewe 5.000 dm 3 oplossing van fosforsuur (H 3 PO 4 ) moet bepaal word. Die volgende prosedure was in die laboratorium uitgevoer: 15.00 cm 3 van die gegewe oplossing was oorgedra met n pipet in leë 250.00 cm 3 volumetriese fles. Gedistilleerde water was bygevoeg tot by die ykmerk van die volumetriese fles. Die inhoud van die fles was goed gemeng. 20.00 cm 3 van die oplossing van die volumetriese fles was oorgedra met n pipet in n koniese fles. Indikator was bygevoeg by hierdie oplossing. Die inhoud van die koniese fles was getitreer met n 0.4051 mol.dm 3 natriumhidroksied-oplossing. By die endpunt van die titrasie was die lesing op die buret 22.22 cm 3. Met hierdie inligting, bereken die konsentrasie van die gegewe oplossing van fosforsuur in mol.dm 3. [10] If necessary, continue on the next page. Indien nodig, gaan voort op die volgende bladsy. CMY 117 2 / 8 University of Pretoria
H3PO4(aq) + 3NaOH(aq) Na3PO4(aq) + 3H2O(l) Balanced equation Plan: titre mol NaOH mol H3PO4 in 25 ml 250/20 mol in 250 ml 5000/15 mol in 5L C=n/V [H3PO4]5L or Balanced equation Plan: titre mol NaOH mol C=n/V H3PO4 [H3PO4]20.00 ml = dilution [H3PO4]250mL [H3PO4]5L Mol NaOH: e = C V = 0.4051 0.02222 = 0.009000 mol NqOH Mol H3PO4 in conical flask: 0.009000 mol 1 = 0.003000 mol (from balanced equation) 3 Mol H3PO4 in 250.0 ml flask: 0.003000 mol 250.00 20.00 = 0.03750 mol Mol H3PO4 in 5.00 dm 3 flask: 0.03750 mol 5000. = 12.50 mol [H3PO4] in 5.00 L flask: C = n V 15.00 = 12.5 mol 5.000 dm 3 = 2.500 mol dm 0 or 2.500 M or C = n 0.003000 mol = V 0.020000 dm3 = 0.1500 mol qm 3 Concentration of solution in 250.00 cm 3 flask is 0.1500 M e = C V = 0.1500 0.2500 = 0.03750 mol in 250.00 cm 3 flask C = n 0.03750 mol = = 2.500 mol V 0.01500 dm 3 dm 0 or 2.500 M or C 1 V 1 = C 2 V 2 C 1 (15.00 qm 3 ) = (0.1500 M)(250.00 qm 3 ) C 1 = 2.500 mol dm 0 or 2.500 M CMY 117 3 / 8 University of Pretoria
Question 2 Bonding and Molecular Structure [10] Vraag 2 Binding e n Molekulê re Geome trie [10] Consider the XeO 2 F 2 molecule. 2.1 By using only single bonds, draw the Lewis Structure of this molecule. Use different symbols for the valence electrons of the various atoms. Do not use dashes for bonds. Show all bonding electrons and lone electron pairs. [2] Beskou die XeO 2 F 2 molekule. 2.1 Deur slegs enkelbindings te gebruik, teken die Lewisstruktuur van hierdie molekule. Gebruik verskillende simbole vir die valenselektrone van die verskillende atome. Moet nie strepies vir bindings gebruik nie. Wys alle bindingselektrone en alleen elektronpare. [2] Atom / Atoom Formal charge / Formele lading O -1 O -1 F 0 F 0 Xe +2 [2] 2.2 Calculate the formal charge of each atom in your structure in question 2.1. Write the values in the table above. 2.3 Modify the Lewis Structure of your answer in 2.1 by taking the tendencies of formal charges into account. [3] 2.2 Bereken die formele lading van elke atoom in jou struktuur in vraag 2.1. Skryf die waardes in die tabel hierbo. 2.3 Verander die Lewisstruktuur van jou antwoord in vraag 2.1 deur die neigings van formele ladings in ag te neem. [3] Atom / Atoom Formal charge / Formele lading O 0 O 0 F 0 F 0 Xe 0 [1] 2.4 Calculate the formal charge of each atom in your Lewis Structure in question 2.3. Write the values in the table above. 2.5 Give the names of the following of the structure in question 2.3: Electron pair geometry: Trigonal bipyramidal [1] Molecular geometry: See-saw [1] 2.4 Bereken die formele lading van elke atoom in jou struktuur in vraag 2.3. Skryf die waardes in die tabel hierbo. 2.5 Gee die name van die volgende in die struktuur in vraag 2.3: Elektronpaargeometrie: Trigonaal bipiramidaal [1] Molekulêre geometrie: Wipplank [1] CMY 117 4 / 8 University of Pretoria
Question 3 Kinetics [10] Vraag 3 Kinetika [10] Consider the following reaction: Beskou die volgende reaksie: 2NO(g) + 2H 2 (g) N 2 (g) + 2H 2 O(g) To obtain the rate law for this reaction, the following Om die tempowet vir hierdie reaksie te bepaal, is die experiments were run. For each experiment the initial volgende eksperimente gedoen. Vir elke eksperiment is rate of reaction for NO was determined. die aanvangstempo van reaksie vir NO bepaal. Experiment no. / nr. [NO] 0 (mol/l) [H 2 ] 0 (mol/l) Initial rate for NO Aanvangstempo vir NO (mol L -1 s -1 ) 1 0.0200 0.0300 0.00276 2 0.0200 0.0900 0.00828 3 0.0350 0.0350 0.00985 3.1 Determine the rate law. [Hint: determine the order for H 2 first] [7] 3.1 Bepaal die tempowet. [Wenk: bepaal eers die orde vir H 2 ] [7] The rate law will have the following format: oqqe = k[no] x [H 2 ] y Comparing experiments 1 and 2: When [H2]0 triples, the rate is 0.00828 = 3 times faster 0.00276 Conclusion: the reaction is first order in H2 and y = 0 or Order with respect to NO: o 2 = k[no]x [H 2 ] y o 1 k[no] x [H 2 ] y = k[0.0200]x (0.0900) y k[0.0200] x (0.0300) y y 0.00828 0.00276 = 0.0900 0.0300 3 = 3 y y = 0 o 3 = k[no]x [H 2 ] y o 2 k[no] x [H 2 ] y = k[0.0350]x (0.0350) 1 k[0.0200] x (0.0900) 1 x 0.00985 0.00828 = 0.0350 0.0200 0.38889 3.059 = 1.75 x x = lol3.059 = 1.9979 2 lol1.75 Rqqe lqw: rate = k[no] 2 [H 2 ] CMY 117 5 / 8 University of Pretoria
3.2 Determine the value and unit of the rate constant. [3] 3.2 Bepaal die waarde en eenheid van die tempokonstante. [3] Carry rate law from 3.1 Use any experiment, e.g. experiment 1: Experimental rate was given for NO rate of reaction = 1 2 From the rate law: [NO ] t = 1 2 (0.00276) = 0.00138 mol L 1 s 1 1 if rate not halved 0.00138 mol L 1 s 1 = k (0.0200 mol L 1 ) 2 (0.0300 mol L 1 ) OR Experiment 2: k = 0.00138 mol L 1 s 1 1.20 10 5 mol 3 L 3 = 005 mol 2 L 2 s 0 oo oo 005 mol 2 dm 6 s 0 005 M 2 s 0 k = 0.00414 mol L 1 s 1 3.60 10 5 mol 3 L 3 = 005 mol 2 L 2 s 0 230 OR Experiment 3: k = 0.004925 mol L 1 s 1 4.2875 10 5 mol 3 L 3 = 005 mol 2 L 2 s 0 CMY 117 6 / 8 University of Pretoria
Question 4 Chemical Equilibrium [10] Vraag 4 Chemiese Ewewig [10] 0.6750 mol of CO 2 (g) and 0.6750 mol CF 4 (g) are 0.6750 mol CO 2 (g) en 0.6750 mol CF 4 (g) word in ʼn mixed in a 5.000 L container at 120. The following 5.000 L houer by 120 gemeng. Die volgende reaksie reaction takes place: vind plaas: CO 2 (g) + CF 4 (g) 2COF 2 (g) 4.1 At equilibrium the COF 2 concentration is 0.0810 M. What is the value of the equilibrium constant? [4] 4.1 By ewewig is die COF 2 konsentrasie 0.0810 M. Wat is die waarde van die ewewigskonstante? [4] CO 2 + CF 4 2COF 2 I (M) 0.1350 0.1350 0 C (M) 0.0405 0.0405 +0.0810 E (M) 0.0945 0.0945 0.0810 K c = [COF 2 ]2 [CO 2 ][CF 4 ] = 0.08102 0.0945 2 = 0.705 or in terms of moles: CO 2 + CF 4 2COF 2 I (mol) 0.6750 0.6750 0 C (mol) 0.2025 0.2025 +0.4050 E (mol) 0.4725 0.4725 0.4050 E (M) 0.0945 0.0945 0.0810 K c = [COF 2 ]2 [CO 2 ][CF 4 ] = 0.08102 0.0945 2 = 0.705 CMY 117 7 / 8 University of Pretoria
4.2 The above equilibrium is disturbed by adding 0.270 mol COF 2 to this equilibrium mixture, still at 120. What will the concentrations of both reactants and product be when equilibrium is reestablished? [6] 4.2 Die bostaande ewewig word versteur deur 0.270 mol COF 2 by die ewewigmengsel te voeg, steeds by 120. Wat sal die konsentrasies van beide reagense en produk wees wanneer die ewewig herstel is? [6] Carry equilibrium concentrations from 4.1 CO 2 + CF 4 2COF 2 E (M) 0.0945 0.0945 0.0810 Disturb (M) 0 0 +0.0540 I (M) 0.0945 0.0945 0.1350 C (M) +x +x 2x E (M) 0.0945+x 0.0945+x 0.1350 2x K c = 0.735 = [ COF 2 ] 2 = (0.1350 2x)2 [CO 2 ][CF 4 ] (0.0945+x)2 0.1350 2x 0.735 = 0.8573 = 0.0945 + x 0.8573(0.0945 + x) = 0.1350 2x 0.8573x + 2x = 0.1350 0.0810 2.8573x = 0.05398 x = 0.01889 oo qqqqqqqqq eeeeeeee: 3.265x 2 + 0.6789x 0.011661 = 0 x = b ± b2 4aa 2a x = 0.18904 (ttt lllll) oo x = 0. 000000 [CO 2 ] = [CF 4 ] = 0.0945 + x = 0.1134 0.000 M [COF 2 ] = 0.1350 2x = 0.09722 0.0072 M AFDELING B / SECTION B Answers / Antwoorde # 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Answer C H C F E D C A A B C B E C I H D E B B B C D F A A F E CMY 117 8 / 8 University of Pretoria