CMY 117 SEMESTERTOETS 2 / SEMESTER TEST 2

Transcription

1 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 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. INSTRUKSIES 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.

2 Question 1 Dilutions and Titrations [10] Vraag 1 Verdunnings en Titrasies [10] The concentration of a given dm 3 solution of phosphoric acid (H 3 PO 4 ) must be determined. The following procedure was carried out in the laboratory: cm 3 of the given solution was transferred with a pipette into an empty cm 3 volumetric flask. Distilled water was added to the graduation mark of this volumetric flask. The content of the flask was well mixed 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 mol.dm 3 solution of sodium hydroxide. At the endpoint of the titration the reading on the burette was 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 dm 3 oplossing van fosforsuur (H 3 PO 4 ) moet bepaal word. Die volgende prosedure was in die laboratorium uitgevoer: cm 3 van die gegewe oplossing was oorgedra met n pipet in leë cm 3 volumetriese fles. Gedistilleerde water was bygevoeg tot by die ykmerk van die volumetriese fles. Die inhoud van die fles was goed gemeng 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 mol.dm 3 natriumhidroksied-oplossing. By die endpunt van die titrasie was die lesing op die buret cm 3. Met hierdie inligting, bereken die konsentrasie van die gegewe oplossing van fosforsuur in mol.dm 3. [10] CMY / 8 University of Pretoria

3 If necessary, continue on the next page. Indien nodig, gaan voort op die volgende bladsy. H 3 PO 4 (aq) + 3NaOH(aq) Na 3 PO 4 (aq) + 3H 2 O(l) Balanced equation Plan: titre mol NaOH mol H3PO4 in 20 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 = = mol NqOH Mol H 3 PO 4 in conical flask: mol 1 = mol (from balanced equation) 3 Mol H 3 PO 4 in ml flask: mol = mol Mol H 3 PO 4 in 5.00 dm 3 flask: mol = mol [H 3 PO 4 ] in 5.00 L flask: C = n 12.5 mol = = mol V dm 3 dm 0 or M or C = n mol = V dm3 = mol qm 3 Concentration of solution in cm 3 flask is M or C A V A C B V B = n A n B = 1 3 C A = 1 3 C BV B V A = M cm3 = M cm 3 e = C V = = mol in cm 3 flask C = n mol = = mol V dm 3 dm 0 or M or C 1 V 1 = C 2 V 2 C 1 (15.00 qm 3 ) = ( M)( qm 3 ) C 1 = mol dm 0 or M CMY / 8 University of Pretoria

4 Question 2 Bonding and Molecular Structure [10] Vraag 2 Binding en Molekulêre Geometrie [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 / 8 University of Pretoria

5 Question 3 Kinetics [10] Vraag 3 Kinetika [10] Consider the following reaction: Beskou die volgende reaksie: To obtain the rate law for this reaction, the following experiments were run. For each experiment the initial rate of reaction for NO was determined. Experiment no. / nr. 2NO(g) + 2H 2 (g) N 2 (g) + 2H 2 O(g) [NO] 0 (mol/l) Om die tempowet vir hierdie reaksie te bepaal, is die volgende eksperimente gedoen. Vir elke eksperiment is die aanvangstempo van reaksie vir NO bepaal. [H 2 ] 0 (mol/l) Initial rate for NO Aanvangstempo vir NO (mol L -1 s -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 [H 2 ] 0 triples, the rate is = 3 times faster Conclusion: the reaction is first order in H 2 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 = = 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 = = 1.75 x x = lol3.059 lol1.75 = Rqqe lqw: rate = k[no] 2 [H 2 ] CMY / 8 University of Pretoria

6 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 ( ) = mol 2 L 1 s 1 1 if rate not halved mol L 1 s 1 = k ( mol L 1 ) 2 ( mol L 1 ) OR Experiment 2: k = mol L 1 s mol 3 L 3 = 005 mol 2 L 2 s 0 oo 005 mol 2 dm 6 s 0 oo 005 M 2 s 0 k = mol L 1 s mol 3 L 3 = 005 mol 2 L 2 s OR Experiment 3: k = mol L 1 s mol 3 L 3 = 005 mol 2 L 2 s 0 CMY / 8 University of Pretoria

7 Question 4 Chemical Equilibrium [10] Vraag 4 Chemiese Ewewig [10] mol of CO 2 (g) and mol CF 4 (g) are mixed in a L container at 120. The following reaction takes place: 4.1 At equilibrium the COF 2 concentration is M. What is the value of the equilibrium constant? [4] mol CO 2 (g) en mol CF 4 (g) word in ʼn L houer by 120 gemeng. Die volgende reaksie vind plaas: CO 2 (g) + CF 4 (g) 2COF 2 (g) 4.1 By ewewig is die COF 2 konsentrasie M. Wat is die waarde van die ewewigskonstante? [4] CO 2 + CF 4 2COF 2 I (M) C (M) E (M) K c = [COF 2] 2 [CO 2 ][CF 4 ] = = or in terms of moles: CO 2 + CF 4 2COF 2 I (mol) C (mol) E (mol) E (M) K c = [COF 2] 2 [CO 2 ][CF 4 ] = = CMY / 8 University of Pretoria

8 4.2 The above equilibrium is disturbed by adding 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 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) Disturb (M) I (M) C (M) +x +x 2x E (M) x x x K c = = [COF 2] 2 = ( x)2 [CO 2 ][CF 4 ] ( x) x = = x ( x) = x x + 2x = x = x = oo qqqqqqqqq eeeeeeee: 3.265x x = 0 x = b ± b2 4aa 2a x = (ttt lllll) oo x = [CO 2 ] = [CF 4 ] = x = M [COF 2 ] = x = M AFDELING B / SECTION B Answers / Antwoorde # 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 / 8 University of Pretoria