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2 THE FIFTEETH ITERATIOAL CHEMISTRY OLYMPIAD 11 JULY 1983, TIMISOARA, ROMAIA THEORETICAL PROBLEMS PROBLEM 1 A) Describe the thermal decomposition of the following ammonium salts in terms of chemical equations: H ClO a) 4 4 t C (H ) SO b) 4 4 (H ) S O c) 4 8 t C t C H O d) 4 t C B) Indicate the right answer: a) Can the molar mass be determined by measuring the density of a gaseous compound at a given temperature and pressure? 1. Yes, under any conditions.. Yes, if the gaseous compound does not dissociate and associate. 3. Yes, if the gaseous compound does not dissociate. 4. Yes, if the gaseous compound does not associate. b) Is a liquid boiling at a constant temperature (at a given pressure) a pure substance? 1. Yes, if the liquid is not azeotropic.. Yes, if the liquid is azeotropic. C) Complete and balance the following equation: (in H O) K Cr O 7 SnCl... CrCl 3... KCl... 69

3 D) The solubility of Hg Cl in water is g/100 ml solution. a) What is the solubility product? b) What is the solubility (in mol dm -3 ) of this substance in a 0.01 M acl solution? c) What is the volume of a 0.01 M acl solution which dissolves the same quantity of mercurous chloride as that dissolved in one litre of pure water? A r (Hg) = A r (Cl) = E) Which of the following groups contains solid compounds at 10 C? a) H O, H 3, CH 4 b) F, Cl, Br c) SO 3, I, acl d) Si, S 8, Hg F) Which of the following salts forms an acidic aqueous solution? a) CH 3 COOa b) H 4 Cl c) a HPO 4 d) a CO 3 e) ahco 3 G) Write the electronic formulas for the following compounds so that the nature of the chemical bonds is evident: a) aclo 3, b) HClO 3, c) SiF 4, d) H 3, e) CaF, f) H O H) Solid perchloric acid is usually written as HClO 4.H O. Based on experimental data showing four equal bonds, suggest a structure accounting for the experimental result. I) The compounds of the second row elements with hydrogen are as follows: LiH, BeH, B H 6, CH 4, H 3, H O, HF. a) Which compounds are solid at room temperature? Explain. b) Which of them are ionic? c) Which are polymeric? d) Which ones do not react with water under normal conditions? e) Give products of the following reactions. 70

4 BeH H O B H 6 H O B H 6 LiH f) Supposing that H 3, H O and HF are acids under some conditions, write their corresponding conjugated bases and arrange them in order of increasing basic strength. J) The following E 0 values are given for the half-reactions: MnO 8 H 5 e = Mn 4 H O E = 1.5 V MnO 4 H 3 e = MnO H O E = 1.69 V Calculate E 0 for the following reaction: MnO 4 H e = Mn H O E =? SOLUTIO t C A) a) 4 H4ClO4 4 HCl 6 H O 5 O b) 3 (H 4 ) SO4 c) (H 4 ) SO8 t C SO 4 H 3 6 H O t C 4 SO 8 H O H O d) 4 t C H O B) a) 1,, 3, 4 b) 1, C) K Cr O 7 3 SnCl 14 HCl CrCl 3 3 SnCl 4 KCl 7 H O D) a) s = g/100 cm 3 = g dm -3 = = g dm = mol dm 47 g mol Hg Cl Hg Cl K s = 4 s 3 = 4 ( ) 3 = b) c(cl ) = 0.01 mol dm -3 71

5 18 K s s = = = [Cl ] (0.01) s = mol dm c) The volume of 0.01 M acl solution in which dissolves the same quantity of Hg Cl as in 1 dm 3 of water, is as follows: V = = dm E) c) SO 3, I, acl F) b) G) a) b) a O Cl O O O Cl O O H c) F d) F Si F H H H F e) F (-) (-) Ca F f) H O H H) O H 3 O ClO 4 - or H 3 O O Cl O O 7

6 I) a) LiH, (BeH ) n polymer b) LiH c) (BeH ) n d) CH 4 e) BeH H O Be(OH) H B H 6 6 H O B(OH) 3 6 H B H 6 LiH Li[BH 4 ] f) H > OH > F J) 4 E MnO 4 H 3 e = MnO H O = 1.69 V MnO 4 H e = Mn H O E =? MnO 8 H 5 e = Mn 4 H O E = 1.5 V E = 3 E E = 5.07 x x = 1.6 V 73

7 PROBLEM In a gaseous mixture of CO and CO, a mass ratio of carbon : oxygen = 1 : was determined..1 Calculate the mass percent composition.. Calculate the volume percent composition..3 Indicate values of the carbon: oxygen ratios for which both gases cannot be present simultaneously. SOLUTIO Write x = number of moles of CO in 100 g y = number of moles of CO in 100 g 8 x 44 y = (xy) 1 = 16(x y) x = mol CO y = 1,389 mol CO = % CO = % CO 100. X = y 50 % CO 50 % CO (by volume).3 The two gases cannot be simultaneously present in the mixture if: carbon mass 1 = which correspeonds to pure CO oxygen mass 16 1 which correspeond s to pure CO 3 74

8 PROBLEM 3 A sample containing a mixture of sodium chloride and potassium chloride weights 5 g. After its dissolution in water 840 ml of AgO 3 solution (c = 0.5 mol dm -3 ) is added. The precipitate is filtered off and a strip of copper weighing g is dipped into the filtrate. After a given time interval the strip weights g. Calculate the mass percent composition of the mixture. SOLUTIO A r (Cu) = 63.5 A r (Ag) = 108 Cu AgO 3 Cu(O 3 ) Ag y x x = the quantity of deposited silver y = the quantity of dissolved copper = y x x y = x = 1.5 y = y 1.5 x Mass of silver nitrate: y = 0.63 x =.15 g Ag = 71.4 g AgO g AgO = 108 g Ag x x = g Ag Silver consumed for participation = 43.1 g Ag 75

9 Total mass of chloride 108 g Ag 43. = g Cl x x = 14. g Cl - M r (acl) = 58.5 M r (KCl) = 74.6 x = mass of acl in the mixture y = mass of KCl in the mixture mass of Cl - in acl: 35.5 x 58.5 mass of Cl - in KCl: 35.5 y x y 74.6 = 14. x y = 5 x = 17.6 g acl y = 7.4 g KCl 70.4 % acl 9.6 % KCl 76

10 PROBLEM 4 The following data were gathered for the alkaline hydrolysis of certain chlorinated compounds: a) A certain volume of a solution of the neutral potassium salt of chlorosuccinic acid is mixed with an equal volume of hydroxide solution. The initial concentration of each solution is 0. mol dm -3. The potassium hydroxide concentration in the reaction mixture was determined at different time intervals at 5 C. The following values were obtained: t (minutes) c(koh) (mol dm -3 ) The experiment was repeated with the same initial solutions at 35 C. The hydroxide concentration is reduced to one half after 1 minutes. b) In the hydrolysis of 3-chloro-3-methylhexane with potassium hydroxide, the concentration of potassium hydroxide was found to have been reduced to one half after 3 minutes at 5 C or 11 minutes at 35 C, r egardless of the initial reactant concentrations (identical). c) In the alkaline hydrolysis of 3-chloro-,4-dimethyl-3-isopropylpentane an identical reaction mechanism as for reaction b was found but the reaction rate was about 100 times faster under the same reaction conditions. Considering the above data answer the following questions: 4.1 What is the reaction order in cases a, b, and c? 4. What is the rate constant at 5 C for reaction a? Indicate the units. 4.3 Calculate the activation energies for reactions a and b. 4.4 If in reaction a dipotassium salt of L-chlorosuccinic acid (which is levorotatory,) is used, what type of optical rotation will be exhibited by the corresponding salt of malic acid formed by hydrolysis? 4.5 If the levorotatory isomer is also used in reaction b, what optical rotation will be exhibited by 3-methyl-3-hexanol formed in the hydrolysis reaction? 77

11 4.6 Why is the rate of reaction c much faster than that of reaction b when both reactions are of the same type and occur under the same temperature and concentration conditions? SOLUTIO 4.1 For reaction a the reaction order is estimated as follows: assuming the first-order reaction: 1 a k = ln t a - x t ( C) k k is not constant, hence the reaction is not of the first-order. for the second-order reaction (with reactant concentrations equal at time zero): 1 a 1 k = t a - x a t ( C) k As k has almost a constant value the condition for a second-order reaction is fulfilled. The half-life of reaction b is independent on the initial concentrations, i. e. it is a firstorder reaction: 1 a 1 a 1 k = ln = ln = ln t a - x t a 1/ a - t1/ Reaction c has the same mechanism as reaction b. Therefore, it will also be a firstorder reaction. 4. The rate constant of reaction a is an average of the above calculated values. k = dm 3 mol -1 min In order to determine the activation energy, the rate constant, k', at 35 C is to be calculated. 78

12 For the second-order reactions the relationship between the rate constants and halflives is as follows: 1 a k = = t a - x a t a = 1/ a - a t1/ a The half-life at 35 C and the initial concentratio n, a = 0.1 mol dm -3, are known. (By mixing equal volumes of the two solutions the concentration of each reacting species is reduced to a half.) Calculation of the rate constant at 35 C: 1 1 k ' = = dm mol min The activation energy of reaction a will be: k ' T ' T Ea = R ln = 8314 ln = J mol k T ' T For reaction b that is a first-order reaction, the rate constants at the two temperatures are calculated from the half-lives: at 5 C: at 35 C: ln 3 1 k = = min ln k ' = = min 11 Hence the activation energy is: E a = 8314 ln = J mol The product of the hydrolysis reaction a will become dextrorotatory as a result of configuration inversion. ( ) CH COO () CH COO ( ) CH COO OH ( ) H C ( ) COO Cl OH C ( ) COO H Cl H C ( ) COO Cl As an S type reaction, it involves a transition state in which the inversion of the configuration of the asymmetric carbon atom occurs. Thus, if the substrate is levorotatory, the product will become dextrorotatory. 79

13 4.5 The reaction b is a unimolecular S 1 reaction and involves the transient formation of an almost stable carbonium ion in the rate-determining step. H 3 C C Cl CH 3 () C ( ) Cl H 5 C H 7 C 3 C 3 H 7 C H 5 The most probable structure of the carbonium ion is planar. The carbonium ion may be attached by the nucleophylic reagent (the OH - ion) on both sides of the plane with the same probability. The product will result as a racemic mixture, with no optical activity, inactive by intermolecular compensation. 4.6 The same is true for the reaction c, the only difference being a more marked repulsion among bulkier substituents. The tendency towards carbonium ion formation with a planar structure and reduced repulsions is increased. C 3 H 7 H 7 C 3 C Cl () C ( ) Cl H 7 C 3 C 3 H 7 H 7 C 3 C 3 H 7 The rate of the carbonium ion formation, and therefore the overall reaction rate, is consequently increased. 80

14 PROBLEM 5 On passing ethanol over a catalyst at 400 K, a dehydration reaction occurs resulting in the formation of ethylene: C H 5 OH(g) C H 4 (g) H O(g) At the above temperature and p 0 = kpa, the conversion of ethyl alcohol is 90.6 mol %. 5.1 Calculate the equilibrium constant K p of the reaction under given conditions. 5. Calculate the values of the equilibrium constants K x and K c at the above temperature. 5.3 Calculate the ethanol conversion at the following pressures: 5 p 0, 10 p 0, 50 p 0, 100 p 0, and 00 p Plot the graph for the variation of conversion vs. pressure. SOLUTIO The reaction: C H 5 OH C H 4 H O Moles: initial: at equilibrium: 1 x x x total: 1 x Ethanol Ethylene Water Molar fraction 1- x 1 x x 1 x x 1 x Partial pressure 1- x 1 x p x 1 x p x 1 x p p' p = p' total pressure, p 0 = kpa p 81

15 K p p x x p p p 1 x 1 x x 1- x p 1- x 1 x CH4 HO = = = p CH5OH p 5.1 p' = kpa x K p = = = 1- x K x = K p p - n ; p' = kpa; n = 1; K x = 4.56 c RT 0 n Kc = K p R = Jmol K ; c = 1mol dm ; T = 400 K p0 K c = x K p x = = p p a) b) c) d) e) x x 5 = = 0.91 x = 0.69 x x 10 = = x = 0.56 x x 50 = = x = 0.9 x x 100 = = x = 0.1 x x 00 = = 0.08 x =

16 5.4 1,0 x 0,5 0, p 83

17 PROBLEM 6 One mole of compound A reacts successively with 3 moles of compound B in aqueous solution in the presence of a basic catalyst (such as Ca(OH) ): A B C C B D D B E Hydrogenation of compound E yields compound F: E H F F has the composition: C = %, H = 8.8 %, O = %. Its molar mass: M = 136 g mol -1 Knowing that 13.6 g of F reacts with 40.8 g acetic anhydride to form product G and acetic acid write down all chemical equations and assign the letters A, B, C, D, E, F, and G to particular formulas of compounds. SOLUTIO The molecular formula of F: C : H : O = : : = 1.5 : 3 : 1 = 5 : 1 : 4 (C 5 H 1 O 4 ) n Since M(F) = 136 and (5 1) (1 1) (4 16) = 136 F = C 5 H 1 O 4 Since F reacts with acetic anhydride it could be a mono- or polyhydroxy alcohol. If it were a monohydroxy alcohol, 136 g of F (1 mol) could react with 10 g (1 mol) of acetic anhydride. In fact 13.6 g of F (i. e. 0.1 mol) reacts with 40.8 g of acetic anhydride (40.8 / 10 = 0.4 mol), i. e. F is a polyol (tetrahydroxy alcohol). F is formed by the reduction of E, so that E has one carbonyl and three OH groups. E is formed from 3 molecules of B and one molecule of A. Since compound E has three OH groups and one CO group and the reaction conditions used are typical for aldol condensation, it is clear that A is acetaldehyde and B 84

18 is formaldehyde. C and D are the products of successive aldol condensation of acetaldehyde with formaldehyde: H 3 C-CH=O H C=O HO-CH -CH -CH=O A B C HO-CH -CH -CH=O H C=O (HO-CH ) CH-CH=O C B D (HO-CH ) CH-CH=O H C=O (HO-CH ) 3 C-CH=O D B E (HO-CH ) 3 C-CH=O H (HO-CH ) 4 C E F (HO-CH ) 4 C 4 (CH 3 CO) O (CH 3 COO-CH ) 4 C 4 CH 3 COOH G 85

19 PROBLEM 7 Knowing that compounds A and B are isomers with the molecular formula C 7 H 7 O and the relative molecular mass of compound M is 93, determine the formulae of compounds A to S taking in account the reactions given in the following reaction scheme: P [O] HOO [H] A H O HOH E HO 3 /H SO 4 F [O] [H] C HOH D HO 3 /H SO 4 G [H] B H O HOH H H aobr HO HCl HOH M L K HO /HCl CO HOH [O] HOH R S J I 86

20 SOLUTIO A B C 6 H 5 -CH=-OH C 6 H 5 -CO-H C C 6 H 5 -C D C 6 H 5 -COOH E C 6 H 5 -CHO F CHO O G COOH H COOH O H I COOH J COOH = Cl - OH K C 6 H 5 -OH L C 6 H 5 -= Cl - M C 6 H 5 -H C 6 H 5 -CH -H P C 6 H 5 -CH -OH R CH=OH S CHO OH OH 87

21 PRACTICAL PROBLEMS PROBLEM 1 (practical) In test tubes A, B, C, and D there are four benzene derivatives containing one or two functional groups of three distinct types. Identify the functional groups of compounds A, B, C, and D using the available reagents. Justify your choice by writing down the identification reactions. Using as reagents the four compounds A, B, C, and D synthesize four organic dyes and write the equations for the reactions performed. SOLUTIO The four compounds are as follows: COOH -H -OH -OH H - -COOH A B C D The identification reactions: a) With H SO 4 : -H H SO 4 H 3 HSO 4 - H 3 COO - H SO 4 H COOH HSO b) With aoh: -OH aoh O a HOH 88

22 COOH COOa -OH aoh OH HOH H 3 COO aoh H COO a HOH c) With ahco 3 : COOH -OH ahco 3 COOa OH CO HOH d) With H SO 3 H O 3 S H 3 ao ahso H 4 HOH SO 4 O 3 S O 3 S OH aoh HO 3 S OH (orange) O 3 S COOH -OH aoh HO 3 S COOH OH (orange) H SO 4 O 3 S H HO 3 S H (orange) II. e) With β-naphthol: H ao H SO 4 HSO 4 89

23 HSO 4 OH aoh OH a SO 4 HOH yellow - orange HOOC H ao H SO 4 OOC ahso 4 HOH OOC OH aoh aooc OH HOH red The following dyes can be obtained: aoh HSO 4 OH OH (red - orange) COOH -OH aoh COOH OH (red - orange) HOOC HSO 4 OH HOOC OH (red - orange) HOOC HSO 4 COOH COOH -OH HOOC OH (red - orange) 90

24 PROBLEM (practical) A solution in a graduated flask contains a mixture of oxalic acid and ammonium oxalate. One of the bottles denoted X, Y, and Z contains a solution of a calibration substance with reducing character at a concentration of mol dm -3. You are required to solve the following tasks: a) Determine the quantity of oxalic acid and of ammonium oxalate in the solution in the graduated flask. (The result will be given in grams.) b) Write the formula for the substance with reducing character and the equations of the chemical reactions which led to its determination. In order to carry out the analyses the following solutions are available: HCl (c = mol dm -3 ), aoh (c = mol dm -3 ), KMnO 4 (c = 0.0 mol dm -3 ), 5 % H SO 4, HO 3 (c = mol dm -3 ), 5 % BaCl, 5 % AgO 3, 5 % Hg (O 3 ), phenolphthalein 0.1 %, methyl red 1 %. c) Describe the procedure used in the individual steps, indicators employed and partial results. M r (H C O 4 ) = M r ((H 4 ) C O 4 ) = SOLUTIO ASWER SHEET: A 1 Identification of the solution with the reducing substance X, Y, Z: Fe(H 4 ) (SO 4 ) A Identification reactions for the ions of the substance - Fe aoh Fe(OH) a - H 4 aoh H 3 H O a - 4 H 3 Hg (O 3 ) H O O(Hg) H.O 3 3 H 4 OH - - SO 4 BaCl BaSO 4 Cl B 1 Preparation of the 0.1 M aoh solution cm 3 in 00.0 cm 3 91

25 B Concentration of the aoh in its solution: M Indicator used: C Concentration of KMnO 4 in its solution M D 1 Mass of oxalic acid in the initial solution g Indicator used D Mass of ammonium oxalate in the initial solution g S o l u t i o n A 1 1- cm 3 of solution X, Y and Z are put into three test tubes. 6 H SO 4 and a drop of KMnO 4 solution are added. The solution which loses colour is the one with reducing character. A Establishment of the formula: aoh greenish white precipitate Fe aoh at the upper end of the test-tube, filter paper with a drop of Hg (O 3 ), black spot H BaCl white precipitate - SO AgO 3 HO 3 Cl is absent Accordingly the substance used is Fe(H 4 ) (SO 4 ). The chemical reactions: Fe a OH Fe(OH) a H 4 a OH H 3 H O a 4 H 3 Hg (O 3 ) O(Hg) H. O 3 Hg 3 H 4 O 3 - SO 4 Ba Cl BaSO 4 Cl B 1 5 cm 3 M solution 100 cm M solution B V cm HCl 0.1 aoh in the presence of phenolphthalein. C V cm 3 solution X 10.0 cm 3 H SO 4 H O is titrated at elevated temperature with KMnO 4. 9

26 D 1 The solution which is to be analyzed is filled to the mark; V cm 3 of this solution is titrated with aoh in the presence of methyl red. The quantity of oxalic acid (moles and g) is calculated. D V cm 3 solution to be analyzed 10.0 cm 3 6 H SO 4 H O are heated and titrated with KMnO 4 solution. The total amount of oxalate is calculated (in mol). The difference gives the amount of ammonium oxalate (moles and g). 93

27 PROBLEM 3 (practical) Six test-tubes contain aqueous solutions of FeSO 4, H SO 4, Mn(O 3 ), H O, Pb(O 3 ), aoh. a) Identify the content of each test-tube without using other reagents. Write the results in tabular form. Write the equations for the chemical reactions used for the identification. b) After identification, perform four reactions each time using three of the identified compounds and write the equations. SOLUTIO FeSO 4 H SO 4 Mn(O 3 ) H O Pb(O 3 ) aoh 1) FeSO 4 Fe(OH)SO 4 yellowish PbSO 4 white Fe(OH) whitegreenish Fe(OH)3 brownredish ) H SO 4 3) PbSO 4 white Mn(O 3 ) 4) HO 5) Pb(O 3 ) 6) aoh Fe(OH)SO 4 yellowish PbSO 4 white Fe(OH) whitegreenish Fe(OH) 3 brownredish Mn(OH) white MnMnO 3 brown black PbSO 4 white Mn(OH) white MnMnO 3 brown black Pb(OH) white Pb(OH) 4 94

28 Reactions Observation (1) (4) FeSO 4 H O Fe(OH)SO 4 Colour change - yellowish (Fe 3 ) (1) (5) FeSO 4 Pb(O 3 ) PbSO 4 Fe(O 3 ) Appearance of a white precipitate. (1) (6) FeSO 4 aoh Fe(OH) a SO 4 Fe(OH) ½ O H O Fe(OH) 3 Appearance of a greenish white precipitate Fe(OH) which after oxidation by air turns into a reddish brown precipitate Fe(OH) 3. () (5) H SO 4 Pb(O 3 ) PbSO 4 HO 3 Appearance of a white precipitate PbSO 4. (3) (6) Mn(O 3 ) aoh Mn(OH) ao 3 Mn(OH) ½ O MnMnO 3 H O Appearance of a white precipitate Mn(OH) which after oxidation by air coverts into a Mn(OH) ½ O MnO H O brown-black precipitate MnMnO 3 which eventually changes into MnO a blackbrown precipitate. (5) (6) Pb(O 3 ) aoh Pb(OH) ao 3 Pb(OH) aoh a Pb(OH) 4 Appearance of a white precipitate Pb(OH) which dissolves in excess reagent. b) (1) () (4) FeSO 4 H O H SO 4 Fe (SO 4 ) 3 H O Colour change yellowish (Fe 3 ) (1) (4) (6) FeSO 4 H O 4 aoh Fe(OH) 3 a SO 4 Appearance of a brown-reddish precipitate Fe(OH) 3 (3) (4) (6) Mn(O 3 ) H O aoh MnO ao 3 H O Appearance of a brown precipitate MnO (5) (4) (6) Pb(O 3 ) H O aoh PbO ao 3 H O Appearance of a brown precipitate PbO. 95

Chem 1A Dr. White Fall Handout 4

Chem 1A Dr. White Fall Handout 4 Chem 1A Dr. White Fall 2014 1 Handout 4 4.4 Types of Chemical Reactions (Overview) A. Non-Redox Rxns B. Oxidation-Reduction (Redox) reactions 4.6. Describing Chemical Reactions in Solution A. Molecular