امتحانات الشهادة الثانوية العامة الفرع: علوم عامة مسابقة في مادة الكيمياء المدة: ساعتان

Transcription

1 وزارة التربية والتعليم العالي المديرية العامة للتربية دائرة االمتحانات الرسمية امتحانات الشهادة الثانوية العامة الفرع: علوم عامة مسابقة في مادة الكيمياء المدة: ساعتان االسم: الرقم: دورة العام 72 العادي ة الخميس 5 حزيران 752 This Exam Includes Three Exercises. It Is Inscribed on 4 Pages Numbered From to 4. The Use of A Non-programmable Calculator is Allowed Answer The Three Following Exercises: Exercise (6 points) Identification of an organic compound Available is an organic compound (A) of molecular formula C 4 H 8 O of a saturated non cyclic carbon chain. The aim of this exercise is to identify compound (A) in order to prepare an ester (E).. Identification of The Compound (A) The compound (A) is subjected to the two tests in document-. Chemical Test Document- Experimental Result Test : (A)+ DNPH Formation of yellow-orange precipitate Test 2: (A)+ Fehling reagent Formation of brick-red precipitate.. Interpret the result of each of these two tests..2. Write the possible condensed structural formulas of the compound (A)... Name compound (A) knowing that the carbon chain is non-branched. 2. Preliminary Study (B) and (C) are two organic compounds used in preparation of the ester (E). (B) is obtained by catalytic hydrogenation of a sample of the compound (A). (C) is obtained by mild oxidation of another sample of the compound (A). 2.. Write, using condensed structural formulas, the equation of the reaction of formation of product (B). Name it Identify the organic compound (C).. Esterification Reaction An equimolar mixture of the compounds (B) and (C) is heated to reflux in the presence of few drops of concentrated sulfuric acid..- Indicate the role of sulfuric acid..2- Give the condensed structural formula and the name of the ester (E) obtained during this reaction. Page of 4

2 .- The carboxylic acid used in the preparation of the ester (E) is replaced by its chlorinated derivative.... Identify the derivative used...2. Choose, from the three following propositions, the one that corresponds to the characteristics of the above reaction: a- complete and athermic b-slow and athermic c- complete and exothermic... Write, using condensed structural formulas of the organic compounds, the equation of the reaction of formation of the ester (E) in this case. Exercise 2 (7points) Sodium Thiosulfate and Hydrochloric Acid In acidic medium, thiosulfate ions (S 2 O ) react slowly and completely with the hydronium ions (H O + ), according to the following equation: S 2 O (aq) + 2 H O + (aq) S (S) + SO 2(aq) +H 2 O (l) In order to study the kinetic of the above reaction, the following experiment is carried out, At the instant t =, a volume V =. ml of a hydrochloric acid solution (H O + + Cl - ) of concentration C = 5. mol.l - is poured into a beaker containing a volume V 2 = 4. ml of a sodium thiosulfate solution (2Na + + S 2 O ) of a concentration C 2 =.5 mol.l -. By an appropriate method the evolution of this reaction is followed and the concentration of the thiosulfate ions is determined at different instants. The results are grouped in the table of document-. t (s) [S 2 O ] mol. L Preliminary Study Document-.. Show that the initial concentration of the thiosulfate ions is [S 2 O ] o = mol.l - and that of hydronium ions is [H O + ] o =. mol.l - in the reactional mixture..2. Identify the limiting reactant. 2. Kinetic Follow-up 2.. Plot the curve representing the variation of the concentration of thiosulfate ions as a function of time [S 2 O ] = f (t) within the time interval: [ - s]. Take the following scales: abscissa: cm for s ordinate: cm for.4 mol.l Determine, graphically, the half-life time t / Show that at instant t = t /2 the concentration of hydronium ions, [H O +, is given by the following relation: [H O + [H O + ] o - [S 2 O ] o 2.4. Deduce the value of [H O + Page 2 of 4

3 2.5. Choose among the three graphs of document-2 the one that corresponds to the shape of the curve that represents the variation of the concentration of H O + ions as a function of time. Justify. [H [H O + ] mol.l - [H O + ] mol.l - O + ] mol.l t / t /2 t(s) t(s) t(s) Graph a Graph b Graph c Document t /2. Kinetic Factors To highlight the effects of the kinetic factors on the duration of this reaction.the three experiments represented in document- are carried out, where t represents the end time of the reaction in each experiment. [S 2 O ] o [H O + ] o Temperature ( o C) Time (t) Experiment mol.l - mol.l - 4 t Experiment 2 mol.l - mol.l - 2 t 2 Experiment mol.l - mol.l - 4 t Document- Compare t 2 and t as well as t and t.justify. Exercice (7 points) Acid-Base Reactions The aim of this exercise is to identify aqueous solutions in order to realize a ph-metric study of an acid-base mixture. Acid/Base pair H O + / H 2 O C 6 H 5 COOH / C 6 H 5 COO NH 4 / NH pka The study is carried out at 25 o C. Document-. Identification of Aqueous Solutions Available are three beakers numbered,2 and. Beaker contains a hydrochloric acid solution (H O + + Cl - ). One of the two other beakers contains an aqueous solution of sodium benzoate (Na + + C 6 H 5 COO ) and the other beaker contains an aqueous ammonia solution NH. Page of 4

4 All of the above three solutions have the same molar concentration C. The ph of each solution is measured. The results are grouped in the table of document-2. Number of 2 beaker ph. 8.5 Document-2.. Show that the concentration C is equal to 5. 2 mol.l Identify, by referring to documents () and (2), the solution contained in each of the two beakers 2 and... The ammonia solution of concentration C is prepared from a commercial solution (S o ) of concentration C o = mol.l -. Choose, by justifying, from the following two sets a and b of document- the appropriate one for the above preparation. - volumetric pipet: 5 ml. - graduated pipet: 5 ml. -volumetric flask: 5 ml. - volumetric flask: 5 ml. Set a 2. ph-metric Follow-up Document- Set b The hydrochloric acid solution of concentration C is added progressively into a beaker containing a volume V b = 2. ml of solution of ammonia of concentration C. 2.. Write the equation of the reaction that takes place between H O + ions and NH Show that this reaction is complete. 2.. Determine the volume, V E, of the acid solution added at equivalence Choose among the three following values: ph = 2 ; ph 2 = 7 ; ph = the one that corresponds to the ph value of the solution obtained after the addition of a volume of the acid equal to ml. Justify without calculation Plot the shape of the curve that represents the variation of ph as a function of the volume of the hydrochloric acid added: V E ph = f (V a ), passing through the points of abscissa : V a = ; V a = ; Va = V E and V a = ml. 2 Take the following scale: abscissa cm for 2 ml and ordinate cm for ph unit. (Knowing that the ph at equivalence is equal to 5.4) Page 4 of 4

5 وزارة التربية والتعلين العالي الوديرية العاهة للتربية دائرة االهتحانات هشروع هعيار التصحيح اهتحانات الشهادة الثانىية العاهة فرع : علىم عاهة دورة العام 72 العادية االسن : الرقن : هسابقة في هادة الكيوياء الودة : ساعتاى Exercise (6 points) Identification of an organic compound Part of Answers the Q... Test : Compound (A) gives with DNPH a yellow orange precipitate ; compound (A) is a carbonyl compound (aldehyde or ketone). Test 2 : Compound (A) gives a red brick precipitate with Fehling reagent ;(A) is an aldehyde..2. The isomers of (A) are: CH - CH 2 - CH 2 - CHO Mark CH - CH - CHO CH.. Compound (A) is Butanal CH - CH 2 - CH 2 - CHO + H 2 CH - CH 2 - CH 2 - CH 2 OH -butanol 2.2. CH - CH 2 - CH 2 - COOH butanoic acid.5.. The role of sulfuric acid is a catalyst 5.2. CH - CH 2 - CH 2 - COO - CH 2 - CH 2 - CH 2 CH Butylbutanoate... CH - CH 2 - CH 2 - COCl Butanoyl chloride c- Complete and exothermic CH - CH 2 - CH 2 - COCl + CH - CH 2 - CH 2 - CH 2 OH HCl + CH - CH 2 - CH 2 - COO - CH 2 - CH 2 - CH 2 CH.5 Exercice 2 (7 pts) Sodium Thiosulfate and Hydrochloric acid Part of Answers the Q... n HO CV 5x. [ HO ] mol. L Vtotal V V2 5. n SO CV 2 2 2,5x4. [ S2O ],4 mol. L V V V 5. total 2 Mark.2. R n,5 n HO S2O R S2O H O Page de

6 S 2 O is the limiting reactant 2.. [S 2 O 2- ] mol.l t(s) 2.2. Half time :is the time needed for the limiting reactant to lose half of its initial value. [ SO 2 ],4,2 mol. L 2 2 Graphically t /2 = 52s 2.2. At each instant of time t : n + HO remained = n + HO (o) - n + HO react n + HO react = 2 n S2O react n + HO remain = n + HO o - 2 n S2O react At t /2 : 2n SO 2 O n n n n HO res tan t HO O HO O S2O O 2 Divide by the volume of the solution : [H O + ] t/2 = [H O + ] o - [S 2 O ] o 2.4. [H O + ] t/2 = [H O + ] o - [S 2 O ] o = =.6 mol.l The graph c corresponds to the shape of the curve representing the variation of the concentration of H O + ions with respect to time, because : At t = we have [H O + ] o = mol.l - At t /2 = 52s we have [H O + ] t/2 =.6 mol.l - At t = s it does not reach the X- axis... The initial concentration of reactants and the temperature are two kinetic factors. Δt 2 > Δt. By comparing the 2 experiments and 2, we found that the initial concentration of reactants is the same in the 2 experiments but the temperature is higher in experiment than that in experiment 2. The rate of the reaction in experiment is higher than that in experiment 2. Δt > Δt. The temperature is the same in the two experiments but the concentration Page 2 de.5

7 of reactant S 2 O is less than that in experiment.the rate of the reaction in experiment is higher than that in experiment..5 Exercice (7 points) Acid-Base Reactions Part of the Q. Answers Mark.. In the beaker, hydrochloric acid is a strong acid : ph = - log C ;. = - log C ; C = -, = mol.l -.2. Ammonia NH and benzoate ion C 6 H 5 COO - are two weak bases. Since the two bases have the same concentration C. The base which has the higher ph is the stronger base. Since pk a (C 6 H 5 COOH/C 6 H 5 COO ) = 4.2 < pka (NH /NH 4 ) = 9.2. then NH is a stonger base than C 6 H 5 COO. Therefore ph of NH > ph of C 6 H 5 COO. Beaker 2 contains solution of NH Beaker contains benzoate ion C 6 H 5 COO -... Upon dilution the number of mole of the solute is conserved : n o = n ; C o V o = C V ; Vo = 5-2 V ; V = 2 V o ; For a volumetric flask of volume V = 5 ml ; Vo = 2,5 ml The appropriate set is b : graduated pipet 5 ml and volumetric flask 5 ml. 2.. H O + + NH NH H 2 O [ NH4 ] 9,2 9 4 KR,58. pka 9,2 [ NH][ HO ] Ka Therefore the reaction is complete 2.. At equivalence : n + HO (added) = n NH present in beaker ; C V E = C V b ; V E = V b = 2 ml V= ml > V E = 2 ml. The hydrochloric acid becomes in excess in the mixture in the beaker which renders the ph acidic (ph < 7). ph = The curve ph = f(v a ) passes through 4 remarkable points initial point: V a = ml ph = half-equivalence point : E /2 : V E/2 = ml ph = pk a = 9.2 equivalence point E : V E = 2 ml ph E = 5.4 Point after equivalence : V = ml ph = 2 ph V a.5 Page de