الشھادة الثانویة العامة فرع علوم الحیاة مسابقة في مادة الكیمیاء المدة: ساعتان

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1 وزارة التربیة والتعلیم العالي المدیریة العامة للتربیة داي رة الامتحانات امتحانات الشھادة الثانویة العامة فرع علوم الحیاة مسابقة في مادة الكیمیاء المدة: ساعتان الاسم: الرقم: دورة سنة ۲۰۰٦ العادیة 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 Following Three Exercises: First Exercise (7 points) Hydrolysis of an Ester The hydrolysis of an ester (E) gives an alcohol (A) and an acid (B). The aim of this exercise is to identify the compounds (A), (B), and (E). I- Study of the Hydrolysis Reaction A mixture of ester (E) and water is heated with few drops of concentrated sulphuric acid solution. After a certain time, the established homogenous equilibrium is represented by the following equation: E + H O A + B - Let x be the number of moles of (B) formed at equilibrium. Rewrite and complete, on the answer sheet, the following table: E H O A B Initial state mol mol 0 0 Equilibrium state - Indicate the role of heating and that of sulphuric acid in performing this hydrolysis reaction. 3- Determine the equilibrium constant K c knowing that x = 0.4 mol. 4- Propose and justify a way to make the hydrolysis of (E) almost complete. II- Identification of Compounds (A), (B) and (E) After an almost complete hydrolysis of ester (E), the obtained alcohol (A) and acid (B) are separated. - A study shows that (A) is a saturated monoalcohol, with an open carbon chain, and the mass of carbon is 3 times that of oxygen. a) Determine the molecular formula of (A). b) In order to identify this alcohol, the following tests are carried out: Test N o Initial mixture Observation and experimental results Test (A) + Acidified potassium permanganate solution (purple) - Final solution is colorless. - Formation of an organic compound (C). Test (C) +,4- D.N.P.H. - Yellow precipitate. Test 3 (C) + Fehling s solution (blue) - Blue solution. Deduce the condensed structural formula of (A) and give its name. - A study shows that (B) is an α-amino acid.

2 a) Write the general formula of an α-amino acid. b) Deduce the condensed structural formula of acid (B) knowing that its group R is composed of carbon and hydrogen atoms and its molar mass is M (B) = 89 g.mol -. Given, in g.mol -. M (C) = ; M (H) =; M (O) =6 and M (N) = 4. 3-Write the condensed structural formula of ester (E) and give its name. Second Exercise (6 points) Kinetics for the Decomposition of C 6 H 5 N Cl C 6 H 5 N Cl is a non-oxygen explosive which is stored at a temperature less than 5 0 C. The decomposition of C 6 H 5 N Cl in an aqueous medium at a temperature of 5 0 C is slow. This decomposition takes place according to the following equation: Given: C 6 H 5 N Cl (aq) C 6 H 5 Cl (aq) + N (g). - Molar volume of gas at the experimental conditions: V m = 5 L.mol -. I- Preliminary Study Consider L of C 6 H 5 N Cl aqueous solution of concentration C =.0x0-3 mol.l -. - Calculate, in ml, the volume of nitrogen gas formed as time tends to infinity. - Show, at each instant t, that the concentration of C 6 H 5 Cl, expressed in mol.l - is given by the following expression: [C 6 H 5 Cl] t = 4x0-5 xv where V is the volume, in ml, of nitrogen gas formed at each instant t. II- Kinetics of this Decomposition The kinetics of this decomposition reaction in solution (S) is studied by measuring the volume V of the liberated nitrogen gas at different instants. The results are given in the following table: t (min) V (ml) [C 6 H 5 Cl] t 0-4 (mol.l - ) - Rewrite and complete, on the answer sheet, the above table. - Plot, on a graph paper, the curve [C 6 H 5 Cl] = f(t). Take the following scales: abscissa: cm for min; ordinate: cm for x0-4 mol.l Determine the half-life of the reaction. 4- One liter of C 6 H 5 N Cl solution, having a higher concentration than solution (S) is prepared. The kinetics of the decomposition of C 6 H 5 N Cl is studied at the same temperature 5 0 C a) The obtained volume of nitrogen gas at t = 6 min exceeds 9 ml. Justify. b) Compare the volume of the liberated nitrogen gas at the end of the reaction (t ) to the volume of the nitrogen gas obtained in question (I- -).

3 Third Exercise (7 points) A Scale Product for Coffee-Pot The main constituent of a scale product is sulfamic acid. It is sold commercially as white small crystals. The aim of this exercise is to determine the percentage of sulfamic acid in the scale product by ph-metry. Given: - Sulfamic acid of formula NH SO 3 H will be represented as HA in this exercise. - Molar mass of sulfamic acid in g.mol - : M = 97.. Experimental Procedure: first step:.05 g of this scale product are dissolved in distilled water to obtain a solution (S) of volume V = 500 ml. second step:a volume V a = 0 ml of solution (S) is put into a beaker. A volume V of distilled water is then added to immerse the electrode of the ph-meter. third step: Titration is carried out by using sodium hydroxide solution of concentration C b = 4x0 - mol.l -. I- Preliminary Study - Sulfamic acid is a strong acid. Write the equation of the reaction of this acid with water. - Indicate the appropriate material used to: o weigh the mass of.05 g ; o prepare the volume V ; o withdraw the volume V a ; o add the sodium hydroxide solution. 3- Write the equation of the titration reaction. 4- Based on the chemical species which are present in the beaker, justify the value of ph = 7 at the equivalence point. II- Make Use of the Titration Results Part of the titration results is given in the following table: V b (ml) ph Draw the shape of the curve ph = f(v b ) for 0 V b 5 ml. Take the following scales: abscissa : cm for ml, ordinate: cm for unit of ph. - Determine the concentration C a of sulfamic acid in solution (S). 3- Deduce the mass percentage of sulfamic acid in this scale product. III- Effect of Dilution on the ph Measurements A solution (S ) of the scale product is prepared by diluting solution (S) ten times. - Among the following kits, choose by justifying, the kit that should be used to carry out the most precise dilution of solution (S). 3

4 Kit Kit Kit ml volumetric pipet ; 00 ml volumetric flask. 0 ml volumetric pipet ; 00 ml beaker. 0 mlgraduated cylinder; 00 ml volumetric flask. - Another titration is carried out with (S ) by repeating the second step of the above experimental procedure and using another sodium hydroxide solution of concentration C ' b = 4x0-3 mol.l -. Give, by justifying, the value of ph that corresponds to the volume V b =0.4 ml. 4

5 Official Exam Marking Scheme of Chemistry L. S. t session 006 First Exercise (7 points) Hydrolysis of an Ester Expected Answer Mark Comment I- - According to the stoechiometric coefficients, we have: nereacting = nh Oreacting na formed = nb formed = = x E H O A B Initial State mol mol Equilibrium State (-x) mol (-x) mol x mol x mol - Heating increases the rate of the reaction (kinetic factor). Sulphuric acid, which is added in a small quantity, has the role of a catalyst. 3- The equilibrium constant of the above reaction is giving by the following relation: K c = [ A][ B] [ E ][ H O. We have then ] K c = V V 4 = = V V 4- In order to have a complete hydrolysis, we add a quantity of water. The reaction is then displaced in the forward direction (direction of hydrolysis), according to Le Chatelier s principle, when a stress is applied to a system at an equilibrium state, the system readjust itself by reducing or opposing this stress. II- - a) The general formula of a saturated monoalcohol of an open carbon chain is C n H n+ O. Since m (C) = 3 m (O), we have then: n = 3x6. n = 4. The molecular formula of (A) is then C 4 H 9 OH. b) According to test, we conclude that (A) is a primary or a secondary alcohol, because it undergoes mild oxidation. According to test, (C) is a carbonyl compound (aldehyde or ketone). According to test 3, we conclude that (C) is a ketone. Then (A) is a secondary alcohol of condensed structural formula: CH 3 CH CHOH CH 3. Its name is -butanol. - a) The general formula of an α-amino-acid is: R CH COOH NH b) M R = 89, then: M R = 5. R formed of carbon and hydrogen should be in the form of CH 3. c) The formula of ester (E) is then: CH 3 CH C O CH CH CH 3. NH O CH 3 It is -butyl --amino propanoate. x Any other suitable proposition is acceptable C n H n+ OH

6 Second Exercise (6 points) Kinetics Of the Decomposition of C 6 H 5 N Cl Expected Answer Mark Comment I- - According to the stoechiometric coefficients: n (N ) formed at t = n (C 6 H 5 N Cl) = CxV = x0-3 mol. The volume of nitrogen is then: V (N ) = nxv m =x0-3 x5=50x0-3 L=50 ml - n (C 6 H 5 Cl) formed = n (N ) formed. V ( N ) Vx [C 6 H 5 Cl] t = V xv = = Vx = 4x0-5 V m solution Vm x 5 II- - We multiply by 0.4, we have then : t (min) [C 6H 5N Cl]x0-4 mol.l The curve [C 6 H 5 Cl] = f(t). La courbe 0 Concentration Series t (min) 3- The half-life of the reaction is the time needed for half the amount of C 6 H 5 N Cl to disappear. So the time that corresponds to the concentration [C 6 H 5 Cl] t/ = 0x0-4 mol.l - is t ½ = 8. min. 4- a) Since the new solution has a higher concentration than solution (S), so the rate of the reaction is greater. Starting from the same volume L, the obtained volume of nitrogen is greater than 9 ml at t = 6 min. b) Because the solution has a higher concentration, starting from the same volume L, the number of moles of the reactant is greater and consequently the volume of nitrogen gas at t is greater than the volume obtained in the first case or the time needed for half the maximum amount of C 6 H 5 Cl to appear.

7 Third Exercise (7 points) Scale Product for Coffee-Pot Expected Answer Mark Comment I- - The equation of the reaction of sulfamic acid with water is: HA + H O H 3 O + + A - To weigh: sensitive balance (/00 ). To prepare V = 500 ml: 500 mlvolumetric flask. 4x To take V a : 0 ml volumetric pipet. To add NaOH: buret. 3- Since, it is a reaction between a strong acid and a strong base, the equation is then: H 3 O + + HO H O. 4- The species that are present at equivalence point, other then water, are Na + and A, which are spectator ions (neutral particles). The ph is then equal to 7. II- - Allure de la courbe représentant la variation du ph en fonction de Vb ph Vb en ml Series - At equivalence point, we have: n H 3 O + (in beaker) = n HO (poured). Knowing that, in a solution: n = CxV, where: C a V a = C b V be ; we 4xo x0.4 conclude: Ca = = 4.6x0 - mol.l n (HA) = 4.6x0 - x =.08x0 - mol ; m (HA) =.08x0 - x97. =.0 g ; and the percentage by.0 mass is: % = x 00 = % III- - When the solution is diluted ten times, the final volume becomes ten times greater than the initial volume. The suitable 3

8 glass-ware that should be used are: 0 ml volumetric pipet and 00 ml volumetric flask, so kit-- is convenient. - Because the titration takes place between a strong acid and a strong base and both have been diluted 0 times, the ph at the equivalence point, where V be = 0.4 ml remains the same, is equal to 7. Ca Or C ' V ' = C ' V ' V ' = 0 4.6x0 b b a a b x0 = x0 = 0.4mL C b 4x0 0 So the ph that corresponds to this volume is 7. 0,5 4

مسابقة في الكیمیاء الرقم:

دورة سنة ۲۰۰٦ العادیة امتحانات شھادة الثانویة العامة وزارة التربیة و التعلیم العالي فرع العلوم العامة المدیریة العامة للتربیة داي رة الامتحانات الاسم: مسابقة في الكیمیاء الرقم: المدة ساعتان This Exam Includes