مسابقة في الكیمیاء الرقم:

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1 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 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 (6 points) Esterification Reaction Two flasks are available. Their labels show the following indications: Flask (): acid (A) of formula R COOH. Flask (): alcohol of formula C 3 H 7 OH. The aim of this exercise is to identify the content of each of the two flasks. Given: - Molar mass in g.mol - : M (H) = ; M (C) = ; M (O) = 6. - The value of the equilibrium constant K c of the esterification reaction * of a carboxylic acid and a primary alcohol is K c = 4.; * of a carboxylic acid and a secondary alcohol is K c =.. I- Identification of the Carboxylic Acid (A) The dehydration of.48 g of (A), in the presence of P O 5 as a dehydrating agent, leads to the formation of the anhydride of acid (A) and 0.0 mol of H O. - Write the equation of the above dehydration reaction. - Determine the number of moles of (A). Deduce its molar mass. 3- Show that the molecular formula of (A) is C 3 H 6 O, knowing that R is an alkyl group. 4- Write the condensed structural formula of (A) and give its name. II- Identification of the Alcohol by Esterification Reaction A mixture of acid (A) and the alcohol is heated for a certain time. A homogenous equilibrium, in the reactional system of volume V, is established. It is represented by the following equation: R COOH + C 3 H 7 OH R COOC 3 H 7 + H O. The remaining amount of acid at equilibrium is titrated. This titration permits to determine the number of moles of each constituent at equilibrium. The result is given in the following table: Constituent R COOH C 3 H 7 OH R COOC 3 H 7 H O n equilibrium (mol) Specify the role of heating on the above reaction and its effect on the equilibrium. - Determine the equilibrium constant K c and deduce the class of the used alcohol. 3- Write the condensed structural formula of this alcohol and give its name. 4- Write the condensed structural formula of the obtained ester in this reaction and give its name.

2 Second Exercise (7 points) Ammonia Solution Ammonia is commercially sold in dark bottles as concentrated solutions. Given: Ammonia is a weak base that corresponds to the pair: NH + 4 / NH 3. I- Titration of a Commercial Solution of Ammonia (S 0 ) A commercial solution of ammonia, (S 0 ), is diluted 00 times to obtain a solution (S). A volume V S = 0 ml of solution (S) is titrated with hydrochloric acid solution of concentration C a = 0. mol.l -, in the presence of a coloured indicator. The volume of the acid solution added to reach the equivalence point is V ae =.6 ml. - Write the equation of the titration reaction. - Determine the concentration of solution (S). Deduce that the concentration of solution (S 0 ) is C 0 = 0.8 mol.l Specify, by referring to the chemical species present in the beaker, the acid-base nature of ph at equivalence. 4- Choose, by justifying, which of the two indicators given in the table below, the one that will be more convenient to detect the equivalence point. Indicator Methyl red Phenolphthalein ph range Red : : yellow colorless : 8. 0 : purple 5- Indicate the change in color observed during the titration. II- Synthesis of Ammonia to Prepare (S 0 ) Ammonia is synthesized from a mixture composed of n mol of N and 3 n mol of H. This synthesis reaction is reversible and is represented by the following equation: N (g) + 3 H (g) NH 3 (g). - Let α be the degree of conversion of N at equilibrium. Rewrite, on the answer sheet, the table below and complete it in terms of n and α. N (g) H (g) NH 3 (g) Total number of moles:n t Initial state (mol) n 3 n 0 Equilibrium state (mol) - P is the total pressure at equilibrium. Give, in terms of α and P, the expression of the partial pressure of each of the constituents at equilibrium. 3- At equilibrium P(NH 3 ) = P(N ). a) Find the value of α. b) Deduce the value of n needed to prepare L of solution (S 0 ) of concentration C 0 = 0.8 mol.l -.

3 Third Exercise (7 points) Sodium Thiosulfate Hydrated sodium thiosulfate, Na S O 3.H O, is a white solid. It undergoes a self oxidation-reduction reaction in a strongly acidic medium. It is a reducing agent when used to titrate iodine solution. Given: Molar mass in g.mol - : M (H) = ; M (O) = 6; M (Na) = 3; M (S) = 3. I- Preparation of a Sodium Thiosulfate Solution A volume of 500 ml of sodium thiosulfate solution of concentration C = mol.l - is to be prepared. - Determine the mass of hydrated sodium thiosulfate needed for this preparation. - Choose, from the list below, the appropriate material needed to perform this preparation: - sensitive balance - 0 and 0 ml volumetric pipets - 50 and 500 ml beakers - spatula - 50 and 500 ml volumetric flasks - watch glass - buret - 50mL and 00mL graduated cylinders - funnel II- Self Oxidation-Reduction of Sodium Thiosulfate At time t = 0, an excess of concentrated hydrochloric acid solution is placed into a beaker containing a volume V = 40 ml of a sodium thiosulfate solution of concentration C = mol.l -. With time, suspended fine solid particles are progressively formed. The equation of this reaction is: S O + H 3 3O + S (s) + SO + 3 H O - Show that the number of moles of the sulfur formed at the end of the reaction is equal to x0 - mol. - Draw the shape of the curve that represents the variation of the number of moles of sulfur formed versus time: n (s) = f(t), that passes through the points of abscissa: t = 0; t / = min and the point of the following coordinates: [ t = 5 min and n (S) = x0 - mol ]. Take the following scale: abscissa cm for min; ordinate cm for 4x0-3 mol 3- Specify how the rate of formation of sulfur varies with time. Deduce the kinetic factor responsible for this variation. 3

4 III- Indirect Titration of H O At time t = 0, hydrogen peroxide solution, an excess of potassium iodide solution, and concentrated sulphuric acid solution, are placed into a beaker. We observe a coloured reactional medium that becomes darker with time. The equation of this reaction is: H O + I + H 3 O + I + 4 H O () In order to determine the amount of H O left in the reactional medium at a given time t, the beaker is immersed in an ice-water bath. The iodine formed is then titrated with sodium thiosulfate solution of concentration C = mol.l - in the presence of starch. The equation of the reaction is: S O + I 3 S 4 O + I 6 () - Specify the reason for immersing the beaker that contains the reactional medium in an ice-water bath. C.V - Show that at instant t, we have the relation: n t = n 0 - E where n 0 = n (H O ) at t = 0; n t = n (H O ) left at instant t; C = concentration of sodium thiosulfate solution used in the titration; V E = volume of sodium thiosulfate solution added to reach equivalence. 3- Calculate V E that would be added to titrate the iodine formed at the end of reaction () when n 0 = 0.0mol. 4- Explain the formation of turbidity (fine solid particles) in the beaker when the volume of sodium thiosulphate solution exceeds V E. 4

5 General Sciences t session 006 Marking Scheme of Chemistry First Exercise (6 points) Esterification Reaction Expected Answer Mark Comments I- - The equation of the dehydration reaction of R COOH is: R COOH (R CO) O + H O - According to the stoechiometric coefficients, we have: n (R COOH) reacting = n (H O) formed = x 0.0 = 0.0 mol. m.48 The molar mass M = = = 74 g.mol -. n In the formula R COOH, R is an alkyl group of formula: C n H n+, its mass is 4n+. Where: M = 4n So n = and the molecular formula of acid (A) is C 3 H 6 O. 4- The condensed structural formula of (A) is: CH 3 -CH -COOH. Its name is propanoic acid. II- - Heating speeds up the reaction or it decreases the time that is needed to reach the equilibrium. It has no effect on the equilibrium because esterification reaction is athermic. - The equilibrium constant is given by the following expression: [ R COOC3H 7 ].[ H O] K c =. With the given values, we have: [ R COOH ].[ C3H 7OH ] x K c = V V =.8. V is the volume of the reactional x V V medium. By comparing the value of K c with the two given values (4. and.) we deduce that the alcohol is a secondary alcohol. 3- The condensed structural formula is CH 3 CHOH CH 3 Its name is -propanol. 4- The condensed structural formula of the ester is: CH 3 CH C O CH CH 3. It is the -propylpropanoate O CH x0.5 x x0.5 Second Exercise (7 points) Ammonia Solution Expected Answer Mark Comment I- - The equation of the titration reaction is: NH 3 + H 3 O + NH + + H 4 O - At equivalence, we have : n (NH 3 ) in 0 ml of (S) = n (H 3 O + ) in.6 ml of acid solution. Then: x.6 C (S) xv (S) = C a V a. We conclude: C (S) = 0 = 0.08 mol.l -.

6 Solution (S) is obtained by diluting 00 times the solution (S 0 ), then, the concentration of (S 0 ) is: C 0 = 00 x C (S) = 00 x 0.08 = 0.8 mol.l The present species at equivalence, other than water, are NH + 4 and Cl. Cl is a spectator ion, but NH + 4 is a weak acid that leads to an acidic medium (ph < 7) at equivalence. 4- Because the ph at equivalence point is less than 7, the convenient indicator to detect the equivalence point is that having ph range included between two values less than 7. This indicator is the methyl red. 5- First the solution is yellow (solution (S) base), it becomes red after the equivalence point. II- N (g) H (g) NH 3(g) n total (mol) Initial state n 3 n 0 4 n Équilibrium state n (-α) 3 n (-α) nα n (-α) n - In a gaseous mixture, the partial pressure P i = i xp. Then: n t n( α) ( α)p P (N ) = xp =, n( α) ( α) 3n( α) 3( α)p P (H ) = xp = and n( α) ( α) nα αp αp P (NH 3 ) = xp = = n( α) ( α) ( α) 3- The value of α that verify the relation P (NH 3 ) = P (N ) is αp ( α) P given by the relation : =. Where α =. ( α) ( α) 4- To prepare L of (S 0 ) the needed number of mol is: x 0.8 = 0.8 mol of ammonia. n (NH 3 ) = nα. We have then: 0.8 = nα = n x so n = 0.8 mol Third Exercise (7 points) Sodium Thiosulfate Expected Answer Mark Comment I- - To prepare this solution, we need n = C x V mol hydrated sodium thiosulfate of mass m = C x V X M, with: M = x3+x3+3x6+x8 = 94 g.mol -. We have: m = x 500 x0-3 x 94 = 48.5 g. - The needed material to carry out this preparation includes: sensitive balance, 500 ml volumetric flask, funnel, watchglass and spatula. II- - n (S) at the end of the reaction = n (S O 3 ) initial = CxV = 0,4x50x0-3 = x0 - mol..5

7 n(s) -At t = 0, n (S) = 0 and at t /, n (S) formed = = x0 - mol mol Shap of the curve n (S) = f(t) number of moles of sulfur t (min) 3- The value of the rate is equal to the slope of the tangent on the curve: n (S) = f(t). An observation shows that this slope decreases when the time increases. We conclude that the rate decreases as time increases. In fact, when time increases the concentration of the reactants decreases that leads to a decrease in the rate of the reaction. The concentration of reactants is then the kinetic factor which is responsible for this variation. III- -The advantage to put the beaker that contains the reactional medium into an Ice-water bath is to block the reaction. - Based on the equation of reaction () n (H O ) reacting = n (I ) formed = n (H O ) at t = 0 n (H O ) remaining at t. n (I ) formed = n 0 n t. Based on the equation of reaction (): n(s O ) n (I ) formed () = n (I ) reacting in () = 3 added = n 0 - CxV Where n t = n 0 - E 3- At t n t = 0 V = n 0 CxV E = 0 VE = n 0 = C CxV E. x0.0 = 0,04 L = 40 ml 4- Where volume of thiosulfate exceeds 40 ml, the excess of thiosulfate in the strongly acid medium undergoes self oxidationreduction that leads to the formation of fine solid particles of sulfur. 3