DEPARTMENT OF CHEMISTRY AND CHEMICAL TECHNOLOGY CHEMISTRY OF SOLUTIONS 202-NYB-05 15/16 TEST 3 MAY 2, 2012 INSTRUCTOR: I. DIONNE.

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1 DEPARTMENT OF CHEMISTRY AND CHEMICAL TECHNOLOGY CHEMISTRY OF SOLUTIONS 202-NYB-05 15/16 TEST 3 MAY 2, 2012 INSTRUCTOR: I. DIONNE Print your name: Answers INSTRUCTIONS: Answer all questions in the space provided. 1. Duration of this test is 75 minutes. 2. No books or extra paper are permitted. 3. Answer the questions in ink in order to preserve the right to grieve. 4. In order to obtain full credit for your answers, you must clearly show your work. Answers to problems involving calculations must be expressed to the correct number of significant figures and proper units. 5. Calculators may not be shared. Programmable calculators are not permitted. 6. Your attention is drawn to the College policy on cheating. This policy will be enforced. 7. A Periodic Table with constants is provided. Problem 1: /3 Problem 6: /3 Sig. Fig.: /1 Problem 2: /3 Problem 7: /3 Units: /1 Problem 3: /4 Problem 8: /3 Problem 4: /5 Bonus /1 Total: /30 Problem 5: /4

2 PROBLEM 1 A saturated aqueous solution of lead(ii) iodate, Pb(IO 3 ) 2, contains 3.1 x 10 5 M Pb 2+ at 25 C. Assuming that the ions do not react with water, calculate the K sp of lead(ii) iodate at that temperature. Pb(IO 3 ) 2 (s) Pb 2+ (aq) + 2IO 3 (aq) K sp = [Pb 2+ ][IO 3 ] 2 K sp = (3.1 x 10 5 )(2 x 3.1 x 10 5 ) 2 = (3.1 x 10 5 )(6.2 x 10 5 ) 2 K sp = x Answer: 1.2 x PROBLEM 2 (a) Which of these salts has the greatest molar solubility? Circle your choice. 1. Ag 2 S 2. CaF 2 3. MgF 2 smallest greatest (b) Will the solubility of the following salts depend on ph? yes no 1. AgCl 2. LiNO 3 3. Fe(OH) 2 4. NaOCl Page 2

3 PROBLEM 3 (4 marks) A solution is prepared by mixing 50.0 ml of 0.10 M Pb(NO 3 ) 2 with 50.0 ml of 1.0 M KCl. Calculate the concentrations of Pb 2+ and Cl at equilibrium. Assume precipitation of PbCl 2 (s) is complete. [K sp for PbCl 2 (s) is 1.6 x 10 5 ] Pb(NO 3 ) 2 (aq) + 2KCl (aq) PbCl 2 (s) + 2KNO 3 (aq) Pb(NO 3 ) 2 : 0.10 M x L = mol KCl: 1.0 M x L = mol After the reaction, we are left with Pb(NO 3 ) 2 : mol mol = 0 KCl: mol (2)(0.0050) = mol which is in ml K sp = [Pb 2+ ][Cl ] 2 K sp = 1.6 x 10 5 = (s)( s) 2 (s)(0.40) 2 s =1.0 x 10 4 Answers: [Pb 2+ ] = 1.0 x 10 4 M and [Cl ] = 0.40 M Page 3

4 PROBLEM 4 (5 marks) (a) If ln[reactant] is plotted versus time, and a straight line of negative slope is observed, what is the order of the reaction? Answer: First-order If 1/[ reactant] is plotted versus time =2 nd order (b) Indicate whether the following statements are true (T) or false (F). 1. T A rate law follows directly from the molecularity of an elementary step. 2. T The value of the rate constant increases as the temperature increases. 3. F A rate law could include the concentration of an intermediate. 4. F In a multi-step reaction, only one step can be fast and rate determining. 5. F The half-life of a first-order reaction depends on reactant concentration. (c) The reaction CO(g) + NO 2 (g) CO 2 (g) + NO(g) is second order in NO 2 and zero order in CO at temperatures less than 500K. 1. Write the rate law for the reaction. Rate = k[no 2 ] 2 2. How will the reaction rate change if the NO 2 concentration is halved? 1/4 original value (d) For the hypothetical reaction A + B C + D, the activation energy is 32 kj/mol. For the reverse reaction, the activation energy is 58 kj/mol. Is the reaction A + B C + D exothermic or endothermic? Answer: exothermic (e) Assume that an A molecule reacts with two B molecules in a one-step process to give AB 2. That is, A + 2B AB 2 If the initial rate of appearance of AB 2 is 2.0 x 10 5 mol/l s when the initial concentrations of A and B are 0.30 M, calculate the rate constant for the reaction. Rate = k[a][b] x 10 5 M/s = k[0.30 M][0.30 M] 2 k = x 10 4 M 2 s 1 Answer: 7.4 x 10 4 M 2 s 1 Page 4

5 PROBLEM 5 (4 marks) HOF decomposes by a first order reaction to HF and O 2 with a half-life of 30.0 minutes at 25 C. HOF(g) HF(g) + ½O 2 (g). If the partial pressure of HOF in a 1.0 L flask is initially 100. mmhg at 25 C, what is the partial pressure of HOF and the total pressure in the flask after 45.0 minutes? 1 st order: t 1/2 = 0.693/k k = / 30 min k = min 1 ln[x] = (0.0231)(45) + ln100 x = 35 HOF (g) HF (g) + ½O 2 (g). After 45 min P HOF = = 35 mmhg P HOF = 35 mmhg, P HF = 65 mmhg, P O2 = 33 mmhg Answers: P HOF = 35 mmhg and P total = 133 mmhg PROBLEM 6 Calculate the activation energy (E a ) for the reaction N 2 O 5 (g) 2NO 2 (g) + ½O 2 (g), from the observed rate constants at the following two temperatures: (a) at 25 C, k = 3.46 x 10 5 s 1 and (b) at 55 C, k = 1.5 x 10 3 s 1. Ln (k 2 /k 1 ) = E a /R(1/T 1 1/T 2 ) ln(3.16 x 10 5 /1.5 x 10 3 ) = E a /8.314(1/328 1/298) E a = x 10 5 J/mol Answer: 1.0 x 10 5 J/mol Page 5

6 PROBLEM 7 A study of the reaction 2A + B C + D gave the following results: Experiment [A] 0, M [B] 0, M R 0, M/s x x x x 10 1 What is the rate law for this reaction and the overall order of the reaction? Exp 1 / exp x 10 3 / 1.2 x 10 2 = k(0.100) x (0.050) y / k(0.200) x (0.050) y 0.50 = 0.50 x x = 1 Exp 2 / exp x 10 2 / 1.1 x 10 1 = k(0.200) x (0.050) y / k(0.200) x (0.150) y = y y = 2 Answers: Rate law = Rate = k[a][b] 2 Overall order = Third order Page 6

7 PROBLEM 8 The ozone decomposes according to the equation 2O 3 (g) 3O 2 (g) The mechanism of the reaction is thought to proceed through an initial fast equilibrium and a slow second step. 1. O 3 (g) O 2 (g) + O(g) (fast) 2. O 3 (g) + O(g) 2O 2 (g) (slow) (a) Which of the steps is the rate-determining step? (b) Write the rate law for the rate-determining step. (c) What is the molecularity of the first step? Step 2 Rate = k[o 3 ] 2 /[O 2 ] unimolecular Rate = k[o 3 ][O] but O is an intermediate k'[o 3 ] = k [O 2 ][O] [O] = k'[o 3 ]/k [O 2 ] Rate = k[o 3 ][O] = k[o 3 ]{k'[o 3 ]/k [O 2 ]} Bonus (no partial mark max of 30 for the test) (1 mark) After five half-life periods for a first order reaction, what fraction of reactant remains? t 1/2 1 1/2 2 1/4 3 1/8 4 1/16 5 1/32 Answer: Page 7