ACS EXAM KEY 11

ACS EXAM KEY 11 - 2006 U S NATIONAL CHEMISTRY OLYMPIAD...

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Not valid for use as an USNCO Olympiad National Exam after April 25, 2006. Page 1 2006 U. S. NATIONAL CHEMISTRY OLYMPIAD NATIONAL EXAM—PART II – ANSWER KEY Prepared by the American Chemical Society Olympiad Examinations Task Force 1. (11%) A(g) + 3B(g) r 2C(g) Use the tabulated data to answer the questions about this reaction, which is carried out in a 1.0 L container at 25°C. Experiment A o , mol B o , mol Initial rate of formation of C, mol . L -1 . min -1 1 0.10 0.10 0.25 2 0.20 0.20 2.0 3 0.10 0.20 2.0 a. For experiment 1, give the initial rate of disappearance of i. A ii. B b. Determine the orders of A and B and write the rate law for the reaction. c. Calculate the value of the rate constant and give its units. d. For the initial amounts of A and B in experiment 1, state the initial rate of formation of C under the following conditions. Justify your answer in each case. i. 0.50 mol of neon gas is added to the 1.0 L container. ii. the volume of the container is increased to 2.0 L. a. rate of formation C = 0.25 mol·L-1·min-1 so i. rate of disappearance of A = 0.25 × 1 2 = 0.13 mol L -1 min - 1 ii rate of disappearance of B = 0.25 × 3 2 = 0.38 mol L -1 min - 1 b. for experiments 2 and 3: [B] is constant while [A] doubles and the rate of the reaction is unchanged. The reaction order with respect to A is zero. for experiments 3 and 1: [A] is constant while [B] doubles and the rate of the reaction increases by a factor of 8. The reaction order with respect to B is 3. Rate = k B [ ] 3 c. k = 0.25 0.10 ( 29 3 = 250 L 2 mol -2 min d. i. The rate of formation of C is 0.25 mol·L-1·min-1. Adding an inert gas does not change the rate law. ii. The rate of formation of C is 0.031 mol·L-1·min-1. Doubling the volume of the container changes the concentration of B. [B] = 0.050 mol·L-1· Rate = k B [ ] 3 = 250 L 2 mol -2 min 0.050 mol L -1 [ ] 3 = 0.031 2. (13%) A 0.472 g sample of an alloy of tin and bismuth is dissolved in sulfuric acid to produce tin(II) and bismuth(III) ions. This solution is diluted to the mark in a 100 mL volumetric flask and 25.00 mL aliquots are titrated with a 0.0107 M solution of KMnO 4 , forming tin(IV) and manganese(II) ions. (The bismuth ions are unaffected during this titration.)
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2 Not valid for use as an USNCO Olympiad National Exam after April 25, 2006. a. Write a balanced equation for the reaction of the MnO 4 - ion with Sn(II) in acid solution. b. If an average titration requires 15.61 mL of the MnO 4 - solution, calculate the number of moles of MnO 4 - used in an average titration. c. Determine the percentage of tin in the alloy. d. State how the end point of the titration is detected. e. Describe and explain the effect on the calculated percentage of tin in the alloy if the same volume of MnO 4 - solution is used with the following differences: i. During the titration the solution pH increases so that MnO 2 is formed rather than Mn(II). ii.
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ACS EXAM KEY 11 - 2006 U S NATIONAL CHEMISTRY OLYMPIAD...

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