Download Chemistry Exam II: Reaction Rates and Mechanisms - Prof. Robert A. Walker and more Exams Chemistry in PDF only on Docsity! Chemistry 153 Hour Exam II 18 March, 2003 Welcome to your second Chem 153 exam. The exam is 1 hour. There are 4 questions for a total of 100 points. Please read each question completely, as many have multiple parts. Be complete in your answers but work quickly - don't spend undue amounts of time on any particular question without going through the exam once. If you are uncertain about what a question is asking, please ask us. Good Luck! Potentially useful information: Integrated rate laws: 1 st order ln[A] = -kt + ln[A]0 2nd order 1/[A] = kt + 1/[A]0 _^a_ = RT AH°f: HCN 135.1 kJ/mol HNC 204.1 kJ/mol rCH =1.06A rNH =1.00A Thermody n ami c s AE = q + w AH = AE + PAY AS = qrev/T AG = AH° - TAS° AG° = -RT lnKeq AG = AG° + RT InQ (Note: Neatness counts! Please label your answers clearly and show all work.) 1. Reaction rates and rate laws (15 points) a. (10 points) Acid rain results from sulfur dioxide (SO2 ) dissolving in water in the presence of molecular oxygen (O2) to form sulfuric acid (H2SO4 or, more accurately, SO4 2 " and 2 H3O+) Balance the following reaction assuming that the stoichiometric coefficient in front of O2aq) is 1. -_S02(aq) + 02(aq) + __ H20(l) -> __ S042 -(aq) + __ H30+(aq) Given your balanced reaction, which reactant disappears at the fastest rate? The slowest? Which product forms at tne fastest rate? The slowest? b. (10 points) Unimolecular reactions (in which A falls apart spontaneously to form B and C) typically have very large activation energies. Propose a reason why this should be so. Also, sketch a reaction coordinate diagram for a unimolecular reaction paying attention to the relative energetics of reactants and products. 3. Reaction mechanisms (30 points) At high temperatures, many molecules that are stable at room temperature fall apart. If the pressure (or concentration in solution) is high, decomposition is described by a rate that: is first order in reactant. However, if pressure (or concentration) is low, decomposition follows a rate law that is second order overall. Let the reactant be designated A and the products P. a. (5 points) Write down the rate in terms of the disappearance of A in both the high and low pressure limits. The net reaction is: A > P Assume that the etiergy for decomposition comes from a collision partner, M. In other words, assume that A + M > A* + M (with rate constant kj (where A* is an "activated" reactant with tremendous excess energy) A* can be deactivated (by M) or can go on to form products according to: A* + M > A + M (with rate constant kd ) A* > P (with rate constant kr) b. (5 points) Write down expressions for the following in terms of [A], [A*], [M]. [P], ka , kd , and d[A] dt d[A*] dt d[P] dt c. (10 points) Using the steady state approximation for A*, show that the rate of product formation can be expressed as follows: d[P] = k r ka [A][M] dt k d. (10 points) Now evaluate this rate expression in the high and low presuure limits and show how at high pressures, decomposition of A is first order in [A] and at low pressures, decomposition is first order in [A]. (At high pressure, kd[M] » kr ; at low pressure, kr »kd[M].) In both limits, identify what kobs equals in terms of ka , kd , and kr .
