CHEM 341. Fall 2000.   Previous PS   Answers   PS#9.doc (Word 97)  Due Tuesday, Dec. 12   Problem Set #9 Note: Please note that due to technical difficulties in creating the accepted notation for an equilibrium reaction, this problem set uses to indicate that equilibrium is established. Rate Law 1. At 25° C and at a constant pH of 5, the inversion of sucrose proceeds with a constant half-life of 500 min. At this same temperature, but at a pH of 4, the half-life is constant at 50 min. The rate law is . What are the values of a and b? 2. The rate of acid catalyzed hydrolysis of ethyl acetate in hydrochloric acid solution obeys the following rate law where k = 0.1 M-1 h-1 at 25° C. Neglecting any back reaction, calculate the time required for half of the ester to be hydrolyzed if the initial concentrations of ester and of HCl catalyst are 0.02 M and 0.01 M, respectively. 3. In the study of a first-order reaction it is found that A/Ao is 0.125 after 1 hour. The system initially consisted of 0.20 mole of gaseous A at STP. Calculate the initial rate of reaction in moles of A reacting per minute. 4. For the following mechanism: where k1 = rate of forward Reaction #1; k-1 = rate of reverse reaction for Reaction #1; k2 = rate of forward Reaction #2. Using the steady-state assumption, find the expression for in terms of PN2O5, PNO, PNO2, k2, k1, and/or k-1. 5. Consider the reaction where k1 = rate of forward Reaction #1; k-1 = rate of reverse reaction for Reaction #1; k2 = rate of forward Reaction #2. Find the rate law of this reaction in terms of [A], [B], [P], k2, k1, and/or k-1. Now suppose that we set [B] = [A]. Under these conditions, what is the overall order of the reaction if k-1 >> k2[A]? What is the overall order of the reaction if k-1 << k2[A]? 6. The reaction obeys the rate law: . It takes 0.5 min for one percent of CH3CH2NO2 to react at 25° C, in the case of a solution 0.002 M in CH3CH2NO2 and 0.3 M in NaOH. Calculate k. Calculate also how long it would take for half of CH3CH2NO2 to react. 7. Complete the following statements: Plot of ln [A]t or vs. time yields a straight line of slope _______ for a first-order reaction. Plot of vs. time gives a straight line of slope ________ for a second-order reaction. 8. Consider the following reaction where k1 = rate of forward reaction and k-1 = rate of reverse reaction. This reaction is first order in both the forward and the reverse direction. Calculate k1 if the initial velocity (i.e. while [B] is still insignificant) is 5.1 ´ 10-4 M s-1 and the initial concentration of A, [A]o is 4.7 ´ 10-2 M. A plot of vs. time has a slope equal to -0.01795 s-1. Find k-1. Hint: The information from part (a) is also true for part (b). Calculate the equilibrium constant and the half-life for this reaction. 9. Sketch approximate plots of the concentration of [A], [B], and [C] vs. time for (on the same graph) if both reactions are first order, and is the rate-limiting step (i.e. k1 << k2). Repeat these plots on a separate graph for the same reaction but if is the rate limiting step (i.e. k1 >> k2). 10. For the following reaction A plot of vs. time has a slope of 5.0 s-1. Calculate the rate constant, k, of the forward reaction if the rate of the reverse reaction is insignificant and if [A]o = 3.1 M and [B]o = 4.7 M. Simple Arrhenius Analysis 11. Suppose a reaction has a rate constant of 1.23 x 103 s-1 at 300 K and of 6.45 x 103 s-1 at 400 K. Calculate the Arrhenius activation energy and the pre-exponential factor A. Then use these values to predict the rate constant at 500 K. Eyring Analysis 12. Explain how a plot of ln (k/T) vs. 1/T can be used to determine DHtransition state and DStransition state if the values of kB, k, and h are known. (kB is Boltzman's constant, k is the tunneling factor, and h is Planck's constant). Chain Reactions 13. The chlorination of an organic molecule (M) with initiation by thermal dissociation of chlorine, may proceed by the following mechanism (in the absence of light). Find the rate law for the overall reaction in terms of [M], [Cl2], and various constants (i.e. k1 and/or k2 and/or k3 and/or k4).