Extra Credit. This problem is worth 5% points toward your final grade and there
ID: 992993 • Letter: E
Question
Extra Credit. This problem is worth 5% points toward your final grade and there is no partial credit. You can work in groups and you can use your notes and textbook, but please do not consult your TAs, ULAs or faculty. Consider the reaction below: Based on the information provided in the table below (all measurements are at 25 degreeC), determine the experimental rate law and calculate the rate constant at this temperature. You can assume ideal gas behavior for both N0_2 and F_2. Contrasting NO_2 and F_2, which one of these two real gases would be closer to ideal gas assumed in ? Draw the Lewis structures of both to support your argument. Based on the rate law determined in , propose a reaction mechanism comprising of two bimolecular elementary steps. F is an intermediate in the overall reaction. Which elementary process in part c is rate-determining? You conduct experiment 1 from the table in . If your initial p(N0_2) is 0.038 atm and p(F_2) is kept at constant flow of 0.060 atm, how long would it take for the pressure of NO_2 to drop to 0.001 atm? The collision frequency and steric factor for the rate-determining step in the overall process at 25 degreeC are 5.9 times 10^10 and 0.16, respectively. Calculate the activation energy (in kJ/mol) for the rate determining step. If the enthalpy of reaction (i.e. standard enthalpy change for the overall process) is -100 kJ/mol, sketch the energy profile of the two-step reaction given that i) the activation energy for the fast step in part c is 5 kJ/mol and ii) equilibrium partial pressures of all species involved in the fast step are 1.0 atm.Explanation / Answer
a.
Going from experiment 1 to 2, p(NO2) doubles, but p(F2) remains same. Rate also gets doubled. Hence rate is first order w.r.t. NO2.
Going from experiment 1 to 2, p(NO2) remains same, but p(F2) halves. Rate also gets halved. Hence rate is first order w.r.t. F2.
Rate = kp(NO2)p(F2)
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