Part A MISSED THIS? Read Section 15.5 (Pages 650 - 656), 15.7 (Pages 661 - 666) .Suppose that a catalyst lowers the activation barrier of a reaction from 123 kJ/mol to 55 kJ/mol. By what factor would you expect the reaction rate to increase at 25.00 °C? (Assume that the frequency factors for the catalyzed and uncatalyzed reactions are identical.) Express your answer to two significant figures. > View Available Hint(s) Πνα ΑΣΦ Xb Vx Vx |X| x•10" =|1.03 • k1 1013 Submit Previous Answers X Incorrect; Try Again; One attempt remaining Since the reaction rate is directly proportional to the rate constant, apply the Arrhenius equation to the two different activation energies. Then divide the rate constant for the catalyzed reaction by the rate constant for the uncatalyzed reaction to determine the factor by which the rate increased. You may want to review Hint 1. Express the ratio of the rate constants at different activation energies.

Part A MISSED THIS? Read Section 15.5 (Pages 650 - 656), 15.7 (Pages 661 - 666) .Suppose that a catalyst lowers the activation barrier of a reaction from 123 kJ/mol to 55 kJ/mol. By what factor would you expect the reaction rate to increase at 25.00 °C? (Assume that the frequency factors for the catalyzed and uncatalyzed reactions are identical.) Express your answer to two significant figures. > View Available Hint(s) Πνα ΑΣΦ Xb Vx Vx |X| x•10" =|1.03 • k1 1013 Submit Previous Answers X Incorrect; Try Again; One attempt remaining Since the reaction rate is directly proportional to the rate constant, apply the Arrhenius equation to the two different activation energies. Then divide the rate constant for the catalyzed reaction by the rate constant for the uncatalyzed reaction to determine the factor by which the rate increased. You may want to review Hint 1. Express the ratio of the rate constants at different activation energies.

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter11: Chemical Kinetics: Rates Of Reactions

Section: Chapter Questions

Problem 11.ACP: (Section 11-5) A rule of thumb is that for a typical reaction, if concentrations are unchanged, a...

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Experiments during a recent summer on a number of fireflies (small beetles, Lampyridaes photinus) showed that the average interval between flashes of individual insects was 16.3 s at 21.0C and 13.0 sat 27.8C. a. What is the apparent activation energy of the reaction that controls the flashing? b. What would be the average interval between flashes of an individual firefly at 30.0C? c. Compare the observed intervals and the one you calculated in part b to the rule of thumb that the Celsius temperature is 54 minus twice the interval between flashes.
Bacteria cause milk to go sour by generating lactic acid. Devise an experiment that could measure the activation energy for the production of lactic acid by bacteria in milk. Describe how your experiment will provide the information you need to determine this value. What assumptions must be made about this reaction?
The decomposition of NH3 to N2 and H2 was studied on two surfaces: Surface Ea (kJ/mol) W 163 Os 197 Without a catalyst, the activation energy is 335 kJ/mol. a. Which surface is the better heterogeneous catalyst for the decomposition of NH3? Why? b. How many times faster is the reaction at 298 K on the W surface compared with the reaction with no catalyst present? Assume that the frequency factor A is the same for each reaction. c. The decomposition reaction on the two surfaces obeys a rate law of the form Rate=k[NH3][H2] How can you explain the inverse dependence of the rate on the H2 concentration?
Experiments have shown that the average frequency of chirping by a snowy tree cricket (Oecanthus fultoni) depends on temperature as shown in the table. Chirping Rate (per min) Temperature (C) 178 25.0 126 20.3 100. 17.3 What is the apparent activation energy of the process that controls the chirping? What is the rate of chirping expected at a temperature of 7.5C?
(Section 11-5) A rule of thumb is that for a typical reaction, if concentrations are unchanged, a 10-K rise in temperature increases the reaction rate by two to four times. Use an average increase of three times to answer the questions below. (a) What is the approximate activation energy of a typical chemical reaction at 298 K? (b) If a catalyst increases a chemical reactions rate by providing a mechanism that has a lower activation energy, then what change do you expect a 10-K increase in temperature to make in the rate of a reaction whose uncatalyzed activation energy of 75 kJ/mol has been lowered to one half this value (at 298 K) by addition of a catalyst?
11.44 A possible reaction for the degradation of the pesticide DDT to a less harmful compound was simulated in the laboratory. The reaction was found to be first order, with k = 4.0 X 10_H s"' at 25°C. What is the half-life for the degradation of DDT in this experiment, in years?
The hydrolysis of the sugar sucrose to the sugars glucose and fructose, C12H22O11+H2OC6H12O6+C6H12O6 follows a first-order rate equation for the disappearance of sucrose: Rate =k[C12H22O11] (The products of the reaction, glucose and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.) (a) In neutral solution, k=2.11011s1 at 27 C and 8.51011s1 at 37 C. Determine the activation energy, the frequency factor, and the rate constant for this equation at 47 C (assuming the kinetics remain consistent with the Arrhenius equation at this temperature). (b) When a solution of sucrose with an initial concentration of 0.150 M reaches equilibrium, the concentration of sucrose is 1.65107M . How long will it take the solution to reach equilibrium at 27 C in the absence of a catalyst? Because the concentration of sucrose at equilibrium is so low, assume that the reaction is irreversible. (c) Why does assuming that the reaction is irreversible simplify the calculation in pan (b)?
11.35 For the reaction 2 NO(g) + 2 H?(g) — N,(g) + 2 H,O(g) at 1100°C, the following data have been obtained: [NOJ [HJ Rate = A(N2]/At (mol L~1) (mol L_1) (mol L-1 s_1) 5.0 X 10’1 0.32 0.012 1.0 X 10~’ 0.32 0.048 1.0 X 10"2 0.64 0.096 Derive a rate law for the reaction and determine the value of the rate constant.