The reaction between nitrogen dioxide and carbon monoxide is NO₂(g) + CO(g) → NO(g) + CO₂(g) The rate constant at 701 K is measured as 2.57 M-1.s and that at 895 K is measured as 567 M-1.8-¹. Part A Use the value of the activation energy (E₁ = 1.50 x 10² kJ/mol) and the given rate constant of the reaction at either of the two temperatures to predict the rate constant at 459 K. 1957 ΑΣΦ xa x₁ = √x √x x x x x.10 k= 3.32 10-4 ● Submit Previous Answers Request Answer X Incorrect; Try Again; 2 attempts remaining SWIX ? SAI M-1.8-1

The reaction between nitrogen dioxide and carbon monoxide is NO₂(g) + CO(g) → NO(g) + CO₂(g) The rate constant at 701 K is measured as 2.57 M-1.s and that at 895 K is measured as 567 M-1.8-¹. Part A Use the value of the activation energy (E₁ = 1.50 x 10² kJ/mol) and the given rate constant of the reaction at either of the two temperatures to predict the rate constant at 459 K. 1957 ΑΣΦ xa x₁ = √x √x x x x x.10 k= 3.32 10-4 ● Submit Previous Answers Request Answer X Incorrect; Try Again; 2 attempts remaining SWIX ? SAI M-1.8-1

Chemistry for Engineering Students

4th Edition

ISBN:9781337398909

Author:Lawrence S. Brown, Tom Holme

Publisher:Lawrence S. Brown, Tom Holme

Chapter11: Chemical Kinetics

Section: Chapter Questions

Problem 11.36PAE: The reaction NO(g) + O,(g) — NO,(g) + 0(g) plays a role in the formation of nitrogen dioxide in...

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The reaction NO(g) + O,(g) — NO,(g) + 0(g) plays a role in the formation of nitrogen dioxide in automobile engines. Suppose that a series of experiments measured the rate of this reaction at 500 K and produced the following data; [NO] (mol L ’) [OJ (mol L 1) Rate = -A[NO]/Af (mol L_1 s-1) 0.002 0.005 8.0 X 10"'7 0.002 0.010 1.6 X 10-'6 0.006 0.005 2.4 X IO-'6 Derive a rate law for the reaction and determine the value of the rate constant.
The following rate constants were obtained in an experiment in which the decomposition of gaseous N2O; was studied as a function of temperature. The products were NO, and NO,. Temperature (K) 3.5 x 10_i 298 2.2 x 10"4 308 6.8 X IO-4 318 3.1 x 10 1 328 Determine Etfor this reaction in kj/mol.
One mechanism for the destruction of ozone in the upper atmosphere is a. Which species is a catalyst? b. Which species is an intermediate? c. Ea for the uncatalyzed reaction O3(g)+O(g)2O2(g) is 14.0 kJ. Ea. for the same reaction when catalyzed is 11.9 kJ. What is the ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at 25C? Assume that the frequency factor A is the same for each reaction.
Why does a catalyst increase the rate of a reaction? What is the difference between a homogeneous catalyst and a heterogeneous catalyst? Would a given reaction necessarily have the same rate law for both a catalyzed and an uncatalyzed pathway? Explain.
Hydrogen iodide, HI, decomposes in the gas phase to produce hydrogen, H2, and iodine, I2. The value of the rate constant, k, fur the reaction was measured at several different temperatures and the data are shown here: Temperature (K) k (M -1 5-1) 555 6.23107 575 2.42106 645 1.44104 700 2.01103 What is the value of the activation energy (in kJ/mol) for this reaction?
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)?