QUESTIONS No. 1) – 3): Consider the 3rd order reaction: 2NOG) + 02 (9) → 2NO2 (9). The specific rates of the reaction, expressed in moles/L-s are: For the forward reaction: @600 K, k = 6.63 x 105; @645 K, k = 6.52 x 105 For the reverse reaction: @600 K, k = 83.9; 1) Calculate the equilibrium constant at 600K? a) 7.6 x 103 @645 K, k = 407 b) 7.9 x 103 c) 7.27 x 104 d) 6.25 x 104
QUESTIONS No. 1) – 3): Consider the 3rd order reaction: 2NOG) + 02 (9) → 2NO2 (9). The specific rates of the reaction, expressed in moles/L-s are: For the forward reaction: @600 K, k = 6.63 x 105; @645 K, k = 6.52 x 105 For the reverse reaction: @600 K, k = 83.9; 1) Calculate the equilibrium constant at 600K? a) 7.6 x 103 @645 K, k = 407 b) 7.9 x 103 c) 7.27 x 104 d) 6.25 x 104
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![QUESTIONS No. 1) – 3): Consider the 3rd order reaction: 2NOG) + O2 (9) → 2NO2 (g).
The specific rates of the reaction, expressed in moles/L-s are:
For the forward reaction: @600 K, k = 6.63 x 105;
For the reverse reaction: @600 K, k = 83.9;
1) Calculate the equilibrium constant at 600K?
a) 7.6 x 103
2) Calculate the activation energy for the forward reaction?
a) 2.699 x 104
3) Calculate the frequency factor for the forward reaction?
a) 8.43 x 105
@645 K, k = 6.52 x 105
@645 K, k = 407
b) 7.9 x 103
c) 7.27 x 10-4
d) 6.25 x 104
b) 2.727 x 104
c) -2.859 x 102 d) 2.859 x 102
b) 3.23 х 104
с) 5.86 х 102
d) 2.59 x 103](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4220efb7-607a-41f9-828b-a5715733ff16%2F9bf1bc54-7021-4bc7-9c5c-52caea941d4d%2Fy781neg_processed.png&w=3840&q=75)
Transcribed Image Text:QUESTIONS No. 1) – 3): Consider the 3rd order reaction: 2NOG) + O2 (9) → 2NO2 (g).
The specific rates of the reaction, expressed in moles/L-s are:
For the forward reaction: @600 K, k = 6.63 x 105;
For the reverse reaction: @600 K, k = 83.9;
1) Calculate the equilibrium constant at 600K?
a) 7.6 x 103
2) Calculate the activation energy for the forward reaction?
a) 2.699 x 104
3) Calculate the frequency factor for the forward reaction?
a) 8.43 x 105
@645 K, k = 6.52 x 105
@645 K, k = 407
b) 7.9 x 103
c) 7.27 x 10-4
d) 6.25 x 104
b) 2.727 x 104
c) -2.859 x 102 d) 2.859 x 102
b) 3.23 х 104
с) 5.86 х 102
d) 2.59 x 103
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