Answered: 13. a) Find the ratio of the rate of…

Introduction to Chemical Engineering Thermodynamics

8th Edition

ISBN:9781259696527

Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Chapter1: Introduction

Section: Chapter Questions

Problem 1.1P

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Answer 13 and 14 ASAP PLEASE THANK YOU

$in this case quite a bit cioser to zero! Since E, is in kJ/mol, use the Gas Constant R. as 8.314 x 10-3 kJ/mol K. The temperature should be in K 39.35kj.mol/8.314 x 10-3 x 311.10-15.21 e-15.21)-2.48e-7 b) Calculate this fraction at the lowest temperature of the experiments, approx. 11.0°C. 39 35kj mol/8 314 x 10-3 x 284.2k-16.65 c) Does the ratio of these firactions reflect the ratio of the k's that were found? In other words, does the dependence of the fraction of collisions with the needed Ea on temperature completely explain the change in rate constant with temperature? No. because while the temperature in k increased, the fraction decreased. Ratio of fraction at high to low temperature 15.21/16.65=D0.91 Ration of k at high to low temperature (2.3e(0.54e)-4.3 6. 13. a) Find the ratio of the rate of reaction with the catalyst to the rate of reaction without the catalyst. Comment on what this ratio tells you about the effect of a catalyst on a reaction rate. 14. Use your result for Ea and the fact that AH for this reaction is -350. kJ to sketch a reaction profile diagram for this system. (This is a diagram similar to those given on page 1 of this handout.) Show the effect of the catalyst on the reaction profile.$

Transcribed Image Text:in this case quite a bit cioser to zero! Since E, is in kJ/mol, use the Gas Constant R. as 8.314 x 10-3 kJ/mol K. The temperature should be in K 39.35kj.mol/8.314 x 10-3 x 311.10-15.21 e-15.21)-2.48e-7 b) Calculate this fraction at the lowest temperature of the experiments, approx. 11.0°C. 39 35kj mol/8 314 x 10-3 x 284.2k-16.65 c) Does the ratio of these firactions reflect the ratio of the k's that were found? In other words, does the dependence of the fraction of collisions with the needed Ea on temperature completely explain the change in rate constant with temperature? No. because while the temperature in k increased, the fraction decreased. Ratio of fraction at high to low temperature 15.21/16.65=D0.91 Ration of k at high to low temperature (2.3e(0.54e)-4.3 6. 13. a) Find the ratio of the rate of reaction with the catalyst to the rate of reaction without the catalyst. Comment on what this ratio tells you about the effect of a catalyst on a reaction rate. 14. Use your result for Ea and the fact that AH for this reaction is -350. kJ to sketch a reaction profile diagram for this system. (This is a diagram similar to those given on page 1 of this handout.) Show the effect of the catalyst on the reaction profile.

This erperiment
has the sae
[H=
0.040
[ S20s ]
0.020
concentrations as #7
M
M
Average time
(sec)
Reaction Rate
Rate Constant k
Temp.(°C)
or other
Еxpt
Reaction Times
(sec)
(Ms)
or
Run
conditions
sec
7a
11.2°C
135
140.5
4.3 x 106
5.4 x 10-3
146
7b
19.5°C
84
93.5
6.4 x 10-6
8.0 x 10-3
102.9
7c
23.0°C
70.2
70.8
8.5 x 10-6
1.0 x 10-2
71.4
7d
29.0°C
53
60
1.0 x 10-5
1.3 х 10-2
67
7e
38.1°C
35
33.5
1.8 x 10-5
2.3 x 10-2
32
(no
cat)
Benchtop,
No catalyst
(w/o
Cu(NO:)2)
(w/cat) Benchtop,
With
102
102
5.9х 104
7.4 x 10-1
27
27
2.2 x 10-5
2.8 x 10-2
catalyst
(w/
Cu(NO:)2)
Cconcentroin

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