The data should fit a linear model of the form log k EA/(2.303RT), where A is the preexponential log A factor, and Ris the gas constant. (a) Fit the data to a straight line of the form log k = a 1000b/ T. - *(b) Find the slope, intercept, and standard error of the estimate. *(c) Noting that E, = -b× 2.303R X 1000, find the activation energy and its standard deviation (Use R= 1.987 cal mol K-'). *(d) A theoretical prediction gave EA = 41.00 kcal mol'K. Test the null hypothesis that EA is this value at the 95% confidence level.
The data should fit a linear model of the form log k EA/(2.303RT), where A is the preexponential log A factor, and Ris the gas constant. (a) Fit the data to a straight line of the form log k = a 1000b/ T. - *(b) Find the slope, intercept, and standard error of the estimate. *(c) Noting that E, = -b× 2.303R X 1000, find the activation energy and its standard deviation (Use R= 1.987 cal mol K-'). *(d) A theoretical prediction gave EA = 41.00 kcal mol'K. Test the null hypothesis that EA is this value at the 95% confidence level.
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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Question
Transcribed Image Text:8-19. A study was made to determine the activation energy
E, for a chemical reaction. The rate constant k was de-
termined as a function of temperature T, and the data
in the table below obtained.
Т, К
k, s-1
599
0.00054
629
0.0025
647
0.0052
666
0.014
683
0.025
700
0.064
The data should fit a linear model of the form log k
log A – EA/(2.303RT), where A is the preexponential
factor, and R is the gas constant.
(a) Fit the data to a straight line of the form log k =
a -
1000b/ T.
*(b) Find the slope, intercept, and standard error of the
estimate.
*(c) Noting that E,
the activation energy and its standard deviation
(Use R = 1.987 cal mol K-').
*(d) A theoretical prediction gave EA
molK. Test the null hypothesis that EA is this
-6 X 2.303R X 1000, find
= 41.00 kcal
value at the 95% confidence level.
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