This preview shows page 1. Sign up to view the full content.
The temperature dependence of K:
S
T
H
G
S
T
H
G
2
2
1
1
assuming
S
and
H
are independent of temperature.
But:
K
RT
G
ln
So we have:
S
T
H
K
RT
S
T
H
K
RT
2
2
1
1
ln
ln
dividing by –RT:
R
S
RT
H
K
R
S
RT
H
K
2
2
1
1
ln
ln
essentially equation 4.18 on page 114.
now subtract:
1
2
2
1
1
2
2
1
1
1
ln
ln
ln
ln
T
T
R
H
K
K
R
S
R
S
RT
H
RT
H
K
K
or
1
2
2
1
1
1
ln
T
T
R
H
K
K
Note that a plot of ln K vs 1/T yields a straight line with:
This is the end of the preview. Sign up
to
access the rest of the document.
Unformatted text preview: Slope = R H and intercept = R S Thus, if rxn is endothermic, slope is negative. As T increases (or 1/T decreases), ln K increases thus favoring the products. If rxn is exothermic, slope is positive. As T increases, ln K decreases thus favoring reactants. Consider this when working Problem 5 of WA Q09....
View
Full
Document
This note was uploaded on 03/29/2009 for the course CH 201 taught by Professor Warren during the Fall '07 term at N.C. State.
 Fall '07
 Warren
 Chemistry

Click to edit the document details