Problem 5.4
Define points on 20C line
> with(geometry):point(p1_20,44e6/(20+273),ln(1e-4)):
> point(p2_20,53.2e6/(20+273),ln(1e-2)):
get slope of line and multiply by 2R to get activation volume
> Digits:=4:'V (m^3/mol)'= 2*8.314*slope(p2_20,p1_20);
V
3
m
mol
=,002440
Compute activation energy by horizontal difference between 20C and -60C lines
> eq:=(V/2)*((69e6/(-60+273)) -
(44e6/(20+273)))=Delta[H]*(1/(-60+273) - 1/(20+273));
80
eq,= 211.9 =?
H
62409
> 'Delta[H] (kJ/mol) ' = solve(eq,Delta[H])/1000;
H
kJ
mol
= 165.3
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