Adventures in Thermochemistry
James S. Chickos*
Department of Chemistry and Biochemistry
University of Missouri-St. Louis
Louis MO 63121
E-mail: jsc@umsl.edu
9
A portion of the Japanese Gardens
at the Missouri Botanical garden
Applications of the Correlation-Gas Chromatographic Method
Objectives: To go where no one else has gone
1. Evaluation of the vaporization enthalpies of large molecules
2. Application of Correlation-Gas Chromatography to a Tautomeric
Mixture - Acetylacetone
3. Evaluation of the Vaporization Enthalpy of Complex Hydrocarbon
Mixtures
4. The Vaporization Enthalpies of Drugs and Related Substances
Amantadine (1-aminoadamantane) and rimantadine (1-adamantylethanamine)
has been used for the prevention of influenza A, for treatment of dementia and other
disorders of the central nervous system,1-3 Use of these materials is no longer
recommended
Phentermine (α,α-dimethylphenethylamine), a substance with a pharmacology similar to
amphetamine, is used medicinally as an appetite suppressant for patients who suffer from
obesity.4
Tranylcypromine or trans-1-phenylcyclopropylamine, developed as an analogue of
amphetamine, is a monoamine oxidase inhibitor and has been used in treatment of mood
and anxiety disorders.5
(1) De Clercq, E. Antiviral drugs in current clinical use. J. Clin. Virol. 2004, 30, 115−133.
(2) Morozov, I. S.; Ivanova, I. A.; Lukicheva, T. A.). Pharm. Chem. J. 2001, 35, 235−8.
(3) Schnell, J. R.; Chou, J. J. Nature 2008, 451, 591−5.
(4) http://www.drugs.com/monograph/phentermine.html (accessed3/12/
(5) Laux, G.; Ulrich, S. Tranylcypromine: A review. Psychopharmakotherapie 2006, 13, 130−141.
The vaporization enthalpies of all these drugs were unavailable A sublimation enthalpy
for amantadine was reported by Bazyleva et al.1
NH2
Δcrg H(298.15 K) = (61.7±0.6) kJ·mol-1
For a crystalline solid melting at T = (479 to 481) K, this value at first glance appeared
remarkably small at least as far as estimates are concerned.
Estimation of Vaporization
Enthalpies: A digression
Bazyleva, A. B.; Blokhin, A. V.; Kabo, A. G.; Kabo,
G. J.;Emel’yanenko, V. N.; Verevkin, S. P. J. Chem.
Thermodyn. 2008, 40, 509−22.
Estimations of Vaporization Enthalpies
Numerous method have been developed to estimate vaporization enthalpies. Some are more
accurate than others. A method we found to be accurate to about 5% is given by:
ΔlgHm(298.15 K)/kJ·mol-1 = 4.69(nC – nQ) + 1.3 nQ + b + 3.0 + C
where nC = number of carbon atoms
nQ = number of quaternary carbons
b = functional group contribution
C = a correction term when branching at sp3 carbons
decreases accessibility to the functional group
NH2
The carbon bearing the –NH2 group is considered a quaternary carbon
C=0
Chickos, J. S.; Acree Jr., W. E.; Liebman, J. F. Estimating Phase Change Enthalpies and Entropies. In
Computational Thermochemistry; Symposiun Series 677; Irikura, K. K., Frurip, D. J., Eds.; ACS:
Washington DC, 1998; Chapter 4.
Estimations of Vaporization Enthalpies
Numerous method have been developed to estimate vaporization enthalpies. Some are more
accurate than others. A method we found to be accurate to about 5% is given by:
ΔlgH(298.15 K) = 4.69(nC – nQ) + 1.3 nQ + b + 3.0 + C
where
nC = 10
nQ = 1
b = 14.8
C= 0
NH2
ΔlgH(298.15 K) = 58.3± 2.9 kJ·mol-1
ΔcrgH(298.15 K) = 61.7±0.6 kJ·mol-1
Est 58.3± 2.9
kJ·mol-1 : 59.9 ± 2.5
Estimate: 58.3 ± 2.9
68.7 ± 3.7
67.7 ± 3.4
60.2 ± 2.6
61.3 ± 3.1
60.3 ± 2.1
60.0 ± 3.0
NH2
Sublimation
ΔcrgH(298.15 K) = 61.7 ± 0.6 kJ·mol-1
∆crlHm(298.15 K)/kJ·mol-1 =
∆crlHm(Tfus)/ kJ·mol-1 + {0.15Cp(cr) – 0.26Cp(l)/J·mol-1·K-1 – 9.83][Tfus-298.15]/1000
In our case, however, our uncertainty permits a fusion enthalpy of 1.5 kJ·mol-1
Lit: Δcrg H(298.15 K) = (61.7±0.6) kJ·mol-1
This work: 59.9 ± 3.5 kJ·mol-1
An attempted X-ray crystal structure determination was unsuccessful because of the lack of
sufficient crystallinity
Crystallinity Approximately 50 %
Is it possible to have a negative fusion enthalpy?
Kauzmann’s parodox
Heat capacites of liquids> heat capacities of solids
If you cool a liquid sufficiently fast can you get a super cooled liquid that is
thermodynamically more stable than the crystal?
Clausius Clapeyron Eq:
dp/dT = (S(l) – S(cr))/(V(l)-V(cr))
At c:
If S(cr)> S(l) and V(l)> V(cr)
V(l) = V(cr)
Are there any examples?
Stillinger, F. H.; Debenedetti, P. G.; Truskett, T. M. The Kauzman Paradox Revisited. J. Phys Chem. B
2001, 105, 11809.
Stillinger, F. H.; Debenedetti, P. G. Phase transitions, Kauzmann Curves and Inverse melting. Biophys.
Chem. 2003, 105, 211.
3He
and 4He
Both are quantum liquids
3He
has a nuclear spin that is free to orient in the crystal but is strongly coupled in
the liquid due to particle exchange causing S(cr)> S(l). Follows Fermi Dirac
statistics.
4He,
S(cr)> S(l) due to the relative number of phonon states of the solid and
liquid. Follows Bose –Einstein statistics.
Somewhat related 4-cyano-4’-octyloxybiphenyl
forms a liquid crystal. At slightly below 350 K
and 2100 bar, the material is a nematic liquid
possessing just molecular orientation order.
Heating at constant pressure the material reorders
to a smectic A phase possessing both orientational
and partial translational order
Cladis, P. E.; Guillon, Bouchet, F. R.; Finn, P. L. Reentrant
nematic transitions in cyano-octylbiphenyl. (80CB) Phys. Rev.
A. 1981, 23, 2594
ln(p/po) = A·T-3 + B·T-2 + C·T -1 + D
A typical correlation between ln(p/po) vs ln(to/ta)
(3)
c
Estimate. (24) Estimation Program Interface EPI Suite, version 4.11; US EPA: Washington, DC, 2012;
available at http://www.epa.gov/opptintr/exposure/pubs/episuite.htm.
NH2
ln(p/po) = – 7.32·107·T-3 – 9.3·104T-2 – 3.93·103·T -1 + 8.506
At the triple point, 477.4 K we calculate:
p477(l) = p477(cr) = 45.8 kPa
Ttp/K = 477.4; ΔlgHm(Ttp)
= 43.8 ± 0.6 kJ·mol-1
ΔlgHm(477.4 K) = 7.3 ± 0.2 kJ·mol-1
ΔcrgHm(477.4 K) = 51.1 ± 0.7 kJ·mol-1
ΔcrgHm(298.15 K) = 58.4 ± 2.3 (61.7±0.6) lit
kJ·mol-1
ln(p(298)cr/Pa)= [ΔcrgH(477.4 K) + ΔcrCp(cr)ΔT ][1/Tfus/K – 1/298.15]/R + ln(p477cr/Pa)]
where ΔcrgCpΔT = [0.75 + 0.15 Cp(cr)][(Tfus/K – 298.15)/2]
Acknowledgement
Chase Gobble
Joe Wilson