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Secondary Metabolites in Plants
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Secondary metabolites are chemicals produced by plants for which no role has yet been
found in growth, photosynthesis, reproduction, or other "primary" functions. These
chemicals are extremely diverse; many thousands have been identified in several major
classes. Each plant family, genus, and species produces a characteristic mix of these
chemicals, and they can sometimes be used as taxonomic characters in classifying
plants. Humans use some of these compounds as medicines, flavorings, or recreational
drugs.
3. Secondary metabolites can be classified on the basis of chemical structure (for example,
having rings, containing a sugar), composition (containing nitrogen or not), their
solubility in various solvents, or the pathway by which they are synthesized (e.g.,
phenylpropanoid, which produces tannins). A simple classification includes three main
groups: the terpenes (made from mevalonic acid, composed almost entirely of carbon
and hydrogen), phenolics (made from simple sugars, containing benzene rings,
hydrogen, and oxygen), and nitrogen-containing compounds (extremely diverse, may
also contain sulfur).
The apparent lack of primary function in the plant, combined with the observation that
many secondary metabolites have specific negative impacts on other organisms such as
herbivores and pathogens , leads to the hypothesis that they have evolved because of
their protective value. Many secondary metabolites are toxic or repellant to herbivores
and microbes and help defend plants producing them. Production increases when a
plant is attacked by herbivores or pathogens. Some compounds are released into the air
when plants are attacked by insects; these compounds attract parasites and predators
that kill the herbivores. Recent research is identifying more and more primary roles for
these chemicals in plants as signals, antioxidants , and other functions, so "secondary"
may not be an accurate description in the future.
Consuming some secondary metabolites can have severe consequences. Alkaloids can
block ion channels, inhibit enzymes , or interfere with neurotransmission, producing
hallucinations , loss of coordination, convulsions, vomiting, and death. Some
phenolics interfere with digestion, slow growth, block enzyme activity and cell division,
or just taste awful.
Most herbivores and plant pathogens possess mechanisms that ameliorate the impacts
of plant metabolites, leading to evolutionary associations between particular groups of
pests and plants. Some herbivores (for example, the monarch butterfly) can store
(sequester) plant toxins and gain protection against their enemies. Secondary
metabolites may also inhibit the growth of competitor plants (allelopathy). Pigments
(such as terpenoid carotenes, phenolics, and flavonoids) color flowers and, together with
terpene and phenolic odors, attract pollinators.
4. Secondary chemicals are important in plant use by humans. Most pharmaceuticals are
based on plant chemical structures, and secondary metabolites are widely used for
recreation and stimulation (the alkaloids nicotine and cocaine; the terpene cannabinol).
The study of such plant use is called ethnopharmacology. Psychoactive plant chemicals
are central to some religions, and flavors of secondary compounds shape our food
preferences. The characteristic flavors and aroma of cabbage and relatives are caused by
Class Example Example Some Effects and Uses
Compounds Sources
NITROGEN-
CONTAINING
Alkaloids nicotine cocaine tobacco coca plant interfere with
theobromine chocolate (cocao) neurotransmission, block
5. Class Example Example Some Effects and Uses
Compounds Sources
enzyme action
NITROGEN-AND
SULFUR-
CONTAINING
Glucosinolates sinigrin cabbage, relatives
TERPENOIDS
Monoterpenes menthol linalool mint and relatives, interfere with
many plants neurotransmission, block ion
transport, anesthetic
Sesquiterpenes parthenolid Parthenium and contact dermatitis
relatives (
Asteraceae )
Diterpenes gossypol cotton block phosphorylation; toxic
Triterpenes, cardiac digitogenin Digitalis (foxglove) stimulate heart muscle, alter
glycosides ion transport
Tetraterpenoids carotene many plants antioxidant; orange coloring
Terpene polymers rubber Hevea (rubber) gum up insects; airplane tires
trees, dandelion
Sterols spinasterol spinach interfere with animal hormone
action
PHENOLICS
6. Class Example Example Some Effects and Uses
Compounds Sources
Phenolic acids caffeic, all plants cause oxidative damage,
chlorogenic browning in fruits and wine
Coumarins umbelliferone carrots, parsnip cross-link DNA, block cell
division
Lignans podophyllin mayapple poison cathartic, vomiting, allergic
urushiol ivy dermatitis
Flavonoids anthocyanin, almost all plants flower, leaf color; inhibit
catechin enzymes, anti- and pro-
oxidants, estrogenic
Tannins gallotannin, oak, hemlock trees, bind to proteins, enzymes,
condensed birdsfoot trefoil, block digestion, antioxidants
tannin legumes
Lignin lignin all land plants structure, toughness, fiber
nitrogen-and sulfur-containing chemicals, glucosinolates, which protect these plants
from many enemies. The astringency of wine and chocolate derives from tannins. The
use of spices and other seasonings developed from their combined uses as preservatives
(since they are antibiotic) and flavorings.
SEE ALSO Flowers ; Herbivory and Plant Defenses ; Metabolism, Cellular ; Poisons
Jack Schultz
Bibliography
7. Agosta, William. Bombardier Beetles and Fever Trees: A Close-up Look at Chemical
Warfare and Signals in Animals and Plants. Reading, MA: Addison-Wesley, 1996.
Bidlack, Wayne R. Phytochemicals as Bioactive Agents. Lancaster, PA: Technomic
Publishers, 2000.
Karban, Richard, and Ian T. Baldwin. Induced Responses to Herbivory. Chicago:
University of Chicago Press, 1997.
Rosenthal, Gerald A., and May R. Berenbaum. Herbivores, Their Interactions with
Secondary Plant Metabolites. San Diego, CA: Academic Press, 1991.
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User Contributions:
1
aarcee varte
Mar 4, 2008 @ 11:23 pm
please explain how 2 extract secondary metabolites which is not secreted out from plants
and remain inside the plants only
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Cellulose[1]
Identifiers
CAS number 9004-34-6
UNII SMD1X3XO9M
EC-number 232-674-9
ChEMBL CHEMBL1201676
Properties
Molecular formula (C6H10O5)n
Appearance white powder
Density 1.5 g/cm3
Melting point decomp.
Solubility in water none
10. Hazards
EU Index not listed
NFPA 704 1
1
0
Related compounds
Related compounds Starch
(verify) (what is: / ?)
Except where noted otherwise, data are given for materials
in their standard state (at 25 °C, 100 kPa)
Infobox references
Cellulose is an organic compound with the formula (C6H10O5)n, a polysaccharide consisting of a
linear chain of several hundred to over ten thousand β(1→4) linked D-glucose units.[2][3]
Cellulose is the structural component of the primary cell wall of green plants, many forms of
algae and the oomycetes. Some species of bacteria secrete it to form biofilms. Cellulose is the
most common organic compound on Earth. About 33% of all plant matter is cellulose (the
cellulose content of cotton fiber is 90%, that of wood is 40–50% and that of dried hemp is
approximately 75%).[4][5][6]
For industrial use, cellulose today is mainly obtained from wood pulp and cotton. Cellulose is
mainly used to produce paperboard and paper; to a smaller extent it is converted into a wide
variety of derivative products such as cellophane and rayon. Converting cellulose from energy
crops into biofuels such as cellulosic ethanol is under investigation as an alternative fuel source.
Some animals, particularly ruminants and termites, can digest cellulose with the help of
symbiotic micro-organisms that live in their guts. Humans can digest cellulose to some
11. extent,[7][8] however it mainly acts as a hydrophilic bulking agent for feces and is often referred
to as "dietary fiber".
Contents
1 History
2 Products
o 2.1 Cellulose source and energy crops
3 Structure and properties
4 Assaying a cellulose-containing material
5 Biosynthesis
6 Breakdown (cellulolysis)
7 Hemicellulose
8 Derivatives
9 References
10 External links
[edit] History
Cellulose was discovered in 1838 by the French chemist Anselme Payen, who isolated it from
plant matter and determined its chemical formula.[2][9][10] Cellulose was used to produce the first
successful thermoplastic polymer, celluloid, by Hyatt Manufacturing Company in 1870.
Hermann Staudinger determined the polymer structure of cellulose in 1920. The compound was
first chemically synthesized (without the use of any biologically derived enzymes) in 1992, by
Kobayashi and Shoda.[11]
Cellulose in a plant cell.
[edit] Products
12. Cotton fibres represent the purest natural form of cellulose, containing more than 90% of this
carbohydrate.
See also: dissolving pulp and pulp (paper)
The kraft process is used to separate cellulose from lignin, another major component of plant
matter. Cellulose is the major constituent of paper, paperboard, and card stock and of textiles
made from cotton, linen, and other plant fibers.
Cellulose can be converted into cellophane, a thin transparent film, and into rayon, an important
fiber that has been used for textiles since the beginning of the 20th century. Both cellophane and
rayon are known as "regenerated cellulose fibers"; they are identical to cellulose in chemical
structure and are usually made from dissolving pulp via viscose. A more recent and
environmentally friendly method to produce a form of rayon is the Lyocell process. Cellulose is
the raw material in the manufacture of nitrocellulose (cellulose nitrate) which is used in
smokeless gunpowder and as the base material for celluloid used for photographic and movie
films until the mid 1930s.
Cellulose is used to make water-soluble adhesives and binders such as methyl cellulose and
carboxymethyl cellulose which are used in wallpaper paste. Microcrystalline cellulose (E460i)
and powdered cellulose (E460ii) are used as inactive fillers in tablets[12] and as thickeners and
stabilizers in processed foods. Cellulose powder is for example used in Kraft's Parmesan cheese
to prevent caking inside the tube.
Cellulose is used in the laboratory as the stationary phase for thin layer chromatography.
Cellulose fibers are also used in liquid filtration, sometimes in combination with diatomaceous
earth or other filtration media, to create a filter bed of inert material. Cellulose is further used to
make hydrophilic and highly absorbent sponges.
Cellulose insulation made from recycled paper is becoming popular as an environmentally
preferable material for building insulation. It can be treated with boric acid as a fire retardant.
Cellulose consists of crystalline and amorphous regions. By treating it with strong acid, the
amorphous regions can be broken up, thereby producing nanocrystalline cellulose, a novel
material with many desirable properties.[13] Recently, nanocrystalline cellulose was used as the
13. filler phase in bio-based polymer matrices to produce nanocomposites with superior thermal and
mechanical properties. [14]
[edit] Cellulose source and energy crops
Main article: Energy crop
The major combustible component of non-food energy crops is cellulose, with lignin second.
Non-food energy crops are more efficient than edible energy crops (which have a large starch
component), but still compete with food crops for agricultural land and water resources.[15]
Typical non-food energy crops include industrial hemp (though outlawed in some countries),
switchgrass, Miscanthus, Salix (willow), and Populus (poplar) species.
Some bacteria can convert cellulose into ethanol which can then be used as a fuel; see cellulosic
ethanol.
A strand of cellulose (conformation Iα), showing the hydrogen bonds (dashed) within and between
cellulose molecules.
[edit] Structure and properties
Cellulose has no taste, is odorless, is hydrophilic with the contact angle of 20–30,[16] is insoluble
in water and most organic solvents, is chiral and is biodegradable. It can be broken down
chemically into its glucose units by treating it with concentrated acids at high temperature.
Cellulose is derived from D-glucose units, which condense through β(1→4)-glycosidic bonds.
This linkage motif contrasts with that for α(1→4)-glycosidic bonds present in starch, glycogen,
and other carbohydrates. Cellulose is a straight chain polymer: unlike starch, no coiling or
branching occurs, and the molecule adopts an extended and rather stiff rod-like conformation,
aided by the equatorial conformation of the glucose residues. The multiple hydroxyl groups on
the glucose from one chain form hydrogen bonds with oxygen atoms on the same or on a
neighbor chain, holding the chains firmly together side-by-side and forming microfibrils with
14. high tensile strength. This strength is important in cell walls, where the microfibrils are meshed
into a carbohydrate matrix, conferring rigidity to plant cells.
A triple strand of cellulose showing the hydrogen bonds (cyan lines) between glucose strands
Compared to starch, cellulose is also much more crystalline. Whereas starch undergoes a
crystalline to amorphous transition when heated beyond 60–70 °C in water (as in cooking),
cellulose requires a temperature of 320 °C and pressure of 25 MPa to become amorphous in
water.[17]
Several different crystalline structures of cellulose are known, corresponding to the location of
hydrogen bonds between and within strands. Natural cellulose is cellulose I, with structures Iα
and Iβ. Cellulose produced by bacteria and algae is enriched in Iα while cellulose of higher plants
consists mainly of Iβ. Cellulose in regenerated cellulose fibers is cellulose II. The conversion of
cellulose I to cellulose II is irreversible, suggesting that cellulose I is metastable and cellulose II
is stable. With various chemical treatments it is possible to produce the structures cellulose III
and cellulose IV.[18]
Many properties of cellulose depend on its chain length or degree of polymerization, the number
of glucose units that make up one polymer molecule. Cellulose from wood pulp has typical chain
lengths between 300 and 1700 units; cotton and other plant fibers as well as bacterial cellulose
have chain lengths ranging from 800 to 10,000 units.[11] Molecules with very small chain length
resulting from the breakdown of cellulose are known as cellodextrins; in contrast to long-chain
cellulose, cellodextrins are typically soluble in water and organic solvents.
Plant-derived cellulose is usually found in a mixture with hemicellulose, lignin, pectin and other
substances, while microbial cellulose is quite pure, has a much higher water content, and consists
of long chains.
Cellulose is soluble in cupriethylenediamine (CED), cadmiumethylenediamine (Cadoxen), N-
methylmorpholine N-oxide and lithium chloride / dimethylformamide.[19] This is used in the
production of regenerated celluloses (such as viscose and cellophane) from dissolving pulp.
[edit] Assaying a cellulose-containing material
Given a cellulose-containing material, the carbohydrate portion that does not dissolve in a 17.5%
solution of sodium hydroxide at 20 °C is α cellulose, which is true cellulose. Acidification of the
extract precipitates β cellulose. The portion that dissolves in base but does not precipitate with
acid is γ cellulose.
15. Cellulose can be assayed using a method described by Updegraff in 1969, where the fiber is
dissolved in acetic and nitric acid to remove lignin, hemicellulose, and xylosans. The resulting
cellulose is allowed to react with anthrone in sulfuric acid. The resulting coloured compound is
assayed spectrophotometrically at a wavelength of approximately 635 nm.
In addition, cellulose is represented by the difference between acid detergent fiber (ADF) and
acid detergent lignin (ADL).
[edit] Biosynthesis
Location and arrangement of cellulose microfibrils in the plant cell wall
In vascular plants cellulose is synthesized at the plasma membrane by rosette terminal complexes
(RTCs). The RTCs are hexameric protein structures, approximately 25 nm in diameter, that
contain the cellulose synthase enzymes that synthesise the individual cellulose chains.[20] Each
RTC floats in the cell's plasma membrane and "spins" a microfibril into the cell wall.
RTCs contain at least three different cellulose synthases, encoded by CesA genes, in an unknown
stoichiometry.[21] Separate sets of CesA genes are involved in primary and secondary cell wall
biosynthesis.
Cellulose synthesis requires chain initiation and elongation, and the two processes are separate.
CesA glucosyltransferase initiates cellulose polymerization using a steroid primer, sitosterol-
beta-glucoside, and UDP-glucose.[22] Cellulose synthase utilizes UDP-D-glucose precursors to
elongate the growing cellulose chain. A cellulase may function to cleave the primer from the
mature chain.
Cellulose is also synthesised by animals, particularly in the tests of ascidians (where the cellulose
was historically termed "tunicine") although it is also a minor component of mammalian
connective tissue.[23]
[edit] Breakdown (cellulolysis)
Cellulolysis is the process of breaking down cellulose into smaller polysaccharides called
cellodextrins or completely into glucose units; this is a hydrolysis reaction. Because cellulose
molecules bind strongly to each other, cellulolysis is relatively difficult compared to the
breakdown of other polysaccharides.[24] Processes do exist however for the breakdown of
16. cellulose such as the Lyocell process [25] which uses a combination of heated water and acetone
to break down the cellulose strands.
Most mammals have only very limited ability to digest dietary fibres such as cellulose. Some
ruminants like cows and sheep contain certain symbiotic anaerobic bacteria (like Cellulomonas)
in the flora of the rumen, and these bacteria produce enzymes called cellulases that help the
microorganism to break down cellulose; the breakdown products are then used by the bacteria
for proliferation. The bacterial mass is later digested by the ruminant in its digestive system
(stomach and small intestine). Similarly, lower termites contain in their hindguts certain
flagellate protozoa which produce such enzymes; higher termites contain bacteria for the job.
Some termites may also produce cellulase of their own.[26] Fungi, which in nature are responsible
for recycling of nutrients, are also able to break down cellulose.
The enzymes utilized to cleave the glycosidic linkage in cellulose are glycoside hydrolases
including endo-acting cellulases and exo-acting glucosidases. Such enzymes are usually secreted
as part of multienzyme complexes that may include dockerins and carbohydrate-binding
modules.[27]
[edit] Hemicellulose
Main article: Hemicellulose
Hemicellulose is a polysaccharide related to cellulose that comprises about 20% of the biomass
of most plants. In contrast to cellulose, hemicellulose is derived from several sugars in addition
to glucose, especially xylose but also including mannose, galactose, rhamnose, and arabinose.
Hemicellulose consists of shorter chains – around 200 sugar units. Furthermore, hemicellulose is
branched, whereas cellulose is unbranched.
[edit] Derivatives
The hydroxyl groups (-OH) of cellulose can be partially or fully reacted with various reagents to
afford derivatives with useful properties like mainly cellulose esters and cellulose ethers (-OR).
In principle, though not always in current industrial practice, cellulosic polymers are renewable
resources.
Ester derivatives include:
Cellulose
Reagent Example Reagent Group R
ester
Organic Organic Acetic acid and acetic
Cellulose acetate H or -(C=O)CH3
esters acids anhydride
Acetic acid and acetic
Cellulose triacetate -(C=O)CH3
anhydride
17. Cellulose propionate Propanoic acid H or -(C=O)CH2CH3
Cellulose acetate Acetic acid and propanoic H or -(C=O)CH3 or -
propionate acid (C=O)CH2CH3
Cellulose acetate H or -(C=O)CH3 or -
Acetic acid and butyric acid
butyrate (C=O)CH2CH2CH3
Inorganic Inorganic Nitrocellulose Nitric acid or another
H or -NO2
esters acids (cellulose nitrate) powerful nitrating agent
Sulfuric acid or another
Cellulose sulfate H or -SO3H
powerful sulfuring agent
The cellulose acetate and cellulose triacetate are film- and fiber-forming materials that find a
variety of uses. The nitrocellulose was initially used as an explosive and was an early film
forming material. With camphor, nitrocellulose gives celluloid.
Ether derivatives include:
Water E
Cellulose Group R = H
Reagent Example Reagent solubilit Application numbe
ethers or
y r
Cold
Halogenoalkan Methylcellulo Chloromethan
Alkyl -CH3 water E461
es se e
soluble
A
commercial
thermoplasti
c used in
Water
coatings,
Ethylcellulose Chloroethane -CH2CH3 insolubl E462
inks, binders,
e
and
controlled-
release drug
tablets
Ethyl methyl Chloromethan -CH3 or - E465
cellulose e and CH2CH3
18. chloroethane
Cold/ho Gelling and
Hydroxyalk Hydroxyethyl Ethylene
Epoxides -CH2CH2OH t water thickening
yl cellulose oxide
soluble agent
Hydroxypropy - Cold
Propylene
l cellulose CH2CH(OH)C water E463
oxide
(HPC) H3 soluble
Chloromethan
Hydroxyethyl Cold Production of
e and -CH3 or -
methyl water cellulose
ethylene CH2CH2OH
cellulose soluble films
oxide
Viscosity
Hydroxypropy Chloromethan modifier,
-CH3 or - Cold
l methyl e and gelling,
CH2CH(OH)C water E464
cellulose propylene foaming and
H3 soluble
(HPMC) oxide binding
agent
Ethyl Chloroethane
-CH2CH3 or—
hydroxyethyl and ethylene E467
CH2CH2OH
cellulose oxide
Often used
as its sodium
Carboxymethy Cold/Ho
Carboxyalk Halogenated Chloroacetic salt, sodium
l cellulose -CH2COOH t water E466
yl carboxylic acids acid carboxymeth
(CMC) soluble
yl cellulose
(NaCMC)
The sodium carboxymethyl cellulose can be cross-linked to give the croscarmellose sodium
(E468) for use as a disintegrant in pharmaceutical formulations.
[edit] References
1. ^ Nishiyama, Yoshiharu; Langan, Paul; Chanzy, Henri (2002). "Crystal Structure and Hydrogen-Bonding
System in Cellulose Iβ from Synchrotron X-ray and Neutron Fiber Diffraction". J. Am. Chem. Soc 124 (31):
9074–82. doi:10.1021/ja0257319. PMID 12149011..
ab
2. ^ Crawford, R. L. (1981). Lignin biodegradation and transformation. New York: John Wiley and Sons.
ISBN 0-471-05743-6.
19. 3. ^ Updegraff DM (1969). "Semimicro determination of cellulose in biological materials". Analytical
Biochemistry 32 (3): 420–424. doi:10.1016/S0003-2697(69)80009-6. PMID 5361396.
4. ^ Cellulose. (2008). In Encyclopædia Britannica. Retrieved January 11, 2008, from Encyclopædia Britannica
Online.
5. ^ Chemical Composition of Wood
6. ^ G. Buschle-Diller, C. Fanter, F. Loth (April 1999). "Structural changes in hemp fibers as a result of
enzymatic hydrolysis with mixed enzyme systems". Textile Research Journal 69 (4): 244-251.
http://www.globalhemp.com/1999/04/structural-changes-in-hemp-fibers-as-a-result-of-enzymatic-
hydrolysis-with-mixed-enzyme-systems.html.
7. ^ Slavin, JL; Brauer, PM; Marlett, JA (1981). "Neutral detergent fiber, hemicellulose and cellulose
digestibility in human subjects.". The Journal of Nutrition 111 (2): 287–97. PMID 6257867.
8. ^ Joshi, S; Agte, V (1995). "Digestibility of dietary fiber components in vegetarian men.". Plant foods for
human nutrition (Dordrecht, Netherlands) 48 (1): 39–44. doi:10.1007/BF01089198. PMID 8719737.
9. ^ A. Payen (1838) "Mémoire sur la composition du tissu propre des plantes et du ligneux" (Memoir on the
composition of the tissue of plants and of woody [material]), Comptes rendus, vol. 7, pages 1052-1056.
Payen added appendices to this paper on December 24, 1838 (see: Comptes rendus, vol. 8, page 169
(1839)) and on February 4, 1839 (see: Comptes rendus, vol. 9, page 149 (1839)). A committee of the
French Academy of Sciences reviewed Payen's findings in : Jean-Baptiste Dumas (1839) "Rapport sur un
mémoire de M. Payen, relatif à la composition de la matière ligneuse" (Report on a memoir of Mr. Payen,
regarding the composition of woody matter), Comptes rendus, vol. 8, pages 51-53. In this report, the word
"cellulose" is coined and author points out the similarity between the empirical formula of cellulose and
that of "dextrine" (starch). The above articles are reprinted in: Brongniart and Guillemin, eds., Annales des
sciences naturelles ..., 2nd series, vol. 11 (Paris, France: Crochard et Cie., 1839), pages 21-31.
10. ^ Young, Raymond (1986). Cellulose structure modification and hydrolysis. New York: Wiley. ISBN 0-471-
82761-4.
ab
11. ^ Klemm, Dieter; Brigitte Heublein, Hans-Peter Fink, Andreas Bohn (2005). "Cellulose: Fascinating
Biopolymer and Sustainable Raw Material". ChemInform 36 (36). doi:10.1002/chin.200536238.
12. ^ Weiner, Myra L.; Lois A. Kotkoskie (2000). Excipient Toxicity and Safety. New York ; Dekker. p. 210.
ISBN 0-8247-8210-0.
13. ^ Peng, B. L., Dhar, N., Liu, H. L. and Tam, K. C. (2011). "Chemistry and applications of nanocrystalline
cellulose and its derivatives: A nanotechnology perspective.". The Canadian Journal of Chemical
Engineering 89 (5): 1191–1206.
http://www.arboranano.ca/pdfs/Chemistry%20and%20applications%20of%20nanocrystalline%20cellulos
e%20and%20its%20derivatives%20A%20nanotechnology%20perspective-2011.pdf.
14. ^ Lawrence Pranger and Rina Tannenbaum "Biobased nanocomposites prepared by in situ polymerization
of furfuryl alcohol with cellulose whiskers or montmorillonite clay" Macromolecules 41 (2008) 8682.
http://dx.doi.org/10.1021/ma8020213
15. ^ Holt-Gimenez, Eric 2007. Biofuels: Myths of the Agrofuels Transition. Backgrounder. Institute for Food
and Development Policy, Oakland, CA. 13:2
16. ^ Charles A. Bishop, ed. (2007). Vacuum deposition onto webs, films, and foils, Volume 0, Issue 8155.
p. 165. ISBN 0-8155-1535-9. http://books.google.com/books?id=vP9E3z7o6iIC&pg=PA165.
17. ^ Deguchi, Shigeru; Tsujii, Kaoru; Horikoshi, Koki (2006). "Cooking cellulose in hot and compressed water".
Chemical Communications (31): 3293. doi:10.1039/b605812d.
18. ^ Structure and morphology of cellulose by Serge Pérez and William Mackie, CERMAV-CNRS, 2001.
Chapter IV.
19. ^ Stenius, Per (2000). "1". Forest Products Chemistry. Papermaking Science and Technology. 3. Finland:
Fapet OY. p. 35. ISBN 952-5216-03-9.
20. ^ Kimura, S; Laosinchai, W; Itoh, T; Cui, X; Linder, CR; Brown Jr, RM (1999). "Immunogold labeling of
rosette terminal cellulose-synthesizing complexes in the vascular plant vigna angularis". The Plant cell 11
(11): 2075–86. doi:10.2307/3871010. JSTOR 3871010. PMC 144118. PMID 10559435.
//www.ncbi.nlm.nih.gov/pmc/articles/PMC144118/.
21. ^ Taylor, N. G. (2003). "Interactions among three distinct CesA proteins essential for cellulose synthesis".
Proceedings of the National Academy of Sciences 100 (3): 1450. doi:10.1073/pnas.0337628100.
20. 22. ^ Peng, L; Kawagoe, Y; Hogan, P; Delmer, D (2002). "Sitosterol-beta-glucoside as primer for cellulose
synthesis in plants". Science 295 (5552): 147–50. doi:10.1126/science.1064281. PMID 11778054.
23. ^ Endean, The Test of the Ascidian, Phallusia mammillata, Quarterly Journal of Microscopical Science, Vol.
102, part 1, pp. 107-117, 1961.
24. ^ David G. Barkalow, Roy L. Whistler, "Cellulose", in AccessScience, McGraw-Hill, doi:10.1036/1097-
8542.118200. Retrieved 11 January 2008.
25. ^ H.Lyocell, "Cellulose" Issue 41, pp 419
26. ^ Tokuda, G; Watanabe, H (22 June 2007). "Hidden cellulases in termites: revision of an old hypothesis".
Biology Letters 3 (3): 336–339. doi:10.1098/rsbl.2007.0073. PMC 2464699. PMID 17374589.
http://rsbl.royalsocietypublishing.org/content/3/3/336.long
27. ^ Brás, Natércia; N. M. F. S. A. Cerqueira, P. A. Fernandes, M. J. Ramos (2008). "Carbohydrate Binding
Modules from family 11: Understanding the binding mode of polysaccharides". International Journal of
Quantum Chemistry 108 (11): 2030–2040. doi:10.1002/qua.21755.
[edit] External links
Structure and morphology of cellulose by Serge Pérez and William Mackie, CERMAV-CNRS
Cellulose, by Martin Chaplin, London South Bank University
Clear description of a cellulose assay method at the Cotton Fiber Biosciences unit of the USDA.
Cellulose films could provide flapping wings and cheap artificial muscles for robots -
TechnologyReview.com
Using cellulase enzymes in the bioethanol process
A list of cellulolytic bacteria
v
t
e
Types of carbohydrates
Aldose
Furanose
General
Ketose
Pyranose
Anomer
Geometry Cyclohexane conformation
Mutarotation
21. Aldodiose
Dioses
o Glycolaldehyde
Aldotriose
o Glyceraldehyde
Trioses
Ketotriose
o Dihydroxyacetone
Aldotetroses
o Erythrose
Tetroses o Threose
Ketotetrose
o Erythrulose
Monosaccharides
Aldopentose
o Arabinose
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o Xylose
Pentoses
Deoxy sugar
o Deoxyribose
Ketopentose
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Aldohexose
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o Allose
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22. o Galactose
o Glucose
o Gulose
o Idose
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o Talose
Deoxy sugar
o Fucose
o Fuculose
o Rhamnose
Ketohexose
o Fructose
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Ketoheptose
Heptoses o Mannoheptulose
o Sedoheptulose
Octose
>7 Nonose
o Neuraminic acid
Cellobiose
Multiple Disaccharides Lactose
Maltose
Sucrose
23. Trehalose
Turanose
Maltotriose
Trisaccharides Melezitose
Raffinose
Tetrasaccharides Stachyose
Acarbose
Fructooligosaccharide (FOS)
Other Galactooligosaccharide (GOS)
oligosaccharides Isomaltooligosaccharide (IMO)
Maltodextrin
Mannan-oligosaccharides (MOS)
Beta-glucan
o Lentinan
o Sizofiran
o Zymosan
Cellulose
Polysaccharides Chitin
Dextrin / Dextran
Fructose / Fructan
o Inulin
Galactose / Galactan
Glucose / Glucan
o Glycogen
24. Levan beta 2→6
Mannan
Starch
o Amylopectin
o Amylose
biochemical families: carbohydrates
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o eicosanoids
o fatty acids / intermediates
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30. Introduction
Some useful constants in thermodynamics:
1 eV = 9.6522E4 J/mol
k Boltzmann's constant = 1.38E-23 J/K
volume: 1 cm3 = 0.1 kJ/kbar = 0.1 J/bar
mole: 1 mole of a substance contains Avogadro's number (N = 6.02E23) of molecules.
Abbreviated as 'mol'.
atomic weights are based around the definition that 12C is exactly 12 g/mol
R gas constant = Nk = 8.314 J mol-1 K-1
Units of Temperature: Degrees Celsius and Kelvin
The Celsius scale is based on defining 0 °C as the freezing point of water and 100°C as the boiling point.
The Kelvin scale is based on defining 0 K, "absolute zero," as the temperature at zero pressure
where the volumes of all gases is zero--this turns out to be -273.15 °C. This definition means that
the freezing temperature of water is 273.15 K. All thermodynamic calculations are done in
Kelvin!
kilo and kelvin: write k for 1000's and K for kelvin. Never write °K.
Units of Energy: Joules and Calories
Joules and calories and kilocalories: A calorie is defined as the amount of energy required to raise the
temperature of 1 g of water from 14.5 to 15.5°C at 1 atm.
4.184 J = 1 cal; all food 'calories' are really kcal.
31. Many times it is easiest to solve equations or problems by conducting "dimensional analysis,"
which just means using the same units throughout an equation, seeing that both sides of an
equation contain balanced units, and that the answer is cast in terms of units that you want. As an
example, consider the difference between temperature (units of K) and heat (units of J). Two
bodies may have the same temperature, but contain different amounts of heat; likewise, two
bodies may contain the same heat, but be at different temperatures. The quantity that links these
two variables must have units of J/K or K/J. In fact, the heat capacity C describes the amount of
heat dQ involved in changing one mole of a substance by a given temperature increment dT:
dQ = CdT
The heat capacity C is then
C = dQ/dT
and must have units of J K-1 mol-1. (The specific heat is essentially the same number, but is expressed per
gram rather than per mole.)
Don't forget significant digits. 1*2=2; 1.1*2=2; 1.1*2.0=2.2; 1.0*2.0=2.0
Why Thermodynamics?
Think about some everyday experiences you have with chemical reactions.
Your ability to melt and refreeze ice shows you that H2O has two phases and that the reaction
transforming one to the other is reversible--apparently the crystallization of ice requires removing some
heat.
Frying an egg is an example of an irreversible reaction.
If you dissolve halite in water you can tell that the NaCl is still present in some form by tasting the water.
Why does the NaCl dissolve? Does it give off heat? Does it require energy?
How is it that diamond, a high-pressure form of C, can coexist with the low pressure form, graphite, at
Earth's surface? Do diamond and graphite both have the same energy? If you burn graphite and
diamond, which gives you more energy?
When dynamite explodes, why does it change into a rapidly expanding gas, which provides the energy
release, plus a few solids?
Chemical thermodynamics provides us with a means of answering these questions and more.
A Few Definitions
A system is any part of the universe we choose to consider.
Matter and energy can flow in or out of an open system but only energy can be added to or subtracted
from a closed system. An isolated system is one in which matter and energy are conserved.
A phase is a homogeneous body of matter. The components of a system are defined by a set of
chemical formula used to describe the system.
The phase rule:
32. F + P = C + 2.
Extensive parameters are proportional to mass (e.g., V, mass, energy).
Intensive parameters are independent of mass (e.g., P, T); these are the "degrees of freedom" F
contained in the phase rule.
Thermodynamics: Power and Limitations
Thermodynamics allows you to predict how chemical systems should behave from a supra-atomic
"black-box" level--it says nothing about how chemical systems will behave. Thermodynamics also
pertains to the state of a system, and says nothing about the path taken by the system in changing from
one state to another.
Chemical Reactions and Equations
How to write chemical reactions; stoichiometry.
Mass and charge balance: e.g.,
2Fe3+ + 3H2O = Fe2O3 + 6H+
Reaction-Produced Change in Mass, Density, Volume
The change in volume rV of a reaction is the volume V of the products minus the volume of the
reactants:
rV = Vproducts - Vreactants
Thus, if the products are smaller than the reactants, rV < 0.
In a generalized reaction such as
aA + bB ... = cC + dD...
rV = cVC + dVD - aVA - bVB
This sort of additive relationship is true for other state variables and is usually stated as
r i i
where i are the stoichiometric coefficients, positive for products and negative for reactants.
What Actually Drives Reactions? Is it Energy? Can We Just Calculate or Measure the Energy
Difference of Reactants and Products and Know Which Way the Reaction Will Go?
33. For many years people felt that chemical reactions occurred because the reactants had some kind of
energy to give up (i.e., use to do work)--and that therefore the energy of the products would be less
than the energy of the reactants. However, we all know that when ice melts it consumes rather than
releases heat, so there must be more to the story behind why chemical reactions occur.
Le Chatelier's Principle
"If a change is made to a system, the system will respond so as to absorb the force causing the change."
Equilibrium
A mechanical analogy for chemical change is that of a ball rolling down a slope with multiple valleys; we
explain the ball's behavior by saying that mechanical systems have a tendency to reduce their potential
energy.
At equilibrium none of the properties of a system change with time. A system at equilibrium returns to
equilibrium if disturbed.
"Stable" describes a system or phase in its lowest energy state.
"Metastable" describes a system or phase in any other energy state.
The figure above shows the mechanical analogy for H2O at -5°C and + 5°C and 1 atm. Left: at -5°C, solid
H2O has the lowest possible energy state. Right: at +5°C, liquid H2O has the lowest possible energy state.
When solid H2O is actually present at +5°C, the difference between the free energy of solid H2O and
liquid H2O is available to drive the reaction to form the stable solid H2O phase, and the reaction will go
to completion if kinetically possible.
34. Energy: How Do We Calculate and Measure Energy and How Can We Use this Knowledge to
Predict Reaction Behavior?
Thermodynamics works equally well to describe any kind of work or energy: magnetic, potential, kinetic,
etc. For geological systems we typically talk about pressure-volume work, which, because mechanical
work is F x, you can imagine might be
P V
or
V P
Because we noted that rV < 0 if the products are smaller than the reactants, we choose to write the P-
V work term as
-P V
so that a decrease in volume - V is seen as positive work or that an increase in volume + V results in
a decrease in crystal energy.
The absolute energy of a body can be calculated from Einstein's equation U=mc2, but the presence of
the c2 term means that the energy of any system is quite large and that measuring this energy is
impractical. It is more practical to measure differences in energy U, and we always discuss or measure
differences relative to some arbitrary standard state. Analogous to this might be if someone in Namibia
asked you to measure the elevation of the crests of waves at Campus Point--without agreement on
some kind of standard, you wouldn't be able to do much more than measure the heights of individual
waves. If however, you could both agree on an equivalent "sea level" at both localities, you could then
compare the absolute elevations of the wave crests.
A typical thermodynamic standard state is normal laboratory conditions: 25°C (298.15 K) and 1
atm (often called STP for standard temperature and pressure).
The internal energy U of a mineral is the sum of the potential energy stored in the interatomic
bonds and the kinetic energy of the atomic vibrations. Thus, you might expect that weakly
bonded minerals have relatively low potential energy and thus low internal energy, and when a
mineral is cold such that its atomic vibrations are slow it will have low kinetic energy and thus
low internal energy. Internal energies are always defined relative to some non-zero standard
state, so we typically talk about changes in internal energy dU.
An Aside on Differences and Differentials
What's the difference among , d , and ?
is used to indicate any kind of difference.
35. d is used to indicate a differential.
is used to indicate a partial differential. For example, the partial differential, with respect to y, of
f(x,y) = x3y4
is
= 4x3y3
First Law of Thermodynamics
Adding heat Q to a crystal increases its internal energy U:
dU dQ
( indicates 'proportional') but if the crystal is allowed to expand, some of the added energy will be
consumed by expansion dV, so the total energy of the crystal is reduced:
dU = dQ - PdV
This is effectively the First Law of Thermo: that total energy (heat + P-V work) is conserved.
Heat Capacity
Heat capacity C describes the amount of heat required to change the temperature of a substance:
C=
By definition, the heat capacity of water at 15°C is 1 cal K-1 g-1 or 18 cal K-1 mol-1 (i.e., the heat required
to heat 1 gram of water from 14.5 to 15.5°C is 1 calorie).
Heat capacities of solids approach zero as absolute zero is approached:
C=0
The heat capacity is written with a subscript P or V depending on whether it obtains for constant
pressure CP or constant volume CV.
As an aside,
2
CP = CV + TV /
36. where and are the expansivity and compressibility--for solids the difference between CP and CV is
minimal and can be ignored as a first approximation. For gases, CP = CV + R, and is quite significant.
Heat capacities are measured by calorimetry and often fit by a function of the form:
CP = a + bT + cT-2 + dT-0.5
but there are other expansions for the heat capacity involving more or fewer terms.
Below are some examples of heat capacities of minerals. Note how silicates have a nearly constant heat
capacity of ~1 J K-1 g-1 above 400K.
37. Enthalpy
We have already talked about the familiar concept of heat as energy.
Let's define another measure of energy called enthalpy H--a kind of measure of the thermal energy of a
crystal. As we will see below,
38. dH = dQ + VdP
Recall that we interpreted
dU = dQ - PdV
to mean that the internal energy change is the heat change minus the energy lost to relaxation of the
crystal. Thus,
dH = dQ + VdP
means that the enthalpy change is the heat change plus the energy the crystal gains by virtue of not
being allowed to expand.
Enthalpy includes the vibrational and bonding energy at absolute zero H0°, plus the energy required to
increase temperature:
H = H0° + CPdT
i.e., we can find the enthalpy change H produced by changing temperature by integrating the heat
capacity CP:
H= CPdT
Integration Reminder
How to integrate the heat capacity (to determine change in enthalpy H):
CP dT = (a + bT + cT-2 + dT-0.5)
=aT + bT2/2 - c/T + 2dT0.5
and is evaluated as
=a(T2 - T1) + b(T22 - T12)/2 - c/(T2-1 - T1-1) + 2d(T20.5 - T10.5)
How to integrate the heat capacity divided by T (to determine entropy S):
dT = (a/T + b + cT-3 + dT-1.5)
39. = a ln T + b T - c T-2/2 - 2 d T -0.5
and is evaluated as
a(ln T2 - ln T1) + b(T2 - T1) - c(T2-2 - T1-2)/2 - 2d(T2-0.5 - T1-0.5)
As an example, let's calculate the change in enthalpy H°298-1000 that results from heating quartz
from 298 K to 1000 K, given the following heat capacity expansion coefficients:
a = 104.35, b = 6.07E-3, c = 3.4E+4, d = -1070
(CP dT = (a + bT + cT-2 + dT-0.5)
=aT + bT2/2 - c/T + 2dT0.5
evaluated from 298 to 1000K
=a*(1000-298) + b*(10002-2982)/2 - c*(1000-1-298-1) + 2d*(10000.5-2980.5)
= 45.37 kJ/mol
Relation Among Enthalpy, Heat, and Heat Capacity ( HP= QP)
An important relationship between enthalpy change H and heat change Q is revealed by
differentiating H = U + PV to obtain the total differential
dH = dU + PdV + VdP
substituting dU = dQ - PdV we get
dH = dQ + VdP
dividing by dT gives
= -V
at constant pressure, = 0, leaving
=
40. which is equal to CP:
= = CP
Determining Enthalpies
Thus, if we want to measure how the internal energy U of a crystal changes U with increasing
temperature at constant pressure, we want to know H, and we can get that by integrating the heat
capacity CP over the temperature range of interest.
There's another way to measure H, though: calorimetry. By dissolving a mineral in acid and measuring
the heat produced by the dissolution, we get a heat of dissolution (usually positive). The enthalpy of
"formation" fH° of the mineral is then just the opposite of the heat of dissolution (usually negative).
Exceptions to the "usually positive/negative" rule include CN, HCN, Cu2+, Hg2+, NO, Ag+, and S2-.
Enthalpies of formation appear in tables of thermodynamic data and are usually referenced to 298 K and
1 atm.
Enthalpy of Reaction
To get an enthalpy of reaction rH° we can measure the enthalpies of formation of the reactants and
products fH° and then take the difference between them as
rH° = fH°products- fH°reactants
For example, we can compute the enthalpy of the reaction
anhydrite + water = gypsum:
CaSO4 + 2H2O = CaSO4 2H2O
from
Ca + S + 2O2 = CaSO4 fH° = -1434.11 kJ/mol
H2 + 0.5O2 = H2O fH° = -285.830 kJ/mol
Ca + S + 3O2 + 2H2 = CaSO4 2H2O fH° = -2022.63 kJ/mol
Thus,
rH° = fH°gypsum - fH°anhydrite - fH°water = -16.86 kJ/mol.
Exothermic vs. Endothermic
41. If rH° < 0 the reaction produces a reduction in enthalpy and is exothermic (heat is given up by the rock
and gained by the surroundings). If rH° > 0 the reaction produces an increase in enthalpy and is
endothermic (heat from the surroundings is consumed by the rock). An easy way to remember this is
that spontaneous reactions produce a decrease in internal energy, and because we know that
UP HP
a decrease in HP is also a decrease in UP.
Calculating fH° at Temperatures Other Than 298 K
So far we know how to calculate the change in enthalpy caused by heating and we know that we can get
enthalpies of formation from tables. What if we want to know the enthalpy of formation of a mineral at
a temperature other than 298 K?
We do this by calculating rCP for the reaction that forms the mineral of interest:
rCP = rCPproducts - rCPreactants
and then integrating. Thus, for example if we want to know fH° for quartz at 1000 K, we get
coefficients for the heat capacities of Si, O2 and SiO2:
compound a b c d
Si 31.778 5.3878E-4 -1.4654E5 -1.7864E2
O2 48.318 -6.9132E-4 4.9923E5 -4.2066E2
SiO2 104.35 6.07E-3 3.4E-4 -1070
for the reaction
Si + O2 = SiO2
and we calculate
a = 24.254
b = 6.2225E-3
c = -3.5E5
d = -470.7
and thus,
fH°1000 - fH°298 = CP dT = a*(1000-298) + b/2*(10002-2982) - c*(1000-1-298-1)
0.5 0.5
+2 d*(1000 -298 ) = 5.511 kJ/mol
42. This is the change in the enthalpy of formation that results from heating. We add this to the enthalpy of
formation at 298 K to get the enthalpy of formation at 1000 K:
fH°1000 =( fH°1000 - fH°298) + fH°298 = 5.511 - 910.700 = -905.2 kJ/mol
In other words, forming quartz from the elements at 1000 K yields slightly less heat than at 298 K.
Compare this with the change in enthalpy H°298-1000 that results from heating quartz from 298 K to
1000 K, which we calculated is 45.37 kJ/mol.
Entropy
We have discussed the intuitive statement that reactions probably proceed because the reactants can
decrease their internal energy by reacting. We also noted that internal energy scales with enthalpy,
suggesting that reactions might 'go' because of a decrease in enthalpy. However, we then noted that not
all reactions give off heat--some, such as the melting of ice, proceed in spite of consuming heat.
Moreover, there are other processes that proceed in the apparent absence of any heat change: e.g.,
mixing of gases or the spreading of dye in water. What is it that causes these reactions to proceed
spontaneously even if the heat change is zero or endothermic?
The answer is entropy S, which is a measure of the order or disorder. Entropy has three sources:
configurational, electronic, and vibrational.
Configurational entropy refers to the entropy resulting from imperfect mixing of different
atoms in the same site in a crystal, and is described by the Boltzmann distribution:
Sconfigurational = k ln (This is engraved on Boltzmann's tomb in Vienna!)
where is the probability that a given number of atoms in a given number of sites will have a particular
configuration.
For N atomic sites that can contain fraction XA A atoms and XB B atoms,
=
43. N is always large where moles of material are concerned, so we can simplify this (using Stirling's
approximation) to
S = - n R (XA ln XA + XB ln XB)
where n is the number of sites per mole. For example in cordierite there are 4 Al atoms and 5 Si atoms
distributed over 9 tetrahedral sites. For a random distribution the entropy is
S = - 9 R (4/9 ln 4/9 + 5/9 ln 5/9) = 51.39 J mol-1 K-1
Note that the form of the configurational entropy equation (and electronic entropy as well) indicates
that if XA or XB are 0 or 1, Sconfig is zero:
Electronic entropy arises when an electron in an unfilled orbital can occupy more than one orbital; e.g.,
for Ti3+, the single 3d electron can occupy one of three possible t2g orbitals and Selectronic = 9 J mol-1 K-1.
Vibrational (or calorimetric) entropy arises because the energy of lattice vibrations can only increase or
decrease in discrete steps and the energy quanta (phonons) can be distributed within the possible
energy steps in different ways. Vibrational entropy is very difficult to calculate from statistical mechanics
but can be calculated easily from heat capacity. Here's why:
The entropy of a system always increases during irreversible processes; i.e., for a reversible process, dS =
0, whereas for irreversible processes dS >0. This is the Second Law of Thermo--better known as "You
can't feed s**t into the rear of a horse and get hay out the front."
If a mineral becomes more ordered during a reaction, reducing its entropy, the heat liberated must
increase the entropy of the surroundings by an even greater amount. Thus, we write
dS >
then
44. >
and recalling that
CP =
then
>
and
S= dT
In other words, the vibrational entropy can be found by integrating the heat capacity divided by
temperature.
In a perfectly ordered, pure crystalline material the entropy is zero. This is a simple statement of the
Third Law of Thermo, which follows from the fact that heat capacities approach zero at zero K:
C=0
However, because the rate of atomic diffusion also goes to zero at 0 K, all compounds have some zero-
point entropy S°0.
Entropy is thus the only thermodynamic potential for which we can calculate an absolute value. What
we typically do is determine the heat capacity from near absolute zero to ambient conditions and then
integrate it to get the (absolute) entropy (in fact this gives us only the vibrational entropy and ignores
configurational and electronic contributions to entropy).
45. Entropy Change of Reaction
Just like rH and rV, we can calculate entropies of reactions by using absolute entropies S and
calculating a difference in entropy rS. For example, if we know that
S°CaSO4 = 106.7 J mol-1 K-1
S°Ca = 41.42 J mol-1 K-1
S°S = 31.80 J mol-1 K-1
S°O2 = 205.138 J mol-1 K-1
then the entropy of the reaction
Ca + S + 2O2 = CaSO4
is rS° = 106.7 - 41.42 - 31.80 - 2 * 205.138 = -376.8 J mol-1 K-1
46. Energy Associated With Entropy
The units of entropy suggest that the energy associated with S scales with temperature:
dU -TS
(The minus sign is there for reasons similar to the -PV we encountered earlier.)
The energy associated with configurational entropy in the Al4Si5 cordierite we talked about earlier looks
like this:
The energy associated with vibrational entropy in tremolite, quartz, and chalcopyrite looks like
this:
(Josiah Willard) Gibbs Free Energy of a Phase
The Gibbs free energy G is the thermodynamic potential that tells us which way a reaction goes at a
given set of physical conditions--neither the enthalpy change nor the entropy change for a reaction
47. alone can provide us with this information. The two measures of energy (enthalpy H and entropic
energy TS) are brought together in the Gibbs free energy equation: (the chemical potential is the
equivalent for a component)
G = U + PV - TS
which says that the Gibbs free energy G is the internal energy of the crystal U plus the energy the crystal
gains by virtue of not being allowed to expand minus the entropic energy TS. Recalling that
H = U + PV
we can write this in a more understandable way
G = H - TS
which says that G is the difference between the heat energy and the entropic energy.
Relationship Among G, S, and V
If we differentiate
G = U + PV - TS
to obtain
dG = dU + PdV + VdP - TdS - SdT
and substitute
48. TdS = dU + PdV
(this comes from dS = dQ/T and dU = dQ - PdV); we are left with
dG = VdP - SdT
meaning that changes in Gibbs free energy are produced by changes in pressure and temperature acting
on the volume and entropy of a phase.
Realize that when we write
dG = VdP - SdT
we are implicitly writing
dG = dP - dT
which means that
=V
and
= -S
49. These relations indicate that the change in Gibbs free energy with respect to pressure is the molar
volume V and the change in Gibbs free energy with respect to temperature is minus the entropy S.
Gibbs Free Energy of Formation
The defining equation for Gibbs free energy
G = H - TS
can be written as
G= H-T S
such that the Gibbs free energy of formation fG° is
fG° = fH° -T fS°
For example, to calculate the Gibbs free energy of formation of anhydrite, we can use
fH°CaSO4 = -1434.11 kJ/mole
S°CaSO4 = 106.7 J mol-1 K-1
S°Ca = 41.42 J mol-1 K-1
50. S°S = 31.80 J mol-1 K-1
S°O2 = 205.138 J mol-1 K-1
and we calculate the entropy of formation of anhydrite
fS° = S°CaSO4 - S°Ca - S°S - 2 * S°O2 = -376.796 J mol-1 K-1
and then use
fG° = fH° -T fS° = -1434,110 - 298.15 * -376.796 = -1321.77 kJ/mol
Gibbs Free Energy of Reaction
We can write the Gibbs free energy of reaction as the enthalpy change of reaction minus the entropic
energy change of reaction
rG = rH -T rS
If the heat energy equals the entropic energy
rH =T rS
then
rG =0
and there is no reaction. Finally we have come to a satisfying point--we can now determine whether a
given reaction will occur if we know H and S, and both of these are measurable or can be
calculated.
51. If rG < 0, the Gibbs free energy of the products is lower than the Gibbs free energy of the reactants
and the reaction moves to produce more products. If rG > 0, the Gibbs free energy of the products is
greater than the Gibbs free energy of the reactants and the reaction moves to produce more reactants.
For example, to calculate rG° at STP for the reaction
aragonite = calcite
we use
rH° = 370 J
rS° = 3.7 J mol-1 K-1
to calculate
rG° = 370 - 298.15 * 3.7 = -733 J/mol
The negative value of G tells us that calcite has lower Gibbs free energy and that the reaction runs
forward (aragonite calcite).
Clapeyron Relation
There is a useful relation between the slope of a reaction in PT space (i.e., dP/dT) and the entropy and
volume changes of the reaction that follows from
rG = VrdP - SrdT
At equilibrium G = 0, such that
52. rVdP = rSdT
or
=
So, the P-T slope of a reaction is equal to the ratio of the entropy change to the volume change.
Alternatively, along the equilibrium curve, the changes in pressure times the volume change are equal to
changes in temperature times the entropy change. This is the Clapeyron Equation.
So, a phase diagram is a kind of free energy map. = along an equilibrium, < at
high P and low T, and > at high T and low P. Along the equilibrium boundary the Gibbs
Free energies of the reactants and products are equal and the Gibbs Free energy of reaction rG,
is zero.
Shortcutting H and S and Finding G Directly
Like other thermodynamic potentials, we can write the change in Gibbs free energy of reaction as
rG° = fG°reactants- fG°products
Instead of using fH° and fS°, it is often possible to obtain fG° values for most compounds from
electronic data bases. For example, if the following Gibbs free energies of formation are known:
fG°CaSO4 = -1707.280 kJ/mol
2H2O
fG°CaSO4 = -1321.790 kJ/mol
fG°H2O = -237.129 kJ/mol
53. then for
CaSO4 + 2H2O = CaSO4 2H2O
rG° = -1.232 kJ/mol
Gibbs Free Energy at Any Pressure and Temperature
We know many ways to determine rG at STP--but how do we calculate rG for other pressures and
temperatures? Recall that the changes in Gibbs free energy with pressure and temperature are given by
two of Maxwell's relations
= rV and =- rS
If we recast these as
= rV P
and
=- rS T
and integrate, we get
rGdP = rGP - rGPref = rVdP
or
rGP = rGPref + rVdP
and
rGdT = rGT - rGTref =- rSdT
or
rGT = rGTref - rSdT
thus
rGPT = rGPrefTref + rVdP - rSdT
Solving the Pressure Integral at Constant Temperature
To a first approximation, we can ignore the expansivity and compressibility of solids and use
54. rVsdP = rVs(P - 1)
as a simplification. Don't forget that this approximation is valid for solids only! An even more common
assumption for P>>1 is
rVsdP = rVsP
For example, calculate the change in Gibbs free energy for the reaction
2 jadeite = albite + nepheline
if pressure increases from 1 bar to 10 kbar, given
nepheline = 54.16 cm3
3
albite = 100.43 cm
3
jadeite = 60.40 cm
First we calculate rV and find
r = nepheline + albite -2 jadeite = 33.79 cm3 = 3.379 J/bar
and thus
rGPT - rG1,T = rVsP = 33.79 kJ/mol
Solving the Temperature Integral at Constant Pressure
Recall that the effect of temperature on the entropy change of reaction rS depends on the heat
capacity change of reaction rCP:
rS = dT
Thus
rGT = rGTref - rSdT
expands to
rGT = rGTref - STref + dT dT
If the form of the heat capacity expansion is
55. CP = a + bT + cT-2 + dT-0.5
then the above double integral is
a(T - T ln T) - bT2/2 - cT-1/2 + 4 dT0.5 - aTref - bTref2/2 + cTref-1 - 2 dTref0.5 +
aTlnTref + bTTref - cTTref-2/2 - 2 dTTref-0.5 - T rSTref + Tref rSTref
Note that this considers only vibrational entropy and ignores configurational entropy. This means of
solving for rG requires that you know rG at the reference temperature.
An alternative path that requires that you know the enthalpy change rH at the reference temperature
is
rGT = rHTref + CPdT - T rSTref + dT
Solving the Temperature and Pressure Integrals for rGP,T
To calculate the Gibbs free energy change of a reaction at any pressure and temperature, we can use
either of the following equations, depending on whether we know rH or rG
rGP,T = rG1,Tref - rSTref + dT dT + rVsP
rGP,T = rH1,Tref + CPdT - T rSTref + dT + rVsP
If you don't have heat capacity data for the reaction of interest, these equations can be roughly
approximated as
rGP,T = rG1,Tref - rS1,Tref(T - Tref) + rVsP
rGP,T = rH1,Tref - T rS1,Tref + rVsP
For example, calculate rG for jadeite + quartz = albite at 800 K and 20 kbar. The data at 298 K and 1
bar are
rH° = 15.86 kJ/mol
-1 -1
rS° = 51.47 J K mol
3
rVs° = 1.7342 J/bar = 17.342 cm /mol
Using
rGP,T = rH1,Tref - T rS1,Tref + rVsP
= 15,860 - 800 * 51.47 + 1.7342 * 20,000 = 9.37 kJ/mol
If we had used the complete equation for solids, integrating the heat capacities, we would have
obtained an answer of 9.86 kJ/mol--not horrifically different.
56. Calculating the PT Position of a Reaction
If we say that rG = 0 at equilibrium, then we can write our solids-only and constant-heat-capacity
approximations as
0= rG1,T - rS1,Tref(T - Tref) + rVsP
0= rH1,Tref -T rS1,Tref + rVsP
and thus we can calculate the pressure of a reaction at different temperatures by
P= rG1,Tref - rS1,Tref(T - Tref) /- rVs
P= rH1,Tref -T rS1,Tref /- rVs
and we can calculate the temperature of a reaction at different pressures by
T = Tref + rG1,Tref + rVsP / rS1,Tref
T = Tref + rH1,Tref + rVsP / rS1,Tref
Let's do this for the albite = jadeite + quartz reaction at T = 400 K and T = 1000 K:
P = (15,860 - 5147 * 400) / -1.7342 = 2.7 kbar
P = (15,860 - 5147 * 1000) / -1.7342 = 20.6 kbar
Assuming that dP/dT is constant (a bad assumption, we know), the reaction looks like this
Introduction to the Equilibrium Constant
A bit farther down the road we will encounter a monster called the equilibrium constant K:
K = exp(- rG°/RT)
57. or
ln K = - rG°/RT
At equilibrium, where rG°= 0, ln K = 0 and K = 1. Let's see what K looks like for jadeite + quartz = albite
at 800 K and 20 kbar:
ln K = - ( rH1,Tref - T rSTref + rVsP)/ RT
= -(15,860 - 800 * 51.47 + 1.7342 * 20,000)/(8.314*800) = -1.4
If we do this for all of PT space, we can contour PT space in terms of lnK:
Solutions
58. Almost no phases are pure, but typically are mixtures of components. For example, olivine varies from
pure forsterite Mg2SiO4 to pure fayalite Fe2SiO4, and can have any composition in between--it is a solid
solution. We need a way to calculate the thermodynamic properties of such solutions.
As a measure of convenience, we use mole fraction to describe the compositions of phases that are solid
solutions. For example, a mix of 1 part forsterite and 3 parts fayalite yields an olivine with 25 mol%
forsterite and 75 mol% fayalite, which can be written as (Mg0.25Fe0.75)2SiO4 or fo25fa75, etc. Mole fractions
are denoted as Xi.
We need a way of splitting up the Gibbs free energy of a phase among the various components of the
phase--how for example do we decide how much of the Gibbs free energy of an olivine is related to the
forsterite component and how much derives from the fayalite component? Likewise, how does the
Gibbs free energy of a phase vary with composition--is the relationship linear between endmembers??
We address these issues by defining a partial Gibbs free energy for each component at constant
pressure and temperature and constant composition of other components, called the partial molar
Gibbs free energy or chemical potential
i =
where n is the amount of substance. For olivine solid solution composed of fayalite and forsterite
components or endmembers, we write
dG = dnfayalite + dnforsterite
Volume of Mixing
Imagine that mole fractions of phase A and phase B with molar volumes VA and VB, are mixed together.
We can describe the volume of the mixture as
V = XAVA° + XBVB°
and it is a linear mixing of the two endmember volumes. We call this ideal mixing or mechanical mixing.
Real solutions, however, do not behave this way, and the mixing is always non ideal, although
sometimes only weakly so. The figure shows mixing that produces a smaller volume than expected, but
it is not possible to predict the shapes and positions of such mixing curves.
59. Partial Molar Volume
The partial molar volume is defined as
i
If you mix two compounds A and B together and find a volume of mixing that is non-ideal, how can you
determine the contribution that A and B each make to the volume? That is, what are the partial molar
volumes of A and B, A and B?? Graphically, the partial molar volumes are the A and B axis intercepts
of the tangent to the mixing curve, and can be described by the simple relationship:
Vmix = XA A + XB B
or
Vmix = Xi i
The behavior of this function is such that when XA is 1, Vmix = VA and when XA is 0, Vmix = VB. Alternatively,
60. A = (Vmix - XB B) / XA
Entropy of Mixing
The entropy of mixing is never zero because mixing increases entropy. As we discussed days ago, the
entropy of mixing (i.e., the configurational entropy) is
Smix = -R (Xi ln Xi)
where i = 1..n is the number of sites over which mixing is occurring.
Enthalpy of Mixing
Enthalpies also do not combined ideally (linearly) in mixtures because the mixture may have stronger
bonds than were present in either of the unmixed phases. The excess enthalpy is
Hmix = 0.5 * N z XAXB [2 AB - AA - BB]
where AB is the interaction energy among A-B atoms, AA is the interaction energy among A-A atoms,
and BB is the interaction energy among B-B atoms.
Gibbs Free Energy of Mixing
Recall that all spontaneous processes/reactions occur because of a decrease in Gibbs free energy. It
should therefore not surprise you that the Gibbs free energy of mixing is always negative--otherwise
mixing would not occur. The fact that A < G°A and B < G°B illustrates why compounds combine
spontaneously--each compound is able to lower its free energy.
61. The above figure is hypothetical because we cannot measure or calculate the absolute Gibbs free
energy of phases. For this reason, is always expressed as a difference from some standard state
measurement, as , - °, or - G°.
The difference between the absolute Gibbs free energy G° per mole ° of a pure compound and
the chemical potential per mole of dissolved compound is
A - G°A = A - °A = RT lnXA
This function has the following shape:
62. implying that when the mineral is pure (X = 1) then = 0, and when the mineral is infinitely
dilute (X = 0) then the chemical potential is undefined. For example, in a two-component
mineral if XA = 0.4, at T = 298 K,
A- °A = 8.314 * 298 ln 0.4 = -2271 J
B- °B = 8.314 * 298 ln 0.6 = -1266 J
The equation of the Gmix line is the sum of the chemical potentials of the endmembers:
Gmix = RT (XA ln XA + XB ln XB) or
Gmix = RT (Xi ln Xi)
which looks like this for two components:
Actually, all this discussion has been predicated on the assumption that Hmix = 0. If this is not
true, Gmix is not a simple function of composition, but has the general form:
63. Depending on the relative values of Hmix and -T Smix, the free energy of mixing may be
negative throughout the whole composition range if the entropic energy contribution outweighs
the enthalpy increase; this is more likely at higher temperature.
The two free energy minima in the above figure indicate that minerals of intermediate
compositions can reduce their free energy by unmixing into two phases. This explains the
appearance and driving force for exsolution. Note that this can only be true if Hmix > 0, i.e., if
2 AB > AA + BB, which makes sense because it means that the A-B bonds have a higher free
energy than the sum of the free energies of separate AA and BB bonds.
Activities
In reality, no phases behave ideally--that is, their chemical potentials are never simple logarithmic
functions of composition as
A - °A = RT lnXA
implies. Instead, we say that the chemical potential is a simple logarithmic function of activity and
define activity as
a = ( X)
where a is the activity of a compound, is the "site occupancy coefficient" (e.g., = 2 for Mg in
Mg2SiO4), and is the activity coefficient that describes the non-ideal behavior. Thus we write
A - °A = RT lnaA
For pure compounds a=1 because X=1. For ideal compounds =1. As a specific example, the chemical
potential of the almandine (Fe3Al2Si3O12) component of a garnet solid solution ((Fe, Mg, Ca,
Mn)3Al2Si3O12) is
alm = °alm + RT lnXalm
64. To be clear, ° is the chemical potential of the component in its pure reference state and varies as a
function of pressure and temperature; this we measure with calorimetry. is the chemical potential as
it actually occurs and varies as a function of phase composition; this we measure with an electron
microprobe. The activity forms a bridge between idealized behavior and real behavior.
The Equilibrium Constant
At equilibrium the sum of the Gibbs free energies of the reactants equals the sum of the Gibbs free
energies of the products. Equally, the sum of the partial molar Gibbs free energies (chemical potentials)
of the reactants equals the sum of the partial molar Gibbs free energies (chemical potentials) of the
products. In other words, for
SiO2 + 2H2O = H4SiO4
at equilibrium,
H4SiO4 = SiO2 + 2 H2O
More generally, for
aA + bB = cC + dD
then
c C+d D=a A+b B
or, at equilibrium
r =0=c C+d D-a A-b B
which we can reformat as
r = i i
where i is the stochiometric coefficient of a product or reactant and is positive if for a product and
negative if for a reactant. If we then remember that
- ° = RT lna
and rewrite it as
= ° + RT lna
we can reformat the earlier equation as
65. r = 0 = c( C° + RT lnaC) + d( D° + RT lnaD) - a( A° + RT lnaA) - b( B° + RT lnaB)
which looks nicer as
r = r ° + RT ln (aCc aDd/aAa aBb)
To be completely general we write
r = r ° + RT ln ai ( means to multiply all i terms)
This equation is invariably simplified to
r = r ° + RT lnQ
and Q is the activity product ratio. The activities in the Q term change as the reaction progresses toward
equilibrium.
To be clear again, r ° is the difference in the Gibbs free energies of the products and reactants when
each is in its pure reference state and varies as a function of pressure and temperature. r is the
difference in the Gibbs free energies of the products and reactants as they actually occur and varies as a
function of phase composition.
At equilibrium, the product and reactant activities have adjusted themselves such that r = 0. We
write this (with K instead of Q, to signify equilibrium) as
0= rG° = -RT ln K
K is called the equilibrium constant. If K is very large (ln K positive), the combined activities of the
reaction products are enormous relative to the combined activities of the reactants and the reaction will
likely progress. On the other hand, if K is small (ln K negative), there is unlikely to be any reaction.
The utility of K is that it tells us for any reaction and any pressure and temperature, what the activity
ratios of the phases will be at equilibrium. For example, for the reaction
albite = jadeite + quartz
let's say that at a particular P and T,
rG° = -20.12 kJ/mol
Using
rG° = -RT lnK
we calculate
66. log K = 3.52
This means that at equilibrium,
(ajadeite aquartz / aNaAlSi3O8) = e3.52
Where ajadeite is the activity of NaAlSi2O6 in clinopyroxene and aalbite is the activity of NaAlSi3O8 in
plagioclase.
Alternative Route to the Equilibrium Constant
When we think of mass balance in a reaction, we can explicitly write
0= iMi
where i are the stoichiometric coefficients and Mi are the masses or the phase components.
Analogously, we can explicitly write a similar balance among the chemical potentials:
0= i i
For each chemical potential we can write
i= °i + RT lnai
Combining these two equations we find
0= i °i + iRTln ai
0= i °i + RTln (ai)
0= i °i + RTln ai
0= i °i + RTln K
and eventually
0= rG° + RT ln K
or
rG° = -RT ln K
The equation
K= ai
67. is the law of mass action (which actually discusses the action of chemical potential rather than mass).
We can also write for 298 K and 1 atm
rH° -T rS° = - RT ln K
and for any P and T of interest:
rH1,Tref + CPdT - T rSTref + dT + rVP = - RT ln K
This has been called "the most important equation in thermodynamics," so you'd better like it(!) The
equilibrium constant K is a function of 1/T
-ln K = ( rG° / RT) = [( rH / RT) - ( rS / R)]
Which looks like
Activity Models (Activity-Composition Relations) for Crystalline Solutions
Garnets are solid solutions of
component abbrev. Formula
pyrope prp Mg3Al2Si3O12
almandine alm Fe3Al2Si3O12
grossular grs Ca3Al2Si3O12
spessartine sps Mn3Al2Si3O12
andradite and Ca3Fe23+Si3O12
Mixing models derive from entropy considerations. In particular the relation
68. Smix = -R Xi ln Xi
although we will not go through the derivation.
Mixing on a Single Site
The simplest type of useful activity model is the ionic model, wherein we assume that mixing occurs on
crystallographic sites. For a Mg-Fe-Ca-Mn garnet with mixing on one site, which we can idealize as
(A,B,C,D) Al2Si3O12, the activities are
3 3
aprp = Mg XMg
3 3
aalm = Fe XFe
3 3
agrs = Ca XCa
3 3
asps = Mn XMn
The pyrope activity is shown in the above figure.
In general, for ideal mixing in a mineral with a single crystallographic site that can contain
ions,
ai = Xj
where a, the activity of component i, is the mole fraction of element j raised to the power. For non-
ideal mixing, we include an activity coefficient
ai = j Xj
Mixing on a Several Sites
For minerals with two distinct sites and the general formula
(A,B) (Y,Z)
69. there are four possible end member components A Y , A Z , B Y , and B Z .
The ideal activities of these components are
aA Y = XA XY
aA Z = XA XZ
aB Y = XB XY
aB Z = XB XZ
For non-ideal garnet activities we write
aprp = XMg3 XAl2 or Mg3 XMg3 Al2 XAl2
aalm = XFe3 XAl2 or Fe3 XFe3 Al2 XAl2
agrs = XCa3 XAl2 or Ca3 XCa3 Al2 XAl2
asps = XMn3 XAl2 or Mn3 XMn3 Al2 XAl2
aand = XCa3 XFe3+2 or Ca3 XMn3 Fe3+2 XFe3+2
where the X3 term describes mixing on the 8-fold trivalent site and the X2 term describes mixing on the
octahedral divalent site.
The pyrope activity is shown in the above figure for Mg from 0 3 and Al from 0 2.
It is common to modify such models that are based on completely random mixing of elements
with models that consider local charge balance on certain sites or the Al-avoidance principle.
Geothermometry and Geobarometry
Exchange Reactions
Many thermometers are based on exchange reactions, which are reactions that exchange elements but
preserve reactant and product phases. For example:
70. Fe3Al2Si3O12 + KMg3AlSi3O10(OH)2 = Mg3Al2Si3O12 + KFe3AlSi3O10(OH)2
almandine + phlogopite = pyrope + annite
We can reduce this reaction to a simple exchange vector:
(FeMg)gar+1 = (FeMg)bio-1
Popular thermometers include garnet-biotite (GARB), garnet-clinopyroxene, garnet-hornblende, and
clinopyroxene-orthopyroxene; all of these are based on the exchange of Fe and Mg, and are excellent
thermometers because rV is small, such that
=
is large (i.e., the reactions have steep slopes and are little influenced by pressure). Let's write the
equilibrium constant for the GARB exchange reaction
K = (aprpaann)/(aalmaphl)
thus
rG = -RT ln (aprpaann)/(aalmaphl)
This equation implies that the activities of the Fe and Mg components of biotite and garnet are a
function of Gibbs free energy change and thus are functions of pressure and temperature.
If we assume ideal behavior ( = 1) in garnet and biotite and assume that there is mixing on only 1 site
aalm = Xalm3 = [Fe/(Fe + Mg + Ca + Mn)]3
aprp = Xprp3 = [Mg/(Fe + Mg)]3
aann = Xann3 = [Fe/(Fe + Mg)]3
aphl = Xphl3 = [Mg/(Fe + Mg)]3
Thus the equilibrium constant is
K = (XMggar XFebio)/(XFegar XMgbio)
When discussing element partitioning it is common to define a distribution coefficient KD, which is just
the equilibrium constant without the exponent (this just describes the partitioning of elements and not
the partitioning of chemical potential):
KD = (XMggar XFebio)/(XFegar XMgbio) = (Mg/Fe)gar /(Mg/Fe)bio = K1/3
71. Long before most of you were playground bullies (1978) a couple of deities named John Ferry and Frank
Spear measured experimentally the distribution of Fe and Mg between biotite and garnet at 2 kbar and
found the following relationship:
If you compare their empirical equation
ln KD = -2109 / T + 0.782
this immediately reminds you of
ln K = - ( rG° / RT) = -( rH / RT) - (P rV / RT) + ( rS / R)
and you realize that for this reaction
rS = 3*0.782*R = 19.51 J/mol K
(the three comes from the site occupancy coefficient; i.e., K = KD3) and
-( rH / R) - (P rV / R) = -2109
or
rH = 3*2109*R -2070* rV
Molar volume measurements show that for this exchange reaction rV = 0.238 J/bar, thus
rH = 52.11 kJ/mol
The full equation is then
72. 52,110 - 19.51*T(K) + 0.238*P(bar) + 3RT ln KD = 0
To plot the KD lines in PT space
Net-Transfer Reactions
Net-transfer reactions are those that cause phases to appear or disappear. Geobarometers are often
based on net-transfer reactions because rV is large and relatively insensitive to temperature. A
popular one is GASP:
3CaAl2Si2O8 = Ca3Al2Si3O12 + 2Al2SiO5 + SiO2
anorthite = grossular + kyanite + quartz
which describes the high-pressure breakdown of anorthite.
For this reaction
rG = -RT ln [(aqtzaky2agrs) / aan3] = -RT ln agrs / aan3
(the activities of quartz and kyanite are one because they are pure phases). A best fit through the
experimental data for this reaction by Andrea Koziol and Bob Newton yields
P(bar) = 22.80 T(K) - 7317
for rV = -6.608 J/bar. Again, if we use
73. ln K = -( rH / RT) - (P rV / RT) + ( rS / R)
and set ln K = 0 to calculate values at equilibrium, we can rewrite the above as
(P rV / R) = -( rH / R) + (T rS / R)
or
P=T rS / rV - rH / rV
if T rS / rV = 22.8 then rS = -150.66 J/mol K
if rH / rV = 7317 then rH = -48.357 kJ/mol
So, we can write the whole shmear as
0 = -48,357 + 150.66 T(K) -6.608 P (bar) + RT ln K
Contours of ln K on a PT diagram for GASP look like this:
Kinetics
Thermodynamics places no constraints on the rate or mechanism of reaction--that is the realm of
kinetics. A popular method for describing the rate at which reactions proceed is to talk of an activated
state through which the reaction must pass:
74. When a system passes from an initial to a final state it must overcome an activation energy
barrier G*.
The advantages of this activation energy barrier paradigm are that it qualitatively explains the
i) persistence of metastable states; ii) effect of catalysts in lowering G*; iii) temperature
dependence of transformation. We can draw a similar diagram for a change in enthalpy induced
by the reaction rH and an activation enthalpy barrier H* (usually called an activation
energy Q*). Of course, unlike rG, rH can be positive or negative:
It is not easy to generalize about the activation entropy S*, however, in general, reactions with
positive entropy change rS are faster. For example, evaporation is faster than condensation,
melting is faster than crystallization, and disordering is faster than ordering.
Because thermal energy dictates whether an atom has sufficient energy to overcome an
activation energy barrier, we will write that the fraction of atoms with thermal energy greater
than H* is
f = exp ( - H*/RT)
