First Law of Thermodynamics
Stated simply; The total energy of the universe does not change. This does
not mean that the form of the energy cannot change. Indeed, chemical energies of
a molecule can be converted to thermal, electrical or mechanical energies.
The internal energy of a system can change only by work or heat exchanges.
From this the change in the free energy of a system can be shown by the
following equation:
DE = q - w Eqn.
1
When q is negative heat has flowed from the system and when q is positive
heat has been absorbed by the system. Conversely when w is negative work has
been done on the system by the surrounding and when positive, work has been done
by the system on the surroundings.
In a reaction carried out at constant volume no work will be done on or by
the system, only heat will be transferred from the system to the surroundings.
The end result is that:
DE = q Eqn.
2
When the same reaction is performed at constant pressure the reaction vessel
will do work on the surroundings. In this case:
DE = q - w Eqn.
3
where w = PDV Eqn.
4
When the initial and final temperatures are essentially equal (e.g. in the
case of biological systems):
DV = Dn[RT/P] Eqn. 5
therefore, w = DnRT Eqn.
6
by rearrangement of equation 3 and incorporation of the statements in
equations 4-6, one can calculate the amount of heat released under constant
pressure:
q = DE + w = DE +
PDV = DE + DnRT Eqn. 7
In equation 7 Dn is the change in moles of gas per
mole of substance oxidized (or reacted), R is the gas constant and T is absolute
temperature.
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Enthalpy
Since all biological reactions take place at constant pressure and
temperature the state function of reactions defined to account for the heat
evolved (or absorbed) by a system is enthalpy
given the symbol, H.
The changes in enthalpy are related to changes in free energy by the
following equation:
DH = DE + PDV Eqn. 8
Equation 8 is in this form because we are addressing the constant pressure
situation. In the biological setting most all reaction occur in a large excess
of fluid, therefore, essentially no gases are formed during the course of the
reaction. This means that the value DV, is extremely
small and thus the product PDV is very small as well.
The values DE and DH are very
nearly equivalent in biological reactions
Stated above was the fact that state functions, like DH and DE, do not depend on the path
taken during a reaction. These functions pertain only to the differences between
the initial and final states of a reaction. However, heat (q) and work (w) are
not state functions and their values are affects by the pathway taken.
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Reversible and Irreversible Reactions
In an idealized irreversible reaction such as one done by expanding an ideal
gas against zero pressure, no work will be done by or on the system so the:
w = 0 Eqn. 9
In the case of an ideal gas (whose molecules do not interact) there will be
no change in internal energy either so:
DE = 0 Eqn.
10
since DE = q - w, in this irreversible reaction q =
0 also.
In a reversible reaction involving an ideal gas, DE
still will equal zero, however, the pressure will be changing continuously and
work (w) is a funtion of P, work done must be determined over the entire course
of the reaction. This result in the following mathematical reduction:
w = RTln[V2/V1] Eqn.
11
Since in this situation DE = 0, q = w. This
demonstrates that some of the heat of the surroundings has to be absorbed by the
system in order to perform the work of changing the system volume.
Reversible reactions differ from irreversible in that the former always
proceeds infinitely slowly through a series of intermediate steps in which the
system is always in the equilibrium state. Whereas, in the irreversible reaction
no equilibrium states are encountered. Irreversible reactions are also
spontaneous or favorable processes. Thermodynamic calculations do not give
information as to the rates of reaction only whether they are favorable or not.
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Second Law of Thermodynamics
Entropy
The second law of thermodynamics states that the universe (i.e. all systems)
tend to the greatest degree of randomization. This concept is defined by the
term entropy, S.
S = klnW Eqn. 12
where k = Boltzmann constant (the gas constant, R, divided by Avagadros'
number) and W = the number of substrates. For an isothermal reversible reaction
the change in entropy can be reduced to the term:
DS = DH/T Eqn. 13
Whereas, enthalpy is a term whose value is largely dependent upon electronic
internal energies, entropy values are dependent upon translational, vibrational
and rotational internal energies. Entropy also differs from enthalpy in that the
values of enthalpy that indicate favored reactions are negative and the values
of entropy are positive. Together the terms enthalpy and entropy demonstrate
that a system tends toward the highest entropy and the lowest enthalpy.
In order to effectively evaluate the course (spontaneity or lack there of)
of a reaction and taking into account both the first and second laws of
thermodynamics, Josiah Gibbs defined the term, free energy. Free energy:
DG = DH - TDS Eqn. 14
Free energy is a valuable concept because it allows one to determine whether
a reaction will proceed and allows one to calculate the equilibrium constant of
the reaction which defines the extent to which a reaction can proceed. The
discussion above indicated that a decrease in energy, a negativeDH, and an increase in entropy, a positive DS, are indicative of favorable reactions. These terms would,
therefore, make DG a negative value. Reactions with
negative DG values are termed exergonic and those with
positive DG values endergonic. However, when a system
is at equilibrium:
DG = 0 Eqn.
15
Gibb's free energy calculations allows one to determine whether a given
reaction will be thermodynamically favorable. The sign of DG states that a reaction as written or its reverse process
is the favorable step. If DG is negative then the
forward reaction is favored and visa versa for DG
values that are calculated to be positive.
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Standard State Conditions in Biological Reactions
To effectively interpret the course of a reaction in the presence of a
mixture of components, such as in the cell, one needs to account for the free
energies of the contributing components. This is accomplished by calculating
total free energy which is comprised of the individual free energies. In order
to carry out these calculations one needs to have a reference state from which
to calculate free energies. This reference state, termed the Standard State, is chosen to be the condition where each
component in a reaction is at 1M. Standard state free energies are given the
symbol:
Go
The partial molar free energy of any component of the reaction is related to
the standard free energy by the following:
G = Go + RTln[X] Eqn.
16
From equation 16 one can see that when the component X, or any other
component, is at 1M the ln[1] term will become zero and:
G = Go Eqn. 17
The utility of free energy calculations can be demonstrated in a
consideration of the diffusion of a substance across a membrane. The calculation
needs to take into account the changes in the concentration of the substance on
either side of the membrane. This means that there will be a DG term for both chambers and, therefore, the total free
energy change is the sum of the DG values for each
chamber:
DG = DG1 +
DG2 =
RTln{[A]2/[A]1}Eqn.
18
Equation 18 tells one that if [A]2 is less than [A]1
the value of DG will be negative and transfer from
region 1 to 2 is favored. Conversely if [A]2 is greater than
[A]1 DG will be positive and transfer from
region 1 to 2 is not favorable, the reverse direction is.
One can expand upon this theme when dealing with chemical reactions. It is
apparent from the derivation of DG values for a given
reaction that one can utilize this value to determine the equilibrium constant,
Keq. As for the example above dealing with transport across a
membrane, calculation of the total free energy of a reaction includes the free
energies of the reactants and products:
DG = G(products) -
G(reactants) Eqn. 19
Since this calculation involves partial molar free energies the DGo terms of all the reactants and products are
included. The end result of the reduction of all the terms in the equation is:
DG =DGo +
RTln{[C][D]/[A][B]} Eqn. 20
When equation 20 is used for a reaction that is at equilibrium the
concentration values of A, B, C and D will all be equilibrium concentrations
and, therefore, will be equal to Keq. Also, when at equilibrium DG = 0. Therefore:
0 =DGo + RTlnKeq Eqn. 21
Keq = e-{DGo/RT}
Eqn. 22
This demonstrates the relationship between the free energy values and the
equilibrium constants for any reaction.
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Coupled Reactions
Two or more reactions in a cell sometimes can be coupled so that
thermodynamically unfavorable reactions and favorable reactions are combined to
drive the overall process in the favorable direction. In this circumstance the
overall free energy is the sum of individual free energies of each reaction.
This process of coupling reactions is carried out at all levels within cells.
The predominant form of coupling is the use of compounds with high energy to
drive unfavorable reactions.
The predominant form of high energy compounds in the cell are those which
contain phosphate. Hydrolysis of the phosphate group can yield free energies in
the range of -10 to -62 kJ/mol. These molecules contain energy in the phosphate
bonds due to:
- 1. Resonance stabilization of the phosphate products
- 2. Increased hydration of the products
- 3. Electrostatic repulsion of the products
- 4. Resonance stabilization of products
- 5. Proton release in buffered solutions
The latter phenomenon indicates that the pH of the solution a reaction is
performed in will influence the equilibrium of the reaction. To account for the
fact that all cellular reactions take place in an aqueous environment and that
the [H2O] and [H+] are essentially constant these terms in
the free energy calculation have been incorporated into a free energy term
identified as:
DGo' =DGo + RTln{[H+]/[H2O]}
Eqn. 23
Incorporation of equation 23 into a free energy calculation for any reaction
in the cell yields:
DG =DGo' +
RTln{[products]/[reactants]} Eqn.
24
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This article has been modified by Dr. M. Javed Abbas.
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20:36 21/12/2002
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