Bioenergetics Course - 2004 - Lecture
3a, review of thermodynamics
Keith D. Garlid
April 16, 2004
ENERGY UNITS AND
CONVERSION FACTORS
Here are some common constants used in
bioenergetic calculations:
F = 96,487 coul / equiv (The Faraday)
R = 8.314 joules / deg (The Gas Constant)
RT/F = 25.688 at 25 oC
Following is a brief review of the
origin of these units, for your information.
newton (N)
The newton is the Standard International (SI) unit of force. One
newton is the force required to cause a mass of one kilogram to
accelerate at a rate of one meter per second squared in the absence of
other force-producing effects. Force (F) in newtons, mass (m) in
kilograms, and acceleration (a) in meters per second squared are
related by a formula well known in physics:
F = ma
joule (J)
When objects interact, i.e. exert forces on each other, then the work
exerted by one object on another is defined as
Work = Force x Distance
One joule is defined as the amount of energy exerted when a force of
one newton is applied over a displacement of one meter. (One
joule is the equivalent of one watt of power radiated or dissipated for
one second.)
Energy
The work done by an object is exactly equal to the loss in energy that
the object experiences while doing that work. It is also equal to
the energy that the object being acted on gains. This discovery (the
conservation of energy) is called the "work-energy theorem" in
physics. Note that this means that Energy must have the same
units as Work, i.e., the joule.
Calories
The calorie is defined as the amount of heat needed to raise the
temperature of 1 gram of water by 1C . But this makes the calorie
depend on the initial temperature of the water. Recall that heat
is a form of energy, so it can also be quantified by the joule.
The joule is superior because it is independent of
temperature.
We may take 1 kcal (1000 cal) = 4.184 kJ
Voltage (V)
Voltage, also called electromotive force, is a quantitative expression
of the potential difference in charge between two points in an
electrical field. The greater the voltage, the greater the flow of
electrical current (that is, the quantity of charge carriers that pass
a fixed point per unit of time) through a conducting or semiconducting
medium for a given resistance to the flow. The standard unit is the
volt. One volt will drive one coulomb (6.24 x 1018) charge
carriers, such as electrons or ions, through a resistance of one ohm in
one second.
A charge (electron or ion) moving through a voltage
is a form of work. Moreover, a charge having the potential to
move through a voltage is a form of energy. The general
conversion between electrical force and energy is:
ΔG = znFΔV
where ΔG is in J/mol, z is the valence, n is the number of charges
moved, and F is the Faraday:
1 Faraday (F) = 96485.3415 Coulombs
Examples
(a) What is the free energy of an electron falling through 1 volt?
For an electron, n = 1 and z = - 1, ΔV = 1 volt, so ΔG = 96.485
kJ/mol
(b) Protonmotive force, Δp ~ 200 mV
Definition: Δp = - ΔμH+/F (n
= 1, the charge on H+)
Therevore, ΔμH+ = - 96.485 x 0.2 V
= - 19.3 kJ/mol
(b) Redox energy, ΔE, for NADH to O (2 electrons, z = -1)
ΔE = -2 x 96.485 x 1.16 = -216 kJ/mol
GIBBS FREE ENERGY,
ΔG, AND CHEMICAL POTENTIAL, Δμ
1. Gibbs Free
Energy, ΔG
Josiah Willard Gibbs introduced the State Function ΔG, which is DEFINED
as follows:
ΔG = ΔH - TΔS
2. General
properties of ΔG:
1) A system is at equilibrium if ΔG = 0 (very
important)
2) A reaction can occur spontaneously ONLY if
ΔG is negative.
3) A reaction CANNOT occur spontaneously if ΔG
is positive.
4) ΔG is a STATE function. Therefore it
depends only on the free energy of the products minus the free energy
of the reactants.
5) ΔG is INDEPENDENT OF THE PATH of the
transformation. (The same whether or not catalyzed by an enzyme)
6) ΔG provides no information about the RATE
of the reaction
3. ΔG for a
chemical reaction
Consider the chemical reaction
A + B ⇄ C + D
ΔG = ΔG o + RT ln [C][D]
[A][B]
ΔG o is the standard free energy change
R is the gas constant = R = 8.314 joules / deg
T is absolute temperature = 298.14 at 25 oC
[A], [B], [C], [D ] are molar concentrations of reactants and products.
4. The chemical
potentials - μ
G is a function of T, P, and the number of mols of each component:
G = G(T, P, N1,....Nn)
We will want to know the free energy associated with a given component
of the system. This is called the chemical potential or partial
molal free energy and is defined thus:
μj ≡ (∂G/∂Nj )T,P,Ni
(This partial derivative means: the derivative of G with respect
to quantity, N, of a particular component, holding all other properties
constant).
The chemical potential takes the same form as the free energy (because
it is a free energy).
μj = μj o + RT ln cj
where cj is the concentration of species “j”, and μj o is the
standard chemical potential, usually defined as 1 molal.
We are normally interested in differences in μ across the inner
membrane. Because Δμo relates to standard conditions in two
(similar) aqeuous phases, we take Δμo = 0. Therefore,
Δμ = RT ln c(in)/c(out)
This is the partial molal free energy of moving a solute
(nonelectrolyte) from outside to inside. Note that if c(out) >
c(in), Δμ is negative, and the reaction will occur spontaneously.
5.
Electrochemical potentials, Δμ
The electrochemical potential difference for an ion takes into account
the electrical energy as well as the chemical energy. For any ion
of valence, z,
Δμ = RT ln c(in)/c(out) + zF ΔΨ
where ΔΨ refers to the electrical potential difference, inside minus
outside.
6. Protonmotive
force, Δp (mV)
Δp is DEFINED as,
Δp = - ΔμH+ / F
Therefore,
Δp = Z ΔpH - ΔΨ
where Z ≡ RT log(e) / F ~ 59 mV at 25 oC
and ΔpH = pHin - pHout
Typical measurements in mitochondria respiring on succinate (but not
phosphorylating) are: ΔpH ~ 0.3 (alkaline inside), and ΔΨ~ -190 mV
(negative inside). Therefore,
Δp ~ 208 mV.
7. Chemical
reactions at equilibrium - the equilibrium constant, Keq
ΔG = 0 at equilibrium. From section 3,
go to ATP
ΔGo = - RT ln [C’][D’]
[A’][B’]
where primes indicate that all concentrations refer to those achieved
at equilibrium. We define the equilibrium constant as
Keq = [C’][D’],
[A’][B’]
so ΔGo = - RT ln Keq
Keq can, in principle, be measured, allowing calculation of ΔGo and
vice versa. For example, take ΔGo for ATP synthesis to be + 30.6
kJ/mol.
Then ln Keq = -30.6/(0.008314 x 298.16) = -12.3442, and Keq ~
4.36 x 10-6
Nicholls and Ferguson introduce a parameter Γ, to describe the actual
concentrations in the free energy equation. This leads to a
simple form:
ΔG = RT ln (Γ / Keq)
8. Partitioning
of a solute between two phases - the partition coefficient, Kp
It is frequently of interest in biology to know how molecules partition
between the aqueous phase and the hydrophobic interior of
membranes. Here, we use the equation given in section 4, but we
must include a standard chemical potential change Δμo, because the two
phases are different. Therefore,
Δμ = Δμo + RT ln c1/c2
where 1 and 2 refer to the lipid and aqueous phases, respectively.
We allow the solute to come to equilibrium between the two phases,
therefore, Δμ = 0, and
Δμo = - RT ln c1/c2 = - RT ln Kp
where Kp = c1/c2
Δμo is referred to as the free energy of transfer of the solute from
phase 2 to phase 1.
Note also that another way of writing the equation is,
c1/c2 = e-Δμo/RT
9. Ions at
equilibrium across a membrane - the Nernst potential
From section 5, we may write the equation for the electrochemical
potential of K+ across the inner membrane of mitochondria.
ΔμK+ = RT ln [K+]in /[K+]out + F ΔΨ
For mitochondria in vivo, K+(in) ~ K+(out) ~ 140 mM. Therefore,
ΔμK+ ~ F ΔΨ ~ -19.3 kJ/mol (for ΔΨ = 0.2
V). This is a very large driving force for inward K+
diffusion!
Experimentally, we may estimate ΔΨ from the K+ equilibrium
potential. It is necessary to lower [K+]out to about 0.14
mM. We then add valinomycin, a K+ ionophore that catalyzes
electrophoretic flux of K+ across the inner membrane, until K+ comes to
equilibrium. Note that mitochondria are respiring, so the system
is NOT at equilibrium. We say that the system is in a state of
partial equilibrium with respect to the process of electrophoretic K+
diffusion. When this is achieved, we know that ΔμK+ =
0.
Therefore, at K+ equilibrium,
Δψeq = - (RT/F) ln [K+]in /[K+]out
(NOTE THAT RT/F = 25.688 at 25 oC)
We frequently observe that K+ is taken up until [K+]out ~ 0.1 mM.
Then [K+]in /[K+]out = 1400 (assuming [K+]in = 140 mM), and
ΔΨeq = -193 mV.
Note that we can apply the preceding equations to mitochondria with
NORMAL K+ concentrations. We define Δψeq as the hypothetical
electrical potential that would obtain if the ACTUAL K+ concentrations
were in electochemical equilibrium. This leads to
ΔμK+ = F(Δψ - Δψeq).
Note that, for [K+]in = [K+]out , Δψeq = 0, and once again,
ΔμK+ = FΔψ for this case.
This kind of equation is used in describing the biophysics of plasma
membrane ion transport, but it is not commonly applied to mitochondria.