Suggestions and Corrections to
Gale Rhodes
Chemistry Department
University of Southern Maine
Revised 2006/07/14
This handout shows how to calculate the free energy available from
A) redox reactions; B) concentration gradients; C) voltage gradients;
and D) proton or other ion gradients, which have two components,
concentration and voltage. I have taken pains in all sections to
employ sign conventions consistently, so take note of how each
process is defined, and how the definition determines the sign of the
free-energy change.
Consider the oxidation of ubiquinone by cytochrome c. How much free
energy is available from this process?
Half reactions |
E^{0}'(V) |
a) UQ (ox) + 2e^{-} + 2H^{+} --> UQH2 (red) |
0.04 |
b) Cytochrome c-Fe^{3+} (ox) + e^{-} --> Cytochrome c-Fe^{2+} (red) |
0.23 |
rev-a): |
UQH_{2} ---> UQ + 2 e^{-} + 2 H^{+} |
2 x b): |
2 Cytochrome c-Fe^{3+} + 2 e^{-} ---> 2 Cytochrome c-Fe^{2+} |
Sum: |
UQH_{2} + 2 Cytochrome c-Fe^{3+} ---> UQ + 2 H^{+} + 2 Cytochrome c-Fe^{2+} |
NOTE: In calculations using this equation, and whenever you have concentration terms inside a logarithm term, remember that the proper terms are really activities and not concentrations. The activity of the solute is its actual concentration divided by its standard concentration, so activities are unitless (which is good, because it's very hard to attach a physical meaning to units like ln[mol/L] ). Click here to learn how to compute activies for solutes, gases, hydrogen ions, and water or other solvents.
Consider the movement of protons from the cytoplasm into the matrix
of the mitochondrion:
H^{+}_{out} <==> H^{+}_{in}
How much free energy is available from the movement of protons down
the concentration gradient created by electron transport?
DG = DG
^{0}' + RT lnQ = DG ^{0}'
+ RT
ln([H^{+}_{in}]/[H^{+}_{out}])
DG ^{0}' = 0 because
K_{eq} for the process is 1.0 (DG
^{0}' = - RT ln K_{eq}),
so DG = RT
ln([H^{+}_{in}]/[H^{+}_{out}])
To express DG in terms of the pH
gradient (rather than the concentration gradient), change ln to log
(that is, log_{10}) and expand the log term:
DG = 2.303 RT
log([H^{+}_{in}]/[H^{+}_{out}])
DG = 2.303 RT
(log[H^{+}_{in}] -
log[H^{+}_{out}]) = -2.303 RT
(pH_{in} - pH_{out})
or
DG = - 2.303 RT
DpH ===> (NOTE:
DpH = pH_{in} -
pH_{out} )
If proton pumping maintains a pH gradient of 1.4 units (lower
outside ), then
DpH = + 1.4 and
DG = - 2.303 (8.315 x 10^{-3}
kJ/mol-K)(298K)(1.4) = - 7.99 kJ/mol
This is the free-energy change attributable to the
concentration gradient. This would be the free energy
available from a gradient of a nonionic solute, or from an ionic
gradient if the movement of other ions maintained equal voltage on
both sides of the membranes, as is true in chloroplasts. The movement
of other ions across the thylakoid membrane maintains electrical
neutrality across the membrane, despite light-driven proton pumping
into the thylakoid lumen. In particular, as H^{+} moves from
the stroma into the lumen, Mg^{2+} moves out of the lumen
into the stroma. So in chloroplasts, the proton gradient is simply a
concentration gradient, and there is no accompanying voltage
gradient.
In mitochondria, electron transport drives proton pumping from the matrix into the intermembrane space. There is no compensating movement of other charged ions, so pumping creates both a concentration gradient and a voltage gradient, the latter resulting from the excess of proton charges outside the inner mitochondrial membrane. This voltage component makes the proton gradient an even more powerful energy source. Here's how to calculate the contribution of voltage to the energy available from such a gradient.
Define the membrane voltage gradient, or membrane potential, as
Dy_{m} = y_{in}
- y_{out}.
How much free energy is available from the movement of protons down
the voltage gradient created by electron transport? NOTE:
Same process, as before: H^{+}_{out} <==>
H^{+}_{in}.
D G = -nF Dy^{0}'
+ nFDy_{m} ,
and Dy^{0}' = 0 (membrane
potential = 0 under standard conditions.)
so
DG = nF Dy_{m}
(NOTE: Dy_{m}
= y_{in} - y_{out}
.)
If proton pumping maintains a voltage gradient of 0.14 V, positive
outside, then Dy_{m}
is negative, as defined here.
Dy_{m} =
y_{in} - y_{out}
= - 0.14 V.
DG = (1)(96.48 kJ/V-mol)(- 0.14 V) = -
13.5 kJ/mol
This is the free-energy change attributable to the voltage
gradient.
"Proton-motive force" (Dp) is a
Dy- or DE-like
term (DE is electromotive force )
that combines the concentration and voltage effects of a proton
gradient such that
DG = - nFDp^{0}'
+ nFDp, and Dp^{0}'
= 0 (proton motive force = 0 under std conds),
so
DG can also be expressed as the sum of the
DpH and Dy_{m}
contributions:
DG = - 2.303 RT DpH
+ nFDy_{m}
so nFDp = - 2.303 RT DpH
+ nFDy_{m}
or
This way of expressing the proton-motive force was probably adopted
because of its elegant resemblance to the Nernst equation, but the
clearest expression of the energy available from a proton gradient is
probably
Remember (this is a recording) that DpH =
pH_{in} - pH_{out }and Dy_{m}
= y_{in} - y_{out}
because we started out by considering the movement of protons from
the cytoplasm to the matrix:
H^{+}_{out} <==> H^{+}_{in}
The total free energy available from the movement of 1 mole of
protons from the cytoplasm to the matrix under cellular conditions
(DpH = 1.4, Dy
= 0.14 V) is the sum of the free energy changes calculated in
sections B and C:
DG = - 2.303 RT DpH
+ nFDy = -7.99 kJ/mol - 13.5 kJ/mol
DG = -21.5 kJ/mol
Estimated consumption of the proton gradient by ATP synthesis is
about 3 moles protons per mole ATP. If DG
= 50 kJ/mol for ATP synthesis in mitochondria, then by Hess's law,
DG = 50 + 3(- 21.5) = - 3.4 kJ/mol, and
the process of synthesis of ATP at the expense of the proton gradient
is spontaneous under mitochondrial conditions.
Estimated proton pumping associated with electron transport is 10
protons per electron pair passed from NADH to O_{2}. In
addition to the 3 protons consumed per ATP synthesized, one proton is
spent in transporting ATP to the cytoplasm, for a total of 4
H^{+} per cytoplasmic ATP. So the yield of cytoplasmic ATP
per electron pair is
(1 ATP/4 H^{+})/(10 H^{+}/electron pair) = 2.5
ATP/electron pair.
Only 6 protons are pumped for each electron pair passed from
FADH_{2 }to O_{2}, so ATP yield is
(1 ATP/4 H^{+})/(6 H^{+}/electron pair) = 1.5
ATP/electron pair.
These are the ATP yields most commonly quoted by careful textbook authors.