Using the expression derived in the last slide (ΔGorxn = –RT ln Keq), we can now work
out conditions involving ΔGorxn that will tell us which side of a reversible reaction is
favored. If the equilibrium constant for a reaction is greater than 1, then ΔGorxn would
be negative (< 0). This tells us that the final state with 100% B has less G compared to
the initial state with 100% A and the reaction will proceed spontaneously with a release
of energy. For a reaction with Keq < 1, ΔGorxn is positive (> 0). Therefore, the final state
with 100% B now has more G compared to the initial state with 100% A. In order to
convert a substantial amount of reactants A to products B, energy is required. The
reaction is disfavored relative to the products. Finally, when K eq = 1, ΔGorxn = 0, and
the amount of G in the initial and final states are equal. Neither the initial state and the
final state is favored, and the reaction proceeds to give equal amounts of products and
reactants.
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The cell is a complex system that must be able to make and break down molecules.
Thermodynamics tells us that when a reaction is favorable in one direction, it is
unfavorable in another direction. How then does Nature accomplish the feat of getting
reactions to go in either directions depending on the requirements of the cell?
Obviously, Nature must use energy to drive reactions in the disfavored direction. The
most common energy currency in a cell is called adenosine triphosphate (ATP). It is a
pretty special molecule and we will take a closer look later in the course. What you
need to know now is that Nature has devised ways to carry out disfavored chemical
reactions, typically by coupling the energy released in a very favorable reaction (such
as ATP hydrolysis as shown, where certain chemical bonds in ATP are broken, and the
ΔGorxn for the process is large and negative) to the disfavored reaction. In this way,
when we consider the coupled processes together, the overall ΔGorxn is negative and
thus favorable.
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This slide summarizes what we have just gone through about chemical equilibrium and
thermodynamics. When we start with a solution with 100% H2CO3, the concentrations
of H2CO3 and CO2 will change with time until the system reaches equilibrium, where
[H2CO3]/[CO2] = Keq[H2O], the Gibbs free energy (G) is at its minimum and where ΔG
= 0. When the system reaches equilibrium, it will stay there for an infinite amount of
time until it is perturbed by some external factor. This is exactly what happens in our
flask analogy. Given infinite time, every reaction in the flask will eventually reach
equilibrium (no matter how slow the process is).
Before we end our discussion of chemical equilibrium and thermodynamics in the next
slide, it is worthwhile to take note that for any reaction, two parameters do not change
in the process of approaching equilibrium (under constant temperature). These are K eq
and ΔGorxn. These parameters are directly related to the G in the state of 100%
reactants relative to the G in the state of 100% products and are only dependent on
the identity of the reaction, and temperature.
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One final point to appreciate: cells are always changing. They grow, they divide, they
respond to environmental stresses in order to survive. You can appreciate that if they
are in a constant state of flux, they can’t be at equilibrium. At equilibrium, there is no
net change in concentrations. A cell at equilibrium is a cell in which nothing is changing
and that is a dead cell. Most chemical processes in a cell are away from equilibrium
where ΔG is not equal to zero, and there is thus always a driving force for change
(towards equilibrium). In the next lecture, we shall try to understand how the cells
prevent themselves from reaching equilibrium.
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1. Understanding thermodynamics and chemical equilibrium
a. First law of thermodynamics
b. The concept of chemical equilibrium and Keq
c. Gibbs free energy, ΔG and ΔG°rxn
d. Relationship between Keq and ΔG°rxn
e. Coupled reactions
f. State of the cell
2. Influence the position of an equilibrium
a. Acidity, Ka and pKa
b. The Henderson-Hasselbalch equation
c. Le Chatelier’s Principle
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Chemical equilibrium is a very important concept that you will see throughout the
course. It can be used to describe any process that is reversible. We have used the
example of hydration of carbon dioxide to illustrate how chemical equilibrium can be
employed to describe chemical reactions. Chemical equilibrium can also be used to
describe processes that involve state changes, such as the dissolution of carbon
dioxide gas into water. Note that this process is the first step that must occur in order
for plants to utilize carbon dioxide during photosynthesis. To be able to use carbon
dioxide efficiently, plants must have a way to ‘trap’ high levels of carbon dioxide from
the atmosphere.
In the equilibrium between gaseous carbon dioxide and aqueous carbon dioxide, K eq =
[CO2(aq.)]/[CO2(g)] = 0.8317. Based on our understanding of chemical equilibrium (Keq
< 1), we know for a fact that the equilibrium favors carbon dioxide in the gaseous state,
if only by a little. How then do plants ‘trap’ high levels of carbon dioxide? Plants need
a stable pool of a CO2 equivalent to use to make sugars.
We just learned that carbon dioxide can react with water to form carbonic acid.
Perhaps this is a way to retain carbon dioxide in water (and hence in plants).
However, we also learned that the Keq for this process is 3.09 x 10-5 M-1, a value that is
far less than 1. Therefore, aqueous carbon dioxide (CO2(aq.)) is favored over carbonic
acid by a significant amount. If the only equilibria that mattered were the dissolution of
carbon dioxide in water and the reaction of carbon dioxide with water to form carbonic
acid, we could say that most carbon dioxide remains in the form of CO 2 gas. However,
there is another equilibrium we must consider: the ionization of aqueous carbonic acid
page 22
in water to form bicarbonate.
page 22
In the ionization of carbonic acid, a H+ ion is transferred to water to form the
hydronium ion (H3O+) and the bicarbonate ion (HCO3-). This form of ionization is
commonly referred to as an acid-base equilibrium because carbonic acid (the acid) is
actually in equilibrium with bicarbonate (the base). The Keq (=[H3O+][HCO3]/[H2CO3][H2O]) for the acid-base equilibrium is found to be 4.46 x 10-6 M-1, which
indicates that even this ionization process is not favorable. As you will see, however,
even though the Keq favors reactants, almost all of the H2CO3 in the cell dissociates to
form HCO3-.
You can appreciate that if most of the H2CO3 is driven towards HCO3-, then more CO2
(aq.) will react to form H2CO3, that in turn, will lead to the dissolution of more CO2 gas
to form CO2 (aq.). This is an example of Le Chatelier’s Principle, which we will
discuss shortly. The example above shows that by coupling equilibria, you can end up
with a significant amount of HCO3- which serves as a source for CO2 for biosynthetic
processes.
In order to appreciate the above example more fully, we need to examine the third
equilibrium in a more detailed manner. First we need to understand the concept of
acidity.
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A molecule is said to be acidic when it is able to give up a proton (H+). Acidic
molecules usually contain highly polar X-H bonds (where X = electronegative atom),
which can be broken easily to form a H+ and an X-. In many acidic molecules, such as
H-F, H-Cl or –COOH, there is only one such X-H bond that can be broken (i.e., only
one H+ can be given up). These are known as monoprotic acids. In the case of a
diprotic acid such as carbonic acid, there are two O-H bonds that are highly polar (due
to the high electronegativity of oxygen). This slide depicts one of the O-H bonds in
carbonic acid being broken to give H+ and HCO3-. Because H+ ions are highly unstable
on their own, they are bonded to water molecules to form hydronium ions (H3O+).
Not shown on this slide is the fact that a second ionization can take place where the
bicarbonate loses a H+ ion to form the dianionic carbonate (CO32-) ion.
When evaluating a molecule’s acidity, the two relevant states are the protonated and
deprotonated forms of the molecule. When an acid gives up a proton (when it is
deprotonated), its covalent structure changes slightly because one of its atoms is no
longer bonded to a hydrogen atom. The deprotonated molecule, also called the
conjugate base form of the molecule, is left with one additional electron that used to
be the hydrogen atom’s valence electron; the hydrogen atom, in turn, becomes the
hydronium ion. Therefore, a neutral acid upon deprotonation forms a negatively
charged conjugate base. Likewise, a positively charged acid becomes a neutral
conjugate base upon giving up a proton. At equilibrium, the concentration of
protonated and deprotonated molecules is constant (but not necessarily equal).
page 24
We can write a Keq equation (= [H3O+][HCO3-]/[H2CO3][H2O]) for the ionization process
of carbonic acid. The concentration of water (in water) is about 55 M and does not
change significantly during ionization. Therefore, a new equilibrium constant K a is
often written where Ka = Keq[H2O] = [H3O+][HCO3-]/[H2CO3]. Ka is known as the acid
dissociation constant and is a measure of whether an acid is a strong or weak acid
(as discussed in the next slide). In general, Ka = [H3O+][A-]/[HA], where HA is the
protonated form of the molecule (the acid) while A- is the deprotonated form of the
molecule (the conjugate base). Since carbonic acid is a diprotic acid, there are two Ka
values; one for the acid-base equilibrium between carbonic acid and bicarbonate (Ka1),
and the other for the equilibrium between bicarbonate and carbonate (K a2).
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For convenience, the acidity of a molecule is usually indicated by its pKa value rather
than by its Ka value. Ka and pKa are related by the simple equation pKa = –log(Ka).
Molecules with lower pKas are more acidic than molecules with higher pKas. By
convention, molecules with pKa values less than 7 are defined as acidic in water,
while molecules with pKas greater than 7 are considered basic in water. The pKa
values of several representative molecules are shown here.
Since we know that pKa = -log(Ka) = -log(Keq[H2O]), a pKa value smaller than -1.75
(which represents Keq > 1) would mean that the ionization process is favorable and
the equilibrium lies towards the conjugate base. Strong acids like HCl (pK a ~ -8.0) fall
into this category and will completely dissociate into H3O+ and A- when placed in
water. As you can see, all pKa values shown in this slide are positive and hence
larger than -1.75 (i.e. Keq << 1). These acids (which are commonly found in the cell)
are therefore considered weak acids and only dissociate into H3O+ and A- to a very
small extent. In other words, the ionization process for these acids is not favorable.
page 26
The combination of acidic or basic molecules dissolved in an aqueous solution
imparts an acidity or basicity in that solution. The acidity or basicity of a solution is
described by a number called its pH. By definition, the pH of a solution is equal to –
log [H+]. Since in water protons actually become H3O+ cations, pH = –log [H3O+] in
water. Therefore, a solution containing a proton concentration of 0.1 M has a pH of
1. As you can deduce, the greater the concentration of protons in solution, the lower
the pH of that solution. Neutral aqueous solutions have a pH of 7.0. Acidic solutions
such as lemon juice have a pH below 7, while basic solutions such as ammoniacontaining window cleaners have a pH above 7.
page 27
The protonation state of a molecule depends on both its pK a as well as the pH of the
surrounding solution. The cell contains a large number of species that ionize and the
pH in a cell is maintained in most compartments at around 7 (actually, 7.4). The
relationship between pKa, pH, and protonation state is given by the HendersonHasselbalch equation: pKa = pH + log ([HA]/[A–]). This equation can be easily
derived from the Ka expression. The Henderson-Hasselbalch equation is very useful
for understanding the molecules of life because it allows you to determine what
fraction of a molecule exists in a protonated versus deprotonated form at any given
pH.
Thus, under physiological conditions (pH ~7), a molecule having a pKa of 7 exists as
an equal mixture of protonated and deprotonated forms. Molecules with pK as less
than 7 are mostly deprotonated and exist largely as their conjugate base forms in pH
7 solutions. Conversely, molecules with pKa values above 7 are mostly protonated
under physiological conditions. From inspecting this equation, you can deduce the
two crucial results that (i) increasing the pH of a solution by one pH unit increases
the fraction of deprotonated molecules by exactly 10-fold, and (ii) when pH = pKa, the
concentrations of protonated and deprotonated forms are equal.
page 28
Let us use the example of the cell to illustrate the usefulness of the HendersonHasselbach equation. First of all, the pH inside a cell is maintained at around 7.4.
There are numerous acid-base equilibria that exist in the cell, each with a different pK a.
We can use the Henderson-Hasselbalch equation to calculate the ratio of protonated
to deprotonated species of each acid in the cell.
The pKa of protonated amines (-NH3+) is around 10.5. At pH ~ 7.4, you will find that the
ratio of [-NH3+]:[-NH2] = 103.1:1 ~ 1260:1. This means that at physiological pH (7.4),
amines are predominantly protonated. Conversely, based on the pK a (4.0) of carboxylic
acids (-COOH), the ratio of [-COOH]:[-COO-] would be 10-3.4:1 ~ 1:2500. Therefore,
carboxylic acids are predominantly deprotonated in the cell. The HendersonHasselbalch equation allows us to easily come to a conclusion that there will almost
always be a positive charge on amines (due to protonation) and a negative charge on
carboxyl groups (due to deprotonation). It also allows us to predict that there are near
equal amounts of H2PO4- (acid, pKa ~ 7.0) and HPO42- (conjugate base) in a solution of
pH 7.4.
If you take a look at the acid-base equilibrium between carbonic acid (pKa ~ 3.6) and
bicarbonate, the ratio of [H2CO3]:[HCO3-] = 10-3.8:1 ~ 1:6300 at pH 7.4. It therefore
seems that the equilibrium now favors the bicarbonate form. Based simply on the
value of Keq for the ionization of carbonic acid combined with the equilibria for
dissolution of gaseous carbon dioxide in water and reaction to form carbonic acid, we
might have concluded that there is no way to trap carbon dioxide at high
concentrations in cells. In fact, because the pH is maintained at 7.4 by other
page 29
processes, virtually all of the carbonic acid is deprotonated.
page 29
The phenomenon that the pH of a solution affects the ratio between two species that
are in an acid-base equilibrium, and thus affects other coupled equilibria, is a
manifestation of Le Chatelier’s Principle.
Le Chatelier’s principle states that if a chemical system at equilibrium experiences a
change in concentration (of any species), temperature, volume or total pressure, the
equilibrium will shift in order to counter-act the change.
page 30
Because the carbonic acid-bicarbonate equilibrium favors the bicarbonate side (because of
pH), the concentration of H2CO3 decreases. The decrease in [H2CO3] affects the equilibrium
for the reaction between carbon dioxide and water. Based on Le Chatelier’s principle, more
carbon dioxide reacts to form H2CO3 in order to attempt to increase the levels of H2CO3.
Similarly, as the levels of aqueous CO2 decrease due to the formation of carbonic acid (which
immediately ionizes to form bicarbonate), Le Chatelier’s principle tells us that more gaseous
CO2 will dissolve into the aqueous environment to make up for that decrease in aqueous CO2.
Through a set of coupled equilibria and the use of Le Chatelier’s principle, carbon dioxide from
the air can thus be easily trapped in the aqueous environment in plant cells in the form of
bicarbonate ions. These bicarbonate ions serve as a source of CO2 which can later be used in
the process of photosynthesis. This is a good thing for plants.
It is not a good thing for the ocean in an age where huge amounts of carbon dioxide are
generated by industrial processes. As more and more CO2 dissolves in the ocean, the pH
begins to drop because the ocean is not as good as a cell at maintaining a constant pH.
Organisms that have calcium carbonate exoskeletons (corals, for example) begin to die
because the concentration of carbonate in seawater decreases, and their exoskeletons begin
to dissolve (or cannot be renewed due to a lack of carbonate ion to use as a building block).
Hence, the three equilibria shown on this slide are coupled to another equilibrium, which is the
ionization of bicarbonate to form carbonate.
Le Chatelier’s principle applies to everything that you see in biology and we will come back to
this principle in the later lectures.
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