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Gibbs Free Energy
Driving
Forces &
Gibbs Free
Energy
The Effect of
Temperature on
the Free Energy of
a Reaction
Standard-State
Free Energies of
Reaction
Interpreting
Standard-State
Free Energy of
Reaction Data
The Relationship Between Free
Energy and Equilibrium
Constants
The Temperature
Dependence of
Equilibrium
Constants
The Relationship
Between Free
Energy and Cell
Potentials
Driving Forces and Gibbs Free Energy
Some reactions are spontaneous because they give off energy in the form of heat ( H < 0). Others are
spontaneous because they lead to an increase in the disorder of the system ( S > 0). Calculations of H and
S can be used to probe the driving force behind a particular reaction.
Practice Problem 5:
Calculate H and S for the following reaction and decide in which direction each of these factors will
drive the reaction.
N2(g) + 3 H2(g)
2 NH3(g)
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What happens when one of the potential driving forces behind a chemical reaction is favorable and the other
is not? We can answer this question by defining a new quantity known as the Gibbs free energy (G) of the
system, which reflects the balance between these forces.
The Gibbs free energy of a system at any moment in time is defined as the enthalpy of the system minus the
product of the temperature times the entropy of the system.
G = H - TS
The Gibbs free energy of the system is a state function because it is defined in terms of thermodynamic
properties that are state functions. The change in the Gibbs free energy of the system that occurs during a
reaction is therefore equal to the change in the enthalpy of the system minus the change in the product of the
temperature times the entropy of the system.
G=
H - (TS)
If the reaction is run at constant temperature, this equation can be written as follows.
G=
H-T S
The change in the free energy of a system that occurs during a reaction can be measured under any set of
conditions. If the data are collected under standard-state conditions, the result is the standard-state free
energy of reaction ( Go).
Go =
Ho - T So
The beauty of the equation defining the free energy of a system is its ability to determine the relative
importance of the enthalpy and entropy terms as driving forces behind a particular reaction. The change in
the free energy of the system that occurs during a reaction measures the balance between the two driving
forces that determine whether a reaction is spontaneous. As we have seen, the enthalpy and entropy terms
have different sign conventions.
Favorable
Unfavorable
Ho < 0
Ho > 0
So > 0
So < 0
The entropy term is therefore subtracted from the enthalpy term when calculating
Go for a reaction.
Because of the way the free energy of the system is defined, Go is negative for any reaction for which Ho
is negative and So is positive. Go is therefore negative for any reaction that is favored by both the
enthalpy and entropy terms. We can therefore conclude that any reaction for which Go is negative should
be favorable, or spontaneous.
Favorable, or spontaneous reactions:
Conversely, Go is positive for any reaction for which
which Go is positive is therefore unfavorable.
Go < 0
Ho is positive and
Unfavorable, or non-spontaneous reactions:
So is negative. Any reaction for
Go > 0
Reactions are classified as either exothermic ( H < 0) or endothermic ( H > 0) on the basis of whether
they give off or absorb heat. Reactions can also be classified as exergonic ( G < 0) or endergonic ( G > 0)
on the basis of whether the free energy of the system decreases or increases during the reaction.
When a reaction is favored by both enthalpy ( Ho < 0) and entropy ( So > 0), there is no need to calculate
the value of Go to decide whether the reaction should proceed. The same can be said for reactions favored
by neither enthalpy ( Ho > 0) nor entropy ( So < 0). Free energy calculations become important for
reactions favored by only one of these factors.
Practice Problem 6:
Calculate
H and
S for the following reaction:
NH4NO3(s) + H2O(l)
NH4+ (aq) + NO3- (aq)
Use the results of this calculation to determine the value of Go for this reaction at 25o C, and explain why
NH4NO3 spontaneously dissolves is water at room temperature.
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The Effect of Temperature on the Free Energy of a Reaction
The balance between the contributions from the enthalpy and entropy terms to the free energy of a reaction
depends on the temperature at which the reaction is run.
Practice Problem 7:
Use the values of H and
spontaneous at 25C:
S calculated in Practice Problem 5 to predict whether the following reaction is
N2(g) + 3 H2(g)
2 NH3(g)
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The equation used to define free energy suggests that the entropy term will become more important as the
temperature increases.
Go =
Ho - T So
Since the entropy term is unfavorable, the reaction should become less favorable as the temperature
increases.
Practice Problem 8:
Predict whether the following reaction is still spontaneous at 500C:
N2(g) + 3 H2(g)
Assume that the values of
Ho and
2 NH3(g)
S used in Practice Problem 7 are still valid at this temperature.
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Standard-State Free Energies of Reaction
Go for a reaction can be calculated from tabulated standard-state free energy data. Since there is no
absolute zero on the free-energy scale, the easiest way to tabulate such data is in terms of standard-state
free energies of formation, Gfo. As might be expected, the standard-state free energy of formation of a
substance is the difference between the free energy of the substance and the free energies of its elements in
their thermodynamically most stable states at 1 atm, all measurements being made under standard-state
conditions.
Interpreting Standard-State Free Energy of Reaction Data
We are now ready to ask the obvious question: What does the value of
reaction?
N2(g) + 3 H2(g)
2 NH3(g)
Go tell us about the following
Go = -32.96 kJ
By definition, the value of Go for a reaction measures the difference between the free energies of the
reactants and products when all components of the reaction are present at standard-state conditions.
Go therefore describes this reaction only when all three components are present at 1 atm pressure.
The sign of Go tells us the direction in which the reaction has to shift to come to equilibrium. The fact that
Go is negative for this reaction at 25oC means that a system under standard-state conditions at this
temperature would have to shift to the right, converting some of the reactants into products, before it can
reach equilibrium. The magnitude of Go for a reaction tells us how far the standard state is from
equilibrium. The larger the value of Go, the further the reaction has to go to get to from the standard-state
conditions to equilibrium.
Assume, for example, that we start with the following reaction under standard-state conditions, as shown in
the figure below.
N2(g) + 3 H2(g)
The value of
2 NH3(g)
G at that moment in time will be equal to the standard-state free energy for this reaction,
When Qp = 1:
G=
Go.
Go
As the reaction gradually shifts to the right, converting N2 and H2 into NH3, the value of G for the reaction
will decrease. If we could find some way to harness the tendency of this reaction to come to equilibrium, we
could get the reaction to do work. The free energy of a reaction at any moment in time is therefore said to be
a measure of the energy available to do work.
The Relationship Between Free Energy and Equilibrium Constants
When a reaction leaves the standard state because of a change in the ratio of the concentrations of the
products to the reactants, we have to describe the system in terms of non-standard-state free energies of
reaction. The difference between Go and G for a reaction is important. There is only one value of Go for
a reaction at a given temperature, but there are an infinite number of possible values of G.
The figure below shows the relationship between G for the following reaction and the logarithm to the base
e of the reaction quotient for the reaction between N2 and H2 to form NH3.
N2(g) + 3 H2(g)
2 NH3(g)
Data on the left side of this figure correspond to relatively small values of Qp. They therefore describe
systems in which there is far more reactant than product. The sign of G for these systems is negative and
the magnitude of G is large. The system is therefore relatively far from equilibrium and the reaction must
shift to the right to reach equilibrium.
Data on the far right side of this figure describe systems in which there is more product than reactant. The
sign of G is now positive and the magnitude of G is moderately large. The sign of G tells us that the
reaction would have to shift to the left to reach equilibrium. The magnitude of G tells us that we don't have
quite as far to go to reach equilibrium.
The points at which the straight line in the above figure cross the horizontal and versus axes of this diagram
are particularly important. The straight line crosses the vertical axis when the reaction quotient for the system
is equal to 1. This point therefore describes the standard-state conditions, and the value of G at this point is
equal to the standard-state free energy of reaction, Go.
When Qp = 1:
G=
Go
The point at which the straight line crosses the horizontal axis describes a system for which G is equal to
zero. Because there is no driving force behind the reaction, the system must be at equilibrium.
When Qp = Kp:
G=0
The relationship between the free energy of reaction at any moment in time ( G) and the standard-state free
energy of reaction ( Go) is described by the following equation.
G=
Go + RT ln Q
In this equation, R is the ideal gas constant in units of J/mol-K, T is the temperature in kelvin, ln represents a
logarithm to the base e, and Q is the reaction quotient at that moment in time.
As we have seen, the driving force behind a chemical reaction is zero ( G = 0) when the reaction is at
equilibrium (Q = K).
0 = Go + RT ln K
We can therefore solve this equation for the relationship between
Go and K.
Go = - RT ln K
This equation allows us to calculate the equilibrium constant for any reaction from the standard-state free
energy of reaction, or vice versa.
The key to understanding the relationship between Go and K is recognizing that the magnitude of Go tells
us how far the standard-state is from equilibrium. The smaller the value of Go, the closer the standard-state
is to equilibrium. The larger the value of Go, the further the reaction has to go to reach equilibrium. The
relationship between Go and the equilibrium constant for a chemical reaction is illustrated by the data in the
table below.
Values of
Go and K for Common Reactions at 25oC
Go (kJ)
Reaction
K
2 SO2(g) + O2(g)
141.7
1.4 x 10-25
H2O(l)
H+(aq) + OH-(aq)
79.9
1.0 x 10-14
AgCl(s) + H2O
Ag+(aq) + Cl-(aq)
55.6
1.8 x 10-10
2 SO3(g)
H+(aq) + OAc-(aq)
27.1
1.8 x 10-5
N2(g) + 3 H2(g)
2 NH3(g)
-32.9
5.8 x 105
HCl(aq) + H2O
H+(aq) + Cl-(aq)
-34.2
1 x 106
Cu(NH3)42+(aq)
-76.0
2.1 x 1013
Zn2+(aq) + Cu(s)
-211.8
1.4 x 1037
HOAc(aq) + H2O
Cu2+(aq) + 4 NH3(aq)
Zn(s) + Cu2+(aq)
Practice Problem 9:
Use the value of
reaction at 25C:
Go obtained in Practice Problem 7 to calculate the equilibrium constant for the following
N2(g) + 3 H2(g)
2 NH3(g)
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The equilibrium constant for a reaction can be expressed in two ways: Kc and Kp. We can write equilibrium
constant expressions in terms of the partial pressures of the reactants and products, or in terms of their
concentrations in units of moles per liter.
For gas-phase reactions the equilibrium constant obtained from Go is based on the partial pressures of the
gases (Kp). For reactions in solution, the equilibrium constant that comes from the calculation is based on
concentrations (Kc).
Practice Problem 10:
Use the following standard-state free energy of formation data to calculate the acid-dissociation equilibrium
constant (Ka) at for formic acid:
Compound
HCO2(aq)
Gfo(kJ/mol)
-372.3
+
H (aq)
0.00
-
HCO2 (aq)
-351.0
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The Temperature Dependence of Equilibrium Constants
Equilibrium constants are not strictly constant because they change with temperature. We are now ready to
understand why.
The standard-state free energy of reaction is a measure of how far the standard-state is from equilibrium.
Go = - RT ln K
But the magnitude of
Go depends on the temperature of the reaction.
Go =
Ho - T So
As a result, the equilibrium constant must depend on the temperature of the reaction.
A good example of this phenomenon is the reaction in which NO2 dimerizes to form N2O4.
2 NO2(g)
N2O4(g)
This reaction is favored by enthalpy because it forms a new bond, which makes the system more stable. The
reaction is not favored by entropy because it leads to a decrease in the disorder of the system.
NO2 is a brown gas and N2O4 is colorless. We can therefore monitor the extent to which NO2 dimerizes to
form N2O4 by examining the intensity of the brown color in a sealed tube of this gas. What should happen to
the equilibrium between NO2 and N2O4 as the temperature is lowered?
For the sake of argument, let's assume that there is no significant change in either Ho or So as the system
is cooled. The contribution to the free energy of the reaction from the enthalpy term is therefore constant, but
the contribution from the entropy term becomes smaller as the temperature is lowered.
Go =
Ho - T So
As the tube is cooled, and the entropy term becomes less important, the net effect is a shift in the equilibrium
toward the right. The figure below shows what happens to the intensity of the brown color when a sealed
tube containing NO2 gas is immersed in liquid nitrogen. There is a drastic decrease in the amount of NO2 in
the tube as it is cooled to -196oC.
Practice Problem 11:
Use values of Ho and So for the following reaction at 25C to estimate the equilibrium constant for this
reaction at the temperature of boiling water (100C), ice(0C), a dry ice-acetone bath (-78C), and liquid
nitrogen (-196C):
2 NO2(g)
N2O4(g)
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The Relationship Between Free Energy and Cell Potentials
The value of G for a reaction at any moment in time tells us two things. The sign of G tells us in what
direction the reaction has to shift to reach equilibrium. The magnitude of G tells us how far the reaction is
from equilibrium at that moment.
The potential of an electrochemical cell is a measure of how far an oxidation-reduction reaction is from
equilibrium. The Nernst equation describes the relationship between the cell potential at any moment in time
and the standard-state cell potential.
Let's rearrange this equation as follows.
nFE = nFEo - RT ln Q
We can now compare it with the equation used to describe the relationship between the free energy of
reaction at any moment in time and the standard-state free energy of reaction.
G=
Go + RT ln Q
These equations are similar because the Nernst equation is a special case of the more general free energy
relationship. We can convert one of these equations to the other by taking advantage of the following
relationships between the free energy of a reaction and the cell potential of the reaction when it is run as an
electrochemical cell.
G = -nFE
Practice Problem 12:
Use the relationship between Go and Eo for an electrochemical reaction to derive the relationship between
the standard-state cell potential and the equilibrium constant for the reaction.
Click here to check your answer to Practice Problem 12.
Free energy is a composite function that balances the influence of energy vs. entropy.
1. Introduction
2. Gibbs Energy
3. How This Equation Was Reached
4. Reasoning Behind the Equation
5. Helmholtz Energy and its Relation to Gibbs Energy
6. References
7. External Links
8. Contributors
Introduction
To first define "free" energy, we shall examine the backgrounds of this term, what definitions carry it, and
which specific definitions we, as chemists, will choose to refer to:
1. Fossil fuels, global warming, and the usual popular controversies in popular science have led people
to cry out for the pursuit of clean and renewable "free" energy (no doubt referring to long-term
monetary cost). A most valiant cause; absolutely disregard it here. Forget you even read this
definition. We'll even start at listing at 1. again!
2. If a system is isothermal and closed, with constant pressure, it is describable by the Gibbs Energy,
known also by a plethora of nicknames such as "free energy", "Gibbs free energy", "Gibbs function",
and "free enthalpy". Because this module is located under "Gibbs Energy", we'll focus on this energy;
the Helmholtz will be briefly mentioned.
3. If a system is isothermal and closed, with constant volume, it is describable by the Helmholtz
Energy, known also by an unnecessary amount of aliases such as "Helmholtz function", "work
function", "Helmholtz free energy", and our favorite, "free energy". It will be mentioned in passing.
4. Having two energies both called "free energy" is like having two brothers named Jack. More
specifically, they'd be twin brothers; the Gibbs and Helmholtz Energies describe situations with
equations easily confused with each other. It's no wonder the IUPAC (the International Union of
Pure and Applied Chemistry) officially refers to the two as Gibbs Energy and Helmholtz Energy,
respectively. This shouldn't be a surprise, because that's what they were originally named in the first
place! Just keep in mind that some outdated or unsophisticated texts might still use the pseudonyms
mentioned above (guised as, say, the title of a module).
Gibbs Energy
The Gibbs Energy is named after a Josiah William Gibbs, an American physicist in the late 19 th century who
greatly advanced thermodynamics; his work now serves as a foundation for this branch of science. This
energy can be said to be the greatest amount of work (other than expansion work) a system can do on its
surroundings,
when
it
operates
at
a
constant
pressure
and
temperature.
First, a modeling of the Gibbs Energy by way of equation:
Where:
= internal energy

Internal Energy
= absolute temperature x final entropy

Entropy
= pressure x volume
Of course, we know that U + PV can also be defined as:
Where:
= enthalpy

Enthalpy
Which leads us to a form of how the Gibbs Energy is related to enthalpy:
All of the members on the right side of this equation are state functions, so G is a state function as well. The
change in G is simply:
How This Equation Was Reached
We will start with an equation for the total entropy change of the universe. Our goal is to whittle it down to a
practical form, like a caveman shaping a unwieldy block of stone into a useful hand held tool!
An equation with variables of such scope is difficult to work with. We want to do away with the vagueness,
and rewrite a more focused equation. We'll consider the case where temperature and pressure is constant.
Here we go:
1.

can be rewritten as
.
The heat, qp, that the system affects the surroundings with is the negative of the
Because
, the change in the entropy of the surroundings will
be
.
for the system.
2. The equation becomes

A simple substitution.
3. The equation becomes

Multiply both sides by T.
4. With the mighty powers of whoever discovering stuff getting to name it, we set
a great big
for the almighty Gibbs. Finally, we achieve the equation,
equal to
.
Reasoning Behind the Equation
As a quick note, let it be said that the name "free energy", other than being confused with another energy
exactly termed, is also somewhat of a misnomer. The multiple meanings of the word "free" can make it seem
as if energy can be transferred at no cost; in fact, the word "free" was used to refer to what cost the system
was free to pay, in the form of turning energy into work.
is useful because it can tell us how a system, when we're given only information on it, will act.
indicates a spontaneous* change to occur.
indicates an absence of spontaneousness.
indicates a system at equilibrium.
The Gibbs Energy reaches the minimum value when equilibrium is reached. Here, it is represented as a
graph, where x represents the extent of how far the reaction has occurred.
The minimum of the function has to be smooth, because G must be differentiable (its first derivative has to
exist at the minimum).
It was briefly mentioned that
evident from the equation.
is the energy available to be converted to work. The definition is self-
Look at
. Recall that
is the total energy that can be made into heat.
energy not available to be converted to work. By a reordering of the Gibbs Energy equation:
Expressed in words: the energy available to be turned into heat =
This lets us see that
is the
- the energy that is not free to do work
MUST be the energy free to do work.
Why constant temperature and pressure? It just so happens that these are regularly occurring factors in the
laboratory, making this equation practical to use, and useful as well, for chemists.
An example of Gibbs Energy in the real world is the oxidation of glucose;
kJ, or 686 Calories. For living cells, this is the primary energy reaction.
Helmholtz Energy and its Relation to Gibbs Energy
The Helmholtz Energy is given by the equation:
It is comparable to G this way:
in this case is equal to 2870
The Helmholtz Energy is used when having a constant pressure is not feasible.
Along with internal energy and enthalpy, the Helmholtz Energy and Gibbs Energy make up the quad group
called the thermodynamic potentials; these potentials are useful for describing various thermodynamic
events. A mnemonic device has been created for your memorizing pleasure:
TS represents energy from surroundings, and PV represents work in expansion. If needed, refer to the links
above to refresh your memory on enthalpy and internal energy.
References
1. Petrucci, et al. General Chemistry: Principles & Modern Applications; Ninth Edition.
Pearson/Prentice Hall; Upper Saddle River, New Jersey 07.
2. Mortimer, Robert G. Physical Chemistry; Third Edition. Elsevier Inc.; 2008.
External Links
1.
Hyperphysics: Georgia State University (accessed June, 2008)
2.Gibbs Free Energy
3. Four quantities called "thermodynamic potentials" are useful in the chemical thermodynamics of
reactions and non-cyclic processes. They are internal energy, the enthalpy, the Helmholtz free energy
and the Gibbs free energy. The Gibbs free energy G is defined by
4.
5. The internal energy U might be thought of as the energy required to create a system in the absence of
changes in temperature or volume. But as discussed in defining enthalpy, an additional amount of
work PV must be done if the system is created from a very small volume in order to "create room"
for the system. As discussed in defining the Helmholtz free energy, an environment at constant
temperature T will contribute an amount TS to the system, reducing the overall investment necessary
for creating the system. This net energy contribution for a system created in environment temperature
T from a negligible initial volume is the Gibbs free energy.
6. The change in Gibbs free energy, ΔG, in a reaction is a very useful parameter. It can be thought of as
the maximum amount of work obtainable from a reaction. For example, in the oxidation of glucose,
the change in Gibbs free energy is ΔG = 686 kcal = 2870 kJ. This reaction is the main energy reaction
in living cells.
7.
Thermodynamics: Chem1 Virtual Textbook by Stephen Lower, Ph.D. Physical Chemistry (accessed
June, 2008)
8.
International Union of Pure and Applied Chemistry (accessed June, 2008)