Gibbs (Free) Energy

Skills to Develop
To get an overview of Gibbs energy and its general uses in chemistry.
Understand how Gibbs energy pertains to reactions properties
Understand how Gibbs energy pertains to equilibria properties
Understand how Gibbs energy pertains to electrochemical properties

Gibbs free energy, denoted G, combines enthalpy and entropy into a single value. The change in free energy, ΔG, is equal
to the sum of the enthalpy plus the product of the temperature and entropy of the system. ΔG can predict the direction of
the chemical reaction under two conditions:
1. constant temperature and
2. constant pressure.
If ΔG is positive, then the reaction is nonspontaneous (i.e., an the input of external energy is necessary for the reaction to
occur) and if it is negative, then it is spontaneous (occurs without external energy input).

Introduction
Gibbs energy was developed in the 1870’s by Josiah Willard Gibbs. He originally termed this energy as the “available
energy” in a system. His paper published in 1873, “Graphical Methods in the Thermodynamics of Fluids,” outlined how
his equation could predict the behavior of systems when they are combined. This quantity is the energy associated with a
chemical reaction that can be used to do work, and is the sum of its enthalpy (H) and the product of the temperature and
the entropy (S) of the system. This quantity is defined as follows:
G = H −TS (1)

or more completely as
G = U +PV −TS (2)

where
U is internal energy (SI unit: joule)
P is pressure (SI unit: pascal)

V is volume (SI unit: m )

3

T is temperature (SI unit: kelvin)

S is entropy (SI unit: joule/kelvin)

H is the enthalpy (SI unit: joule)

Gibbs Energy in Reactions

Spontaneous - is a reaction that is consider to be natural because it is a reaction that occurs by itself without any external
action towards it. Non spontaneous - needs constant external energy applied to it in order for the process to continue and
once you stop the external action the process will cease. When solving for the equation, if change of G is negative, then
it's spontaneous. If change of G if positive, then it's non spontaneous. The symbol that is commonly used for FREE
ENERGY is G. can be more properly consider as "standard free energy change"
In chemical reactions involving the changes in thermodynamic quantities, a variation on this equation is often
encountered:
ΔG = ΔH − T ΔS (3)
change in free energy change in enthalpy (temperature) change in entropy

Example 1.1
Calculate ∆G at 290 K for the following reaction:
2 NO(g) + O (g) → 2 NO (g) (4)
2 2

Given
∆H = -120 kJ

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 ∆S = -150 JK -1
Solution
now all you have to do is plug in all the given numbers into the above equation that was introduced earlier. expect you
have to convert ΔS so the same unit was ΔH .

1 kJ
ΔS = −150 J /K ( ) = −0.15 kJ/K
1000 J

looks like:

ΔG = −120 kJ − (290 K )(−0.150 kJ/ K )

= −120 kJ + 43 kJ

= −77 kJ

Exercise 1.1: The Haber Process

What is the ΔG for this formation of ammonia from nitrogen and hydrogen gas.

N +3 H ⇌ 2 NH (5)
2 2 3

The Standard free energy formations: NH3 =-16.45 H2=0 N2=0

Answer
−1
ΔG = −49.35 kJ mol

Since the changes of entropy of chemical reaction are not measured readily, thus, entropy is not typically used as a
criterion. To obviate this difficulty, we can use G. The sign of ΔG indicates the direction of a chemical reaction and
determine if a reaction is spontaneous or not.
ΔG < 0 : reaction is spontaneous in the direction written (i.e., the reaciton is exergonic)
ΔG = 0 : the system is at equilibrium and there is no net change either in forward or reverse direction.

ΔG > 0 : reaction is not spontaneous and the process proceeds spontaneously in the reserve direction. To drive such a

reaction, we need to have input of free energy (i.e., the reaction is endergonic)
The factors affect ΔGof a reaction (assume ΔH and ΔS are independent of temperature):
ΔH ΔS ΔG Example

at low temperature: + , at high

+ + 2HgO(s) -> 2Hg (l) + O2 (g)
temperature: -
+ - at all temperature: + 3O2 (g) ->2O3 (g)
- + at all temperature: - 2H2O2 (l) -> 2H2O (l) + O2 (g)
at low temperature: - , at high
- - NH3 (g) + HCl (g) -> NH4Cl (s)
temperature: +

Note:
1. ΔG depends only on the difference in free energy of products and reactants (or final state and initial state). ΔG is
independent of the path of the transformation and is unaffected by the mechanism of a reaction.
2. ΔG cannot tell us anything about the rate of a reaction.
The standard Gibbs energy change ΔG (at which reactants are converted to products at 1 bar) for:
o

aA + bB → cC + dD (6)

o o o o o
ΔrG = c Δf G (C ) + dΔf G (D) − aΔf G (A) − b Δf G (B) (7)

0 0 0
Δf G = ∑ vΔf G (products) − ∑ vΔf G (reactants) (8)

The standard-state free energy of reaction ( ΔG ) is defined as the free energy of reaction at standard state conditions:
o

o o o
ΔG = ΔH − T ΔS (9)

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 Note
If |ΔH | >> |T ΔS|: the reaction is enthalpy-driven
If ΔH << T ΔS : the reaction is entropy-driven

Standard-State Free Energy of Formation

The partial pressure of any gas involved in the reaction is 0.1 MPa.
The concentrations of all aqueous solutions are 1 M.
Measurements are generally taken at a temperature of 25° C (298 K).
The standard-state free energy of formation is the change in free energy that occurs when a compound is formed from its
elements in their most thermodynamically stable states at standard-state conditions. In other words, it is the difference
between the free energy of a substance and the free energies of its constituent elements at standard-state conditions:
o o o
ΔG = ∑ ΔG − ∑ ΔG (10)
fproducts frea cta nts

Example 1.2
Used the below information to determine if N H 4N O3(s) will dissolve in water at room temperature.

Compound ΔH
f
o
ΔS
f
o

N H4 N O3(s) -365.56 151.08

+
NH
4(aq ) -132.51 113.4

−
NO
3(aq ) 205.0 146.4

SOLUTION
This question is essentially asking if the following reaction is spontaneous at room temperature.
H O
2
+ −
NH NO (s) ⟶ NH (aq) + NO (aq) (11)
4 3 4 3

This would normally only require calculating ΔG and evaluating its sign. However, the ΔG values are not
o o

tabulated, so they must be calculated manually from calculated ΔH and ΔS values for the reaction. o o

Calculate ΔH :o

o o o
ΔH = ∑ nΔH − ∑ mΔH (12)
fproducts frea cta nts

o
kJ −
kJ
ΔH = [(1 mol N H3 ) (−132.51 ) + (1 mol N O ) (−205.0 )] (13)
3
mol mol

kJ
− [(1 mol N H4 N O3 ) (−365.56 )] (14)
mol

o
ΔH = −337.51 kJ + 365.56 kJ = 28.05 kJ (15)

Calculate ΔS :o

o o o
ΔS = ∑ nΔS − ∑ SΔH (16)
f f
products rea cta nts

o
J −
J
ΔS = [(1 mol N H3 ) (113.4 ) + (1 mol N O ) (146.6 )] (17)
3
mol K mol K

J
− [(1 mol N H4 N O3 ) (151.08 )] (18)
mol K

o
ΔS = 259.8 J/K − 151.08 J/K = 108.7 J/K (19)

Calculate ΔG :o

These values can be substituted into the free energy equation

o
TK = 25 C + 273.15K = 298.15 K (20)

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 1 kJ
o
ΔS = 108.7 J /K ( ) = 0.1087 kJ/K (21)
1000 J

o
ΔH = 28.05 kJ (22)

Plug in ΔH , ΔS and T into Equation 1.7

o o

o o o
ΔG = ΔH + T ΔS (23)

o
ΔG = 28.05 kJ − (298.15 K )(0.1087 kJ/ K ) (24)

o
ΔG = 28.05 kJ − 32.41 kJ (25)

o
ΔG = −4.4 kJ (26)

This reaction is spontaneous at room temperature since ΔG is negative. Therefore N H

o
4N O3(s) will dissolve in water
at room temperature.

Example 1.3
Calculate ΔG for the following reaction at 25 o
C . Will the reaction occur spontaneously?

N H3(g) + H C l(g) → N H4 C l(s)

given for the reaction

ΔH = −176.0 kJ

ΔS = −284.8 J/K

Solution
calculate ΔG from the formula

ΔG = ΔH − T ΔS (27)

but first we need to convert the units for ΔS into kJ/K (or convert ΔH into J) and temperature into Kelvin

1 kJ
ΔS = −284.8 J /K ( ) = −0.284.8 kJ/K
1000 J
o
T = 273.15 K + 25 C = 298 K

The definition of Gibbs energy can then be used directly

ΔG = ΔH − T ΔS (28)

ΔG = −176.0 kJ − (298 K )(−0.284.8 kJ/ K ) (29)

ΔG = −176.0 kJ − (−84.9 kJ) (30)

ΔG = −91.1 kJ (31)

Yes, this reaction is spontaneous at room temperature since ΔG is negative.

Gibbs Energy in Equilibria

Let's consider the following reversible reaction:
A+B ⇋ C +D (32)

The following equation relates the standard-state free energy of reaction with the free energy at any point in a given
reaction (not necessarily at standard-state conditions):
o
ΔG = ΔG + RT ln Q (33)

ΔG = free energy at any moment

ΔG
o
= standard-state free energy
R is the ideal gas constant = 8.314 J/mol-K
T is the absolute temperature (Kelvin)
ln Q is natural logarithm of the reaction quotient

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At equilibrium, ΔG = 0 and Q=K. Thus the equation can be arranged into:
[C ][D]
o
ΔG = ΔG + RT ln (34)
[A][B]

with
ΔG
o
= standard free energy change
R = gas constant = 1.98 * 10-3 kcal mol-1 deg-10
T = is usually room temperature = 298 K
[C ][D]
K =
[A][B]

The Gibbs free energy ΔG depends primarily on the reactants' nature and concentrations (expressed in the ΔG term and o

the logarithmic term of Equation 1.11, respectively).

At equilibrium, ΔG = 0 : no driving force remains
′ [C ][D]
o
0 = ΔG + RT ln (35)
[A][B]

[C ][D]
o
ΔG = −RT ln (36)
[A][B]

The equilibrium constant is defined as

[C ][D]
Keq = (37)
[A][B]

When K eq is large, almost all reactants are converted to products. Substituting K eq into Equation 1.14, we have:
o
ΔG = −RT ln Keq (38)

or
o
ΔG = −2.303RT log10 Keq (39)

Rearrange,
o
−Δ G /(2.303RT )
Keq = 10 (40)

This equation is particularly interesting as it relates the free energy difference under standard conditions to the properties
of a system at equilibrium (which is rarely at standard conditions).
Table 1.1: Converting K ea to ΔG
Keq ΔGo (kcal/mole)

10
−5
6.82
10
−4
5.46
10
−3
4.09
10
−2
2.73
10
−1
1.36
1 0
10
1
-1.36
10
2
-2.73
10
3
-4.09
10
4
-5.46
10
5
-6.82

Example 1.4
What is ΔG for isomerization of dihydroxyacetone phosphate to glyceraldehyde 3-phosphate?
o

If at equilibrium, we have K eq = 0.0475 at 298 K and pH 7. We can calculate:

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 o −3
ΔG = −2.303 RT log10 Keq = (−2.303) ∗ (1.98 ∗ 10 ) ∗ 298 ∗ (log10 0.0475) = 1.8 kcal/mol (41)

Given:
The initial concentration of dihydroxyacetone phosphate = 2 × 10 −4
M

The initial concentration of glyceraldehyde 3-phosphate = 3 × 10 −6

M

SOLUTION
From equation 2:
ΔG = 1.8 kcal/mol + 2.303 RT log10(3*10-6 M/2*10-4 M) = -0.7 kcal/mol

Note
Under non-standard conditions (which is essential all reactions), the spontaneity of reaction is determined by ΔG, not
′

ΔG .
o

Gibbs Energy in Electrochemistry

The Nernst equation relates the standard-state cell potential with the cell potential of the cell at any moment in time:
RT
o
E =E − ln Q (42)
nF

with
E = cell potential in volts (joules per coulomb)
n = moles of electrons
F = Faraday's constant: 96,485 coulombs per mole of electrons

By rearranging this equation we obtain:

o
RT
E =E − ln Q (43)
nF

multiply the entire equation by nF

o
nF E = nF E − RT ln Q (44)

which is similar to:

o
ΔG = ΔG + RT ln Q (45)

By juxtaposing these two equations:

o
nF E = nF E − RT ln Q (46)

o
ΔG = ΔG + RT ln Q (47)

it can be concluded that:

ΔG = −nF E (48)

Therefore,
o o
ΔG = −nF E (49)

Some remarks on the Gibbs "Free" Energy

Free Energy is not necessarily "free": The appellation “free energy” for G has led to so much confusion that
many scientists now refer to it simply as the Gibbs energy. The “free” part of the older name reflects the steam-
engine origins of thermodynamics with its interest in converting heat into work: ΔG is the maximum amount of
energy which can be “freed” from the system to perform useful work. By "useful", we mean work other than that
which is associated with the expansion of the system. This is most commonly in the form of electrical work
(moving electric charge through a potential difference), but other forms of work (osmotic work, increase in surface
area) are also possible.
Free Energy is not energy: A much more serious difficulty with the Gibbs function, particularly in the context of
chemistry, is that although G has the units of energy (joules, or in its intensive form, J mol–1), it lacks one of the

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 most important attributes of energy in that it is not conserved. Thus although the free energy always falls when a
gas expands or a chemical reaction takes place spontaneously, there need be no compensating increase in energy
anywhere else. Referring to G as an energy also reinforces the false but widespread notion that a fall in energy must
accompany any change. But if we accept that energy is conserved, it is apparent that the only necessary condition
for change (whether the dropping of a weight, expansion of a gas, or a chemical reaction) is the redistribution of
energy.The quantity –ΔG associated with a process represents the quantity of energy that is “shared and spread”,
which as we have already explained is the meaning of the increase in the entropy. The quotient –ΔG/T is in fact
identical with ΔStotal, the entropy change of the world, whose increase is the primary criterion for any kind of
change.
Free Energy is not even "real": G differs from the thermodynamic quantities H and S in another significant way:
it has no physical reality as a property of matter, whereas H and S can be related to the quantity and distribution of
energy in a collection of molecules (e.g., the third law of thermodynamics). The free energy is simply a useful
construct that serves as a criterion for change and makes calculations easier.

References
1. Chang, Raymond. Physical Chemistry for the Biosciences. Sansalito, CA: University Sciences, 2005.
2. Atkins, Peter and de Paula, Julio. Physical Chemistry for the Life Sciences. New York, NY: W. H. Freeman and
Company, 2006. Page 153-163, 286.
3. Stryer, Lubert. Biochemistry (Third Edition). New York, NY: W.H. Freeman and Company, 1988. Page 181-184.

Contributors
Cathy Doan (UC Davis), Han Le (UC Davis)
Stephen Lower, Professor Emeritus (Simon Fraser U.) Chem1 Virtual Textbook

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