Skip to main content

# Gibb's Energy

Discussion Questions

• How does energy affect the direction and extend of a reaction?
• What are reaction quotients, equilibrium constants, and the mass action law?
• What are the relationships between the Gibb's free energies and the equilibrium constants?
• What is Gibb's free energy in terms of enthalpy and entropy?
• How does Gibb's free energy drive a chemical reaction? What is the tendency for Gibb's free energy to vary or change?

## Gibb's Free Energy and Equilibrium

Chemical and physical reactions can be represented by equations, using a few symbols to represent actually what is going on. For examples:

H2O(l) = H2O(g), liquid to gas, evaporation
2 NO2 = N2O4 dimerization
H2O = H+ + OH- ionization
C(s) + H2O(g) = H2(g) + CO(g) water gas production

Reactions are driven by energy as entropy, S, or as Gibbs energy, G. This document introduces the concept of Gibb's energy.

### How does energy affect the direction and extend of a reaction?

To answer this question, we have to introduce the concept of chemical equilibrium for reversible reactions. This concept is formulated as the mass action law, which defines the equilibrium constant K. The equilibrium constant K is related to the Gibbs free energy, G, or simply free energy. Furthermore, the Gibb's energy is actually derived from the enthapy (H) and entropy (S) of the reaction.

#### Mass Action Law

Since the mass action law is valid for many reactions, we discuss it using a general reaction with no specific reactants or products. A general reaction of a moles of A and b moles of B to produce c moles of C and d moles of D can be represented by

a A + b B = c C + d D

Furthermore, we represent activities of A, B, C, and D by {A}, {B}, {C} and {D} respectively. In any system, we can always define a reaction quotient, Q

     {C}c {D}d
-----------  = Q
{A}a {B}b


Experience shows that, over time, the quotient Q tends to approach a constant K for a given reaction in a closed system. Such a state is called equilibrium,

Q -> K

The constant K depends on temperature and the nature of the reactants and products. Thus, K is called the equilibrium constant. This is known as the mass action law. In other words, there is a tendency for the reaction to reach a equilibrium such that

     {C}c {D}d
-----------  = K
{A}a {B}b


Using activities covers a wider range of concentrations than using concentrations to define the reaction quotient or equilibrium constant K. Concentrations were used in freshman chemistry for simplicity.

#### Standard Gibbs Free Energy Change, DGo

Energy is the driving force for reactions. The tendency for a reaction to reach a equilibrium is driven by the Gibbs free energy, symbol DGo and the relationship has been defined by the relationship

DGo = - R T ln K

At the standard condition, activities of all reactants and products are unity (all equal to 1). In this system, Q = 1. If K > 1, the the forward reaction is spontaneous,

Q -> K.

The Gibb's free energy DGo is a negative quantity for such a system. Since this is always true, we can generalize the condition. When the Gibb's energy is negative, the reaction is spontaneous. A reaction (change) is the result of a system trying to minimize its Gibb's free energy, DGo changes from 0 to - R T ln K as the system changes from standard condition to an equilibrium state.

In contrast to the decrease in Gibb's energy, the entropy increases as an isolated system undergoes an spontaneous reaction.

#### Gibbs Free Energy Change, DG

The generalized statement can be represented by a generalized Gibb's free energy change, DG, for a system not at standard condition, but whose reaction quotient is Q. Obviously, the formulation is

DG = DGo + R T ln Q.

As the system strives to reach an equilibrium state, (no longer any net change),

Q -> K

we have the following results,

DGo = - R T ln K
DGo + R T ln K = 0
DG = 0.

The previous discussion leads to the following conclusion. When DG is positive, the reverse reaction is spontaneous. When DG is negative, the forward reaction is spontaneous, when DG is zero, the system has reached an equilibrium state.

#### Gibb's Energy in Terms of Enthalpy and Entropy

So far, the Gibb's free energy is defined as the driving force for a system to reach a chemical equilibrium. The energy comes from the enthalpy and entropy of reaction in the system, and DG has been define in terms of enthalpy and entropy changes, DH and DS, at temperature T as:

DG = DH - T DS.

Since Gibb's energy, enthalpy, and entropy are state functions, they have been treated as the functions in thermodynamics, and as a result, the delta D is omitted. The relation is simply,

G = H - T S

The generalized equation is very useful, and it can be differentiacted with respect to other thermodynamical variables. However, we will not discuss it any further at this point.

#### Gibb's Energy and Electric Energy

For redox reactions, Gibb's energy is the electric energy, which, when properly setup in an electric cell, is the charge transferred (q in Coulomb) times the potential E (in V). Each mole of electron has a charge of 1 faraday (1 F = 96458 C), and n moles of electron have a charge of n F. Since the voltage is usually a positive value, we have,

DG = - n F E

Example 1

What are the standard enthalpies of formation and standard (or absolute) entropies for H2(g), O2(g), I2(s), C(graphite), C(diamond), H2O, and HI?

SOLUTION
The data are usually listed in the thermodynamic data table. However, you need not use this table to find the standard enthalpies for elements at their standard-condition states. These are arbitrarily set as zero. Thus, DHfo = 0 for H2(g), O2(g), I2(s or cr) and C(graphite). Note that diamond is not a stable phase from the thermodynamic view point.

However, lines from the Thermodynamic Data Table are reproduced below for you to compare: Note that data for diamond come from another source.

Substance   DHfo          DSo
kJ/mol       J/(mol K)
H2(g)        0           130.680
I2(s)        0           116.14
O2(g)        0           205.152
C(graphite)  0             5.74
C(diamond)   1.985         2.377
H2O(l)    -285.83         69.95
HI(g)       26.50        206.590


DISCUSSION
Note units for the two quantities. The standard entropies in J/(mol K) are also the absolute entropies.

Example 2

Find the enthalpy and entropy changes for the reaction:

H2(g) + I2(s) -> 2 HI(g)

SOLUTION
For simplicity, we write the standard enthalpies below the chemical formula of the equation:

H2 (g) + I2 (s) -> 2 HI (g)
0 0 2*26.5 kJ/mol

DHoreaction = S Hfo(products) - S Hfo(reactants)
= 2*26.5 - (0 + 0) kJ/equation.
= 53.0 kJ/equation.

The standard entropy of reaction at standard conditions. DSoreaction can be evaluated in a similar fashion. For convenience, we again write the standard entropies below the chemical formula of the equation:

H2 (g) + I2 (s) -> 2 HI (g)
130.68 116.14 2*206.59 J mol-1 K-1

DSoreaction = SSo(products) - So(reactants)
= 2*206.59 - (130.68 + 116.14) J eqn-1 K-1
= 166.36 J eqn-1 K-1

DISCUSSION
This is an endothermic reaction, and 53 kJ is required for the formation of 2 moles of HI. The decomposition of HI is exothermic.

Note that DSoreaction is different from standard entropy DSo, which is NOT entropy of formation.

Since the reaction is the formation of HI, the entropy so calculated is twice the entropy of formation of HI. Thus, we have
DSof(HI) = (166.36/2) = 83.18 J mol-1 K-1

Example 3

Calculate the standard Gibb's energy of formation of HI,

½ H2(g) + ½ I2(g) -> HI(g),

and the equilibrium constant K for the above reaction.

SOLUTION
We will use the results from the examples given above to calculate the standard Gibb's energy according to the definition:

DGof = DH - T DS
= (26.5 - 298*0.08318) kJ/mol
= 1.71 kJ/mol
= - R T ln K

Therefore,

         1710 J mol-1
ln K = ------------------------- = 0.6904
8.312 J (mol K)-1 * 298 K


Thus, K = e0.6904 = 2.0

DISCUSSION
Examples 1, 2, and 3 illustrate how DGf can be derived from a table of thermodynamic data. In some tables, the value of DGf are also given.

Example 4

Calculate the standard entropy of formation of H2O(l), its standard Gibb's energy of formation, and the equilibrium constant K for the reaction.

H2(g) + ½ O2(g) -> H2O(l)

SOLUTION
All the required data have already been found in Example 1. Again, we write the standard entropies below the formula

H2(g) + ½ O2(g) -> H2O(l)
130.680 ½(205.152) 69.95

DSof = SSo(products) - SSo(reactants)
= 69.95 - (130.680 + &189; (205.152) J mol-1 K-1
= - 163.306 J/(mol K)
The standard Gibb's free energy of formation is,
DGof = DHof - T DSof
= -285.83 kJ/mol - (298 K)(-0.163306) kJ/(mol K)
= -237.16 kJ/mol

The equilibrium constant is evaluated by

            237160 J mol-1
ln K = ------------------------- = 95.746
8.312 J (mol K)-1 * 298 K


Thus, K = e95.746
= 13.82*1041,
a very large number indeed indicating a reaction to almost exhaust at least one of the reactants.

DISCUSSION
This example illustrates how you may use a thermodynamic data table.

Example 5

Estimate the standard Gibb's free energy of formation for amonia.

SOLUTION
The data required are:

Substance   DHfo          DSo
kJ/mol       J/(mol K)
H2(g)        0           130.680
N2(g)        0           191.609
NH3(g)       -45.94           192.77


Again, we write the standard entropies below the formula

3/2 H2(g) + ½ N2(g) -> NH3(g)
3/2*130.680 ½*191.609 192.77

DSof = SSo(products) - SSo(reactants)
= 192.77 - (3/2*130.680 + ½*191.609)
= -99.125 kJ/mol
The standard Gibb's free energy of formation is,
DGof = DHof - T DSof
= -45.94 - 298*(-0.099125) kJ/mol = -16.40 kJ/mol

DISCUSSION
Results from the previous and this examples are used in the next example.

Example 6

Evaluate DGoreaction for the reaction:

4 NH3(g) + 3 O2(g) -> 2 N2 + 6 H2O(g)

SOLUTION
Let us write the standard Gibb's energies of formation below the formula

4 NH3(g) + 3 O2(g) -> 2 N2 + 6 H2O(g)
4*(-16.40) 3*0 2*0 6*(-237.16) kJ/mol

DGoreaction = = (2*0 - 6*237.16) - (3*0 -4*16.40) kJ/mol
= -1357.36 kJ/mol

DISCUSSION
Note the general rule for the evaluation of the standard Gibb's free energy of reaction introduced in this example.

Equilibrium in chemical reactions gives a slightly different view on Gibb's energy and equilibrium from Brown University.

Example 7

A zinc copper battery has a voltage of 1.10 V. How much is the Gibb's energy for the redox reaction:

Zn + Cu2+ -> Zn2+ + Cu

SOLUTION
Since the number of reaction transferred per Zn or Cu atom is 2, Gibb's energy is evaluated by

DGo = - 2*96485*1.10 J
= 212.2 kJ

DISCUSSION
Gibb's energy is the maximum electric energy derived from a battery.

### Summary

Enthalpies and entropies are important thermodynamic data, and their values have been tabulated due to their requirement for a wide variety of applications. For example, you can apply the Hess's law to derive values for. reactions whose changes of enthalpy, entropy or both cannot be measured. The application of thermodynamic data is expanded due to the introduction of the Gibb's energy, because of its link to the equilibrium constant, K.

In answering the question "How does energy affect the direction and extend of a reaction?", we have introduced the equilibrium constant, K, and its relationship with Gibb's energy, G, entropy S at temperature T. As a summary, here are the key relationships:

DGo = - R T ln K
DGo + R T ln K = 0
DG = DGo + R T ln Q
DG = DH - T DS
G = H - T S
DG = - n F E

These key relationships enables us to derive many other forms, which illustrate how nature behave. Some simple rules of thumb are

Spontaneous reactions (Q/K < 1): G < 0 and/or S > 0
Equilibrium conditions (Q/K = 1): G = 0 and/or S = 0
(Spontaneous reverse reactions (Q/K > 1): G > 0 and/or S < 0)

In other words, Gibb's free energy tends to decrease and entropy tends to increase in a spontaneous reaction.

Three more useful rules have been used in the examples. They are:

DHoreaction = SHof(products) - SHof(reactants)
DSoreaction = SSof(products) - SSof(reactants)
DGoreaction = SGof(products) - SGof(reactants)

State functions DHo, DSo and DGo have similar properties, and the above relationship is due to the conservation of energy.

## Questions

1. What is the standard enthalpy of formation for H?

2. The facts that DHo is 0 and 217.998 kJ/mol for H2 and H respectively allows you to evaluate the bond energy for H-H. What is the bond energy for H-H?

3. The equilibrium constant for the reaction at 298 K,

CH3COOH(aq) + H2O -> CH3COO-(aq) + H3O+

is 1.74*105 M. Calculate the value of DHo.

4. What data are required in order to calculate the DGo for the reaction

Hg(l) + S(s) -> HgS(s)

5. The reaction between hydrogen and fluorine gives a flame reaching a temperature of 6000 K. What is the standard enthalpy of reaction for

H2(g) + F2(g) -> 2 HF(g)

6. Evaluate the DGof for the reaction:

H2(g) + F2(g) -> 2 HF(g)

7. Is the reaction

H2O(l) -> H2O(g)

spontaneous at 298 K?

## Solutions

1. 217.998 kJ/mol

Skill -
Search the thermodynamic data table.

2. 534.9 kJ/mol

Skill -
Explain the meaning of thermodynamic data. The standard enthalpy of formation given above is for the reaction
H2(g) -> 2 H(g), DH = 534.9

3. 27.2 kJ/mol

Skill -
Choose the appropriate formula (DHo = - R T ln K) for the problem.

Discussion -
A positive value for G indicates that the reverse reaction is spontaneous. At standard condition, {H+} = 1.0. The anions is protonated.

4. Standard entropies for Hg(l), S(s), and HgS(s)

Results
So are 75.90, 32.05, ??
DHo for HgS(s) = ??
The signs ?? indicate no data from Thermodynamic Data Table.

Discussion
Sulfur is usually used to cover spilled mercury base on the reaction given in this question. What is the sign of DGo for the reaction? (Using data from other sources gives DGof = -48.8 kJ/mol)

5. (2 mol)*(-273.30 kJ/mol)

Skill -
Use the proper data from the Thermodynamic Data Table, DHof = -273.30 kJ/mol for HF.

6. -550.8 kJ/mol

Skill -
Recognize an intermediate step in the calculation:
DSof
= 2*173.779 - (130.680 + 202.791) J/(mol K)
= 14.078 J/(mol K)

7. Hint: ??

Discussion
Support your argument by calculating DGoreaction