CHEM 2011.03 Introduction to Thermodynamics Fall 2003

Gibbs Free Energy Changes and Chemical Equilibrium Constants

For chemists, the most important use of the Gibbs free energy (which I shall usually shorten to

"Gibbs energy") is due to its close connection to chemical equilibrium constants. Here we will look into

the thermodynamic foundation of that connection. We will use the standard enthalpy and standard

entropy changes for a reaction to compute the standard Gibbs energy change; that is, the Gibbs energy

change if all the reactants and products were in their standard states. The actual Gibbs energy change for

a reaction is generally different from the standard Gibbs energy change since the reactants and products

are rarely in their standard states. We will show that when the reactants and products are in equilibrium

at constant temperature and pressure, the Gibbs energy change for the reaction is zero. We will learn

how the Gibbs energy change depends on the concentrations of the reactants and products. This will

enable us to predict the conditions for which a reaction is at equilibrium. Specifically, we will obtain the

expression for the equilibrium constant and a method to compute the equilibrium constant from

thermochemical data. From the temperature dependence of the Gibbs energy, we will deduce the

temperature dependence of equilibrium constants. Finally, we will use the relationship between Gibbs

free energy and non-expansion work to understand the voltages produced by electrochemical cells.

Throughout the following we shall assume that the chemical reaction takes place at constant

temperature and pressure. This is because these conditions are used most often in practice. We could, if

we wished, derive similar results for other sets of constraints (such as constant temperature and volume).

But it turns out that once we derive the equilibrium conditions for constant temperature and pressure, it

is a fairly straight-forward matter to determine how they vary with temperature and pressure. So the

results derived here are useful even if temperature and pressure are not constant.

1. The standard Gibbs free energy of reaction

We start with the definitions of Gibbs free energy:

, (1.1)

the standard enthalpy change of a chemical reaction:

, (1.2)

and the standard entropy change of a chemical reaction:

(1.3)

where the sums are over all products and all reactants. In these equations, and are the standard

molar enthalpies and entropies of the various reactants and products and the <i and <j are the

stoichiometric coefficients of the products and the reactants. One mole of reaction consumes <j moles of

each reactant and produces <i moles of each product. For example, in the reaction

N2(g) + 3H2(g) W 2NH3(g)

the stoichiometric coefficients of N2(g), H2(g), and NH3(g) are respectively 1, 3, and 2. One mole of

reaction consumes one mole of N2(g) and three moles of H2(g) while producing two moles of NH3(g).

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Note that standard enthalpy, entropy, and Gibbs energy changes are always per mole of any reactant or

product that has a stoichiometric coefficient of unity (that is, <=1).

We now define the standard Gibbs energy of reaction as

(1.4)

From the definition of Gibbs free energy (equation 1.1), we have for each individual species

. (1.5)

By substituting equation (1.5) into equation (1.4) for each species, and using equations (1.2) and (1.3),

we obtain

. (1.6)

This is just what we might expect to get from equation (1.1) for a change in G in a process at constant

temperature.

Equation (1.6) is our first important result. It tells us how to calculate the standard Gibbs energy of

reaction from the standard enthalpy and entropy changes. In computing the standard enthalpy change we

use the enthalpies of formation since we do not know absolute enthalpies. This is permissible since the

zero point chosen for enthalpy does not affect the change in H during a reaction. Since we can determine

absolute entropies, we use them in computing the standard entropy of reaction.

Because absolute values of enthalpies can not be fixed, we also can not fix absolute values of Gibbs

energies. Therefore, instead of absolute Gibbs energies we used standard Gibbs energies of formation in

equation (1.4). In other words, we replace equation (1.4) with

. (1.7)

The )fG° are defined to be zero for elements in their reference states, just like standard enthalpies of

formation. This is equivalent to adding a constant to Gm° for each element. The constants cancel out

when changes in Gibbs energy are calculated; so they don't affect the Gibbs energy of reaction.

Strictly speaking, the molar quantities in equations (1.2), (1.3) and (1.4) are partial molar quantities.

Partial molar quantities are a generalization of molar quantities that are suitable for use with non-ideal

solutions. For ideal solutions, ideal gases, pure solids, and pure liquids, molar quantities and partial

molar quantities are identical. The arguments presented here are identical whether we use molar or

partial molar quantities. It should also be remembered that the partial molar Gibbs free energy is

indentical with the chemical potential.

Example 1.1 - Using entropy and enthalpy data to compute a free energy of reaction

Consider the reaction used to manufacture ammonia:

N2(g) + 3H2(g) W 2NH3(g). (1.8)

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In tables of thermodynamic data (such as Tables 2.5 and 2.6 in Atkins), we find that the standard

enthalpy of formation of NH3(g) at 298.15 K is -46.11 kJ mol-1 and the standard absolute molar entropies

for N2(g), H2(g), and NH3(g) are 191.61 J K-1 mol-1, 130.684 J K-1 mol-1, and 192.45 J K-1 mol-1,

respectively. Since the reaction forms two moles of NH3(g) and the reactants are elements in their

reference states, we have

)rH° = 2)fH°(NH3(g)) = 2(-46.11 kJ mol-1) = -92.22 kJ mol-1.

However, in computing )rS° the absolute entropies of all the species are non-zero; thus,

)rS° = (2(192.45) - 191.61 - 3(130.684)) J K-1 mol-1

)rS° = -198.76 J K-1 mol-1

Now we can compute the Gibbs energy change using equation (1.6):

)rG° = -92.22 kJ mol-1 - (298.15 K)(-198.76 J K-1 mol-1)(10-3 J kJ-1)

)rG° = -32.96 kJ mol-1

Note that we must be very careful about units in these calculations since enthalpies normally use

kilojoules while entropies usually use joules. Since our example reaction forms two moles of NH3(g)

from its elements in their reference states, the standard Gibbs free energy of formation of NH3(g) is one

half the standard Gibbs free energy change of this reaction.

2. The condition for equilibrium in terms of the Gibbs energy change

In the following argument, it will help to refer to a particular reaction. Consider the reaction:

CO(g) + H2O(g) W CO2(g) + H2(g). (2.1)

The equilibrium for this reaction lies to the right, but measureable amounts of the reactants remain at

equilibrium. Imagine that we start with a system containing a mixture of CO(g) and H2O(g); since these

react very slowly at room temperature, we can easily mix them. Then we add a catalyst that enables the

reaction to proceed to equilibrium spontaneously and irreversibly. As the reaction occurs, we exchange

heat and work with the surroundings so that the temperature and pressure remain constant. We assume

that the surroundings are at the same temperature as the system and that the transfers of heat and work

are reversible. In other words, the only irreversible part of the process is the chemical reaction itself.

Let q be the heat added to the system. The heat transfer is reversible and isothermal, thus for the

surroundings

. (2.2)

The chemical reaction is irreversible. Therefore, from the Second Law we have for the system plus the

surroundings (the universe),

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. (2.3)

Since the process takes place at constant pressure, the enthalpy change is equal to the heat added.

Therefore, making use of equation (2.2), we have

. (2.4)

Since the process is isothermal, we have

so we can use equation (2.4) to write the change in Gibbs energy of the system as

. (2.5)

If we substitute equation (2.3) into (2.5) we see that

. (2.6)

In other words, the Gibbs energy of the system decreases.

Although we used a particular reaction as an example, nothing in this derivation is specific to any

particular reaction. We can therefore conclude that equation (2.6) applies to any spontaneous reaction at

constant temperature and pressure. In fact, this is an example of a more general principle:

In any spontaneous process at constant temperature and pressure, the Gibbs free energy of the

system decreases.

Since this is a consequence of the increase in the entropy of the universe during a spontaneous process, it

is, in effect, just another statement of the Second Law of Thermodynamics. A spontaneous process is

one that proceeds towards equilibrium. In other words, the sign of the Gibbs energy change for a reaction

tells us whether it will go in the forward direction or the reverse direction. Here we must be careful to

distinguish between the Gibbs energy change and the standard Gibbs energy change; it is the former that

tells us the direction in which a reaction proceeds under a specific set of conditions. Gibbs free energy

depends on concentration and so the Gibbs energy of reaction changes as the reaction proceeds. The

standard Gibbs energy change is for one particular set of conditions for the reactants and products; we

shall soon see that it is closely related to the value of the equilibrium constant.

It turns out that reaction (2.1) does not go to completion; significant concentrations of both reactants

and products are present at equilibrium. We could have formed our system using CO2 and H2. Then the

addition of the catalyst would cause the reaction to proceed in the reverse direction towards equilibrium.

This would also be a spontaneous process and would also be accompanied by a decrease in the Gibbs

energy of the system. Since Gibbs energy decreases as we approach equilibrium from either direction, it

must be a minimum at the position of equilibrium. Another way of saying this is that a system is at

equilibrium only if no spontaneous change can occur; that is, only if G can not decrease. But if G can not

decrease, it must be at a minimum. This gives yet another way of stating of the Second Law:

At constant temperature and pressure, the Gibbs free energy of the system is a minimum at

equilibrium.

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There is a huge advantage in restating the Second Law in terms of Gibbs free energy. We now only

need to know the properties of the system in order to determine whether a process is spontaneous. This is

in contrast to what we do when we use the entropy to assess spontaneity; then we must also consider the

entropy change of the surroundings. The advantage of using Gibbs energy comes with a price; we can

only use the Gibbs energy to assess spontaneity if the process takes place at constant pressure and

temperature. This condition is not quite as restrictive as it first seems. Since Gibbs energy is a state

function, it is sufficient to have the final state at the same pressure and temperature as the initial state.

In order to apply these results to chemical reactions, we must be able to compute the Gibbs energy

change for any combination of concentrations of reactants and products. Formally, we can write out an

equation like equation (1.4) but with the standard molar Gibbs energies replaced by molar Gibbs

energies under the conditions for the reaction; this gives

. (2.7)

The difference between equations (1.4) and (2.7) is that the former is the Gibbs energy change when all

reactants and products are in their standard states (for example, each gas at a partial pressure of one bar)

while the latter is for the actual conditions under which the reaction occurs.

The use of molar Gibbs energy is only correct when we have a system that consists only of ideal

gases, pure liquids, and pure solids. We now introduce a change in notation to put this in a form in

which it applies to solutions, including non-ideal solutions. To do this, we replace the molar Gibbs free

energy with the chemical potential, :; defined by

. (2.8)

The chemical potential is the partial molar Gibss free energy; for ideal gases, pure liquids, and pure

solids it is the same as the molar Gibbs free energy. But the chemical potential is applicable in all

systems, including liquid solutions and mixtures of non-ideal gases. Thus, the chemical potential can be

thought of as being a generalization of the molar Gibbs free energy.

When we use :i instead of Gm,i, equation (2.7) is written

(2.9)

and equation (1.4) is written

(2.10)

where the various :i° and :j° are the standard chemical potentials of the reactants and products. These

are just the chemical potentials of the species in their standard states. It is important to realize that the :i°

correspond to the rather than the . Because of this, we can not determine absolute values of the

chemical potentials; we will always express the chemical potential of a species in terms of its difference

from the standard chemical potential for that species.

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We are now in a position to take the next step in applying the Second Law to chemical reactions.

Equation(2.9) gives the Gibbs energy change per mole of reaction; that is, for an amount of reaction that

produces one mole of any product that has a stoichiometric coefficient of unity. Now imagine that a

reaction proceeds to an infinitesimal extent, dn. This reaction consumes <idn moles of reactant i and

produces <jdn moles of product j. The reaction has an infinitesimal change in Gibbs energy; we obtain

this by multiplying the Gibbs energy change per mole (equation 2.9) by the number of moles, dn, of

reaction that occurs. Thus,

. (2.11)

If the reactants and products are at equilibrium, then the Gibbs free energy is a minimum. Then for

an infinitesimal change from the equilibrium position we have dG=0. Thus, at equilibrium equation

(2.11) becomes

. (2.12)

Equation (2.12) is the condition for chemical equilibrium at constant temperature and pressure; it applies

only at equilibrium. This is our second important result; we can combine it with equation (2.10) to get a

more convenient form for the equilibrium condition. We shall do this in the following section.

Equation (2.12) is similar to, but more complicated than, other expressions for equilibrium that we

have seen. For example, the condition for thermal equilibrium between two systems is T1-T2 = 0, the

condition for mechanical equilibrium is p1-p2 = 0, and the condition for equilibrium of a substance

between two phases is :1-:2 = 0. Equation (2.12) is different only in that it involves the equivalence of

certain sums over the reactants and products rather than single quantities.

Example 2.1 - The direction of spontaneous chemical change

In example 1.1 we found a standard reaction free energy of -32.96 kJ mol-1 for reaction (1.8). From this

we can say that if we started with a mixture of the three gases each at a partial pressure of one bar, then

the direction of spontaneous change would be to form more of the product, NH3(g) since that would

reduce the Gibbs free energy of the system. As the reaction proceeds, the partial pressures of the

reactants and products will change from the standard pressure of one bar. As a result, )rG will change

from the standard value. Eventually the reaction will reach a point at which )rG = 0; then it will have

reached equilibrium and will not proceed any further. In the next section we will determine the point at

which this occurs.

3. The equilibrium constant for reactions of ideal gases

Equation (2.12) provides the condition for chemical equilibrium in terms of the chemical potentials

of the reactants and products. Equation (2.10) provides the standard Gibbs energy of reaction in terms of

the standard chemical potentials of the reactants and products. If we are dealing with gases, the former

set of chemical potentials are for the various species at their equilibrium partial pressures; while the

standard chemical potentials are for each species at the standard pressure (usually one bar). All are at the

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same temperature. To connect them, we need to discover how the chemical potentials vary with pressure.

This will tell us how the partial pressures of the reactants and products at equilibrium are different from

the standard pressure. The argument in the remainder of this section will be restricted to ideal gases. In

the following section we will see how to generalize the result to reactions involving other substances.

Our first step is to determine how the chemical potential changes with pressure; to do this we first

determine how enthalpy and entropy change with pressure. To compute the entropy change, we consider

the reversible, isothermal volume change; for ideal gases we get

. (3.1)

Therefore, the molar entropy change for this isothermal process is

. (3.2)

For the isothermal compression or expansion of an ideal gas, we have )H=0, therefore the molar free

energy change is

.

For ideal gases, the molar Gibbs free energy is the same as the chemical potential. Thus,

. (3.3)

It is usual to adopt the convention that the units of pressure are chosen so that the standard pressure

is numerically equal to unity. In other words, if the standard state is one bar, then we use bar as our unit

of pressure, but if the standard state is one pascal then we use pascal as the unit of pressure. Then if we

apply equation (3.3) to the change in chemical potential in going from the standard state, which has

chemical potential :° and partial pressure p°=1, to a state in which the gas has chemical potential : and

partial pressure p, we get

. (3.4)

We now use equation (3.4) for each species in the equilibrium condition, equation (2.12). The

subscripts on the pressures will indicate the individual reactants or products. For each species, the final

pressure is just the equilibrium partial pressure. Thus, the equilibrium condition becomes

.

The first two terms on the right hand side are just the standard Gibbs energy of reaction (see equation

2.10). In other words, this equation is just

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(3.5)

where I have put the subscript eq on the brackets to emphasize that this applies at equilibium. The

quantity in brackets can be simplified by recalling that a sum of logarithms is the logarithm of a product.

Just as the symbol G is used to indicate a sum of terms, the symbol A is used to indicate a product of

terms; that is

. (3.6a)

so that

. (3.6b)

We define the equilibrium constant, Kp, as

(3.7)

where the subscript eq on the right hand side is to remind us that this is for a system at equilibrium. To

illustrate this notation, for reaction (2.1) equation (3.7) becomes

while for reaction (1.8) we get

This is something that you should recognize from earlier chemistry courses.

We can rearrange equation (3.7) by taking the logarithm of both sides (as in equation 3.6b) to get

. (3.8)

The right hand side of equation (3.8) is just the quantity in brackets in the equilibrium condition,

equation (3.5). We can now substitute equations (2.10) and (3.8) into equation (3.5) to obtain the very

simple result

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. (3.9)

This is one specific example (for reactions of ideal gases) of the Lewis Equation relating the standard

Gibbs energy change of reaction and the equilibrium constant. We can easily rearrange it to get Kp as a

function of )rG°,

. (3.10)

We now have the following procedure for calculating equilibrium constants from thermochemical data:

(1) Calculate )rH° from the enthalpies of formation of the reactants and products (equation 1.2).

(2) Calculate )rS° from the standard molar entropies of the reactants and products (equation 1.3).

(3) Calculate )rG° from )rG° = )rH° + T)rS° (equation 1.5).

(4) Calculate Kp from equation (3.10). The units of pressure used in Kp are those in which the pressure

used to define the standard states are unity. In other words, if the standard states are defined at a pressure

of one bar, then the pressures used in equation (3.7) should be in bar.

If standard Gibbs energy of formation data are available for all species in the reaction, we could use

them in equation (1.4) to compute )rG° and skip steps (1) and (2). But we can only do this to obtain

)rG° at the temperature for which the data are tabulated. Although it is often a reasonable approximation

to assume that )rH° and )rS° are independent of temperature, it is never reasonable to assume that )rG°

is independent of temperature.

Example 3.1 - Calculation of an equilibrium constant

Here we shall calculate the equilibrium constant for reaction (1.8) at 25° C. In example 1.1, we have

already carried out the first three steps of the above procedure and obtained )rG° = -32.96 kJ mol-1. Now

we use equation (3.10) to get Kp (being careful about joules and kilojoules):

Since )rG° < 0, we get Kp > 1. Here I have added units of bar-2 to remind us that the partial pressures

must use units of bar since the standard states were for a pressure of one bar.

4. Generalization of the equilibrium constant

The derivation in section 3 applies only to ideal gases. For reactions involving solids, liquids, real

gases, or species in solution we need a more general result. We accomplish this by defining the general

equilibrium constant, Keq, so that the Lewis equation holds for all reactions. This is done by defining a

quantity called the activity, ai, of species i so that the equation

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(4.1)

always applies. Notice that equation (4.1) is very similar to equation (3.4).

Equation (4.1) does not give us any physical sense of what the activity is; it just replaces one

unfamiliar quantity, the chemical potential, with new unfamiliar quantity. We can better understand the

activity by first thinking about what the chemical potential represents. Qualitatively, we can think of the

chemical potential as a measures of how chemically active a substance is. For example, the most stable

phase of a pure substance is the one with the lowest chemical potential. We can only have equilibrium

between two phases if they have the same chemical potential; otherwise, the phase with the higher

chemical potential changes into the phase with the lower chemical potential. Similarly, from equation

(2.9) we see that increasing the chemical potential of reactants increases their tendency to turn into

products and increasing the chemical potential of products increases their tendency to turn into reactants.

The activity is also a measure of how chemically active a substance is. From equation (4.1), we see

that an increase in the chemical potential, :i, implies an increase in the activity, ai. So the qualitative

meaning of activity is the same as the qualitative meaning of chemical potential; if we use the same

standard state (that is, the same value of :i°) for a substance no matter what phase it is in, then the most

stable phase is the one with the lowest activity and we can have two phases in equilibrium only if the

species has the same activity in both phases. Also, increasing the activities of the reactants will cause a

chemical reaction to proceed in the forward direction and increasing the activities of the products will

cause it to proceed in the reverse direction.

So why use the activity instead of the chemical potential? The answer is that the activity turns out to

be more conveniently related to familiar quantities such as partial pressure, mole fraction, and

concentration. For example, if you compare equation (4.1) to equation (3.4) you will see that for ideal

gases the activity must be proportional to the partial pressure of the gas. On the other hand, the chemical

potential depends on the logarithm of the partial pressure; that is much less convenient. We shall see that

in ideal solutions, the activity is directly proportional to mole fraction and in dilute solutions activity is

directly proportional to concentration. Thus, activity has a closer relation to the quantities that we use to

describe composition while chemical potential has a closer relation to fundamental thermodynamic

quantities. Equation (4.1) provides the link between the two.

Now we will learn how to write the chemical equilibrium constant in terms of activities and how to

relate activities to measures of composition for various systems. In section 2 we worked out expressions

for the Gibbs energy change per mole (equation 2.9) and the standard Gibbs energy change per mole

(equation 2.10). The former is for reactants and products in arbitrary states; the latter is for reactants and

products in standard states. If we subtract equation (2.10) from equation (2.9) we get

.

This is very like equation (3.5). Now we can use equation (4.1) to replace each of the :i and :j; all of the

standard chemical potentials cancel out and we are left with

. (4.2)

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We now introduce the activity quotient, Q, which is defined by means of the equation

. (4.3)

This is very similar to the definition of the equilibrium constant (equation 3.7) for ideal gases. It differs

from equation (3.7) in two respects. First, it uses activities instead of partial pressure; we shall see that

the two are essentially the same for ideal gases. Second, equation (4.3) can be used for all conditions, not

just at equilibrium.

We proceed as in section 3 and take the natural logarithm of both sizes of equation (4.3). This gives

. (4.4)

Substituting this into equation (4.2) yields

. (4.5)

This is another form of the Lewis equation. We will use it in section 8 when we derive the Nernst

equation. However, the most commonly used form of this equation is for equilibrium conditions. As we

saw in section 2, the condition for equilibrium at constant T and p is that, for an infinitesimal amount of

reaction, )rG=0. Then equation (4.5) becomes

. (4.6)

Where Qeq is the equilibrium value of the activity quotient; it has a unique value at any given

temperature since the standard free energy change has a unique value at any given temperature. To

emphasize the uniqueness of this particular value of the activity quotient we call it the equilibrium

constant and give it its own symbol, Keq . It is defined in terms of activities by the equation

(4.7)

where the subscript eq on the brackets indicates that the quantity is to be evaluated at equilibrium. This

is just like equation (3.7) except that we have replaced the partial pressures with activitites. In terms of

the equilibrium constant, equation (4.6) is written

. (4.8)

This is the Lewis equation for a general equilibrium constant, Keq. Equation (3.9) was a specific example

of this for the case of ideal gases.

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Qualitatively, we can think of the activity as measuring how chemically active a substance is. The

higher the activity of a substance, the greater is its tendency to change into something else. From

equation (4.7) we see that increasing the activities of the reactants causes the position of equilibrium to

shift towards the products; so the reactants have a greater tendency to turn into products. Similarly,

increasing the activities of the products gives them a greater tendency to turn into reactants.

To use these equations we must be able to compute the activities. By comparing equations (4.1) and

(4.7) with equations (3.4) and (3.7), we see that

for ideal gases (4.9)

where pi° is the standard pressure. Normally, we choose our units of pressure so that the standard

pressure is unity, then ai is numerically equal to pi. We may, if we choose, put units on the activity to

remind ourselves of which standard state was used (that is, we use give the activity of a gas units of bar

if the standard state is one bar).

Other simple cases are those of pure solids and pure liquids. These are in their standard states when

the total pressure is equal to the standard pressure. If a substance is in its standard state, then :=:°.

Thus, from equation (4.1), we see that ai=1 for any substance in its standard state. If the total pressure is

different from the standard pressure, then the activity will be slightly different from unity since the

chemical potential, :, will be slightly different from its standard value. However, for liquids and solids :

depends only very slightly on pressure, typical changes are of the order of a joule per mole for each one

bar change in pressure (gases have a much stronger dependence of : on pressure). It is rare that we will

need to worry about errors in the chemical potential of less than 100 J mol-1. Thus, unless we are

working at very high pressures, we can usually ignore the pressure dependencies of : and a for any

substances other than gases. Then we can make the approximation

for pure solids and pure liquids. (4.10)

But we must remember that this is only a good approximation if the pressure is not too high. The

activities of pure solids and pure liquids are always dimensionless.

Of course, the most interesting case for chemists is species in solution. There is no universal formula

for activities of species in solution. To understand why, we must look carefully at the thermodynamics of

solutions; we will do this when we get to chapter 7 of Atkins. It turns out that there are two cases that are

simple: ideal solutions and dilute solutions. The results for these cases are given here; they be justified

when we get to Chapter 7.

For ideal solutions, the activity is proportional to the mole fraction of the species in solution. The

constant of proportionality depends on the choice of standard state. If the standard state is taken to be the

pure liquid and if the mole fraction of substance i is Xi, then we have

ideal solution, pure liquid standard state. (4.11)

This gives an activity of unity for the pure liquid, in agreement with equation (4.10). Since mole

fractions are dimensionless, activities using this standard state are also dimensionless. If the standard

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state is taken to be the vapor and if the vapor can be treated as an ideal gas, then the activity is defined in

terms of the vapor pressure, pi*, of the pure liquid. In this case we have

ideal solution, ideal gas standard state. (4.12)

In this case we give the activity the same units that we would use for the gas since we are using the same

standard state as for the gas. These are not the only choices of standard state, but they are the most

commonly used ones for ideal (or nearly ideal) solutions. The choice of standard state can be made

entirely on the basis of convenience. However, once a choice is made for a particular species it must be

used consistently for that species. For example, if the pure liquid standard state is used to compute )rG°

by means of equation (1.4) then, for that species, the activity must be defined in terms of the pure liquid

standard state.

For sufficiently dilute solutions the activity of species i is equal to its concentration in solution.

Strictly speaking, the molal concentration, bi, should be used; but for aqueous solutions we can usually

approximate the molal concentration with the molar concentration when the solution is dilute. When we

define the activity in this manner we say that we are using a dilute solution standard state, also called a

solute standard state. Thus we have

solute standard state. (4.13)

In this case we may, if we choose, give the activity units of mol kg-1 (the units of molal concentration) to

indicate the standard state used. Note that equation (4.13) can only be used for the solute in a dilute

solution. When we study the properties of solutions, we shall see that if a solution is dilute, then the

solvent behaves as if the solution were ideal; thus, either equation (4.11) or (4.12) can be used for the

solvent.

How dilute does a solution have to be to use equation (4.13)? The answer depends on the nature of

the solution and on how accurate a result is needed. For ideal solutions, equation (4.13) can be used at

any concentration. For solutions of univalent electrolytes (such as NaCl or HNO3) in water, equation

(4.13) is in error by about 10% for a concentration of only 0.01 mol kg-1. The errors are even larger when

divalent or trivalent ions are involved. However, the deviations from ideal behavior do not increase

uniformly with increases in concentration; in fact, the deviations usually decrease at higher

concentrations. As a result, if one is willing to accept errors of a factor of two or three, one can usually

assume dilute solution behavior even for fairly high concentrations. If we want more accurate results, we

must use measured activity data on the specific solutions of interest. We will learn more about the

sources of such data when we look into the thermodynamics of solutions in detail.

Example 4.1

Write an approximate expression for the equilibrium constant, KW, for the auto-ionization of pure

water in terms of molar concentrations.

Solution:

The auto-ionization reaction is

H2O(l) W H+(aq) + OH-(aq).

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From equation (4.7) we have for this reaction

.

We know that in pure water the concentrations of H+ and OH- are of the order of 10-7 M, so we can

reasonably assume dilute solution behavior. We can also assume that molarity and molality are

essentially the same. Using the solute standard states for the ions, we have

and

For the solvent, water, we use equation (4.11). Thus, if the water is very nearly pure

.

The expression for the equilibrium constant then becomes

.

This is a very good approximation since the solution is so dilute.

5. Some examples of using the Lewis equation

In these examples we will use the thermodynamic data in Tables 1 and 2 below. Tables of similar

data can be found in the Data Section in Atkins (Tables 2.5 and 2.6), in the Handbook of Chemistry and

Physics, and in many other sources.

Example 5.1 - Calculating an Equilibrium Constant

Use the data in Table 1 to calculate , , , and Kp at 298.15 K for the reaction

CO(g) + H2O(g) W CO2(g) + H2(g)

Solution:

We follow the four step procedure outlined at the end of section 3, being careful about J and kJ.

(1) Calculate from the enthalpies of formation of the reactants and products:

Since )rH° is negative, the reaction is exothermic; that is, heat is released by the reaction.

(2) Calculate )rS° from the entropies of the reactants and products:

15

Table 1: Some thermochemical data at 298.15 K for ideal gas standard states at one bar.

)fH° (kJ mol-1) )rG° (kJ mol-1) Sm° (J K-1 mol-1)

Cl2(g) 0 0 223.07H2(g) 0 0 130.684N2(g) 0 0 191.61O2(g) 0 0 205.138CO(g) -110.53 -137.17 197.67CO2(g) -393.51 -394.36 213.74H2O(g) -241.82 -228.57 188.83NH3(g) -46.11 -16.45 192.45CH3CHO(g) -166.19 -128.86 250.3

(3) Calculate )rG° from )rG° = )rH° - T)rS°:

We could also have computed )rG° directly from the data in Table 1 in the same way that we computed

)rH° and )rS°.

(4) Calculate Kp from equation (3.10):

.

Since the standard states are at one bar, the units of pressure used in Kp are bar. However, since all of the

units of pressure will cancel out in this case, it does not matter what units we use for pressure in applying

Kp for this particular reaction.

Example 5.2 - Calculating thermochemical data from equilibrium constants

The equilibrium constant, Kp, for the reaction

H2(g) + Cl2(g) W 2HCl(g)

16

Table 2: Some thermochemical data at 298.15 K for pure liquid or pure solidstandard states at one bar.

)fH° (kJ mol-1) )fG° (kJ mol-1) Sm° (J K-1 mol-1)

H2O(l) -285.83 -237.13 69.91Br2(l) 0 0 152.23Hg(l) 0 0 76.02CCl4(l) -135.44 -65.21 216.40CH3COOH(l) -484.5 -389.9 159.8C(s, graphite) 0 0 5.740Ag(s) 0 0 42.55AgBr(s) -100.37 -96.90 107.1AgCl(s) -127.07 -109.79 96.2HgCl2(s) -224.3 -178.6 146.0

is 2.48×1033 at 298 K and the standard molar entropy of HCl(g) is 186.91 J K-1 mol-1 at 298 K. Use these

data together with data in Table 1 to calculate )fG°(HCl(g)) and )fH°(HCl(g)).

Solution:

We start with equation (3.9)

Since this reaction represents the formation of two moles of HCl(g) from the elements in their reference

states, we have

)fG°(HCl(g)) = ½)rG° = -95.3 kJ mol-1.

Now to find )fH°(HCl(g)), we rearrange equation (1.6) to get

where we have changed the subscripts to indicate that the reaction is the formation of one mole of

HCl(g) from its elements (one half of the reaction as written above). Thus, using data from Table 1 and

the given entropy of HCl(g), we have

Thus,

17

Example 5.3 - Equilibrium involving condensed phases

Use the data in Tables 1 and 2 to calculate the equilibrium constant at 298 K for the reaction

C(s,graphite) + 2Cl2(g) W CCl4(l)

and use the equilibrium constant to determine the conditions that would obtain at equilibrium if Cl2(g) is

allowed to react with an excess of graphite.

Solution:

The reaction forms one mole of CCl4(l) from its elements in their reference states, so the standard Gibbs

energy of reaction is equal to the Gibbs energy of formation of CCl4(l). Thus, )rG° = -65.21 kJ mol-1 and

equation (4.8) gives

.

At equilibrium, this is equal to the activity quotient for the reaction:

.

The activity of graphite, aC, is unity since it is a pure solid. The activity of CCl4(l) will be equal to its

mole fraction in the liquid phase; this will be somewhat less than unity, depending on the amount of Cl2

that dissolves in the liquid at equilibrium. For the moment, let's assume that the mole fraction of

dissolved Cl2 is small so that aCCl4 . 1. If we treat Cl2(g) as ideal, then aCl2 = pCl2. So at equilibrium we

have

Since this partial pressure of Cl2(g) is so low, our assumptions that the gas is ideal and that the mole

fraction of dissolved gas in the liquid is small are both probably valid. The reaction will proceed until

either all the graphite is consumed or until the partial pressure of Cl2(g) has dropped to 1.94×10-6 bar,

which ever comes first.

6. Temperature dependence of the equilibrium constant

It is a common observation that changes in temperature change the position of chemical equilibria.

An everyday example is the evaporation of water and other liquids; higher temperature shifts the

equilibrium towards the vapor and away from the liquid. But the rules for how equilibria change with

18

temperature are not always so simple. For example an increase in temperature increases the solubility of

some salts in water but reduces the solubility of other salts in water. Here we will derive the general

expression for the temperature dependence of equilibrium constants.

Our starting point is the Lewis equation (equation 4.8):

. (6.1)

If we can determine the temperature dependence of )rG° then we can discover the temperature

dependence of Keq. To do this, we use the fundamental equation written in terms of the Gibbs energy:

(6.2)

from which we can immediately obtain

. (6.3)

We can apply this to each term in equation (1.4) and obtain

(6.4)

where equation (1.3) has been used to get the second equality. If we rearrange equation (1.6) we get

.

Substituting this into equation (6.4) yields

. (6.5)

The next step requires a trick. Instead of seeking the temperature dependence of )rG° we seek the

temperature dependence of ()rG° /T). From basic calculus, we can write

. (6.6)

Now we substitute equation (6.5) into equation (6.6) and end up with

. (6.7)

This is the Gibbs-Helmholtz equation written for a chemical reaction. The Gibbs-Helmholtz equation

can also be written for the temperature dependence of )G in any process.

19

Now we use the Lewis equation (6.1) to substitute for )rG° in the left hand side of equation (6.7).

The result is

. (6.8)

As explained in the next section, we account for the effect of pressure on chemical equilibria by taking

pressure into account in computing the activities. Then the equilibrium constant does not depend on

pressure; so it does not matter if we keep pressure constant when we change the temperature. Then

equation (6.8) can be written more succinctly as

. (6.9)

We see that if the reaction is endothermic ()rH° < 0), then Keq increases when the temperature increases

while if the reaction is exothermic ()rH° > 0), then Keq decreases when the temperature increases.

Equation (6.9) is the differential form of the van't Hoff equation. It is a perfectly general expression

for the temperature dependence of an equilibrium constant. It is a direct consequence of applying the

laws of thermodynamics to the definition of the equilibrium constant given by equation (4.7). It does not

depend upon any approximations or assumptions about ideal behavior. However, in applying equations

(6.9) and (4.7) we often do need to make assumptions and approximations.

If we know how )rH° dependends on temperature, then we can integrate equation (6.9) to get Keq as

a function of temperature. This is easiest if )rH° is independent of temperature. If we assume that this is

so, then we can rewrite equation (6.9) as

(6.10)

where K1 and K2 are the equilibrium constants at temperatures T1 and T2, respectively. The integrals are

easily carried out to give

. (6.11)

This is the integral form of the van't Hoff equation which you may remember from introductory

chemistry. It is only valid if )rH° does not change much between the temperatures T1 and T2. This will

only be strictly true if the heat capacity of the reactants is identical to the heat capacity of the products. In

practice, equation (6.11) can usually be used over temperature ranges of a few tens of degrees even if

)rH° does change with temperature. As temperature changes get larger, the validity of equation (6.11)

20

gets more and more suspect. For large temperature changes, heat capacity data can be used to determine

how )rH° changes with temperature, but we will not go into that level of detail here.

One consequence of equation (6.11) is that a plot of ln(Keq) versus 1/T (called a van't Hoff plot) will

often give a straight line. If )rH° does not change with temperature, the slope of the line in a van't Hoff

plot is -)rH°/R. However, getting a straight line in a van't Hoff plot does not mean that )rH° is

independent of temperature. This is because when )rH° does depend on T, the main effect is to change

the slope of the line rather than causing it to become a curve. In such a case, the slope of the line is

approximately equal to the value -)rH°/R at the midpoint of the range of temperatures used.

Example 6.1

Estimate the equilibrium constant and )rG° for the reaction in Example 5.1 at a temperature of

250° C.

Solution:

Since we are not given any information on the temperature dependence we have little choice but to

assume that )rH° is independent of temperature. Fortunately, since the both the reactants and the

products consist of two moles of gas, we might reasonably expect that the heat capacities are not much

different and that )rH° does not change much with temperature. Assuming that )rH° is constant, we use

the integral form of the van't Hoff equation to calculate Keq at the higher temperature. Taking T1=298 K

and T2=523 K we have

Since the reaction is exothermic, the equilibrium constant is smaller at the higher temperature. To get

)rG°(523 K) we use the Lewis equation to get

Since we are assuming that )rH° is independent of temperature, we could have approached this

another way. )rH° will be independent of temperature only if )rCp° is independent of temperature. If

that is so, then )rS° will also be independent of temperature. Then we can get )rG°(523 K) by

evaluating )rG° = )rH° - T)rS° at 523 K. Thus,

and, from the Lewis equation,

21

.

Notice how the values vary slightly due to slightly different rounding errors.

7. Pressure dependence of chemical equilibria

In principle, we could approach this in a manner very similar to that used for the temperature

dependence. By differentiating the fundamental equation we find that

. (7.1)

In other words, the change in molar Gibbs energy with pressure is given by the molar volume of the

species. One consequence of this can be seen without any further mathematics: when the pressure is

increased the position of equilibrium changes to favor the side of the reaction with the smallest volume.

Therefore, for the reaction

N2(g) + 3H2(g) W 2NH3(g)

an increase in pressure shifts the equilibrium to the right since two moles of gas occupies less volume

than four moles.

In practice, we usually do not calculate the pressure dependence of equilibrium constants. Instead,

we take pressure into account in computing the activities that we use in the expression for the

equilibrium constant. This is what we did in section 3 when we discussed the equilibrium constant for

reactions of ideal gases.

It turns out that pressure has little effect on activities unless we are dealing with either gases or very

high pressures. As an example, let's see how a one bar (105 Pa) change in pressure changes the Gibbs

energy and activity of H2O(l). If we assume that the molar volume does not change much (a good

approximation for liquids and solids) then equation (7.1) yields

.

If we recall that the molar Gibbs energy is the same thing as the chemical potential and assume that we

started from the standard state, then we have :-:o = )Gm = 1.8 J. Substituting this into equation (4.1)

yields

.

In other words, the change from the standard state activity is less than 0.1% for each one bar increase in

pressure. Even at 100 bar the effect would be only 7%. So we usually ignore the effect of pressure on the

activities of liquids and solids, unless the pressure is very high.

For gases the effect of pressure on activity is much greater. For example, at 298 K, the vapor

pressure of H2O is 3170 Pa and one mole of vapor occupies a volume of 782 liters. Since this is a factor

22

of 4×104 greater than the volume of a mole of liquid water, the effect of pressure is a factor of 4×104

greater for the vapor. So we must account for it. This is what we did in section 3 where we wrote the

equilibrium constant for ideal gases in terms of the partial pressures of the gases. In section 4 we showed

that the activity of an ideal gas is directly proportional to its partial pressure (equation 4.9).

8. Electrochemical cells

Electrochemical cells are devices in which a chemical reaction can occur provided that electrons are

allowed to flow through an external connection between the two electrodes in the cell. Oxidation occurs

at one electrode (the anode) and reduction occurs at the other (the cathode). Electrochemical cells are of

practical importance as a means of storing electricity (batteries), as devices used in analytical chemistry

(pH electrodes are the most common example), and as a means of carrying out certain industrial

processes, such as electroplating. They are also a very important source of thermochemical data since

they provide a means of determining )G for reactions by measuring the voltage produced by an

electrochemical cell. Our interest here is in this last application.

A simple example of an electrochemical cell is a fuel cell; this is a device that produces electricity

directly from H2 and O2. The two electrodes, typically made of nickel, are immersed in water. H2(g) is

continuously bubbled over the anode where it is oxidized according to the reaction

H2(g) + 2OH-(aq) 6 2H2O(l) + 2e-. (8.1)

O2(g) is continuously bubbled over the cathode where it is reduced according to the reaction

½O2(g) + H2O(l) + 2e- 6 2OH-(aq) (8.2)

These two reactions can only occur if the electrons, e-, can flow freely through an external circuit

between the anode and the cathode. Since they can not occur in isolation, they are called half reactions.

The overall reaction that occurs is the sum of the two half reactions:

H2(g) + ½O2(g) 6 H2O(l). (8.3)

If we have pure water in the cell and if the H2 and O2 are supplied at pressures of one bar, then the

reactants and products are all in their standard states. Then )rG for the reaction is the same as )rG°.

From the data in Tables 2 and 3 we see that )rG° = )fG°(H2O(l)) = -237.13 kJ mol-1. Under standard

conditions, this reaction is spontaneous and may be used as a source of energy (more correctly, a source

of work). However, reactions are usually carried out under non-standard conditions. For example, the

gases may be provided at high pressure. Or the O2 might be provided in air at a total pressure of one bar

so that the partial pressure of O2 would be 0.21 bar. In such cases, )rG would be different from )rG°.

In the following, we shall see how to relate the voltage produced by an electrochemical cell (this

voltage is called the cell potential) to the Gibbs free energy change for a reaction. There is a standard cell

potential that is directly related to the standard free energy change. We shall see how the cell potential

varies with the activities of reactants and products. This permits us to use standard cell potentials to

calculate cell potentials under non-standard conditions. Standard cell potentials can also be determined

for half reactions, such as those given by equations (8.1) and (8.2); these are called half-cell potentials.

Tables of half-cell potentials can be used to calculate Gibbs energy changes and equilibrium constants

for any chemical reaction that can be written as a combination of the half reactions that are available.

23

We begin by showing that the Gibbs energy change is equal to the maximum non-pV work that can

be done by an electrochemical cell. The maximum work that can be done on the surroundings in

transferring an electrical charge, Q, through a change in electrical potential, E, is given by the product of

Q and E. The maximum electrical work, wE, done on the system is the negative of the work done on the

surroundings; thus,

. (8.4)

If Q is in coulombs and E is in volts, then wE is in joules. However, for chemical applications it is more

natural to express charge as the number of moles of electrons that are transferred. One mole of electrons

carries F coulombs of charge where F is Faraday's constant; its value is

(8.5)

where it is understood that this refers to moles of electrons. If n moles of electrons are transferred, we

can rewrite the maximum electrical work as

. (8.6)

This is the work that would be done in a reversible process.

We can relate this to the Gibbs energy change as follows. From the definition of Gibbs energy,

G = H - TS = U + pV - TS. (8.7)

Differentiating this yields

dG = dU + pdV + Vdp - TdS - SdT. (8.8)

If we consider a process at constant temperature and pressure and change from differentials to

differences we have

)G = )U + p)V - T)S. (8.9)

From the First Law, we have

)U = q + wpV + wE. (8.10)

Where wpV is the pressure-volume work. For a reversible process at constant temperature and pressure

we have q = T)S and wpV = -p)V. Substituting these into equation (8.10) gives us

)U = T)S - p)V + wE. (8.11)

If we substitute this into equation (8.9), we find that the change in Gibbs energy is simply

)G = -wE. (8.12)

This provides our relationship between the maximum non-expansion work (done on the surroundings)

and the Gibbs free energy change. If this is electrical work, then we can use equation (8.6) for wE.

If the Gibbs energy change is due to a chemical reaction and if we take < to be the number of moles

of electrons transferred per mole of reaction, then for one mole of reaction we replace n in equation (8.6)

with <. Then substituting equation (8.6) into equation (8.12) gives us the Gibbs free energy change of

reaction:

24

. (8.13)

If the system is at equilibrium, )rG=0 and E=0. We can not expect to get any work out of a system at

equilibrium.

If all the reactants and products are in their standard states, then we have a standard Gibbs energy

change and a standard cell potential, E°, which are related by

. (8.14)

In these equations, < is the stoichiometric coefficient of the electrons in the reaction. As an example,

consider the fuel cell reaction in equation (8.3). One mole of the reaction as written consumes one mole

of H2 and one half mole of O2. But two moles of electrons are transferred from the H2 to the O2 (see the

two half reactions, equations (8.1) and (8.2)). So for this reaction, the correct value to use for < in

equations (8.13) and (8.14) is <=2. If instead of equation (8.3) we had written the reaction as

2H2(g) + O2(g) 6 2H2O(l)

then we would be transferring four moles of electrons for each mole of reaction and we would have <=4.

As a result, )rG° would be twice as large as for reaction (8.3) although E° would be the same as for

reaction (8.3). Since the cell potential is for one mole of electrons, it does not depend on how we write

the reaction. But )rG° does depend on how we write the reaction, since it is for one mole of reaction.

To find how the cell potential varies with the activities of the reactants, we substitute equations

(8.13) and (8.14) into equation (4.5), which was one form of the Lewis equation. This yields

(8.15)

which can be rearranged to give

(8.16)

where Q is the activity quotient. This is known as the Nernst equation. You may have previously

encountered it with concentrations replacing activities in calculating Q. In this equation E° is the

potential of a cell in which all reactants and products are in their standard states. In such a cell, all the

activities are unity and therefore Q is also unity so that lnQ=0 and E=E°. In an actual cell some of the

species may not be in their standard states. We can compute the observed cell potential, E, of such a cell

from the activities of the reactants and products for the conditions that obtain in the cell. The activities

are used to compute Q which is then used in the Nernst equation to compute E. In practice, it is more

common to use measured cell potentials to determine Q. With suitably chosen cells, this can be used to

determine the activities of species in solution as a function of concentration or to measure the

concentrations of species in solution. A pH electrode is an example of the latter.

25

Example 8.1

Compute the standard cell potential for a fuel cell operating according to reaction (8.2) and compute

the actual cell potential when H2 is at a partial pressure of 5.0×10-7 bar (the partial pressure of H2 in air)

and O2 is at a partial pressure of 0.21 bar (the partial pressure of O2 in air).

Solution:

From the data in Tables 1 and 2 we obtain )rG° = )fG°(H2O(l)) = -237.13 kJ mol-1. Then by using

equation (8.14) with F from equation (8.5) and <=2 for the reaction as written, we have

To get the actual cell potential, we note that the activity of H2O(l) is unity since it is in its standard

state. From equation (4.9), the activities of H2(g) and O2(g) are 5.0×10-7 and 0.210, respectively. Thus,

we have for the activity quotient

Now we substitute into the Nernst equation and get the actual cell potential:

Thus we could, in principle, get work from the normal chemical components of air; in other words, the

chemical components of air are not at equilibrium. In practice, a ridiculous volume of air would be

needed to get a reasonable amount of work since the concentration of H2(g) in air is so small.

9. Summary

The key to applying thermodynamics to chemical reactions is to understand how to use the Gibbs

free energy. In section 1 we saw how to compute the Gibbs energy change in a chemical reaction. In

section 2 we learned how to restate the Second Law of Thermodynamics in terms of the Gibbs energy

change of a system; then we used this to get the condition for chemical equilibrium. We also introduced

the chemical potential as a sort of shorthand for the partial molal Gibbs energy. In section 3 we used the

condition for chemical equilibrium to derive the equilibrium constant for reactions of ideal gases. We

also derived the Lewis equation for this special case. In section 4 we introduced the activity as a way of

extending the results of section 3 to less simple systems. We then defined the activity quotient and the

equilibrium constant in terms of activities and derived the Lewis equation for a general equilibrium

constant. We also learned how to define activities for ideal gases, pure solids and liquids, and for the

components of ideal and dilute solutions. In section 5 we used these results to solve some simple

problems.

26

In section 6 we derived the differential and integral forms of the van't Hoff equation; these equations

give the temperature dependence of equilibrium constants. The differential form is general but the

integral form only applies over small temperature intervals or when the heat of reaction is independent of

temperature. In section 7 we saw that the effect of pressure on chemical reactions is usually negligible

except in the case of gases. For gases the effect is accounted for by the fact that the activity of an ideal

gas is directly proportional to its partial pressure. Finally, in section 8 we applied the results to

electrochemical cells and derived the relationship between the cell potential and the Gibbs energy of

reaction. This permitted us to express the standard cell potential in terms of the standard Gibbs energy of

reaction and to derive the Nernst equation for the dependence of the cell potential on the activities of the

reactants and products.

© Michael Mozurkewich, October, 2003