# LEARNING OUTCOMES

1. The students should be able to derive the equation for dependence of free energy change on the presurre of the gas.

2.The students should be able to carry out calculations of free energy change with change of presuure of gas at constant temperature.

# LESSON DEVELOPMENT

From the definitions of Gibbs energy, eq.(1.9),and the enthalpy(1st law of thermodynamics lesson 2), eq.(2.3) we have:

G=U-TS+PV (3.1)

In the differential form the equation is:

dG = dU - TdS - SdT + PdV + VdP (3.2)

From the first law we have dU = dq + dw. If the process is reversible, dqrev = TdS, and if dw is only pressure - volume work, dw = - PdV. Hence the change in internal energy can be written:

dU= TdS -PdV (3.3)

Since U, S and V are state functions, the changes dU, dS and dV are independent of the path. This neans that eq. (3.3) is valid for both reversible and irreversible processes. This equation combines the first and the second law, and it is called the fundamental equation for a closed system.

Combining eqs (3.2, 3.3) we obtain:

dG = - SdT + VdP (3.4)

The independent variables for G are T and P. The partial derivatives of G are:

(∂G/∂T)p = - S (3.5)

(∂G/∂P)T= V (3.6)

Equation (3.5) is an expression for the temperature dependence of Gibbs energy at constant pressure, while eq. (3.6) is an expression for the pressure dependence of Gibbs energy at constant temperature. We shall treat the pressure dependence first. From eq. (3.6) we can obtain the change in Gibbs energy for a substance, solid, liquid or gas, as a function of pressure at constant temperature. When the pressure of the substance changes from Po to P, the change in Gibbs energy is:

For solids and liquids the change in volume with pressure is very small, and we can write:

ΔG = VdP = V dP = V (P - Po) (3.8)

The volume of a solid or a liquid is very small compared to the volume of a gas, and the change in Gibbs energy with pressure is very small. Thus the Gibbs energy for a solid or liquid at a pressure P is practically equal to the Gibbs energy at Po, the standard state:

ΔGP,T ≈ ΔGPo,T = ΔGoT (3.9)

For a gas the change in Gibbs energy with pressure is significant. When changing the pressure from r to P, the change in Gibbs energy for 1 mole of an ideal gas is:

ΔG = VdP = RT dP/P = RT ln (P/Po) (3.10)

If the gas is one component, i, of a mixture of ideal gases, it contributes with a pressure Pi = niRT/V (see eq. (2.9). Since there is no force between the molecules of an ideal gas. we can use the equation also for a component:

ΔGi = RT ln (Pi/Po) (3.11)

We have seen earlier how we can find ΔrGo for a chemical reaction from the values of ΔrG° for roducts and reactants, eq. (2.3). Since ΔrG = ΣiViΔfGi we can use eqs (3.9, 3.10) to find ΔrG at a iven temperature when products and reactants are not in standard state. The mixture of products and reactants that represent the equilibrium situation is of particular lterest. The criterion for equilibrium given by eq. (1.11) is that dG = 0 for an infinitesimal change I the system. We shall investigate the condition that gives dG = 0 by means of an example.

### Example 1

A system consists of a mixture of CO(g), 02(g) and CO2(g). The equilibrium:

CO (g) + 1/2 O2 (g) = CO2 (g)

is represented by one particular mixture. We allow our system to undergo an infinitesimal change by a reaction according to the chemical equation. A small quantity of CO, dnco, reacts with a small quantity of 02' dnO2, to form a small quantity of CO2, dnCO2. The quantities can be expressed by the stoichiometric coefficients, Vi' and the extent of reaction:

dnco = Vco dξ = - dξ

dno2 = Vo2 dξ = - 1/2 dξ

dnCO2 = V CO2 dξ = dξ

The corresponding change in Gibbs energy is:

dG = ΣividξxΔfGi=dξ{ΔfG(CO2,g) - ΔfG(CO,g) - 1/2ΔfG(O2,g)}


The condition that dG = 0, that equilibrium is established, is:

dG/dξ ={ΔfG(CO2,g) - ΔfG(CO,g) - 1/2ΔfG(O2,g)}= 0


For a reaction in general we have:

dG =ΔrGxdξ={ΣΔfG(produces)-ΣΔfG(reactants)}dξ (3.12)

At equilibrium we have dG = 0, which implies that:

ΔrG = ΣΔfG(produces)-ΣΔfG(reactants) = 0 (3.13)

We shall use this equation as the equilibrium criterion for a chemical reaction.

Let us return to the example. For the mixture of CO(g), O2(g) and CO2(g) at arbitrary pressure we have:

ΔrG = ΔfG(CO2,g) - ΔfG(CO,g) - 1/2ΔfG(O2,g)

fG(CO2,g)= +ΔfG°(CO2,g) + RT In PCO2/P°

- ΔfG(CO2,g) = - ΔfG°(CO,g) + RT In PCO/P°

- 1/2ΔfG(O2,g)= - 1/2(0 + RT In PO2/P°)

This gives:

ΔrG = ΔrG° + RT{ln(PCO2/P°) - In (PCO/P°) - 1/2 In (P O2/P°)}

or:

ΔrG = ΔrG° + RT ln(PCO2/P°)/(PCO/P°)x(PO2/P°)1/2

When ΔrG = 0, we have equilibrium, and thus:

ΔrG = ΔrG° + RT ln(PCO2/P°)/(PCO/P°)x(PO2/P°)1/2 = -RT ln K

where K is called the thermodynamic equilibrium constant.

For a general reaction between ideal gases one may write:

The chemical equation: 0 = ΣiviBi (3.14)

The change in Gibbs energy: ΔrG = ΔrG° + RT ln IIi(Pi/P°)vi (3.15)

The equilibrium constant: K = IIi(Pi/P°)vi (3.16)

Relation between Gibbs energy and equilibrium constant: ΔrG° = - RT In K (3.17)

The ratio Pi/P° is without dimension, and the thermodynamic equilibrium constant is a dimensionless number. The P° is a standard pressure. In the SI system, the standard pressure is 1 bar. Earlier 1 atm was chosen as the standard, and still many data are based on that standard. If Pi is given in atm and we want to use bar as the standard, we write P⇃(O.9869 atm). H PI is given in bar and we want to use atm as standaJd, we write P⇃(1.013 bar). The use of antilogarithms on eq.(3.17) gives:

K = erG°/RT=10rG°/23RT (3.18)

We can see now that if ΔrG < 0, the exponent will be positive and K will.be greater than unity. The more negative the value of ΔrG°, the higher is the value of K, and thus the more the equilibrium will be shifted towards the products. When ΔrG° > 0, K will be less than unity, and the equilibrium will be shifted towards the reactants.

According to eq.(1.9) we have ΔrG° = ΔrH° - TΔrS°, and thus we can expand eq. (3.18):

K = erH°/RT - ΔrS°/R=erH°/RT (3.19)

We see that the greater the ΔrS°, the greater is K. Recalling that entropy was interpreted in terms of disorder, we can see that the value of K is influenced by the tendency towards maximum disorder. We can also see that the more negative the value of ΔrH°, the greater is K. The value of K is thus also influenced by the tendency towards low enthalpy.

### Heterogeneous Equilibrium

The change in Gibbs energy for a liquid or a solid with pressure, is very small as long as pressure changes are moderate, see eq. (3.9).

For the heterogeneous reaction:

C (graph) + C02 (g) = 2 CO (g)

we have:

ΔrG = 2 ΔfG(CO,g) - ΔrG(C01,g) - ΔrG(C,graph)

Using eqs (3.8, 3.11) we have:

ΔfG(CO,g) = ΔfG°(CO,g) + RT In PCO/P°

ΔfG(C02 g) = ΔfG°(C02,g) + RT In PCO2/P°

ΔfG(C,graph) = 0 + VC,graph x (ΣP - P°)

where V C,graph,is the volume of one mole graphite, and ΣP is the total pressure.

From this we obtain the Gibbs energy change for the reaction:

ΔrG = ΔrG°+2RT ln PCO/P°-RT In PCO2/P°-VC,graph x (ΣP - P°)(3.20)

where:

ΔrG =2ΔrG°(CO2,g)-0


The last term in eq. (3.20) is negligible at moderate pressures.

At equilibrium we have ΔrG = 0, which gives:

ΔrG° = -RT In (PCO/P°)2/(PCO2/P°)=-RT In K

where:

K= (PCO/P°)2/PCO2/P°