FreeEnergyLesson3

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, dq_rev = 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 P^o to P, the change in Gibbs energy is:

ΔG =VdP (3.7)

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

ΔG =VdP = VdP = V (P - P^o) (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 P^o, the standard state:

ΔG_P,T ≈ ΔG_P^o,T = ΔG^o_T (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/P^o) (3.10)

If the gas is one component, i, of a mixture of ideal gases, it contributes with a pressure P_i = n_iRT/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:

ΔG_i = RT ln (P_i/P^o) (3.11)

We have seen earlier how we can find Δ_rG^o for a chemical reaction from the values of Δ_rG° for roducts and reactants, eq. (2.3). Since Δ_rG = Σ_iV_iΔ_fG_i we can use eqs (3.9, 3.10) to find Δ_rG 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:

dn_co = V_co dξ = - dξ

dn_o2 = V_o2 dξ = - 1/2 dξ

dn_CO2 = 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 =Δ_rGxdξ={ΣΔ_fG(produces)-ΣΔ_fG(reactants)}dξ (3.12)

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

Δ_rG = ΣΔ_fG(produces)-ΣΔ_fG(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:

Δ_rG = Δ_fG(CO₂,g) - Δ_fG(CO,g) - 1/2Δ_fG(O₂,g)

+Δ_fG(CO₂,g)= +Δ_fG°(CO₂,g) + RT In P_CO2/P°

- Δ_fG(CO₂,g) = - Δ_fG°(CO,g) + RT In P_CO/P°

- 1/2Δ_fG(O₂,g)= - 1/2(0 + RT In P_O2/P°)

This gives:

Δ_rG = Δ_rG° + RT{ln(P_CO2/P°) - In (P_CO/P°) - 1/2 In (P _O2/P°)}

or:

Δ_rG = Δ_rG° + RT ln(P_CO2/P°)/(P_CO/P°)x(P_O2/P°)^1/2

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

Δ_rG = Δ_rG° + RT ln(P_CO2/P°)/(P_CO/P°)x(P_O2/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 = Σ_iv_iB_i (3.14)

The change in Gibbs energy: Δ_rG = Δ_rG° + RT ln II_i(P_i/P°)^v_i (3.15)

The equilibrium constant: K = II_i(P_i/P°)^v_i (3.16)

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

The ratio P_i/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 P_i 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 = e^-Δ_rG°/RT=10^{-Δ_rG°/23RT} (3.18)

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

According to eq.(1.9) we have Δ_rG° = Δ_rH° - TΔ_rS°, and thus we can expand eq. (3.18):

K = e^{-Δ_rH°/RT - Δ_rS°/R}=e^-Δ_rH°/RT (3.19)

We see that the greater the Δ_rS°, 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 Δ_rH°, 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:

Δ_rG = 2 Δ_fG(CO,g) - Δ_rG(C01,g) - Δ_rG(C,graph)

Using eqs (3.8, 3.11) we have:

Δ_fG(CO,g) = Δ_fG°(CO,g) + RT In P_CO/P°

Δ_fG(C0₂ g) = Δ_fG°(C0₂,g) + RT In P_CO2/P°

Δ_fG(C,graph) = 0 + V_C,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:

Δ_rG = Δ_rG°+2RT ln P_CO/P°-RT In P_CO2/P°-V_C,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 Δ_rG = 0, which gives:

Δ_rG° = -RT In (P_CO/P°)²/(P_CO2/P°)=-RT In K

where:

K= (P_CO/P°)²/P_CO2/P°