Gibbs Free Energy

Chemists often call this the free energy, while physicists often call it the thermodynamic potential.The most important property of the Gibbs free energy is that it is a minimum for a system in equilibrium at constant pressure when in thermal contact with a reservoir.In order to see this, consider the differential dG,

If the system, S, is in thermal contact with a heat reservoir, R₁, at temperature t and in mechanical contact with a pressure reservoir, R₂, that can maintain the pressure p but cannot exchange heat, then dt = dp = 0.So dG becomes

From the thermodynamic identity

we see that

but dN = 0, so dG = 0, which is the condition of an extremum.The fact that G is a minimum follows directly from the fact that the entropy has a minus sign associated with it.Also, from the derivation, we see that G = G(t,p,N).The general differential of the Gibbs free energy is

Comparing this with (15.2) and using the thermodynamic identity, we can immediately see that

(15.4)

(15.5)

(15.6)

Intensive and Extensive Quantities

Thus we see that

But we already saw that , so j(p,t) = m, and we get

Thus, the chemical potential for a system is equal to the Gibbs free energy per particle.If more than one particle species is present, (15.7) becomes

(15.8)

With this change in definition, the differential dG becomes

(15.9)

Gibbs Free Energy and Helmholtz Free Energy

From the definition of the Helmholtz free energy we know that

Holdingt and V constant and integrating, we get

Thus, F is not directly proportional to N if we keep the temperature and volume constant.From the definition of the Gibbs free energy and the definition of the Helmholtz free energy we can immediately see that

since pV = Nt for an ideal gas.Thus

(15.10)

From this we see that the chemical potential as a function of t and p is

(15.11)

Chemical Reactions

(15.12)

where A_j is the j^th chemical species, and n_j is the coefficient of the j^th species in the reaction.For example, in the reaction

we have A₁ = H₂, A₂ = C1₂, A₃ = HCl, n₁ = 1, n₂ = 1 and n₃ = -2.We usually discuss chemical equilibrium for reactions that occur under conditions of constant pressure and temperature.Recalling that the change in the Gibbs free energy is

this reduces to

The change dN_j is proportional to the coefficient n_j.This allows us to write

where dx indicates how many times the reaction takes place.Thus dG becomes

But in equilibrium dG = 0, so thus reduces to the condition

(15.13)

Equilibrium Reactions in an Ideal Gas

where c_j = n_q,jZ_int,j.Substituting this into the equilibrium condition, we get

This can be rewritten as

or

(15.14)

whereis called the equilibrium constant.Exponentiating both sides of (15.14), we get the law of mass action,

(15.15)

Example:

Consider the reaction

The law of mass action yields

where F_H(int) = -I = -13.6 eV and I is the ionization potential.Here [A] is the concentration of the molecule A.The quantum concentration for H and H + is the same, n_q(H) = n_q(H⁺), so the mass action equation reduces to

If charge neutrality is preserved, [e^-] = [H⁺], so this finally becomes

(15.16)

This is called the Saha equation.

Reaction Speed

with concentrations n_A, n_B and n_C = n_AB.What is dn_AB/dt?We can write

where C(t) and D(t) are constant with respect to rate.In equilibrium, dn_AB/dt = 0, so

In general, a graph of the potential energy required for a reaction looks like

HereDE is called the activation energy.This is the potential energy required for the reaction to take place.DH is called the heat of reaction and is the energy that evolves out of the reaction.

Isotherms

Different regions of the isotherm correspond to different forms of the matter, e.g. solid, liquid, or gaseous.The phase is a portion of a system that is uniform in composition.Two phases may coexist, with a definite boundary between them.Liquid and vapor (a gas that is in equilibrium with its liquid or solid form) may coexist on a section of an isotherm only if the temperature of the isotherm lies below a critical temperature, t_c.Above that critical temperature only a single phase - the fluid phase - exists, no matter how great the pressure.Consider a system that is originally only a liquid at a constant temperature.

Raise the piston.As the piston is raised, more vapor is formed until there is only vapor in the chamber.Plotting the pressure verses volume for this transformation we get

The thermodynamic conditions for the coexistence of two phases are the conditions for the equilibrium of two systems that are in thermal, diffusive and mechanical contact.These conditions are that

(15.17)

At a general point in the p-t plane the two phases do not coexist: if m₁ < m₂, the first phase alone is stable, and if m_l > m₂ the second phase alone is stable.

Consider a small segment of the curve.Then the condition for coexistence is that

(15.18)

and

Since the changes are small, we can expand the second condition to get

(15.19)

Subtracting (15.18) from (15.19) and rearranging the terms,

(15.20)

Now recall that the Gibbs free energy could be written as

If we define the volume and entropy per molecule as v = V/N and s = s/N respectively, then

and

so (15.20) becomes

(15.21)

Note that this derivative refers to the very special interdependent change of p and t in which the gas and liquid continue to coexist.The number of molecules in each phase will vary as the volume is varied, subject only to the condition that N_l + N_g = N.

Enthalpy

Thus the quantity of heat added by the transfer of one molecule is

(15.22)

where L is called the latent heat of vaporization.If we write v_g - v_l = Dv, then (15.21) can be rewritten as

(15.23)

This is known as the Clausius-Clapeyron equation, or the vapor pressure equation.Finally, the latent heat of a phase transition is, as we have seen, equal to t times the entropy difference of the two phases at constant pressure.It is also equal to the difference in the enthalpy, H, of the two phases, where H = U + pV.To see this, consider the differential

by the thermodynamic identity.But at constant pressure this becomes

Considering the change across the coexistence curve, we see that the last term in dH becomes (m_g - m_l)DN, while and .But m_g = m_l on the curve, so

(15.24)

Using the definition of the heat capacity at constant pressure, this can be written in a more useful form.The heat capacity at constant pressure is given by

Integrating this yields

(15.25)