Free Energy Functions

 

What are free energy functions?

Significance of Gibbs free energy and Helmholtz free energy.

Free Energy Functions:

In natural processes there are two factors involved when a system approaches equilibrium. One is the tendency to minimum energy and other is the tendency to maximum entropy. In many cases these two tendencies are in competition. If entropy or energy changes were only involved in deciding the direction of change, some strange phenomena would be observed.

For example, if the approach to equilibrium were governed solely by the criterion of lowest energy, all liquids would simultaneously freeze at any temperature, since energy is liberated on freezing. On the other hand, criterion of approach to equilibrium as the state of maximum disorder or entropy would result in all solids spontaneously melting or even evaporating since both processes involve an increase in disorder.

In practice, two tendencies often reached a compromise, which usually depends on temperature. For example solids are the stable form of a substance below the melting point, whereas above the melting point, the liquid is more stable. The factors which relate the energy changes, entropy changes and temperature to the direction of spontaneous change are the free energy functions.

These functions are the Helmholtz free energy or the work function denoted by A (from German Arbeit which means work) and Gibbs free energy represented by F.

These are defined as

        A = E – TS

        G = H – TS

Where E and H are the energies and enthalpy content of system respectively and S is the entropy.

Helmholtz Free Energy:

       A = E- TS

Since E, T and S are functions of state , A will be state function. Hence dA is an exact differential. Moreover E and S are extensive properties, it follows that A must also be an extensive property. State function A is more useful when the process takes place at constant volume and constant temperature.

Change in Helmholtz free energy for an Isothermal process:

The change in A for an isothermal process is given by

            ∆A = ∆E - T∆S

Significance of Helmholtz Free Energy:

For an isothermal infinitesimal change (T = constant)

            dA = dE – TdS

            dA = dE – δQ_rev (TdS = δQ_rev )

But      δQ_rev = dE + W_rev (first law of thermodynamics)

Hence dA = dE – (dE + W_rev)

             dA = - W_rev

Or         - dA = W_rev

We know that work done in reversible process is maximum hence

              W_rev= W_max

And       - dA = W_max

Where   - dA = decrease in work function

W_max includes both mechanical and non mechanical work. In fact it is for this reason that this thermodynamic quantity has been termed as work function.

Work Function Formula

From the definition of thermodynamic work function,

A = E − TS

For a small change in the system,

dA = dE− TdS − SdT

When the work is mechanical,

TdS = q (heat) = dE + PdV

∴ dA = − PdV − SdT

The above equation is the basic thermodynamic work function equation.

Reversible Isothermal Process

For the reversible isothermal process,
dT = 0

Therefore, dA = − PdV (W_max=PdV)

If the system contains ideal gas,

dA = − nRT/V

Integrating within the limits,

ΔA = nRT ln(V₁/V₂)

Therefore, with the increase of the volume of the system, work function decreases.

Isochoric Process

For the reversible isochoric system,

dV = 0
∴ dA = − SdT

Since entropy is always a positive quantity. It implies that Helmholtz’s free energy or work function A decreases with the increasing temperature of the isochoric process.

 

Gibbs Free EnergyGibbs Function:

F = H – TS

H, T and S are functions of state, it follows F will also be a state function. Moreover, since H and S are extensive properties, F will also be an extensive property. That is the value of F will be based on the mass of the substance. This function is useful in explaining the process at constant P and constant T.

Change in Gibbs Free Energy for an Isothermal Process:

The change in Gibbs free energy for an isothermal process is given by            

             ∆G = ∆H -T∆S

Significance of Gibbs Free Energy as useful work:

For an infinitesimal change

           dG= dH – TdS

Or       dG = dE +PdV +VdP –TdS –SdT  (H = E + PV)

For a reversible process occurring at constant temperature and pressure we have

            dT = 0 and dP = 0

Hence dG = dE + PdV – TdS

Also    TdS = δQ_rev  and δQ_rev = dE + W_rev (first law of thermodynamics)

            δQ_rev = dE + W_max           (W_rev= W_max)

            dG = dE + PdV – (dE + W_max)

            dG = PdV -–  W_max  

            -dG = W_max -–  PdV

 

W_max = maximum work obtained in a reversible process

PdV = work wasted against constant pressure

Useful work = maximum work – work wasted

            W_useful = W_max – PdV

Hence definition of F is as follows

 “ The function of total energy which is isothermally available for converting into useful work is called Gibbs free energy of the system.”Or decrease in free energy is a measure of all kinds of work except the PV work or work of expansion.

Gibbs Free Energy Equation

From the definition of Gibbs free energy,

G = H − TS = U + PV − TS
or, dG = dU + PdV + VdP − TdS − SdT

When the work is mechanical work only,
TdS = q = dU + PdV

Combining the above two equations,

dG = dU + PdV + VdP − dU − PdV − SdT
∴ dG = VdP − SdT

The above equation provides another basic thermodynamic Gibbs free energy formula.

For Isothermal Process

For an isothermal process,
dT = 0

∴ dG = VdP

For n moles ideal gas,
V = nRT/P

∴ dG = nRTdP/P

Integrating the above energy equation,

G = nRTlnP + G⁰ (integration constant)

When dividing by n,

μ = μ⁰ + RTlnP

Where μ = G/n = free energy per mole = chemical potential

For Isobaric Process

For the reversible isobaric process,
dP = 0

∴ dG = − SdT

Since the entropy of the system is always positive. Therefore, free energy decreases with increasing the temperature of the system at constant pressure. The rate of decrease of G with temperature is highest for gases and lowest for solids.

 

Some related questions:

Q: For the process H₂O(l) (1 bar, 373) H₂O (g) (1 bar, 373), the correct set of thermodynamic parameters is

(a) ∆G = 0,∆S = +ve

(b) ∆G = 0,∆S = -ve

(c) ∆G = +ve,∆S = 0

(d) ∆G = -ve,∆S = 0

Ans: a

At 373 H₂0(l) is in equilibrium with H2O(g). Hence ∆G =0. Since liquid molecules are converted into gaseous molecules, ∆S = +ve.

Q: ∆H vapour is equal to 30 kJ mol^-1 and , ∆S = 75 kJ mol^-1. Find the temperature of vapour at 1 atmosphere

(a) 400K

(b) 350K

(c) 298K

(d) 250K

Ans: a

∆G = ∆H - T∆S; at equilibrium ∆G = 0; T =  ∆H/∆S = 30 × 10³/75 = 400K