Free Energy and Temperature
Free Energy and the Equilibrium Constant

How is the change in free energy (i.e. DG ) as we go from state 1 to state 2 (or reactants to products) affected by temperature?

DG = DH - T*DS

Exothermic reactions release heat to the surroundings and have a negative value for DH

• If there were no entropic considerations, all exothermic reactions would result in negative values of DG , and would define spontaneous reactions

The value of DS (the entropic change) may be either positive (products have a greater degree of disorder than reactants) or negative (products have a lesser degree of disorder than reactants)

• For those reactions where DS is positive, the -TDS term will be negative. This will contribute to the overall negative value (i.e. spontaneity) of DG
• For those reactions where DS is negative, the -TDS term will be positive. This decrease in entropy will oppose spontaneity. This will reduce the magnitude of a negative value of DG and may even cause DG to be positive (indicating a non-spontaneous process)

• Generally, DH and DS values do not change much with temperature
• Although the DS term may not change much with temperature, the magnitude of the -TDS term is obviously dependent upon the value of T (temperature). For an equivalent change in disorder, greater energy is absorbed at higher temperatures.

If DS is negative (i.e. unfavorable entropy change), the value of the -TDS term is positive, and increasing the temperature will increase the magnitude of the positive value of the -TDS term

• At some high temperature, the magnitude of the (positive) -TDS term can overwhelm the magnitude of an exothermic (i.e. negative) DH term. And even though the reaction is exothermic it will not occur spontaneously (i.e. DG = positive in value)

The opposite type of situation may occur: a reaction may be endothermic (DH is positive), and have a positive value for DS

• With a positive value for DS, the -TDS term favors spontaneity. However, at low temperatures, its contribution will be small (magnitude of -TDS at low temperatures is small). In this case DH predominates and the reaction will be non-spontaneous (DG = positive)
• At higher temperatures, the magnitude of the -TDS term increases and can overwhelm the (positive) DH term. In this case the -TDS predominates and the reaction is spontaneous (DG = negative at high temp)

Summary:

Consider the following possible states for two different types of molecules with some attractive force:

• There would appear to be greater entropy on the left (state 1) than on the right (state 2). Thus the entropic change for the reaction as written (i.e. going to the right) would be (-) in magnitude, and the energetic contribution to the free energy change would be (+) (i.e. unfavorable) for the reaction as written.
• In going to the right, there is an attractive force and the molecules adjacent to each other is a lower energy state (heat energy, q, is liberated). To go to the left, we have to overcome this attractive force (input heat energy) and the left direction is unfavorable with regard to heat energy q. The change in enthalpy is (-) in going to the right (q released), and this enthalpy change is negative (-) in going to the right (and (+) in going to the left).
• This reaction as written, is therefore, enthalpically favorable, and entropically unfavorable. It is enthalpically driven.
• From the above table, it would appear that we might be able to get the reaction to go to the right at low temperatures (lov temperature would minimize the energetic contribution of the entropic change).

Looking at the same process from an opposite direction:

• This reaction as written, is entropically favorable, and enthalpically unfavorable. It is entropically driven.
• From the above table, it would appear that we might be able to get the reaction to go to the right at high temperatures (high temperature would increase the energetic contribution of the entropic change).

Recall that standard free-energy of formation values from tables can be used to calculate the standard free energy change associated with a reaction:

DG⁰ = S n DG_f⁰(products) - S m DG_f⁰(reactants)

• These tabulated DG_f⁰ values are for standard conditions and a defined temperature
• Often we are interested in the value of DG⁰ for a reaction that is not under standard conditions (e.g. other concentrations than standard conditions of 1 atm for a gas, or 1M for a solution)

The general relationship between the free-energy change under standard conditions (i.e. DG⁰) and the free-energy change under any other conditions, (i.e. DG), is defined as:

DG _{(e.g. under non-standard conditions of conc.)} = DG⁰ + RT ln Q

• R is the gas constant, 8.314 J/mol K
• T is absolute temperature (K)
• Q is the calculated reaction quotient

Calculate DG at 298K for the Haber reaction where you start with 1.0 atm N₂(g), 3.0 atm H₂(g) and 1.0 atm NH₃(g). DG⁰_298K = -33.3 kJ

N₂(g) + 3H₂(g) ó 2NH₃(g)

DG = DG⁰ + RT ln Q

DG = -33.3 kJ + 8.314 J/K * 298K * ln (3.7 x 10^-2)

DG = -33.3 kJ - 8.17 kJ = -41.5 kJ

Q reflects the ratios of the various components under starting conditions. K reflects their ratios at equilibrium.

• If the starting system is at equilibrium, then Q = K
• At equilibrium DG = 0 (no spontaneous reaction in either direction)

Therefore:

DG = DG⁰ + RT ln Q

at equilibrium becomes

DG = DG⁰ + RT ln K

and

0 = DG⁰ + RT ln K

or

DG⁰ = -RT ln K

• If K>1 it means the equilibrium favors the products (i.e. spontaneity to the right). In this case lnK will be positive and therefore DG⁰ will be negative (again spontaneity to the right)
• If K<1 it means the equilibrium favors the reactants (i.e. spontaneity to the left). In this case lnK will be negative and therefore DG⁰ will be positive
• If K = 1, lnK will be zero and so will DG⁰. The reaction is at equilibrium

The above equation also allows us to calculate K if we know DG⁰: