For the reaction below, Δ S °

The Temperature Dependence of EquilibriumConstants

Equilibrium constants are not strictly constant because theychange with temperature. We are now ready to understand why.

The standard-state free energy of reaction is a measure of howfar the standard-state is from equilibrium.

G^o = –RT ln K

But the magnitude of G^o depends on thetemperature of the reaction.

G^o =H^o –TS^o

As a result, the equilibrium constant must depend on thetemperature of the reaction.

Given data ,

Δ S ° = − 269 J/(mol · K), Δ H ° = − 103.8 kJ/mol.T=25^oC =298 K

G^o =H^o –TS^o

= -103.8 – 298x (- 269 )

= 80265.8

G^o = –RT ln K

Kc =e^{– Go  / RT}

   ⁼ 8.5572×10^-15

The Gibbs free energy of a systemat any moment in time is defined as the enthalpy of the systemminus the product of the temperature times the entropy of thesystem.

G = H – TS

The Gibbs free energy of the system is a state function becauseit is defined in terms of thermodynamic properties that are statefunctions. The change in the Gibbs free energy of the system thatoccurs during a reaction is therefore equal to the change in theenthalpy of the system minus the change in the product of thetemperature times the entropy of the system.

G = H – (TS)

If the reaction is run at constant temperature, this equationcan be written as follows.

G = H – TS

The change in the free energy of a system that occurs during areaction can be measured under any set of conditions. If the dataare collected under standard-state conditions, the result is thestandard-state free energy of reaction(G^o).

G^o =H^o –TS^o

The beauty of the equation defining the free energy of a systemis its ability to determine the relative importance of the enthalpyand entropy terms as driving forces behind a particular reaction.The change in the free energy of the system that occurs during areaction measures the balance between the two driving forces thatdetermine whether a reaction is spontaneous. As we have seen, theenthalpy and entropy terms have different sign conventions.

Favorable		Unfavorable
Ho < 0		Ho > 0
So > 0		So < 0

The entropy term is therefore subtracted from the enthalpy termwhen calculating G^o for areaction.

Because of the way the free energy of the system is defined,G^o is negativefor any reaction for which H^o is negativeand S^o is positive.G^o is thereforenegative for any reaction that is favored by both the enthalpy andentropy terms. We can therefore conclude that any reaction forwhich G^o is negativeshould be favorable, or spontaneous.

Favorable, or spontaneous reactions:

Go < 0

Conversely, G^o is positivefor any reaction for which H^o is positiveand S^o is negative.Any reaction for which G^o is positive istherefore unfavorable.

Unfavorable, or non-spontaneous reactions:

Go > 0

Reactions are classified as either exothermic(H < 0) orendothermic (H > 0) on the basis ofwhether they give off or absorb heat. Reactions can also beclassified as exergonic (G < 0) orendergonic (G > 0) on the basis ofwhether the free energy of the system decreases or increases duringthe reaction.

When a reaction is favored by both enthalpy (H^o < 0) andentropy (S^o > 0), thereis no need to calculate the value of G^o to decidewhether the reaction should proceed. The same can be said forreactions favored by neither enthalpy (H^o > 0) norentropy (S^o < 0). Freeenergy calculations become important for reactions favored by onlyone of these factors.