For the reaction below, Δ S °

For the reaction below, Δ S ° = − 269 J/(mol · K) and Δ H ° = −103.8 kJ/mol. Calculate the equilibrium constant at 25 °C.

3C(s) + 4H_2 (g) –>C_3+ H_8 (g)


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.

Go = –RT ln K

But the magnitude of Go depends on thetemperature of the reaction.

Go =HoTSo

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=25oC =298 K

Go =HoTSo

= -103.8 – 298x (- 269 )

= 80265.8

Go = –RT ln K

Kc =eGo  / 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.


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.


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(Go).

Go =HoTSo

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 Go for areaction.

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

Favorable, or spontaneous reactions: Go < 0

Conversely, Go is positivefor any reaction for which Ho is positiveand So is negative.Any reaction for which Go 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 (Ho < 0) andentropy (So > 0), thereis no need to calculate the value of Go to decidewhether the reaction should proceed. The same can be said forreactions favored by neither enthalpy (Ho > 0) norentropy (So < 0). Freeenergy calculations become important for reactions favored by onlyone of these factors.

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