Question

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)

Answer 1

The Temperature Dependence of Equilibrium Constants

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

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

G^o = - RT ln K

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

G^o = H^o - TS^o

As a result, the equilibrium constant must depend on the temperature 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}

   ⁼ 8.5572x10^-15

The Gibbs free energy of a system at any moment in time is defined as the enthalpy of the system minus the product of the temperature times the entropy of the system.

G = H - TS

The Gibbs free energy of the system is a state function because it is defined in terms of thermodynamic properties that are state functions. The change in the Gibbs free energy of the system that occurs during a reaction is therefore equal to the change in the enthalpy of the system minus the change in the product of the temperature times the entropy of the system.

G = H - (TS)

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

G = H - TS

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

G^o = H^o - TS^o

The beauty of the equation defining the free energy of a system is its ability to determine the relative importance of the enthalpy and entropy terms as driving forces behind a particular reaction. The change in the free energy of the system that occurs during a reaction measures the balance between the two driving forces that determine whether a reaction is spontaneous. As we have seen, the enthalpy 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 term when calculating G^o for a reaction.

Because of the way the free energy of the system is defined, G^o is negative for any reaction for which H^o is negative and S^o is positive. G^o is therefore negative for any reaction that is favored by both the enthalpy and entropy terms. We can therefore conclude that any reaction for which G^o is negative should be favorable, or spontaneous.

Favorable, or spontaneous reactions:

Go < 0

Conversely, G^o is positive for any reaction for which H^o is positive and S^o is negative. Any reaction for which G^o is positive is therefore unfavorable.

Unfavorable, or non-spontaneous reactions:

Go > 0

Reactions are classified as either exothermic (H < 0) or endothermic (H > 0) on the basis of whether they give off or absorb heat. Reactions can also be classified as exergonic (G < 0) or endergonic (G > 0) on the basis of whether the free energy of the system decreases or increases during the reaction.

When a reaction is favored by both enthalpy (H^o < 0) and entropy (S^o > 0), there is no need to calculate the value of G^o to decide whether the reaction should proceed. The same can be said for reactions favored by neither enthalpy (H^o > 0) nor entropy (S^o < 0). Free energy calculations become important for reactions favored by only one of these factors.