Gibbs Free Energy (G) - The energy associated with a chemical reaction that can be used to do work.  The free energy of a system is the sum of its enthalpy (H) plus the product of the temperature (Kelvin) and the entropy (S) of the system:

Free energy of reaction (G)

• The change in the enthalpy (H) of the system minus the product of the temperature (Kelvin) and the change in the entropy (S) of the system:

Standard-state free energy of reaction (G)

• The free energy of reaction at standard state conditions:

Standard-state conditions

• The partial pressures of any gases involved in the reaction is 0.1 MPa.
• The concentrations of all aqueous solutions are 1 M.

Measurements are also generally taken at a temperature of 25C (298 K)

Standard-State Free Energy of Formation (G_f)

• The change in free energy that occurs when a compound is formed form its elements in their most thermodynamically stable states at standard-state conditions.  In other words, it is the difference between the free energy of a substance and the free energies of its elements in their most thermodynamically stable states at standard-state conditions.

• The standard-state free energy of reaction can be calculated from the standard-state free energies of formation as well.  It is the sum of the free energies of formation of the products minus the sum of the free energies of formation of the reactants:

Recall from the enthalpy notes that reactions can be classified according to the change in enthalpy (heat):

• Endothermic - absorbs heat, H > 0
• Exothermic - releases heat,H > 0

Reactions can also be classified according to the change in the free energy of the reaction:

• Endergonic - NON-SPONTANEOUS, G > 0
• Exergonic - SPONTANEOUS, G < 0

Summary

Favorable Conditions		Unfavorable Conditions
H < 0		H > 0
S > 0		S > 0

SPONTANEOUS:  G is negative (G< 0)

NON-SPONTANEOUS:  Gis positive (G > 0)

EQUILIBRIUM:  G = 0

• If a reaction is favorable for both enthalpy (H < 0 ) and entropy (S > 0),  then the reaction will be SPONTANEOUS (G < 0 ) at any temperature.
• If a reaction is unfavorable for both enthalpy (H > 0 ) and entropy (S < 0 ), then the reaction will be NONSPONTANEOUS (G > 0 ) at any temperature.
• If a reaction is favorable for only one of either entropy or enthalpy, the standard-state free energy equation must be used to determine whether the reaction is spontaneous or not.

Sample free energy calculation

Compound		Hf		S
NH4NO3(s)		-365.56		151.08
NH4+(aq)		-132.51		113.4
NO3-(aq)		-205.0		146.4

Calculate H, S, and G for the above reaction to determine whether the reaction is spontaneous or not.

First let's calculate H_f.  Note that in the above reaction, one mole of NH₄NO₃ dissociates in water to give one mole each of NH₄⁺ and NO₃^-:

Next, let's calculate S:

Now we can plug in these values we've calculated into the free energy equation.

• NOTE:  The units of H_f is kJ and the units of S is J/K.  Since G is generally reported in kJ, we can divide S by  1000 to convert it to units of kJ/K
• NOTE: The temperature in the free energy equation must be in Kelvin, so we must convert the given temperature in Celsius to Kelvin by adding 273.15.

Temperature and Free Energy

• If a reaction is favorable for enthalpy (H < 0 ), but unfavorable for entropy (S < 0 ), then the reaction becomes LESS SPONTANEOUS as temperature increases.
  • WHY? - The standard-state free energy equation states that:

    

    If entropy is unfavorable, the S is negative.  Subtracting a negative number is the same as adding the respective positive number.  As the temperature increases, the TS factor (which is ADDED to the enthalpy if the entropy is unfavorable) increases as well. Eventually, the TS factor becomes larger than  H and G  becomes positive,  i.e. the reaction is no longer spontaneous.

Sample Calculations

Compound		Hf		S
N2(g)		0		191.61
H2(g)		0		130.68
NH3(g)		-46.11		192.45

1) Calculate H and S for the above reaction.  Explain what each of the signs mean.

H is negative which is favorable.

S is negative which is unfavorable.

2) Predict whether the above reaction is spontaneous at 25C.

G is negative, so the reaction is SPONTANEOUS.

3) Predict whether the above reaction is spontaneous at  500C.

Gis positive, so the reaction is NOT SPONTANEOUS. 

Free energy and Equilibrium Constants

The following equation relates the standard-state free energy of reaction with the free energy of reaction at any moment in time during a reaction (not necessarily at standard-state conditions):

• G = free energy at any moment
• G = standard-state free energy
• R = ideal gas constant = 8.314 J/mol-K
• T = temperature (Kelvin)
• lnQ = natural log of the reaction quotient

Reaction quotient (Q_c or Q_p) - The mathematical product of the concentrations (or partial pressures) of the products of a reaction divided by the mathematical product of the concentrations (or partial pressures) reactants of a reaction AT ANY MOMENT IN TIME.

Note: When Q_c = K_c (or when Q_p = K_p), a reaction is at equilibrium.

It was stated earlier that when G = 0, a reaction is at equilibrium.  Let's consider the above reaction at equilibrium:

If we move RTlnK to the opposite side by subtracting it from both sides, we get the following reaction which relates the free energy of a reaction to the equilibrium constant of a reaction:

Summary

The magnitude of G measures how far a reaction is from equilibrium.  The larger the value of G, the further the reaction is from equilibrium and the further the reaction must shift to reach equilibrium.  In reactions in which enthalpy is favorable and entropy is unfavorable, the reaction becomes less spontaneous (G increases) until eventually the reaction is not spontaneous (when G > 0).  As the magnitude of Gchanges, so does the equilibrium constant. K.

Free energy and Cell potentials

Cell potential - A measure of the driving force behind an electrochemical reaction, reported in volts.  The potential of an electrochemical cell measures how far an oxidation-reduction reaction is from equilibrium.

The Nernst equation relates the standard-state cell potential with the cell potential of the cell at any moment in time:

If we rearrange the equation, we get:

This equation is very similar to the equation that relates the standard-state free energy of reaction with the free energy of reaction at any moment in time during a reaction:

We can convert these equations to get the following:

This shows that the free energy of a oxidation-reduction reaction is directly proportional to the cell potential of the reaction.

See notes for cell potentials and the Nernst equation.

Next:  "Electrochemistry:  Oxidation and Reduction"