Thermodynamics : Gibbs Free Energy 
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: Standardstate free energy of reaction (G)
The free energy of reaction at standard state conditions: Standardstate conditions
Tabulated standardstate thermodynamic data are generally for a temperature of 25C (298 K)The partial pressures of any gases involved in the reaction is 0.1 MPa. The concentrations of all aqueous solutions are 1 M. StandardState Free Energy of Formation (G_{f})
The change in free energy that occurs when a compound is formed from its elements in their most thermodynamically stable states at standardstate 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 standardstate conditions.
Recall from the enthalpy notes that reactions can be classified according to the change in enthalpy (heat):
Conditions 
Conditions 
H < 0  H > 0 
S > 0  S < 0 
SPONTANEOUS: G is negative (G< 0, K_{eq} > 1)
NONSPONTANEOUS: Gis positive (G > 0, K_{eq} < 1)
EQUILIBRIUM: G = 0 (K_{eq} = 1)
Compound  H_{f}  S 
NH_{4}NO_{3(s)}  365.56  151.08 
NH_{4}^{+}_{(aq)}  132.51  113.4 
NO_{3}^{}_{(aq)}  205.0  146.4 
CalculateH,S, andG for the above reaction to determine whether the reaction is spontaneous or not.First let's calculateH_{f}. Note that in the above reaction, one mole of NH_{4}NO_{3} dissociates in water to give one mole each of NH_{4}^{+} and NO_{3}^{}:
Next, let's calculateS:
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.
Sample Calculations (standardstate conditions):
 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 standardstate 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.
Compound  H_{f}  S  
N_{2(g)} 



H_{2(g)} 



NH_{3(g)} 


1) CalculateH andS 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 standardstate free energy of reaction with the free energy of reaction at any moment in time during a reaction (not necessarily at standardstate conditions):
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.G = free energy at any moment G = standardstate free energy R = ideal gas constant = 8.314 J/molK T = temperature (Kelvin) lnQ = natural log of the reaction quotient Note: When Q_{c} = K_{c} (or when Q_{p} = K_{p}), a reaction is at equilibrium.
It was stated earlier that whenG = 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 ofG measures how far a reaction is from equilibrium. The larger the value ofG, 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 (whenG > 0). As the magnitude ofGchanges, 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 oxidationreduction reaction is from equilibrium.
The Nernst equation relates the standardstate 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 standardstate 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 oxidationreduction reaction is directly proportional to the cell potential of the reaction.
See notes for cell potentials and the Nernst equation.