Gibbs Free Energy
The value of Gibbs Free Energy for a reaction is based upon its enthalpy and entropy. Both of these factors can have either a negative or positive effect on whether or not a reaction will occur spontaneously or at all. We will even be able to predict at what temperature a nonspontaneous reaction will become spontaneous.

It is called free energy because it is the energy that will be released or freed up to do work. Gibbs Free energy is defined as.
GoΔHo - TΔSo
It is also define as ΔGo = Gproducts - Greactants.   But this equation is only valuable to you of the reaction is carried out at standard temperature and pressure. Any other temperature requires the use of the first equation.

Take a look again at ΔGo = ΔHo  -  T ΔSo

A change can only be spontaneous if it is accompanied by a decrease in free energy. In other words, for a change to be spontaneous, G must be negative. What does this mean in terms of H and S?

When a change is exothermic and is also accompanied by an increase in entropy, both factors favour spontaneity.

ie.     H is negative (-ve)
S is positive (+ve)
G = H - TS
= (-) - T(+)
In such a change, G will be negative regardless of the value of the absolute temperature, T (which can only have positive values). Therefore, the change will occur spontaneously at all temperatures.

On the other hand, if a change is endothermic and is accompanied by a decrease in entropy, both factors work against spontaneity.

i.e.  H is positive (+ve)
S is negative (-ve)
ΔG = ΔH - TΔS
= (+) - T(-)

In this case, G will be positive at all temperatures and the change will always be nonspontaneous.

When H and S have the same sign, the temperature becomes critical in determining whether or not an event is spontaneous.

If H and S are both positive,
G = (+) - T(+)

Only at relatively high temperatures will the value of TS be larger than the value of H so that their difference, G, is negative. A familiar example is the melting of ice.

H2O(s) ----> H2O(l)
Here is a change that we know is endothermic and occurs with an increase in entropy. At temperatures above 0oC (when the pressure is 1 atm), ice melts because the TS term is bigger than the H term. At lower temperatures, ice doesn't melt because the smaller value of T gives a smaller value for TS and the difference H-TS, is positive.

For similar reasons, when H and S are both negative, G will be negative only at relatively low temperatures. The freezing of water is an example.

H2O(l) ----> H2O(s)

Energy is released as the solid is formed and the entropy decreases. You know, of course, that water freezes spontaneously at low temperatures, that is, below 0oC.

 ΔH +ve -ve ΔS +ve Spontaneous     only at      high T Spontaneous at all T -ve Non- Spontaneous      at all T Spontaneous only at low T
 Standard Free Energies Standard free energies of formation can be used to obtain standard free energies of reaction. When G is determined at 25oC and 1 atm, we call it the standard free energy change, Go. There are a number of ways of obtaining Go for a reaction. One of them is to computer Go from Ho and So using: Sample Problem Compute Go for the hydrolysis of urea, CO(NH2)2, CO(NH2)2(aq) + H2O(l) ----> CO2(g) + 2 NH3(g) Solution We can calculate Ho from standard heats of formation from Hess's Law. ΔHo = [CO2 + (2)NH3] - [CO(NH2)2 + H2O]                            = [ -393.5 + (2)-46.19 ] - [ -319.2 - 285.9 ]                            = (-485.9 kJ) - (-605.1)                            = +119.2 kJ We have already calculated the So in a previous example. But we must be careful to express Ho and So in the same units. ΔGo = -119.2 kJ - (298.15 K)(0.3548 kJ/K)                          = +119.2 kJ - 105.8 kJ                          = +13.4 kJ The value obtained for Go means that at room temperature 25oC urea will not spontaneously decompose in water. It is also useful to calculate the Standard Free Energies of Formation, Gfo. The equation to use is a variation of the Hfo and Sfo equations you've already used. ΔGfo = (sum of Gfo of products) - (sun of Gfo of reactants) Sample Problem What is Go for the combustion of ethyl alcohol (C2H5OH) to give CO2(g) and H2O(g)? Solution First we need a balanced equation C2H5OH(l) + 3 O2(g) ----> 2 CO2(g) + 3 H2O(g) Using the information in the databook found next to the enthalpy and entropy data you will find the Gibbs Free energy data. ΔGo = [(2)CO2(g) + (3)H2O(g)] - [C2H5OH(l) + (3)O2(g)] As with Hfo, Gfo is zero for an element in its standard state. ΔGo = [2 mol(-394.4 kJ/mol) + 3 mol(-228.6 kJ/mol)]                         = [1 mol(-174.8 kJ/mol) + 3 mol(0 kJ/mol)]                         = (-1474.6 kJ) - (-174.8 kJ)                         = -1299.8 kJ Go to the Gibb's Free Energy Worksheet Thermochemistry Unit Review Entropy and Gibbs Free Energy Review