Gibbs Free Energy is a criterion for spontaneity and takes into account the entropy change of the surroundings.
The Gibbs Free Energy Function is defined by the equation,
G = H - TS
For change at constant temperature
G = H - T S
Based on G reaction behavior can be predicted accurately.
G negative reaction proceeds spontaneously.
0 reaction is at equilibrium
positive reaction will not proceed at all
Free energy or reaction can be calculated as follows :
G reaction = G products - G reactants
Standard Free Energy
A standard free energy change, given the symbol G0 applies to a reaction run at 250C and 1 atm pressure with
reactions and products in their standard states (that state in which the substance occurs under these conditions).
The standard free energy of formation G0f is defined as the change in standard free energies when 1 mole of a
substance is prepared from its constituent elements.
An element in its standard state at 250C and 1 atm is assigned a standard free energy of zero.
The table lists the standard free energy formation of different compounds.
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.
Go = delta Ho - T delta So
It is also define as Go = 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 Go = delta Ho - T delta So
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)
G = H - TS
= (+) - 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 0 oC.
delta H
+Spontaneous Spontaneous
+ only at at all
high T T
delta S
Non- Spontaneous Spontaneous
at all T only at
- low T
SCH4U Gibb's Free Energy Entropy Worksheet
1. Phosgene, COCl2, was used as a war gas during World War I. It reacts with the moisture in the lungs to
produce HCl, which causes the lungs to fill with fluid, leading to death of the victim. COCl2 has a standard
entropy, So =284 J/mol K and Ho =-223 kJ/mol. Use this information and appropriate tables to calculate G o
for COCl2(g) in kJ/mol at body temperature!
2. Aluminum oxidizes rather easily, but forms a thin protective coating of Al2O3 that prevents further
oxidation of the aluminum beneath. Use the data for H fo and So to calculate Go for Al2O3(s) in kJ/mol.
3. Calculate Go in kJ for the following reactions, using the appropriate data tables.
Use Go = Gproducts - Greactants
(a) SO3(g) + H2O(l) ----> H2SO4(l)
(b) 2 NH4Cl(s) + CaO(s) ----> CaCl2(s) + H2O(l) + 2 NH3(g)
(c) CaSO4(s) + 2 HCl(g) ----> CaCl2(s) + H2SO4(l)
(d) C2H4(g) + H2O(l) ----> C2H5OH(l)
(e) Ca(s) + 2 H2SO4(l) ----> CaSO4(s) + SO2(g) + 2 H2O(l)
4. Calculate the Go in kJ for the following reactions at 100 oC using the appropriate data tables.
(a) C2H4(g) + H2(g) ----> C2H6(g)
(b) 5 SO3(g) + 2 NH3(g) -----> 2 NO(g) + 5 SO2(g) + 3 H2O(g)
(c) N2O(g) + NO2(g) ----> 2 NO(g)