Post on 21-Jan-2021
Gibbs Free Energy (G) is a thermodynamic state
function that combines enthalpy and entropy.
For a process at constant temperature & pressure, the sign of ΔG relates to the spontaneity of the
process.
ΔG = negative, the process is spontaneous
ΔG = positive, the process is not spontaneous, but
the reverse process is spontaneous.
At equilibrium, the process is reversible and ΔG is
zero.
Free energy is the amount of useful work that
can be obtained from a process at constant
temperature and pressure.
p. 8 of handout
Potential energy and Free
energy – an analogy
between the gravitational-PE
change in a boulder rolling
down a hill (position a) and
the free energy change in a
spontaneous reaction (b).
The equilibrium position in (a)
is given by the minimum
gravitational PE available to
the system. The equilibrium
position in (b) is given by the
minimum free energy
available to the system.
In any spontaneous process at constant
temperature and pressure, the free energy
always decreases.
Calculating Free Energy Change (G)
(constant temperature and pressure)
ΔG = ΔH – TΔS
ΔG is the change in free energy measure in kJ/mol)
Δ H is the change in enthalpy (kJ/mol)
ΔS is the change in entropy (J/mol K)
T is Kelvin Temperature
Quantity Change in
Enthalpy
Change in
Entropy
Change in Free
Energy
Symbol ΔH ΔS ΔG
Unit kJ/mol J/mol K kJ/mol
Definition Heat gained by a
system
Change in
randomness of a
system
Available useful work
Comments + for endothermic
- for exothermic
+ for increasing
randomness
- for decreasing
randomness
+ for nonspontaneous
- for spontaneous
-ΔG = 0 at equilibrium
Summary of Important Thermodynamic Quantities
ΔG = ΔG° + RT ln Q
R is the ideal gas constant, 8.314 J/mol-K
T is the absolute temperature
Q is the reaction quotient that corresponds to the
reaction mixture of interest.
Under standard conditions the concentrations of all the
reactants and products are equal to 1. Under standard
conditions Q = 1 and therefore, ln Q = 0.
Therefore, ΔG = ΔG° under standard conditions.
The standard free energy for any reaction is
related to the equilibrium constant.
At equilibrium, ΔG = 0 and Q = K, the
equilibrium constant.
At equilibrium: ΔG = -RT ln K
The standard free energy change for the following reaction at 25°C is -118.4 kJ/mol:
KClO3(s) → KCl(s) + 3/2 O2(g)
Calculate Kp for the reaction at 25°C and the equilibrium
pressure of O2 gas.
ΔG° = -RT ln K
-118.4 kJ/mol = -(8.314 J/mol K)(298 K)(1kJ/1000 J) ln Kp
Ln Kp = 47.8
Kp = e47.8 = 5.68 x 1020
Kp = p3/2 O2
PO2 = Kp2/3 = (5.68 x 1020 )2/3 = 6.9 x 1013 atm
ΔG° for any process can be calculated from
tabulations of standard free energies of
formation – see Tables in Appendix C.