Predicting Reactions with Gibbs Free Energy
Name: Sebastian G.
Date: October 2007
When you put values in for Gibbs free energy (enthalpy,
entropy and temperature) and you want to find out if two substances
mixed together will react, how do you put in the two values? Do you
find the average?
If you write the balanced equation for pure unmixed reactants and
products, you can look up the Gibbs free energy in appropriate tables
found in texts, handbooks, etc. These are expressed in (per mol) terms
at 1 atm pressure, and the stated temperature (usually 298.15 K). If
you do the calculation of "products" minus "reactants" taking the
coefficients in the balanced equation into account and the resultant
Gibbs free energy is negative, that reaction (as written) is spontaneous
(in the thermodynamic sense). These cautions (in parentheses) are
mentioned because some thermodynamically favored reactions can be prevented
from "coming to equilibrium". A simple example is a balloon. Here there
is a difference in pressure -- the higher pressure being inside the
balloon -- and the pressures inside and outside the balloon should be
the same. But the barrier of the balloon prevents that from occurring,
at least very rapidly.
Your perceptive question about what happens with mixed substances is much
large to be derived concisely in a forum such as this. In fact the proper
answer to your question makes up a significant part of a semester course
in chemical thermodynamics!! But the result is: DGo = -RTln(Keq), where
DGo is the standard free energy change for the balanced chemical reaction.
R is the gas constant in consistent energy units, T is the absolute
temperature, and ln(Keq) is the "natural log" of the equilibrium constant
For a reaction of gases, say aA + bB = cC + eE,
Keq is defined as: Keq = (PC)^c x (PE)^e / (PA)^a x (PB)^b.
The pressures of the various gases being:
(PC), ..., (PB) etc., c, e, a, b being the stoichiometric coefficients. Of
course the pressures have to be be in consistent pressure units. In the
case of solutions the P's become concentrations, that is, (MC), ..., (MB),
Pretty messy, but in a course, or a textbook, there are techniques
presented to show you how to solve such equations systematically. What
is "surprising" is that while the Gibbs free energy of pure solid and
liquid reactants or products appears in the calculation of DGo, those
numbers do not appear in Keq. This is not a mistake, that is how the
derivation works out.
Again, your question is perceptive, but the answer is too long to be
answered in a short paragraph.
Note that both enthalpy and entropy values have "delta" or change in
front of the variable. While the 3rd law of thermodynamics states that
we can measure the actual entropy of a system, there is no way to
measure the actual enthalpy of the system. All we can do is measure the
*change* in the enthalpy. In that sense, we are not using the
single-event, one molecule reaction, in describing the change in
enthalpy or entropy of the system. What we are reporting is the over-all
enthalpy and entropy change of the whole system. In the case of enthalpy
change - we measure how much heat is absorbed or released, in the case
of entropy we can statistically obtain its entropy. Temperature, of
course, is a measured value that is related to the average molecular
speeds of the whole system.
Greg (Roberto Gregorius)
No, this is not correct. You do not use averages.
You need to also have Gibbs free energies
for the products. Then you calculate delta G
for the reaction, using the Gibbs free energies
in the same way that you would handle enthalpies
of formation to calculate the heat of reaction
(see your textbook). Once this is done,
if delta G < 0, products are favored at equilibrium
if delta G > 0, reactants are favored at equilibrium.
However, even if delta G < 0 the reaction still might
not be observed, or might be very slow, if there is
a significant energy barrier between reactants and products
(this is called the activation energy, Ea). You would need
to know Ea for the particular reaction.
If delta G < 0 and Ea is small or zero, the reaction will
quickly proceed to products. However, if Ea is big it might
be a very slow reaction.
Hope this helps!
If you have values for the change in enthalpy and entropy for a reaction
(there will be one value for the reaction of two substances together), you
can plug it into the Gibbs free energy equation to see if the reaction will
be spontaneous. Since DG = DH - TDS (where D should be the Greek symbol
delta, symbolizing the change in one of these values), where DH is the
enthalpy change and DS is the change in entropy. If the overall value of DG
is negative the reaction will be spontaneous.
For example, when water changes phases from liquid water to ice, this change
has a negative DH value (since more hydrogen bonds are formed in the
structure of ice this will release heat) and also a negative DS value since
ice is more structured than water. You can see why at higher temperatures
when -TDS is large and positive, DG will be positive and water will not
freeze. At lower temperatures -TDS will be small, and the negative DH
term will dominate leading to a negative DG value and spontaneous freezing.
Keep in mind that just because a reaction is spontaneous however does not
factor into how fast the reaction will occur. If there is a large
activation barrier a spontaneous reaction may take a very long time to
proceed. For example, iron rusting or (more morbidly) us reaching
equilibrium (kicking the bucket) are spontaneous but slow processes.
Click here to return to the Chemistry Archives
Update: June 2012