Endergonic vs. Exergonic

Name: Jacqueline
Status: educator
Grade: 9-12
Location: FL

Question: I have confused myself in teaching endergonic versus
exergonic reactions when my students tested baking soda in vinegar
and antacid tablets in water. I initially thought that the cold
tube/canister showed a great example of an endergonic reaction, but
I was thinking of endoTHERMIC. Do these two go together? Can one
assume that if the substance produced is cold that it is both an
absorption of heat and a positive free energy change? How could it
be explained from the free energy equation?
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Jacqueline,

As you already know, endergonic and exergonic describe processes that
allow the system to absorb or release energy respectively through work.
In contrast, endothermic and exothermic describe processes that allow
the system to absorb or release energy respectively through heat. As
such, since heat and work are two independent processes they do not
necessarily go in the same direction. Notice, for example that internal
energy is defined as the sum of the heat and work processes done on the
system.

It is also important to realize that temperature can be independent of
energy absorption. While in most cases, when work is done on a gas -such
as the quick compression of a bicycle pump- the gas increases in
temperature, an increase in internal energy may be manifested in various
ways not related to temperature. For example, heating water at 100degC
does not result in an increase in temperature but rather a change of
phase (entropy increases).

Greg (Roberto Gregorius)
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Dear Jacqueline,

"Endergonic vs. Exergonic" refers to whether
the change in Gibbs free energy , delta G,
is positive or negative. delta G is related to
the equilibrium constant K of a reaction through the
equation

delta G = - RT ln K

Thus an exergonic reaction has K > 1, and favors products,
whereas an endergonic reaction has K < 1 and favors reactants.

"Endothermic vs. Exothermic" refers to whether the change in
enthalpy, delta H, is postive or negative. Delta H equals the
heat flow caused by the reaction at constant pressure. The relationship
between the two is given by the equation

delta G = delta H - T * delta S

where T is the absolute temperature in kelvin and delta S is the
entropy of reaction. Delta S can be positive or negative depending
on the reaction of interest. As an approximation we can think of
delta H and delta S as being constants (independent of temperature)
for a particular reaction. Thus we have four possibilities:

(1) delta S > 0 and delta H < 0 always gives delta G < 0.
(2) delta S < 0 and delta H > 0 always gives delta G > 0.
(3) delta S > 0 and delta H > 0 will only give delta G < 0 at
    sufficiently high temperatures.
(4) delta S < 0 and delta H < 0 will only give delta G < 0 at
    sufficiently low temperatures.

So, an exothermic reaction can be either endergonic or exergonic.
The same is also true for an endothermic reaction, as in your
example. You'd have delta H > 0 (endothermic)
but delta G < 0 (exergonic).

I hope this helps!

best, Dr. Topper
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Yes, there is. You need a table
which contains thermodynamic information
for each reactant and each product.
The easiest is if you have a table of
Gibbs free energies of formation (dGf)
for all reactants and products. This is
usually in the back of any college-level
general chemistry textbook.


Once you have the dGf values, delta G for
the reaction is obtained as follows:


delta G = (sum over products' dGf) - (sum over reactants' dGf)


You have to be careful when looking things up because
the dGf value depends on the chemical (H2O) and the
phase (solid, liquid, gas, aqueous).


In the sum, you multiply each dGf by the stoichiometric
coefficient for each chemical. So for the combustion of ethanol,


C2H5O5 (l) + 3 O2(g)  ---> 2 CO2 (g)  + 3 H2O (g)


(sum over products' dGf)  = 2 * dGf(CO2) + 3 * dGf(H2O,g)
(sum over reactants' dGf) = dGf(C2H5O5,l) + 3 * dGf(O2)


delta G =  2 * dGf(CO2) + 3 * dGf(H2O,g) - [dGf(C2H5O5,l) + 3 * dGf(O2)]


From Ebbing and Gammon's General Chemistry (8th ed., App C)
delta G = 2*(-394.4) + 3*(-228.6) - [-174.90 + 3 * 0.00] = -1299.7 kJ /
mole


which is very strongly spontaneous / exergonic.


In your case you will need to write the balanced chemical reaction
and know the free energy of formation data for all
of your own reactants and products.


The other way to get at delta G is to find delta H and delta S for your
reaction. These either can be looked up as stand-alone numbers or
calculated from dHf data or dS data using the same prods-reacts sum.


Once you have delta H and delta S,


delta G = delta H - T * delta S


where T is in kelvin and delta S is in kJ / (mole kelvin).


Good luck, and best wishes!


Dr. Topper
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