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Name: Jacqueline
Status: educator
Grade: 9-12
Location: FL
Date: December 2007


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?



Replies:
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)


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


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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