Structure & Reactivity in Organic, Biological and Inorganic Chemistry by Chris Schaller is licensed under a Creative Commons Attribution-NonCommercial 3.0 Unported License.
Reactivity in Chemistry
Thermodynamics
TD8. Hess' Law
Sometimes we can use available information about the energetics of known reactions to predict the energetics of a new reaction. That's because energy is a state function. That means it doesn't matter how we arrive at a particular state, such as a set of molecules under a specific set of conditions. Their energy will always be the same, no matter how they were made.
We can use this rule to calculate the energy change associated with a reaction. The idea is pretty simple. Suppose we are interested in a reaction in which A is turned into B:
A → B
We don't know the energy change of that reaction, but we have a couple of other reactions that we do know something about. In one, A reacts to make a different product, C:
A → C
In another, C reacts to make B:
C → B
Suppose we do know the energy changes for both of those reactions, A going to C and C going to B. Then we can imagine the reaction in which A goes to B takes place in two steps, stopping first at C, even if it doesn't really happen that way. The energy to go from A to B is just the sum of the energy changes of those two reactions.
To see how this works, we'll look at a relatively simple reaction: the combustion of carbon.
A couple of things could happen when we burn some carbon. Burning typically results in the combination of the elements in a material with oxygen. So we are just talking about combining carbon with oxygen to make a new compound.
There are two possibilities. Either the reaction forms carbon monoxide, or it forms carbon dioxide.
The first reaction is:
2 C + O2 → 2 CO
That reaction is exothermic. It releases about 25 kcal per mol of CO produced.
Note that the reaction has been balanced to keep track of the exact numbers of atoms involved in the transaction. Because only one oxygen atom is needed, and the oxygen molecules comes with two oxygen atoms, the oxygen molecule can actually convert two carbon atoms into carbon monoxide.
The second reaction is:
C + O2 → CO2
That reaction is also exothermic. It releases about 90 kcal per mol of CO2 produced.
Because energy is released, or produced, by each reaction, we can think about the energy as another product of the reaction. We'll just list it on the product side of the equation with the other products.
2 C + O2 → 2 CO + 50 kcal/mol
and
C + O2 → CO2 + 90 kcal/mol
There's a third reaction that is related to these two reactions. It's the combustion of carbon monoxide.
2 CO + O2 → 2 CO2
Again, the reaction is exothermic, releasing about 65 kcal per mole of CO2 produced. It would release 130 kcal for the reaction as written, because we are showing the production of two moles of CO2. Rewriting the equation to include the energy produced:
2 CO + O2 → 2 CO2 + 130 kcal/mol
What if we conducted this reaction in stages? What if we combusted the carbon to carbon monoxide, then took the carbon monoxide and allowed it to react further to get carbon dioxide?
2 C + O2 → 2 CO + 50 kcal/mol
2 CO + O2 → 2 CO2 + 130 kcal/mol
Imagine this is a pair of algebraic equations. What would happen if we added them together?
2 C + O2 = 2 CO + 50 kcal/mol
2 CO + O2 = 2 CO2 + 130 kcal/mol
Sum: 2 C + 2 CO + 2 O2 = 2 CO + 2 CO2 + 180 kcal/mol
Note that the CO appears on both sides and would cancel.
2 C + 2 O2 = 2 CO2 + 180 kcal/mol
We can drop the factor of 2:
C + O2 = CO2 + 90 kcal/mol
So two reactions, one after the other, would add up to a third. In addition, the energies of those two reactions, added together, give the energy of the third.
This result is a pretty important aspect of thermodynamics. Enthalpy is a state function. that means it doesn't matter how a reaction is performed. Whether we convert carbon directly into carbon dioxide or we convert it to carbon monoxide, then continue, the e nergy involved is the same overall. That's because the energy of the reaction is a property of the products and the reactants only. It is independent of how we get from one to the other.
One more note on the reactions above. The enthalpies for these reactions, if measured in the correct way, are sometimes called the heats of formation of the compounds. The heat of formation refers to the energy change when the compounds are formed from the elements under standard conditions. We're not going too deeply into what those standard conditions are here. However, because C is the elemental form of carbon and O2 is the elemental form of oxygen, we would loosely consider the energies listed above to be heats of formation.
When you hear the phrase "heat of formation", we're just talking about the formation of the compound from the elements.
Problem TD6c.1.
a) If the heat of formation of potassium chloride, KCl, is -104 kcal/mol, and the heat of formation or potassium chlorite, KClO2, is -95 kcal/mol, then what is the heat of reaction when potassium chloride reacts with oxygen to produce potassium chlorite?
b) If the heat of formation of tantalum(IV) oxide, TaO2, is -40 kcal/mol, and the heat of formation of tantalum(V) oxide, Ta2O5, is -490 kcal/mol, then what is the heat of reaction for the combustion of TaO2 to Ta2O5?
c) If the heat of formation of carbon monoxide, CO, is -25 kcal/mol and the heat of formation of tetracarbonyl nickel, Ni(CO)4, is -145 kcal/mol, then what is the heat of reaction for the formation of tetracarbonyl nickel from nickel and carbon monoxide?
This site was written by Chris P. Schaller, Ph.D., College of Saint Benedict / Saint John's University (retired) with other authors as noted on individual pages. It is freely available for educational use.
Structure & Reactivity in Organic, Biological and
Inorganic Chemistry by
Chris Schaller is licensed under a
Creative
Commons Attribution-NonCommercial 3.0 Unported License.
Send corrections to cschaller@csbsju.edu
This material is based upon work supported by the National Science Foundation under Grant No. 1043566.
Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.
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