Week of May 6th Chemistry Problem: Analysis of a Common Redox Reaction

The thermite reaction is one of the most spectacular chemical reactions one can demonstrate in a classroom or laboratory (for a controlled example of this reaction, see http://www.youtube.com/watch?v=a8XSmSdvEK4). It is the reaction of aluminum metal with iron(III) oxide to yield iron metal and aluminum oxide.

How would one describe this reaction and calculate the change in free energy for an Introductory Chemistry course?

For starters, the reaction is clearly a single replacement reaction that is also described as a redox reaction. The reducing agent is clearly aluminum metal, which provides the electrons to reduce the iron in iron(III) oxide as shown below (the oxidation states are written above each of the metals)

Each aluminum atom provides 3 electrons to each iron atom. Note that since the reaction releases so much heat (see video above), this reaction is clearly exothermic (and therefore the change in enthalpy (∆G°) is negative, about -848 kJ/mol). Since the reaction does occur spontaneously, this must mean the free energy change (∆G°) is negative also. To calculate the change in free energy at room temperature (knowing the enthalpy change to be -848 kJ/mol) we must first calculate the entropy (∆S°) for the reaction, by summing the entropy of the products and subtracting the entropy of the reactants (values are listed in the Appendix of all good general chemistry textbooks).

Change in entropy = [2(27.3) + (50.9)] J/(mol K) - [2(28.3) + 87.4] J/(mol K) =  -38.5 J/(mol K)

Since the entropy change is negative, this is clearly an enthalpy-driven reaction according to the equation

∆G° = ∆H° - T∆S°

Using our calculated values for enthalpy and entropy change for the reaction at room temperature (298 K), we calculate

∆G° = -837 kJ/mol

This very negative value indicates the thermite reaction is highly-favored energetically (under standard conditions), with products that are much more thermodynamically stable relative to reactants.

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