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This document addresses the question of how the Gibbs Free Energy of an equilibrating chemical system approaches its minimum value. This minimum is achieved at constant T and constant P. The extent of reaction adjusts itself until the derivative of the Gibbs Free Energy with respect to the extent of reaction is zero. This defines the position of the minimum in the Gibbs Free Energy (technically an extremum). The example chosen concerns dinitrogentetroxide decomposition, and complements the discussion in J. Chem. Ed., 65,4071988. For the reaction

the molar Gibbs free energy of the two components are

so , the Gibbs Free Energy of a mixture of moles of and moles of would be

 

We define the extent of reaction, , as

 

and

 

where the denominators are the stoichiometric coëfficients taken from the balanced chemical equation. These equations can be inverted to solve for the number of moles of each component as a function of the starting number of moles of that component and the extent of reaction. One has from Equation 4

 

and, from Equation 5, one has

 

so, the mixture's Gibbs Free Energy must be (substituting Equations 6 and 7 inot Equation 3)

whicis

We wish to take the derivative of with respect to , with the intent of setting the result equal to zero, searching for an extremum in (which we suspect is a minimum). Wishing to do this at constant pressure, we need to write each partial pressure in terms of the total pressure and the mole fraction, using Dalton's Law. We have:

where we recognize that each mole fraction is itself a function of the extent of reaction, . From here, the calculus becomes a bit messy, but the result is worth it. What we know is that

and

which is a little harder than the equimolar cases where is zero. We will do this work in parts, so that the differentiation can be explicitly followed, line by line. First, we attempt taking ther derivative of with respect to , i.e.,

 

where we need to just evaluate the remaining partial derivatives.

and

so, doing the dirty deed, we have

Bringing these two equations separately over a common denominator, one has

which become, again sequentiually,

 

 

Expanding into Equation 12

which is, upon substitution of Equations 19 and 20 yields,

which is

Q.E.D.




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Thu Nov 6 13:28:59 EST 1997