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Composition of an Equilibrium Mixture

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  • Students often wonder why many chemical reactions yield an equilibrium mixture in which a significant amount of the reactants are present, even though the products have a lower standard free energy than the reactants. One might at first think that as long as any reactants are present, the free energy could be reduced if conversion of reactants to products were complete.

    The short answer is that by "contaminating" some of the product with reactants, the free energy of the system (including both reactants and products) can be reduced below that of the pure products alone. This additional drop in the free energy is due to the free energy of mixing of reactants with products.

    Unless you are enrolled in a more advanced course, you are probably not expected to know how to calculate free energies of mixing. All you really need to know is that it is formally equivalent to the expansion of gases (or to the dilution of a solute) into a larger volume. See here for more details.

    This example illustrates how the free energies of the reaction components combine with the free energies of mixing reactants with products to minimize the Gibbs function in the equilibrium mixture. To keep things as simple as possible, we will deal with the isomerization equilibrium between the two butanes C4H10 at 298 K:

    n-butane iso-butane
    S°, J mol–1 310 295
    ΔGf°, kJ mol–1 –15.71 –17.97

    Contributors and Attributions

    Stephen Lower, Professor Emeritus (Simon Fraser U.) Chem1 Virtual Textbook