CHM 3400 – Fundamentals of Physical Chemistry

Chapter 7 - Supplementary Material

1. Free energy and the equilibrium constant

As discussed in class, for a general chemical reaction of the form

aA + bB ® cC + dD (1.1)

the relationship for the change in free energy when one mole of the reaction is carried out under conditions of constant temperature and pressure is

DGrxn = DG°rxn + RT ln Q (1.2)

where Q, the reaction quotient, is defined as

Q = (aC)c (aD)d (1.3)

(aA)a (aB)b

and aA, aB, aC, and aD are the activities of the reactants and products.

At equilibrium, DGrxn = 0. If we define the value for the reaction quotient for a system at equilibrium as K, the equilibrium constant, then it follows from eq 1.2 that

DG°rxn = - RT ln K (1.4)

or

ln K = - DG°rxn (1.5)

RT

The significance of eq 1.5 is that it makes it possible to find the numerical value for the equilibrium constant for any chemical reaction from thermochemical data, so long as such data are available for the reaction products and reactants.

2. The temperature dependence of the equilibrium constant

Values for equilibrium constants are typically calculated at 25. °C, the temperature at which thermodynamic data are often given. However, we are sometimes interested in the value for the equilibrium constant at temperatures other than 25. °C.

There are two general methods that can be used to model the temperature dependence of the equilibrium constant.

Method 1 – Assume that DH°rxn and DS°rxn are independent of temperature

For most chemical reactions, particularly for relatively small changes in temperature, it is a good first approximation to assume that both DH°rxn and DS°rxn (but not DG°rxn) are independent of temperature. In that case, the following equation can be derived

ln(K2/K1) = - DH°rxn [ (1/T2) - (1/T1) ] (2.1)

R

where K1 is the value for the equilibrium constant at T1, and K2 is the value at T2.

For cases where values for the equilibrium constant are available over a range of temperatures, these data can be used to find an experimental value for DH°rxn by plotting ln K vs (1/T). The slope of the best fitting line to the data for such a plot is

m = - DH°rxn/R (2.2)

The Clausius-Clapeyron equation is an example of the use of eq 2.1 and 2.2 for the special case of a phase transition from a condensed phase (solid or liquid) to the gas phase.

Method 2 – Assume that DH°rxn and DS°rxn depend on temperature

For precise work there is a better, although longer, method that can be used to find more precise values for K at different temperatures. The method works as follows:

1) Find DH°rxn and DS°rxn at the temperature of interest. We can do this using the general relationships

DH°(T2) = DH°(T1) + òT1T2 DCp dT (2.3)

and

DS°(T2) = DS°(T1) + òT1T2 (DCp/T) dT (2.4)

where DCp, the difference between the constant pressure heat capacities of the products and the reactants, is

DCp = S Cp(products) - S Cp(reactants) (2.5)

We also assume that values for DH°rxn and DS°rxn can be found at some reference temperature T1 (often taken as 25. °C). Note that if we can assume that DCp is independent of temperature, then DCp can be brought outside of the integrals in eq 2.3 and 2.4, simplifying the final results.

2) Find DG°rxn at the temperature of interest. If we know values for DH°rxn and DS°rxn at some temperature T2 (by using eq 2.3 and 2.4, for example), then it follows from the definition of free energy that

DG°rxn(T2) = DH°rxn(T2) – T2 DS°rxn(T2) (2.6)

where DG°rxn(T2), DH°rxn(T2), and DS°rxn(T2) are the change in free energy, enthalpy, and entropy for the reaction at temperature T2.

3) Find the value for K

If we know the value for DG°rxn(T2), then the corresponding value for K, the equilibrium constant at temperature T2, can be found using eq 1.5.

Which of the above procedures should be used to find the value of K at a particular temperature? Usually Method 1 should be used, because it is simple and reasonably accurate. However, for a more accurate value for K, and assuming the data needed are available, Method 2 is preferred.