P

where Ge is the free energy of 1 mole of gas at a pressure of 1 bar.

Applying equation (3.16) to each component of the reaction gives aGA = a(GA + RT In Pa) bGB = b(Ge + RT In pB)

etc.

A Ge is the standard free energy change of the reaction, given by

As we have noted previously, the free energy change for systems at equilibrium is zero, and hence equation (3.17) becomes

Substituting from equation (3.15) gives

Substituting equation (3.19) into equation (3.17) gives

Equation (3.20) gives the change in free energy when a moles of A at a partial pressure pA and b moles of B at a partial pressure pB react together to yield c moles of C at a partial pressure pC and d moles of D at a partial pressure pD. For such a reaction to occur spontaneously, the free energy change must be negative, and hence equation (3.20) provides a means of predicting the ease of reaction for selected partial pressures (or concentrations) of the reactants.

section 3.3.1) of the components rather than the partial pressures. A Ge values can readily be calculated from the tabulated data and hence equation (3.19) is important because it provides a method of calculating the equilibrium constants without resort to experimentation.

A useful expression for the temperature dependence of the equilibrium constant is the van'tHoff equation (equation 3.23), which may be derived as outlined in Box 3.3. A more general form of this equation is log K =

-AHe 2.303RT

+ constant

Plots of log K against 1/T should be linear with a slope of -AHe/2.303R, from which AHe may be calculated.

Equations (3.19) and (3.24) are fundamental equations which find many applications in the broad area of the pharmaceutical sciences: for example, in the determination of equilibrium constants in chemical reactions and for micelle formation; in the treatment of stability data for some solid dosage forms (see section 4.4.3); and for investigations of drug-receptor binding.

Box 3.3 Derivation of the van't Hoff equation

From equation (3.19),

-AGe RT Since

AGe=AH®-TASe then at a temperature T1

In K

If we assume that the standard enthalpy change A H® and the standard entropy change AS® are independent of temperature, then subtracting equation (3.21) from equation (3.22) gives ln - ln = (1 -1

0 0

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