**2.1 Thermodynamic driving force and** <sup>o</sup> *G*<sup>r</sup> **vs. T diagrams**

Equation (6) can be used to formulate a mathematical definition of the thermodynamic driving force for a chlorination reaction. If the reaction proceeds in the desired direction, then *d* must be positive. Based on the fact that by fixing *T*, *P*, *n*(O), *n*(Cl), and *n*(M) the total Gibbs energy of the system is minimum at the equilibrium, the reaction will develop in the direction of the final equilibrium state, if and only if, the value of *G* reduces, or in other words, the following inequality must then be valid:

$$-g\_{\mathrm{M}\_2\mathrm{O}\_5}^s - 5\mu\_{\mathrm{Cl}\_2}^g + \frac{5}{2}\mu\_{\mathrm{O}\_2}^g + 2\mu\_{\mathrm{MCl}\_5}^g < 0\tag{14}$$

The left hand side of inequality (14) defines the thermodynamic driving force of the reaction (*<sup>r</sup>* ).

$$
\Delta\mu\_{\rm r} = -g\_{\rm M\_2O\_5}^{\rm s} - 5\,\mu\_{\rm Cl\_2}^{\rm g} + \frac{5}{2}\,\mu\_{\rm O\_2}^{\rm g} + 2\,\mu\_{\rm MCl\_5}^{\rm g} \tag{15}
$$

If <sup>r</sup> is negative, classical thermodynamics says that the process will develop in the direction of obtaining the desired products. However, a positive value is indicative that the reaction will develop in the opposite direction. In this case, the formed products react to regenerate the reactants. By using the mathematical expression for the chemical potentials (Eq. 8), it is possible to rewrite the driving force in a more familiar way:

$$
\Delta\mu\_{\rm r} = \Delta G\_{\rm r}^{\rm o} + RT\ln\left(\frac{P\_{\rm MC1\_3}^2 P\_{\rm O\_2}^{5/2}}{P\_{\rm C1\_2}^5}\right) = \Delta G\_{\rm r}^{\rm o} + RT\ln Q \tag{16}
$$

According to Eq. (16), the ratio involving the partial pressure of the components defines the so called *reaction coefficient* (*Q*). This parameter can be specified in a given experiment by injecting a gas with the desired proportion of O2 and Cl2. The partial pressure of MCl5, on the other hand, would then be near zero, as after the formation of each species, the fluxing gas removes it from the atmosphere in the neighborhood of the sample.

At a fixed temperature and depending on the value of *Q* and the standard molar Gibbs energy of the reaction considered, the driving force can be positive, negative or zero. In the last case the reaction ceases and the equilibrium condition is achieved. It is important to note, however, that by only evaluating the reactions Gibbs energy one is not in condition to predict the reaction path followed, then even for positive values of <sup>o</sup> *G*<sup>r</sup> , it is possible to find a value *Q* that makes the driving force negative. This is a usual situation faced in industry, where the desired equilibrium is forced by continuously injecting reactants, or removing products. In all cases, however, for computing reaction driving forces it is vital to know the temperature dependence of the reaction Gibbs energy.

On the Chlorination Thermodynamics 791

Further, for a reaction defined by Eq. (1) the number of moles of gaseous products is higher than the number of moles of gaseous reactants, which, based on the ideal gas model, is indicative that the chlorination leads to a state of grater disorder, or greater entropy. In this

The same can not be said about the molar reaction enthalpy. In principle the chlorination reaction can lead to an evolution of heat (exothermic process, then <sup>o</sup> *H*<sup>r</sup> < 0) or absorption of heat (endothermic process, then <sup>o</sup> *H*<sup>r</sup> > 0). In the first case the linear coefficient is positive, but in the later it is negative. Hypothetical cases are presented in Fig. (2) for the chlorination of two oxides, which react according to equations identical to Eq. (1). The same molar reaction entropy is observed, but for one oxide the molar enthalpy is positive, and for the

Finally, it is worthwhile to mention that for some reactions the angular coefficient of the straight line can change at a particular temperature value. This can happen due to a phase transformation associated with either a reactant or a product. In the case of the reaction (1), only the oxide M2O5 can experience some phase transformation (melting, sublimation, or ebullition), all of them associated with an increase in the molar enthalpy of the phase. According to classical thermodynamics, the molar entropy of the compound must also

> t t

*<sup>H</sup> <sup>S</sup>* (18)

<sup>t</sup> *T*

Where <sup>t</sup> *S* , *H* <sup>t</sup> and *T*t represent respectively, the molar entropy, molar enthalpy and temperature of the phase transformation in question. So, to include the effect for melting of M2O5 at a temperature *T*t, the molar reaction enthalpy and entropy must be modified as

<sup>r</sup> *S* < 0), as

particular case then, the straight line must have negative linear coefficient (- <sup>o</sup>

Fig. 2. Endothermic and exothermic reactions

depicted in the graph of Figure (1).

other it is negative.

increase (Robert, 1993).

follows.
