**2. Chemical reaction equilibria**

The equilibrium state achieved by a system where a group of chemical reactions take place simultaneously can be entirely modeled and predicted by applying the principles of classical thermodynamics.

Supposing that we want to react some solid transition metal oxide, say M2O5, with gaseous Cl2. Lets consider for simplicity that the reaction can result in the formation of only one gaseous chlorinated specie, say MCl5. The transformation is represented by the following equation:

$$\rm{M\_2O\_s(s) + 5Cl\_2(g) \to 2MCl\_s(g) + \frac{5}{2}O\_2(g)}\tag{1}$$

In this system there are only two phases, the pure solid oxide M2O5 and a gas phase, whose composition is characterized by definite proportions of Cl2, O2, and MCl5. If temperature, total pressure, and the total molar amounts of O, Cl, and M are fixed, the chemical equilibrium is calculated by finding the global minimum of the total Gibbs energy of the system (Robert, 1993).

$$\mathbf{G} = \eta\_{\mathrm{M}\_2\mathrm{O}\_5}^s \mathbf{g}\_{\mathrm{M}\_2\mathrm{O}\_5}^s + \mathbf{G}^\mathrm{\mathcal{S}} \tag{2}$$

Where <sup>s</sup> OM <sup>52</sup> *g* represents the molar Gibbs energy of pure solid M2O5 at reaction's temperature and total pressure, <sup>s</sup> OM <sup>52</sup> *<sup>n</sup>* the number of moles of M2O5 and <sup>g</sup> *<sup>G</sup>* the molar Gibbs energy of the gaseous phase, which can be computed through the knowledge of the *chemical potential* of all molecular species present ( <sup>g</sup> MCl g O g Cl <sup>22</sup> <sup>5</sup> , , ):

$$\begin{aligned} \mathbf{G}^{\mathcal{S}} &= n\_{\mathrm{Cl}\_{2}}^{\mathcal{S}} \mu\_{\mathrm{Cl}\_{2}}^{\mathcal{S}} + n\_{\mathrm{O}\_{2}}^{\mathcal{S}} \mu\_{\mathrm{O}\_{2}}^{\mathcal{S}} + n\_{\mathrm{MCl}\_{5}}^{\mathcal{S}} \mu\_{\mathrm{MCl}\_{5}}^{\mathcal{S}}\\ \mu\_{\mathrm{Cl}\_{2}}^{\mathcal{S}} &= \left(\frac{\partial \mathbf{G}^{\mathcal{S}}}{\partial n\_{\mathrm{Cl}\_{2}}^{\mathcal{S}}}\right)\_{T, P, n\_{\mathrm{O}\_{2}}, n\_{\mathrm{MCl}\_{5}}} \end{aligned} \tag{3}$$

The minimization of function (2) requires that for the restrictions imposed to the system, the first order differential of *G* must be equal to zero. By fixing the reaction temperature (*T*) and pressure (*P*) and total amount of each one of the elements, this condition can be written according to equation (4) (Robert, 1993).

for a review on chlorination thermodynamics which can combine its basic aspects with a

The present chapter will first focus on the thermodynamic basis necessary for understanding the nature of the equilibrium states achievable through chlorination reactions of metallic oxides. Possible ways of graphically representing the equilibrium conditions are discussed and compared. Moreover, the chlorination of V2O5, both in the absence as with the presence of graphite will be considered. The need of such reducing agent is clearly explained and discussed. Finally, the equilibrium conditions are appreciated through the construction of graphics with different levels of complexity, beginning with the well known <sup>o</sup> *G*r x *T* diagrams, and ending with gas phase speciation diagrams, rigorously calculated

The equilibrium state achieved by a system where a group of chemical reactions take place simultaneously can be entirely modeled and predicted by applying the principles of classical

Supposing that we want to react some solid transition metal oxide, say M2O5, with gaseous Cl2. Lets consider for simplicity that the reaction can result in the formation of only one gaseous chlorinated specie, say MCl5. The transformation is represented by the following

gO2

<sup>52</sup> <sup>2</sup> <sup>5</sup> gMCl2gCl5sOM <sup>2</sup>

In this system there are only two phases, the pure solid oxide M2O5 and a gas phase, whose composition is characterized by definite proportions of Cl2, O2, and MCl5. If temperature, total pressure, and the total molar amounts of O, Cl, and M are fixed, the chemical equilibrium is calculated by finding the global minimum of the total Gibbs energy of the

25 25

the gaseous phase, which can be computed through the knowledge of the *chemical potential*

MCl

g g g gg g g Cl Cl O O MCl MCl

<sup>2</sup> 2 5

The minimization of function (2) requires that for the restrictions imposed to the system, the first order differential of *G* must be equal to zero. By fixing the reaction temperature (*T*) and pressure (*P*) and total amount of each one of the elements, this condition can be written

*Cl TPn n*

,, , *O MCl*

g O

*g g*

*G n*

 

Cl <sup>22</sup> <sup>5</sup> , , ):

*Gn n n*

g

2

g Cl

OM <sup>52</sup> *g* represents the molar Gibbs energy of pure solid M2O5 at reaction's temperature

2 2 22 5 5

OM <sup>52</sup> *<sup>n</sup>* the number of moles of M2O5 and <sup>g</sup> *<sup>G</sup>* the molar Gibbs energy of

 

5

s s <sup>g</sup> *Gn g G* MO MO (2)

(1)

(3)

through the minimization of the total Gibbs energy of the system.

now available new kind of approach.

**2. Chemical reaction equilibria** 

thermodynamics.

system (Robert, 1993).

and total pressure, <sup>s</sup>

of all molecular species present ( <sup>g</sup>

according to equation (4) (Robert, 1993).

equation:

Where <sup>s</sup>

$$d\mathbf{G}\_{\rm T\_{+}P,\mu\_{\rm O},\mu\_{\rm Cl},\mu\_{\rm M}}=\mathbf{0}$$

$$d\mathbf{G}=\mathbf{g}\_{\rm M\_{2}O\_{5}}^{s}\,d\eta\_{\rm M\_{2}O\_{5}}+\mu\_{\rm Cl\_{2}}^{g}\,d\eta\_{\rm Cl\_{2}}+\mu\_{\rm O\_{2}}^{g}\,d\eta\_{\rm O\_{2}}+\mu\_{\rm KCl\_{5}}^{g}\,d\eta\_{\rm MCl\_{5}}\tag{4}$$

$$\mathbf{g}\_{\rm M\_{2}O\_{5}}^{s}\,d\eta\_{\rm M\_{2}O\_{5}}+\mu\_{\rm Cl\_{2}}^{g}\,d\eta\_{\rm Cl\_{2}}+\mu\_{\rm O\_{2}}^{g}\,d\eta\_{\rm O\_{2}}+\mu\_{\rm KCl\_{5}}^{g}\,d\eta\_{\rm MCl\_{5}}=0$$

The development of the chlorination reaction can be followed through introduction of a reaction coordinate called *degree of reaction* ( ),whose first differential is computed by the ratio of its molar content variation of each specie participating in the reaction and the stoichiometric coefficient(Eq. 1).

$$d\mathcal{E} = \frac{dn\_{\text{Cl}\_2}}{\left(-\text{S}\right)} = \frac{dn\_{\text{M}\_2\text{O}\_5}}{\left(-1\right)} = \frac{dn\_{\text{O}\_2}}{\left(+\text{S}/2\right)} = \frac{dn\_{\text{MCl}\_5}}{\left(+2\right)}\tag{5}$$

The numbers inside the parenthesis in the denominators of the fractions contained in equation (5) are the stoichiometric coefficient of each specie multiplied by "-1" if it is represented as a reactant, or "+1" if it is a product. The equilibrium condition (Eq. 4) can now be rewritten in the following mathematical form:

$$\begin{cases} -g\_{\mathrm{M}\_{2}\mathrm{O}\_{5}}^{s}\,d\varepsilon - 5\,\mu\_{\mathrm{Cl}\_{2}}^{g}\,d\varepsilon + \frac{5}{2}\,\mu\_{\mathrm{O}\_{2}}^{g}\,d\varepsilon + 2\,\mu\_{\mathrm{MCl}\_{5}}^{g}\,d\varepsilon = 0\\ \left(-g\_{\mathrm{M}\_{2}\mathrm{O}\_{5}}^{s} - 5\,\mu\_{\mathrm{Cl}\_{2}}^{g} + \frac{5}{2}\,\mu\_{\mathrm{O}\_{2}}^{g} + 2\,\mu\_{\mathrm{MCl}\_{5}}^{g}\right)\mathrm{d}\varepsilon = 0 \end{cases} \tag{6}$$

At the desired equilibrium state the condition defined by Eq. (6) must be valid for all possible values of the differential *d* . This can only be accomplished if the term inside the parenthesis is equal to zero. This last condition is the simplest mathematical representation for the chemical equilibrium associated with reaction (1).

$$-g\_{\mathrm{M}\_2\mathrm{O}\_5}^s - 5\,\mu\_{\mathrm{Cl}\_2}^g + \frac{5}{2}\,\mu\_{\mathrm{O}\_2}^g + 2\,\mu\_{\mathrm{MCl}\_5}^g = 0\tag{7}$$

The chemical potentials can be computed through knowledge of the molar Gibbs energy of each pure specie in the gas phase, and its chemical activity. For the chloride MCl5, for example, the following function can be used (Robert, 1993):

$$
\mu\_{\text{MCl}\_3}^{\text{g}} = \text{g}\_{\text{MCl}\_3}^{\text{g}} + RT \ln a\_{\text{MCl}\_3}^{\text{g}} \tag{8}
$$

Where <sup>g</sup> MCl5 *a* represents the chemical activity of the component MCl5 in the gas phase. By introducing equations analogous to Eq. (8) for all components of the gas phase, Eq. (7) can be rewritten according to Eq. (9). There, the activity of M2O5 is not present in the term located at the left hand side because, as this oxide is assumed to be pure, its activity must be equal to one (Robert, 1993).

$$\ln\left(\frac{a\_{\rm MC1\_5}^2 a\_{\rm O\_2}^{5/2}}{a\_{\rm Cl\_2}^5}\right) = -\frac{\left(2\,\mathrm{g}\_{\rm MC1\_5}^\mathrm{g} + \frac{5}{2}\,\mathrm{g}\_{\rm O\_2}^\mathrm{g} - 5\,\mathrm{g}\_{\rm Cl\_2}^\mathrm{g} - \mathrm{g}\_{\rm M\_2\mathrm{O\_5}}^\mathrm{s}\right)}{RT} = -\frac{\Delta\mathrm{G}\_r}{RT} \tag{9}$$

On the Chlorination Thermodynamics 789

situation defines a process where in the achieved equilibrium state, the atmosphere tends to be richer in the desired products. The second situation characterizes a reaction where the reactants are present in higher concentration in equilibrium. Finally, the third possibility defines the situation where products and reactants are present in amounts of the same order

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,

> 2 5 22 5 s g gg M O Cl O MCl

The left hand side of inequality (14) defines the thermodynamic driving force of the reaction

 

<sup>2</sup> *<sup>g</sup>*

 

 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

<sup>5</sup> 5 20

2 5 22 5 s g gg r MO Cl O MCl <sup>5</sup> 5 2 2

<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

> 5 2 2

ln ln

2 5/2 o o MCl O r r 5 r Cl

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

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

*P P G RT G RT Q <sup>P</sup>*

 

*g* (15)

(14)

(16)

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

words, the following inequality must then be valid:

(Eq. 8), it is possible to rewrite the driving force in a more familiar way:

gas removes it from the atmosphere in the neighborhood of the sample.

know the temperature dependence of the reaction Gibbs energy.

of magnitude.

then *d*

(*<sup>r</sup>* ).

If 

The numerator of the right side of Eq. (9) represents the molar Gibbs energy of reaction (1). It involves only the molar Gibbs energies of the species participating in the reaction as pure substances, at *T* and *P* established in the reactor. The molar Gibbs energy of a pure component is only a function of *T* and *P* (Eq. 10), so the same must be valid for the reactions Gibbs energy (Robert, 1993).

$$d\lg\left(T,P\right) = -s dT + \upsilon dP \tag{10}$$

Where *s* and denote respectively the molar entropy and molar volume of the material, which for a pure substance are themselves only a function of *T* and *P*.

It is a common practice in treating reactions involving gaseous species to calculate the Gibbs energy of reaction not at the total pressure prevailing inside the reactor, but to fix it at 1 atm. This is in fact a reference pressure, and can assume any suitable value we desire. The molar Gibbs energy of reaction is in this case referred to as the *standard molar Gibbs energy of reaction*. According to this definition, the standard Gibbs energy of reaction must depend only on the reactor's temperature.

By assuming that the total pressure inside the reactor (*P*) is low enough for neglecting the effect of the interactions among the species present in the gas phase, Eq. (9) can be rewritten in the following form:

$$\frac{P\_{\text{MCl}\_5} \,^2 P\_{\text{O}\_2} \,^{5/2}}{P\_{\text{Cl}\_2} \,^5} = \exp\left(-\frac{\left(2\,\text{g}\_{\text{MCl}\_5}^6 + \frac{5}{2}\,\text{g}\_{\text{O}\_2}^6 - 5\,\text{g}\_{\text{Cl}\_2}^6 - \text{g}\_{\text{M}\_2\text{O}\_5}^8\right)}{RT}\right) \tag{11}$$

The activities were calculated as the ratio of the partial pressure of each component and the reference pressure chosen (*P* = 1 atm). This proposal is based on the thermodynamic description of an ideal gas (Robert, 1993). For MCl5, for example, the chemical activity is calculated as follows:

$$a\_{\rm MC1\_5} = \frac{P\_{\rm MC1\_5}}{1} = P\_{\rm MC1\_5} = \chi\_{\rm MC1\_5}^{\rm g} P \tag{12}$$

Where <sup>g</sup> MCl <sup>5</sup> *x* stands for the mol fraction of MCl5 in the gas phase. Similar relations hold for the other species present in the reactor atmosphere. The activity is then expressed as the product of the mol fraction of the specie and the total pressure exerted by the gaseous solution.

The right hand side of Eq. (11) defines the equilibrium constant (*K*) of the reaction in question. This quantity can be calculated as follows:

$$K = \exp\left(-\frac{\Delta G\_r^\circ}{RT}\right) \tag{13}$$

The symbol "o" is used to denote that The molar reaction Gibbs energy ( <sup>o</sup> *G*<sup>r</sup> ) is calculated at a reference pressure of 1 atm.

At this point, three possible situations arise. If the standard molar Gibbs energy of the reaction is negative, then *K* > 1. If it is positive, *K* < 1 and if it is equal to zero *K* = 1. The first

The numerator of the right side of Eq. (9) represents the molar Gibbs energy of reaction (1). It involves only the molar Gibbs energies of the species participating in the reaction as pure substances, at *T* and *P* established in the reactor. The molar Gibbs energy of a pure component is only a function of *T* and *P* (Eq. 10), so the same must be valid for the reactions

It is a common practice in treating reactions involving gaseous species to calculate the Gibbs energy of reaction not at the total pressure prevailing inside the reactor, but to fix it at 1 atm. This is in fact a reference pressure, and can assume any suitable value we desire. The molar Gibbs energy of reaction is in this case referred to as the *standard molar Gibbs energy of reaction*. According to this definition, the standard Gibbs energy of reaction must depend

By assuming that the total pressure inside the reactor (*P*) is low enough for neglecting the effect of the interactions among the species present in the gas phase, Eq. (9) can be rewritten

g gg <sup>s</sup> 2 5/2 MCl O Cl M O MCl O

The activities were calculated as the ratio of the partial pressure of each component and the reference pressure chosen (*P* = 1 atm). This proposal is based on the thermodynamic description of an ideal gas (Robert, 1993). For MCl5, for example, the chemical activity is

*g g gg P P*

*P RT*

*P a* <sup>g</sup>

MCl MCl <sup>5</sup> <sup>5</sup> 5

of the mol fraction of the specie and the total pressure exerted by the gaseous solution.

<sup>5</sup> 2 5 2

*PxP*

<sup>5</sup> <sup>1</sup> (12)

(13)

MCl MCl

o

*RT* 

*x* stands for the mol fraction of MCl5 in the gas phase. Similar relations hold for the

other species present in the reactor atmosphere. The activity is then expressed as the product

The right hand side of Eq. (11) defines the equilibrium constant (*K*) of the reaction in

<sup>r</sup> exp *<sup>G</sup> <sup>K</sup>*

The symbol "o" is used to denote that The molar reaction Gibbs energy ( <sup>o</sup> *G*<sup>r</sup> ) is calculated

At this point, three possible situations arise. If the standard molar Gibbs energy of the reaction is negative, then *K* > 1. If it is positive, *K* < 1 and if it is equal to zero *K* = 1. The first

which for a pure substance are themselves only a function of *T* and *P*.

exp

denote respectively the molar entropy and molar volume of the material,

5 22 2 5

(11)

*dgT P sdT vdP* , (10)

Gibbs energy (Robert, 1993).

only on the reactor's temperature.

5 2 2

question. This quantity can be calculated as follows:

5 Cl

in the following form:

calculated as follows:

MCl <sup>5</sup>

at a reference pressure of 1 atm.

Where <sup>g</sup>

Where *s* and

situation defines a process where in the achieved equilibrium state, the atmosphere tends to be richer in the desired products. The second situation characterizes a reaction where the reactants are present in higher concentration in equilibrium. Finally, the third possibility defines the situation where products and reactants are present in amounts of the same order of magnitude.
