# De Donder's Treatment of Chemical Equilibrium

## De Donder's Treatment of Chemical Equilibrium

Let us consider a general equation in a closed system-aA + bB ⇌ cC +dD

The Gibbs free energy, G = ƒ(T,P,n)

Hence, the gibbs free energy change is given by-

or, dG = −SdT + VdP + µ

_{A}dn

_{A}+ µ

_{B}dn

_{B}+ µ

_{C}dn

_{C}+ µ

_{D}dn

_{D}---Equation-1

If the reaction takes palce at constat T and P, we have-

(dG)

_{T,P}= µ

_{A}dn

_{A}+ µ

_{B}dn

_{B}+ µ

_{C}dn

_{C}+ µ

_{D}dn

_{D}---Equation-2

For a closed system, the reaction stoichiometry requires that a change in the number of moles of one of the reactants or the products must be accompanied by an equivalent change in all other reactants and the products.

If, in the balanced equation for the reaction, all the moles of the reactants get converted into products, the progress of the reaction is said to be 100%. This is also expressed by saying that one unit of reaction has occurred.

Obviously, the progress of the reaction would be less than 100% if all the moles of the reactants do not get converted into products. It is thus convenient to define a variable ξ to express the progress or advancement of the reaction. Introduced by de Dondre, ξ is called the degree of advancement of a chemical reaction.

If the reaction advanced by ξ units, then-

n

_{A}= n

^{o}

_{A}− aξ ---Equation-a

n

_{B}= n

^{o}

_{B}− bξ ---Equation-b

n

_{C}= n

^{o}

_{C}+ cξ ---Equation-c

n

_{D}= n

^{o}

_{D}+ dξ ---Equation-d

where n

^{o}

_{A}, n

^{o}

_{B}, n

^{o}

_{C}and n

^{o}

_{D}are the number of moles before the reaction is advanced by ξ units.

Minus sign for the number of moles for reactants denotes the consumption and plus sign for the number of moles for products denotes the formation.

Differentiating equation-a,b,c and d assuming n

^{o}

_{A}, n

^{o}

_{B}, n

^{o}

_{C}and n

^{o}

_{D}are constant, we get-

dn

_{A}= − adξ − aξ ---Equation-e

dn

_{B}= − bdξ − aξ ---Equation-f

dn

_{C}= + cdξ − aξ ---Equation-g

dn

_{D}= + ddξ − aξ ---Equation-h

The quantity dξ is called differential advancement or the increment of the degree of advancement of the reaction.

Now putting the dn

_{A}, dn

_{B}, dn

_{C}and dn

_{D}from equation-e,f,g and h in equation-2 we get-

(dG)

_{T,P}= [(cµ

_{C}+ dµ

_{D}) − (aµ

_{A}+ bµ

_{B})]dξ ---Equation-3

(dG/dξ)

_{T,P}= [(cµ

_{C}+ dµ

_{D}) − (aµ

_{A}+ bµ

_{B})] ---Equation-4

Equation-4 is De Donder's Treatment of Chemical Equilibrium.

The partial derivative (dG/dξ)

_{T,P}gives the rate of change of the Gibbs free energy with the advancement of the reaction. The expression on RHS in equation-4 is called

**reaction potential**.

So, reaction potential = ΔG = (dG/dξ)

_{T,P}

Thus, reaction potential is the rate change of total Gibbs free energy per unit advancement of the chemical relation at constant T and P.

Reaction potential is identical with the free energy change of a reaction. The decrease in the reaction potential is defined as the

**chemical affinity (A**.

_{f})A

_{f}= − (dG/dξ)

_{T,P}= − ΔG

At equilibrium, ΔG = (dG/dξ)

_{T,P}= 0

So, equation-4 becomes-

[(cµ

_{C}+ dµ

_{D}) − (aµ

_{A}+ bµ

_{B})]= 0 ---Equation-5

If the derivative (dG/dξ)

_{T,P}is negative, this means that the free energy of the reaction mixture decreases as the reaction advances in the forward direction and the reaction is spontaneous and affinity is positive. If the derivative is positive, the reverse reaction is spontaneous and the affinity is negative. If the derivative is zero, the reaction is in equilibrium and affinity is zero.

Equation-5 is general and is applicable whether the reactants and products are solids, liquids or gases.

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