Derive law of mass action thermodynamically?

 

Question: Derive law of mass action thermodynamically?

The law of mass action is a thermodynamic principle that states that the rate of a chemical reaction at a given temperature is proportional to the product of the active masses of the reactants, each raised to its stoichiometric coefficient. To derive this law thermodynamically, we can use the concept of chemical potential and equilibrium constant. 

Consider a reversible reaction involving ideal gases:

aA + bB ⇌ cC + dD

The chemical potential of each component is given by:

μ_A = μ_A^0 + RT ln p_A

μ_B = μ_B^0 + RT ln p_B

μ_C = μ_C^0 + RT ln p_C

μ_D = μ_D^0 + RT ln p_D

where μ_i^0 is the standard chemical potential of component i, R is the gas constant, T is the temperature, and p_i is the partial pressure of component i.

At equilibrium, the total change in chemical potential is zero, so we have:

aμ_A + bμ_B = cμ_C + dμ_D

Substituting the expressions for the chemical potentials, we get:

a(μ_A^0 + RT ln p_A) + b(μ_B^0 + RT ln p_B) = c(μ_C^0 + RT ln p_C) + d(μ_D^0 + RT ln p_D)

Rearranging and simplifying, we obtain:

(p_C)^c (p_D)^d / (p_A)^a (p_B)^b = exp[(cμ_C^0 + dμ_D^0 - aμ_A^0 - bμ_B^0) / RT]

The right-hand side of this equation is a constant at a given temperature, and it is called the equilibrium constant K_p. Therefore, we have:

(p_C)^c (p_D)^d / (p_A)^a (p_B)^b = K_p

This is the law of mass action expressed in terms of partial pressures. If we use concentrations instead of partial pressures, we get a similar expression with a different equilibrium constant K_c:

(C)^c (D)^d / (A)^a (B)^b = K_c

where [i] is the concentration of component i. The law of mass action can be used to calculate the equilibrium composition of a reaction mixture or the extent of reaction at a given temperature.

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