Dicarboxylic acid is mixed with ethanol in a molar ratio 1 : x (x > 1) in the presence of a catalyze — Organic Chemistry Chemistry Question

x = 1.05.

Solution:
Denote A = acid, E = ethanol, M = monoester, D = diester. Consider two equilibria:
A + E →← M + H₂O K₁ = [M][H₂O] / ([A][E]) = 20
M + E →← D + H₂O K₂ = [D][H₂O] / ([M][E]) = 20
(here water is not a solvent but a product, therefore, it enters the expressions for equilibrium constants).

The equilibrium yield of monoester is:
η = [M] / ([A]+[M]+[D]) = K1[E] / ([H₂O] + K1[E] + K₁K2[E]^2/[H₂O])

Denote [H₂O] / [E] = x', then η(x') = 1 / (x'/K₁ + 1 + K2/x').
By differentiating with respect to x', we find that this function has maximum value at x' = sqrt(K₁K₂).

Substituting the optimal ratio [H₂O] / [E] into the equilibrium constants, we find the relations:
[M] = sqrt(K1/K₂)[A], [D] = [A].

From the material balance with respect to water we get:
[H₂O] = [M] + 2[D] = [A](2 + sqrt(K1/K₂)),
[E] = [H₂O] / x' = [A](2/sqrt(K₁K₂) + 1/K₂)

The initial concentrations of ethanol and acid are:
[E]0 = [E] + [M] + 2[D] = [A](2/sqrt(K₁K₂) + 1/K₂) + [A]sqrt(K1/K₂) + 2[A]
[A]0 = [A] + [M] + [D] = [A] + [A]sqrt(K1/K₂) + [A]

And the optimal ratio is:
x = [E]0 / [A]0 = (1 + 2/sqrt(K₁K₂) + 1/K₂ + sqrt(K1/K₂)) / (2 + sqrt(K1/K₂)).
At K₁ = K₂ = 20, the optimal ratio is x = 1.05.

η max = 1/3.

Solution:
From the derived equation, the maximum value is:
ηmax = 1 / (1 + 2*sqrt(K2/K₁)).
At K₁ = K₂ = 20, the maximum yield of monoester is: ηmax = 1/3.

For 7.1: x = (1 + 2/sqrt(K₁K₂) + 1/K₂ + sqrt(K1/K₂)) / (2 + sqrt(K1/K₂)).
For 7.2: ηmax = 1 / (1 + 2*sqrt(K2/K₁)).