Poly(lactic acid). Группа авторовЧитать онлайн книгу.

target="_blank" rel="nofollow" href="#ulink_663bf57d-8300-5b57-9a00-c9d4090f1293">(2.10) p equals StartFraction n Subscript normal upper L 1 Baseline upper K Over n Subscript normal upper W Baseline EndFraction

p equals StartFraction n Subscript normal upper L 1 Baseline upper K Over n Subscript normal upper W Baseline EndFraction

The variable p represents the probability of bond formation. Although the value of p depends on concentration, the quantity is the same for all oligomers at each concentration. Vu et al. [27] used the variable r, which is equivalent to the variable p used here. Using recursion, we recognize that Equation 2.9 can be written as:

(2.11) n Subscript normal upper L 3 Baseline equals n Subscript normal upper L 2 Baseline p equals n Subscript normal upper L 1 Baseline p squared semicolon n Subscript normal upper L 4 Baseline equals n Subscript normal upper L 3 Baseline p equals n Subscript normal upper L 1 Baseline p cubed

(2.12) n Subscript normal upper L Sub Subscript j Baseline equals n Subscript normal upper L Sub Subscript left-parenthesis j minus 1 right-parenthesis Baseline p equals n Subscript normal upper L 1 Baseline p Superscript j minus 1 Baseline normal j greater-than-or-equal-to 2

Each oligomer of length j contains j lactic acid molecules, so the apparent moles of lactic acid are given by the balance found by the closed form of the sum:

(2.13)

(2.14) n Subscript normal upper L 1 Baseline equals n Subscript normal upper L Superscript i Baseline left-parenthesis 1 minus p right-parenthesis squared

Equation 2.14 can be inserted into Equation 2.12 to give the Flory‐Schulz distribution:

(2.15) n Subscript normal upper L Sub Subscript j Baseline equals n Subscript normal upper L Superscript i Baseline p Superscript j minus 1 Baseline left-parenthesis 1 minus p right-parenthesis squared j greater-than-or-equal-to 2

The water in an equilibrated solution is the sum of the apparent water plus the water from the condensation reaction. Each step during the condensation releases a water molecule, so an oligomer of length j releases (j − 1) moles of water (n _W):

(2.16)

Recognizing Equation 2.13, we insert it into Equation 2.16 to obtain:

(2.17) n Subscript normal upper W Baseline equals n Subscript normal upper W Superscript i Baseline plus n Subscript normal upper L Superscript i Baseline p

Inserting Equations 2.14 and 2.17 into Equation 2.10, we develop a relation between the apparent number of moles and K that can be solved to find p

(2.18) upper K equals StartFraction p left-parenthesis n Subscript normal upper W Superscript i Baseline plus n Subscript normal upper L Superscript i Baseline p right-parenthesis Over n Subscript normal upper L Superscript i Baseline left-parenthesis 1 minus p right-parenthesis squared EndFraction

(2.19) n Subscript normal upper L Superscript i Baseline left-parenthesis 1 minus upper K right-parenthesis p squared plus left-parenthesis n Subscript normal upper W Superscript i Baseline plus 2 n Subscript normal upper L Superscript i Baseline upper K right-parenthesis p minus n Subscript normal upper L Superscript i Baseline upper K equals 0

(2.20)

For a given K and apparent moles n Subscript normal upper L Superscript i and n Subscript normal upper W Superscript i Baseline comma Equation 2.20 provides a value of p. Then Equations 2.12 and 2.14 can be used to find the equilibrium moles of lactic oligomers and monomer, while Equation 2.17 provides the equilibrium moles of water. The titratable acidity is a measure of the true number of oligomers in solution because each oligomer has one free carboxylic acid group. The titratable acidity is

Schematic illustration of left axis—total titratable acidity tabulated by Holten [28] from various workers (♢) and measurements by Vu et al.

FIGURE 2.4 Left axis—total titratable acidity tabulated by Holten [28] from various workers (♢) and measurements by Vu et al. [27] (▪) compared to the model. Right axis—value of p for the model as a function of apparent wt% using K = 0.2023.

(2.21)Скачать книгу