Consider a free-radical polymerization carried out isothermally in a batch react

Question

Consider a free-radical polymerization carried out isothermally in a batch reactor with bimolecular combination termination chemically-controlled. Given the following data, calculate the initial initiator concentration required to reach a limiting conversion of monomer of 80%: f = 0.5; k_d = 10^-4/sec; [M]_0 = 10 mol/L; rho_m = rho_p = 10^3 g/L; k_P = 10^3 L mol sec; k_tc = 10^7 L/mol sec; monomer molecular weight: 100 g/mol (b) For the free-radical polymerization in Part (a) calculate the instantaneous number- and weight-averages (M_n and M_w) at the start of the polymerization (at zero monomer conversion). How will the averages change during the course of the polymerization? How will the index for the instantaneously produced polymer change with monomer conversion.

Explanation / Answer

Solution:

(a) Given 80% conversion, means 80% of the monomers converted to polymer with 20% left un-converted.

Therefore, [M] = 20%[Mo] = 0.2[Mo]

Again, we know that the degree of polymerization for a free radical reaction, termination by combination can be given by:

x = 2 = 1/(1-p) --------> (1);

x = degree of polymerization, = kinetic chain length, p = % monomer conversion

or, x = 2 = 1/(1-0.8)=5

or, = 2.5

Also we know, = kp[M]/(2kdktf[I])1/2

Or, [I]1/2 = kp[M]/.(2fkdkt)1/2

Again we know, [I] = [Io]e-kd.t

Therefore,

[Io]1/2e-kd.t/2 = kpx0.2[Mo]/.(2fkdkt)1/2

Or, [Io]1/2 = kpx0.2[Mo]x ekd.t/2/.(2fkdkt)1/2

Or, [Io]1/2 = 1000x0.2x10x ekd.t/2/2.5(2x0.5x10-4x107)1/2

Or, [Io]1/2 = 2000x ekd.t/2/79

Or, [Io]1/2 = 25.3xekd.t/2; squaring both the sides we have,

[Io] = 640.1xe0.0001t

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