Rate constants for termination kt may be of the order of 1x108 L/mol s in free r

Question

Rate constants for termination kt may be of the order of 1x108 L/mol s in free radical polymerizations. Consider the polymerization of styrene initiated by ditert -butyl peroxide at 60. C. For a solution of 0.01 M peroxide and 1.0 M styrene in benzene, the initial rate of polymerization is 1.5x10-7 mol/L s and <Mn> of the polymer produced is 138,000. (a) From the above information estimate kp for styrene at 60. C. (b) What is the average lifetime of a macroradical during initial stages of polymerization in this system?

Explanation / Answer

(a) Estimation of Kp for styrene-

It is proposed that a small chain length dependence of kp (propagation rate constant) (overall) may, in part, be a consequence of a marked chain length dependence of kp for the first few propagation steps [i.e. kp(1) > kp(2) < kp(3) ? kp(?4)].

In kp = 16,47 ? 30084/(RT) (chain length ? 4).

here Mn of the polymer = 138000

T = 273 + 60 = 333 K

R = 8.314 J/Kmol

ln Kp = 138000/(8.314*333) = 49.84 mol/Ls

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(b) Macroradicals form macromolecules during termination step (this step (second order) also runs with initiation and propagation (zero order) simultaneously. This step follow second order kinetics as it is clear from the unit of Kt. For a second-order reaction each half-life is twice as long as the life span (average life) of the one before. The equation for the half-life of a second order reaction:

t1/2=1/kt[A]o

Here [A]o = initial concentration of macoradical (formed moles of macroradicals during initial stages of polymerization will be of the order of 10-7. So conc, can be assumed = 10-7 mol/L

So  t1/2= 1/ (108*10-7) = 0.1 sec so average lifetime of a macroradical may be = 0.1/2 = 0.05 sec.

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