Chain-growth polymerization of methyl methacrylate is being performed in a batch
ID: 1043733 • Letter: C
Question
Chain-growth polymerization of methyl methacrylate is being performed in a batch reactor. 100 liters of the liquid monomer are added together with 10 gmol of an initiator whose half-life at the reaction temperature is 50 hours. Assume that the initiator dissolves in the monomer and does not lead to any appreciable change of volume. The efficiency of the initiator, f, is about 0.3 and the density of the liquid monomer is 940 g L at the reaction temperature. The propagation rate constant is 55 L/s.gmol. Termination is primarily by the combination mechanism and the termination rate constant, k, is 25.5 x 106 L/s.gmol Calculate: The molecular weight of the polymer being formed initially, i.e., when the concentrations are those listed above at the beginning of the reaction; The mass of polymer produced after 10 hours of reaction time. (i) (ii)Explanation / Answer
a) We are given with values as follows:
Kp = 55 L/s.gmol
Kt = 25.52*106 L/s. gmol
[I] = 10 gmol
f = 0.3
And finally the initial concentration of the monomer, that is 940 g/L.
We can calculate the rate of polymerization Rp, which is ultimately the amount of monomer consumed with time by the following equation,
Rp = kp [M] (f kd [I] / kt ) 1/2
But we don't know the value of kd, which is the decomposition constant of the initiator. So, we can calculate kd by half-life of the initiator given to us, which is 50 hr.
t1/2 = 0.693 / kd
Then, kd = 0.693/ ( 50*3600)
= 3.85 * 10-6
Now, putting values in the equation for Rp, we get
= 55 * 940 [ (0.3 * 3.85 * 10-6 * 10 ) / (25.5 * 106)] 1/2
= 0.0347 gmol L-1s-1
Hence, 0.0347 gmol polymer is formed per liter, initially. Since, the amount of monomer added is 100 L. We get the initial mass formed as 3.47gmol per second.
b) after 10 hours, we can use
Rp * 10 hr = > Rp * 36000
Hence, after 10 hours, the amount of polymer formed will be
124.92 kgmol .
I hope this helps!