Question

Methylmethacrylate (MMA) is polymerized at 65^oC using this initiator. The rate constants of propagation and of termination for MMA are k_p = 700 l/mole/s and k_t = 5.1x10⁷ l/mole/s. The efficiency factor is assumed to be 0.5 and the initiator concentration is 2x10^-2 mol/l. What is the conversion to polymer attained after 45 minutes?

Answer 1

Hi
Let us write the reaction first
MMA+initiator = IP( intermediate product)
Kp=700 L/mole/s
IP+MMA= Polymeised Product (PP)
Kt=5.1*10^7 L/mole/s

Initiator concentration =2*10^-2 mole/s
From the unit of the rate constant we may see the reaction is second order.

We may assume that MMA is abundantly available( as in the case of the polymerization reaction ).

rate of the reaction 1 = Kp*[initiator]^2
=700*C^2*(1-X)^2=CdX/dt
Here C is the initial concentration of the initiator.
X is the conversion obtained for the initiator.

=(1/1-X1 - 1/1-X2 )= -700*C*t
X1 is the initial conversion.
X2 is the final conversion.
t is the time for this conversation.
t = 45 minutes = -45*60=2600 S
1-1/1-X2 = -700*2600*2*10^-2
=36401=1/1-X2
1-X2 =0.0000274
X2=0.99997252

Concentration of the inintermediate product =CX2
=0.01999945


For the termination reaction 2
rate of the reaction 2 = Kp*[IP]^2
=5.1*10^7*C1^2*(1-X")^2=CdX'/dt
Here C1 is the initial concentration of the IP
X' is the conversion obtained for the IP


=(1/1-X'1 - 1/1-X'2 )= -700*C1*t
X'1 is the initial conversion.
X'2 is the final conversion.
t is the time for this conversation.
t = 45 minutes = -45*60=2600 S
1-1/1-X'2 = -5.1*10^7*2600*2*10^-2
=2.625*10^7=1/1-X2
1-X2 =3.77*10^-8
X2=1-3.77*10^-8

Concentration of the final product =C1X2
C1=0.01999945
Concentration of final product =0.0199944

Since the efficiency is 0.5, the actual concentration product will have half concentration than the expected so the concentration = 0.5*0.0199944=0.0099972

Conversion in the formation of the polymer = final concentration of the product / initial initiator concentration
=0.0099972/0.02
=0.49986

Hope this helps
Thanks