For acrylamide kp2kt 22 lmol s at 25 c and termination is by...

For acrylamide, kp2/kt = 22 L/mol s at 25 ?C and termination is by coupling alone. At this temperature the half-life of isobutyryl peroxide is 9.0 h and its efficiency in methanol can be taken to be equal to 0.3. A solution of 100 g/L acrylamide in methanol is polymerized with 10-1 M isobutyryl peroxide. (a) What is the initial steady state rate of polymerization? (b) How much polymer has been made in the first 10 min of reaction in 1 L of solution?

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0:00 Hello.

00:01 So in this question we have to find out the rate of polymerization and the weight of polymers, that means number of polymers which has been made in the first 10 minutes or we can say that weight of polymers.

00:16 So let us find out the first rate of polymerization here.

00:21 So we know that that rate of polymerization is equal to k -e concentration of m is there and f that is the initial initial initiator efficiency multiply by k -a -t divided by k -t and this is your half -life.

00:40 Right, or 1 by 2 we should write here.

00:43 So now, let us see here that kd or k -e, it is nothing but 0 .693 divided by t 1 by 2 or we have the value it is given here that is 0 .693 divided by 9 .93 divided by 9.

01:00 Hours it is given here we have to convert it into second so for this we have the value by solving this that 2 .1 into 10 raised to the power minus 5 so this is this value now let us see the for the concentration m it is there as we know that that 100 is there 100 gram per l it is given here therefore it will become your 100 by 70 1 moles per next, the concentration i, it is given here that is 0 .1 moles per liter it is there.

01:41 Now we know that the concentration, sorry, for the rate of polymerization that is kp square, it is also given here is equal to kt is equal to 22 or we can just write it as this kp by kt or 1 by 2 we can write it as it will become your under root 22 yes or you can just do the multiplication here by 22 multiply by k t and then do the under root so that is why we can write it in this form also and we know that that f is nothing but your initial concentration whose value it is given here as 0 .3...

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