Question

Please solve following Q with complete step by step solution

cru la mi hert2e0 iso btyryl loo of auylami de in metha nol nt 25 is used and ho initator Con Centrahon is o.M AL es foinpera turhe halflfe of isobntyr rde af what i the inital stea d-stato rate b Ho i tech poly men, in grams has beem e frst lo minutes of eachon in ts one leter much ymer,ih enams

Answer 1

Initiator undergoes a first order reaction to produce a radical that initiates the reaction. [ M l - i - 0 - ^
Half life of a first order reaction = ln2/k where k is the rate constant.

In this case 9 hours = 0.693/k_i
                                k_i = .077/hour = 2 x 10^-5 /second

rate of polymerisation = kp(fk_i/k_t)^_1/2[inititaor]^_1/2[Monomer]

Molecular Mass of acrylamide = 71 gram/mol

kp²/k_t = 22 L/mol/sec f = 0.3 [inititator] = 0.1M [Monomer] = 100/71 M = 1.41 M

Substituing these values,

= ( 2 2 ) ^{1 / 2}( 1 . 4 1 ) [ ( 0 . 3 ) ( 2 χ 1 0 ^{~ 5}) ( 1 0 ^-1) ] ^{1 /2}
= 5.1 x10^-3 moles monomer/I/sec
= (5.1 x10^-3) ( 7 1 )

rate (initially) = 0.36 g polymer/L/sec.

(b) IM 2-17 一一2-17 TH]。 一2-17 IM]0-114 [M]. n lo minuts

Please leave a feedback if you are satisfied with the solution.