Question

In a polymerization of methyl methacrylate, a polymerization rate of 8.0x10-5 M/s was observed when the monomer and initiator concentrations were 9.0 M and 0.010 M, respectively. a. Calculate the value of the observed rate constant, k’, for this reaction. b. How should the monomer concentration be changed to achieve a polymerization rate of 3.2x10-5 M/s? c. How should the initiator concentration be changed to achieve a polymerization rate of 2.4x10-4 M/s?

Answer 1

Answer. Rate =k[monomer][initiator]

a). [Monomer] = 9.0 M

[ initiator] =0.010. M

Rate = 8.0 × 10-5 M/s

K = rate/ [[momomer][initiator]]

K = (8.0×10-5)/((9.0×0.010))

K= 88.8×10-5 M-1s-1

b). If rate of reaction is 3.2 ×10-5 M/s and concentration of initiator is 0.010 M then concentration of monomer ?

Rate = k [ monomer][initiator ]

   ( 3.2 ×10-5) /((88.8×10-5)(0.010))= [ Monomer]

[Monomer]= 3.6 M

Therefore concentration of monomer will decrease .

C). If rate = 2.4 ×10-5 M/s

[Monomer]= 9.0 M

k = 88.8 M-1s-1

[ initiator] = ?

[ Initiator]= (rate)/((k)[monomer])

[Initiator ]= (2.4 ×10-5)/((88.8×10-5)×(9.0))

[ Initiator]= 0.003 M

Concentration of initiator decrease to 0.003M