In a reaction vessel containing pure monomer and no initiator, cations were gene

Question

In a reaction vessel containing pure monomer and no initiator, cations were generated using UV radiation and the process involved two molecules of monomers according to the following mechanism. Initiation: M + M rightarrow X^Propagation: X^+ M rightarrow XM^Termination: XM^+ H_2O rightarrow XMH + OH a Find expressions for the rates of initiation, propagation and termination and use steady-state approximations to show that the rate of polymerization is third-order in monomer concentration. b What is the concentration of the monomer if R_p = 2.49 times 10^3 mol/L.s, k_d = 1.45 times 10^-6 L/mol.s, k_p = 0.175 times 10^-3 L/mol.s and k_t = 7.2 times 10^-7. (Assume [H_2O] is 1)

Explanation / Answer

a)

Initiation:

M + M    --------ki------>    X+

Rate of Initiation Ri = ki [M][M]

Propagation:

X+ + M    ------kp------>     XM+

Rate of Propagation Rp =kp[X+][M]

Termination:

XM+   + H2O -----kTr---->   XMH + OH-

Rate of Termination RTr = kTr [XM+][H2O]

At steady state rate of initiation equals the rate of termination,

Ri = RTr

ki [M][M] = kTr [XM+][H2O]

[XM+] = ki [M]2/kTr[H2O]

Substituting this in RP we have,

Rp =kpki [M]2[M]/kTr[H2O]

Rp = kpki/kTr [M]3/[H2O] = k'[M]3/[H2O]

So rate of polymerization is third order in monomer concentration.

b)

Given,

Rp = 2.49 x 103 mol/L.s

ki = 1.45 x 10-6 L/mol.s

kp = 0.175 x 10-3 L/mol.s

kt = 7.2 x 10-7 L/mol.s

Rp =kpki/kTr [M]3/[H2O]

2.49 x 103=[(0.175 x 10-3 x 1.45 x 10-6 ) / 7.2 x 10-7 ] x [M]3

0.0003524 x [M]3 = 2.49 x 103

[M]3 = 2.49 x 103 / 0.0003524

[M]3 = 7.065 x 106

[M] = 191.885 M

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