This set of Polymer Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Non – Ideal Kinetics in Radical Polymerization”.
1. Which of the following is not a reason for non-ideal kinetics of radical polymerization?
a) initiator-monomer complexation
b) unimolecular termination
c) degradative initiator transfer
d) none of the mentioned
Explanation: All the three mechanisms cause non-ideality and kinetic deviations in the radical polymerization.
2. Consider initiator-monomer complexation in polymerization reaction, and a straight line is plotted between [M]3/Rp2 and [M]. What is the value of quotient of slope and the intercept?
a) equilibrium rate constant of monomer complexation reaction, K
b) termination rate constant, kt
c) propagation rate constant, kp
d) initiator concentration
Explanation: The transformed rate equation, considering initiator monomer complexation, is given by-
[M]3/Rp2 = (kt/kp2 kd K[I0]) + (kt[M]/ kp2 kd [I0])
Thus, the slope of the straight line plotted is, kt/ kp2 kd [I0] and the intercept is kt/kp2 kd K[I0]. Therefore, the quotient is calculated as K.
3. What is the order of reaction with respect to initiator, considering exclusive case of unimolecular termination?
Explanation: For exclusive unimolecular termination,
Rt = kt1 [M∙] Thus, the overall rate of polymerization is expressed as
Rp = kp/kt1 (2fkd) [I] [M] Therefore, the initiator order is equal to 1.
4. Which of the following can restrict the bimolecular termination?
a) high monomer concentration
b) insolubility of polymer formed in monomer
c) presence of solvents, additives
d) none of the mentioned
Explanation: Bimolecular termination is restricted in heterogeneous polymerization, where the polymer formed is insoluble in the monomer solvent media and unimolecular termination shows prominence.
5. What are the initiator and monomer orders, in limiting case of exclusive primary radical termination?
a) 0, 2
b) 1, 2
c) 1, 1
d) 2, 0
Explanation: The rate of polymerization, taking into account exclusive primary radical termination mechanism, is given by
Rp = kp (ki/kprt)[M]2
where kprt is the rate constant for primary radical termination.
Thus, the initiator and monomer orders are 0 and 2, respectively.
6. What is the order of dependence of monomer concentration on Rp, in the exclusive case of degradative initiator transfer process?
Explanation: The overall rate maintains first order dependence on monomer concentration and the rate equation is given by-
Rp =( kp/kt’)(2fkd) [M].
7. How does the addition of solvents and additives, due to their abnormal effects, can affect the process of polymerization?
a) increase initiator efficiency
b) increase radical generation
c) enhance chain initiation rate
d) all of the mentioned
Explanation: The solvents, while behaving abnormally, sometimes can induce an increase in initiator efficiency or radical generation, which consequently increases the rate of chain initiation as well as polymerization.
8. What is the value of intercept of the line plotted between the terms (2kt/kp2)(Rp2/[I][M]2) and Rp/[M], when there is simultaneous occurrence of bimolecular termination and termination by degradative process?
Explanation: For simultaneous occurrence of bimolecular termination and termination by degradative process, under steady-state condition, the equation is given by-
2kt[M∙]2 + kt’[I][M∙] = 2fkd [I].
On eliminating the radical concentration term, [M∙], we get
(2kt/kp2)(Rp2/[I][M]2) = 2fkd – (kt’/kp) Rp/[M] .
So, from the above equation, when straight line is plotted, it has a negative slope and the intercept value gives the value of 2fkd.
9. Viscosity of the polymerization medium has a great influence on the termination process. State true or false.
Explanation: Viscosity of the polymerization medium ofcourse has a influence on termination reaction, for termination being a diffusion based process. The solvent modification of termination reaction causes different kt values depending on the viscosity medium.
10. With the introduction of solvents in modifying the initiator species and affecting the degradative chain transfer, the value of parameter kp2/kt , changes with the solvent concentration. State true or false.
Explanation: The kinetic parameter kp2/kt practically remains unchanged over a whole range of dilution, when the solvents play an important role in modifying the initiator species and bring into effect noticeable degradative chain transfer.
Sanfoundry Global Education & Learning Series – Polymer Engineering.
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