Engineering Questions with Answers - Multiple Choice Questions

# Polymer Engineering MCQ’s – Non – Ideal Kinetics in Radical Polymerization

1 - Question

Which of the following is not a reason for non-ideal kinetics of radical polymerization?
a) initiator-monomer complexation
b) unimolecular termination
d) none of the mentioned
Explanation: All the three mechanisms cause non-ideality and kinetic deviations in the radical polymerization.

2 - Question

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 - Question

What is the order of reaction with respect to initiator, considering exclusive case of unimolecular termination?
a) 1
b) 0
c) 2
d) 0.5

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 - Question

Which of the following can restrict the bimolecular termination?
a) high monomer concentration
b) insolubility of polymer formed in monomer
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 - Question

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: Bimolecular termination is restricted in heterogeneous polymerization, where the polymer formed is insoluble in the monomer solvent media and unimolecular termination shows prominence.

6 - Question

What is the order of dependence of monomer concentration on Rp, in the exclusive case of degradative initiator transfer process?
a) 1
b) 0
c) 2
d) 0.5
Explanation: The overall rate maintains first order dependence on monomer concentration and the rate equation is given by- Rp =( kp/kt’)(2fkd) [M].

7 - Question

How does the addition of solvents and additives, due to their abnormal effects, can affect the process of polymerization?
a) increase initiator efficiency
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 - Question

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?
a) 2fkd
b) fkd
c) kt’/kp
d) [I]

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 - Question

Viscosity of the polymerization medium has a great influence on the termination process. State true or false.
a) True
b) False