1. What is the effect of polymer transfer on the polymer product formed by radical polymerization?
a) becomes less crystallizable
b) becomes mechanically weak
c) becomes less resistant to foreign agents
d) all of the mentioned
Explanation: The polymer transfer leads to the chain branching which makes it less crystalizable, weak in mechanical properties and less resistant to heat, solvents and chemicals.
2. For polymerization of vinyl acetate with hydrogen peroxide at 60ᵒC, the reciprocal of number average degree of polymerization is given by,
1/Xn = 1.75*10-4 + 14.0Rp + 5.4*104Rp2
What is the value of monomer transfer constant?
a) 1.75*10-4
b) 0.875*10-4
c) 5.4*104
d) 14
Explanation: According to the expression for reciprocal of average degree of polymerization, we have
1/Xn = CM + (kt/kp2)(Rp/[M]2) + CI(kt/kp2 fkd)(Rp2/[M]3)
Comparing the given equation with this one, we get CM as 1.75* 10-4.
3. Consider the expression for reciprocal of average degree of polymerization, when styrene is polymerized with t-butyl hydroperoxide at 60ᵒC-
1/Xn = 0.6*10-4+10.0Rp+1.73*104 Rp2
Calculate the value of initiator transfer constant, if the monomer concentration is 8.35 mol/l and the value of the term kp2 fkd/kt= 3.29*10-9.
a) 0.035
b) 0.07
c) 0.063
d) 0.0175
Explanation: According to the expression for reciprocal of average degree of polymerization, we have
1/Xn = CM + (kt/kp2)(Rp/[M]2) + CI(kt/kp2 fkd)(Rp2/[M]3)
Comparing the given equation,
CI(kt/kp2 fkd)(Rp2/[M]3) = 1.73*104 Rp2
CI = 1.73*104 * 3.29*10-9*(8.35)3
CI = 0.035.
4. The transfer constant to the solvent for polymerization of of styrene in toluene at 60ᵒC is 0.125*10-5. How much dilution is required to decrease the molecular weight by 4 times when 1/Xn0 is 3.5*10-4?
a) 75
b) 42
c) 84
d) 7
Explanation: To decrease molecular weight by 4 times, 1/Xn becomes 4/Xn0,
4/Xn0 = 1/Xn0 + CI [S]/[M] Substituting the values, [S]/[M] = 84. Thus, dilution factor is 84.
5. The molecular weight of the polymer, when styrene is polymerized at 60ᵒC, is halved by diluting it with ethyl benzene. What is the value of transfer constant to the solvent, if the dilution factor is 7 and the term 1/Xn0 is 4.8*10-4?
a) 0.335
b) 0.68
c) 1.5
d) 0.82
Explanation: We have the expression for solvent transfer,
1/Xn = 1/Xn0 + CI [S]/[M] When molecular weight is halved, 1/Xn becomes 2/Xn0, then substituting the values and [S]/[M]= 7, we get CI as 0.68.
6. Which of the following is not dependent on result of equilibrium condition of polymerization reaction?
a) degree of polymerization
b) chain length distribution
c) limiting monomer conversion
d) none of the mentioned
Explanation: The result of equilibrium condition is felt on degree of polymerization, chain length distribution and also on the conversion of monomer to polymer.
7. How the increase in temperature to higher values, does affect the equilibrium state of propagation reaction?
a) shifts forward
b) shifts backwards
c) remains constant
d) none of the mentioned
Explanation: The shifting of equilibrium takes place in reverse reaction as the propagation reaction is exothermic. Initially the effect of reverse reaction rate constant, kdp is not felt, but it increases progressively with higher temperatures and at a critical temperature, the overall polymerization reduces to zero
8. What is the value of overall rate of polymerization at ceiling temperature?
a) exceptionally high
b) exceptionally low
c) zero
d) cannot be determined
Explanation: When the temperature is initially increased from a low level, rate increases due to gain in the value of kp. Initially the effect of reverse reaction rate constant, kdp is not felt, but it increases progressively with higher temperatures and at a critical temperature, the rate of propagation is equal to rate of de-propagation and the overall polymerization reduces to zero.
9. What is the critical temperature Tc called, if ∆Hᵒ and ∆Sᵒ are both negative in radical polymerization?
a) ceiling temperature
b) floor temperature
c) theta temperature
d) all of the mentioned
Explanation: If both ∆Hᵒ and ∆Sᵒ are negative, then the critical temperature is called the ceiling temperature, above which it is thermodynamically impossible to convert monomer to polymer.
10. What is the critical temperature called, if ∆Hᵒ and ∆Sᵒ are both positive in radical polymerization?
a) ceiling temperature
b) floor temperature
c) theta temperature
d) all of the mentioned
Explanation: If both ∆Hᵒ and ∆Sᵒ are positive, then the critical temperature is called the floor temperature, below which it is thermodynamically impossible to convert monomer to polymer.