Question

A particular free radical polymerization process yields a polymer with a number averaged degree of polymerization, \( \overline{x_n} = 100 \). The monomer concentration is doubled and the initiator concentration is increased by four times. Assuming that all rate coefficients and other parameters remain unchanged, the value of \( \overline{x_n} \) (rounded off to the nearest integer) is \(\underline{\hspace{2cm}}\).

Hint:

In free radical polymerization, changes in the concentrations of monomer and initiator directly affect the number-averaged degree of polymerization.

Answer 1

Correct Answer: 100

For a free radical polymerization process, the number-averaged degree of polymerization is given by:
\[ \overline{x_n} = \frac{k_p [M]}{k_i [I]} \] where:
- \( k_p \) is the propagation rate constant,
- \( k_i \) is the initiation rate constant,
- \( [M] \) is the monomer concentration,
- \( [I] \) is the initiator concentration.
If the monomer concentration is doubled and the initiator concentration is increased by four times, the new degree of polymerization \( \overline{x_n}' \) is:
\[ \overline{x_n}' = \overline{x_n} \times \frac{[M]'}{[M]} \times \frac{[I]}{[I]'} \] Substitute the given values:
\[ \overline{x_n}' = 100 \times 2 \times 4 = 800. \] Thus, the value of \( \overline{x_n} \) is \( 800 \).