Question

A solution of styrene (1.5 M) in THF is polymerized at 25oC by sodium naphthalene at a concentration of 3.2x10-5 M. Calculate the initial polymerization rate and the average degree of polymerization at the completion of the reaction. What fractions of the polymerization rate are due to free ions and contact ion pairs, respectively

Answer 1

the initial polymerization rate
`R_p = 0.218 \ mol/L.s`

fraction of polymerization rate that is due to free ions = 0.982

fraction of polymerization rate due to contact ion pairs = 0.0176

the average degree of polymerization at the completion of the reaction. =
`9.4*10^4`

Step-by-step explanation:

From the table of Polymerization of tetrahydrofuran below; we obtain the following data for Na⁺ counterion

`K_d = 1.5 *10^(-7) \ mol/L\\\\K_{\bar{p}} = 6.5*10^4 \ L /mol.s\\\\K_{\bar{{p^+}}}= 80\ L/mol.s`

Sodium Naphthalene = 3.2 × 10⁻⁵ M

Concentration of styrene solution = 1.5 M

Now;

`[M^-] = [K_d(M^-C^+])^(1/2)\\ \\\ [M^-] = [(1.5*10^(-7) \ mol/L )(3.20*10^(-5) \ mol/L ) ] ^(1/2)\\\\\ [M^-] = 2.19*10^(-6) \ mol/L`

`R_p = K_{\bar{p}} [M^-][M^+] + K_{\bar{p^+}}[M^-C^+][M^-]\\\\R_p = (6.5*10^4 \ L/mol.s)(2.19*10^(-6) \ mol/L)(1.5 \ mol/L)+(80 L/mol.s)(3.2*10^(-5) \ mol/L)(1.5 \ mol/L)\\\\R_p = 0.224 \ mol/ L.s + 0.00384 \ mol/L.s\\\\`

`R_p = 0.218 \ mol/L.s`

Therefore; the initial polymerization rate
`R_p = 0.218 \ mol/L.s`

Fraction of polymerization rate due to free ions :

`= (0.224)/(0.218)\\\\`

= 0.982

Fraction of polymerization rate due to contact ion pairs .

=
`(0.00384)/(0.218)`

= 0.0176

the average degree of polymerization at the completion of the reaction is :

`\bar{DP_n}= 2 ([M])/([M^- \ C^+])`

`= 2* (1.5 \ mol/L)/(3.2*10^(-5) \ mol/L)`

=
`9.4*10^4`