Answer:
the initial polymerization rate [tex]R_p = 0.218 \ mol/L.s[/tex]
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. = [tex]9.4*10^4[/tex]
Explanation:
From the table of Polymerization of tetrahydrofuran below; we obtain the following data for Na⁺ counterion
[tex]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[/tex]
Sodium Naphthalene = 3.2 × 10⁻⁵ M
Concentration of styrene solution = 1.5 M
Now;
[tex][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[/tex]
[tex]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\\\\[/tex]
[tex]R_p = 0.218 \ mol/L.s[/tex]
Therefore; the initial polymerization rate [tex]R_p = 0.218 \ mol/L.s[/tex]
Fraction of polymerization rate due to free ions :
[tex]= \frac{0.224}{0.218}\\\\[/tex]
= 0.982
Fraction of polymerization rate due to contact ion pairs .
= [tex]\frac{0.00384}{0.218}[/tex]
= 0.0176
the average degree of polymerization at the completion of the reaction is :
[tex]\bar{DP_n}= 2 \frac{[M]}{[M^- \ C^+]}[/tex]
[tex]= 2* \frac{1.5 \ mol/L}{3.2*10^{-5} \ mol/L}[/tex]
= [tex]9.4*10^4[/tex]
