The values given below have been taken from my lecture hand out, and they weren't cited so I don't know where my instructor got them from

The $k_\mathrm{p}$ (rate constant for propagation in radical polymerisation ) step for the following monomers is given below:

$$\begin{array}{cc} \hline \text{Monomer} & k_\mathrm{p}~/~\mathrm{mol^{-1}~s^{-1}} \\ \hline \text{Styrene} & 106 \\ \text{Vinyl ester} & 3000\mathrm{-}4000 \\ \text{Ethylene} & 16 \\ \hline \end{array}$$

And the instructor explained the difference between the first two as follows:

the benzyl radical is stabilised by conjugation, however the radical produced in a vinyl ester lacks stabilisation by conjugation and thus reacts more rapidly.

Seems like a fair explanation, however in the same table the $k_\mathrm{p}$ for ethylene is given as $16~\mathrm{mol^{-1} s^{-1}}$, the slowest propagation step of the three. Thus his explanation is not consistent with this observation I believe, because similar to the vinyl ester radical this one too is not stabilised by conjugation and should be similarly eager to react.

My instructor did not offer a satisfactory explanation and I was wondering if people could weigh in on this here.

  • 2
    $\begingroup$ Perhaps its due to the C-H bond energy in ethylene being quite a bit larger than a C-O bond and possibly a C-H in the styrene? $\endgroup$ – porphyrin Oct 20 '16 at 14:11
  • 1
    $\begingroup$ I feel like I remember a radical $\mathrm{\alpha}$ to a carbonyl as being moderately stabilized via conjugation even though it is destabilized due to the electron withdrawing effect from the carbonyl... $\endgroup$ – Zhe Oct 20 '16 at 21:49

The argument presented by your instructor is based on one factor: the stability of the radical terminus of the polymer. But for bimolecular polymerization reactions we need to consider 4 things:

  1. Stability of the radical
  2. Stability of the monomer
  3. Stability of the transition state
  4. The probability of the radical and the monomers being correctly oriented

The first three items affect the activation energy ($E_\mathrm{a}$), while the final item affects the frequency factor ($A$), or pre-exponential factor in the Arrhenius equation:

$$ k = A \exp{\left(\frac{-E_\mathrm{a}}{RT}\right)} $$

For styrene, vinyl acetate (a vinyl ester), and ethylene, I have found the following values for radical polymerization processes.[1,2] These have been calculated to provide an apples-to-apples comparison:

$$\begin{array}{cc} \hline \text{Monomer} & E_\mathrm{a}~/~\mathrm{kJ~mol^{-1}} & A~/~\mathrm{L~mol^{-1}~s^{-1}} & T~/~^\circ\mathrm{C} & k_\mathrm{p}~/~\mathrm{L ~mol^{-1}~s^{-1}} \\ \hline \text{Styrene} & 32.5 & 4.3 \times 10^7 & 60 & 341 \\ \text{Vinyl acetate} & 20.4 & 1.5 \times 10^7 & 60 & 9460 \\ \text{Ethylene} & 27.7\mathrm{-}32.8 & 2.0 \times 10^3 \mathrm{-} 3.0 \times 10^4 & 60 & 0.1\mathrm{-}0.2 \\ \hline \end{array}$$


"the propagation rate coefficient for most monomers is strongly pressure-dependent with large negative activation volumes; thus an increase in pressure leads to an increased $k_\mathrm{p}$ value."[1]

This is particularly relevant for the ethylene example, which is done at $100\ \mathrm{bar}$. At $60\ ^\circ\mathrm{C}$, this should be about $0.15\ \mathrm{g~mL^{-1}}$,[3] which is a factor of 7 less dense than the styrene and vinyl acetate systems.

With those caveats noted, the argument of the vinyl ester having a lower activation energy ($E_\mathrm{a}$) due to the reactants being less stable relative to the transition state seems valid. However, the data for ethylene shows that the pre-exponential factor ($A$) is likely responsible for the lesser rate of propagation displayed by ethylene.


  1. Krzysztof Matyjaszewski & Thomas P. Davis, Handbook of Radical Polymerization, 2003, p 199.
  2. Etienne Grau, Jean-Pierre Broyer, Christophe Boisson, Roger Spitz, and Vincent Monteil, Macromolecules 2009, 42, 7279–7281. DOI: 10.1021/ma901622u
  3. Thomas, W. & Zander, M. Int J Thermophys (1980) 1: 383. DOI: 10.1007/BF00516565
  • $\begingroup$ Thanks for the great answer, and the references were especially helpful :) $\endgroup$ – getafix Oct 26 '16 at 6:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.