deltaS of a decomposition reaction Suppose we had the following reaction:

$$\ce{Polymer ($n$~units)_{(aq)} -&gt; Polymer($a$~units)_{(aq)} + Polymer($n-a$~ units)_{(aq)}}$$

Without knowing the specific monomer unit of the polymer, and assuming the polymer is straight chain, is there anyway to determine the ΔS of the reaction? I mean, I know we can determine the sign (which will almost certainly be positive), but is there a way to make even a rough approximation of magnitude? We can assume the conformational entropy is negligible, i.e. the polymers are fairly rigid.
The largest contribution to the entropy change will be the translational component, which is easy to evaluate using the Sackur-Tetrode equation
$$S^\circ_{\rm{trans}}=R\ln(bM^{3/2}T^{3/2})$$
where $b$ is a collection of constants with a value of 3.75 (mol/gK)$^{3/2}$ $M$ is the mass in g/mol, $T$ is temperature in Kelvin, and the standard state is 1 M.  

Due to the slowly varying logarithmic dependence on $M$ and $T$ this contribution is usually on the order of 100's of J/molK.  

While the translational contribution is the largest, rotational and vibrational contributions can be sizable so this will give you an order of magnitude estimation at best.



Derivation starting from the wikipedia equation
$$S^\circ=kN \left( \ln \left( \frac{V^\circ}{N} \left( \frac{4\pi mU}{3h^2N}\right)^{3/2} \right)+\frac{5}{2}\right) $$ 
$V^\circ$ = 1 L = 0.001 $m^3$, $N=N_A$, $U=\tfrac{3}{2}N_AkT$, $M=mN_A$
$$S^\circ=R  \ln \left( \left[e^{5/2}\frac{V^\circ}{N_A} \left( \frac{4\pi \tfrac{3}{2}k}{3h^2N_A}\right)^{3/2}\right]M^{3/2}T^{3/2} \right) $$
The term in square brackets corresponds to $b$ and using SI units $b=1.20\times10^5$ mol(kg/K)$^{3/2}$ which corresponds to 3.75 mol(g/K)$^{3/2}$