# deltaS of a decomposition reaction

Suppose we had the following reaction:

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

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.

• I am not a polymer theorist, but I think it would be very hard to estimate this. Consider that you don't know how the units were linked, e.g., what type of bond, how much it can rotate, etc. – Geoff Hutchison Jun 17 '15 at 14:18
• Assume it's a polymer small enough that it can be modeled as a rigid Rod. – Breaking Bioinformatics Jun 17 '15 at 14:34

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)$^{3/2}$

• Isn't the Sackur-Tetrode Equation derived from an ideal gas? Is it reasonable to make that approximation for a macromolecule in solution? – Breaking Bioinformatics Jun 18 '15 at 11:56
• Since the thermodynamic quantities are state functions you can rewrite the aqueuos phase process as a thermodynamic cycle: 1. De-solvation of the reactants, 2. Reaction in the gas phase. 3. Solvation of the products. The largest contribution to $\Delta S^\circ$ is likely the translation contribution to the entropy change in step 2. Also, note that the standard state in aq soln is 1 M ideal solution – Jan Jensen Jun 18 '15 at 13:00
• Oooooh. Different question: the equation you have, while simple, does not immediately seem identical to the equation you linked. Is there anyway you can show the derivation from the more complicated one to the simpler one? – Breaking Bioinformatics Jun 18 '15 at 13:02
• I'll try to find time to write it down. In the mean time some hints if you want to have a go: $U=\tfrac{3}{2}kT$, $\ln(x)+\tfrac{5}{2}=\ln(xe^{5/2})$, $N=N_A$, and $V=1$ L. The rest is just units. – Jan Jensen Jun 18 '15 at 13:15
• I've taken a crack at it and got as far as $R*ln(e^{5/2}*(\frac{2 \pi m}{h^2N_A}k_bT)^{3/2})$ – Breaking Bioinformatics Jun 19 '15 at 10:11