This is Figure 10.1 from Dill and Bromberg's Molecular Driving Forces, which is my favorite book on statistical mechanics. It is a beautiful example from a beautiful book.

The Boltzmann distribution says that, at equilibrium, the probability of being in a microstate (also sometimes referred to as an energy state) is $$p_i=\frac{e^{-\varepsilon_i/kT}}{q}\text{ where }q=\sum_i^{\begin{array}{ c }\text{micro} \\

\text{states} \\

\end{array}} e^{-\varepsilon_i/kT}$$ Let's apply this formula to the simple four-bead polymer example shown in the figure above. Each of the five conformations represents a

\text{states} \\

\end{array}} e^{-\varepsilon_i/kT}$$ Let's apply this formula to the simple four-bead polymer example shown in the figure above. Each of the five conformations represents a

**. There is one low-energy conformation with a bead-bead contact and four higher-energy conformations where the bead-bead contact is lost. Because higher energy conformations have a common feature (the lack of bead-bead contact) and the same energies we group them into one common***micro*state**(a state comprised of more than one microstate). I'll refer to this macrostate at the "***macro*state**unfolded**" state and the to ground state the "**folded**" state.
According to the Boltzmann distribution the probability of the folded state is $$p_f=\frac{1}{q}\text{ where }q=1+4e^{-\varepsilon_0/kT}$$ The probability is the unfolded state is the sum of the (equal) probabilities of being in one of the four unfolded

*micro*states $$p_{uf}=p_{uf1}+p_{uf2}+p_{uf3}+p_{uf4}=\frac{4e^{-\varepsilon_0/kT}}{q}$$
This figure shows a plot of the probabilities of the folded and unfolded state, together with the probability of being in one of the four unfolded

*micro*states, as a function of temperature and for $N_A\varepsilon_0=2.5$ kJ/mol. The polymer can be said to unfold at 217 Kelvin where the unfolded state starts to become more probable than the folded state.**Why does the polymer unfold beyond 217 K**?

*Answer 1.*The plot shows that if there is only

*one*unfolded microstate then the polymer would never unfold (i.e. unfolded state would never be more probable than the folded state). Therefore,

**the polymer unfolds because there are more unfolded microstates than folded microstates**(4 compared to 1).

*Answer 2.*The number of microstates with the same energy is known as the

**degeneracy**, so one can also say that

**the polymer unfolds because the unfolded state has a higher degeneracy than the folded state**

*(4 compared to 1).*

*Answer 3.*

**The ratio of probabilities can be written as $$\frac{p_{uf}}{p_f}=4e^{-\varepsilon_0/kT}=e^{-( \varepsilon_0-kT\ln(4))/kT}$$The latter form clearly shows that the unfolded state will become more probable for temperatures were $T(k\ln(4))\gt\varepsilon_0$. The similarly of $k\ln(4)$ to the entropy $S=k\ln(W)$ is no accident. The entropy of the unfolded state is given by $$S_{uf}=k\ln\left(\frac{N_{uf}!}{N_{uf1}!N_{uf2}!N_{uf3}!N_{uf4}!}\right)$$where $N_{uf1}=N_{uf2}=N_{uf3}=N_{uf4}=\frac{1}{4}N_{uf}$ is the number of polymers in each of the unfolded microstates. For large $N_{uf}$, where Stirling's approximation [$x!\approx(x/e)^x$] holds, it is easy to show that this expression reduces to $$S_{uf}=N_{uf}k\ln(4)$$So one can also say that**

**the polymer unfolds because the unfolded state has a higher entropy than the folded state**.

A simulation of the unfolding of a slightly longer polymer can be found here

This work (except the first figure which is © by Garland Science) is licensed under a Creative Commons Attribution 3.0 Unported License.

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