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.
A system is at equilibrium when its free energy is a minimum
dA=0 Here I use the Helmholtz free energy A=U-TS but the same is true for the Gibbs free energy. Let's explore this for the simple four-bead polymer shown in the figure above. Since I have already defined the probabilities of the folded (p_f) and unfolded (p_{uf}) macrostates in a previous post it is easiest to start with the expression for the internal energy (U) and entropy (S) in terms if probabilities. U=N\left<\varepsilon\right>=N\sum_i^{\begin{array}{ c }\text{micro} \\ \text{states} \\
\end{array}} \varepsilon_ip_i S=-Nk\left<ln(p)\right>=-Nk\sum_i^{\begin{array}{ c }\text{micro} \\ \text{states} \\ \end{array}} p_i\ln(p_i)This results in U=N\varepsilon_0 p_{uf}\text{ and }S=-Nk\left( p_f\ln(p_f)+p_{uf}\ln\left(\frac{p_{uf}}{4}\right) \right)Combining these two terms and using the fact that probabilities must sum to 1 (p_f+p_{uf}=1), the free energy can be written as A=N\varepsilon_0 (1-p_{f})+TNk\left( p_f\ln(p_f)+(1-p_{f})\ln\left(\frac{(1-p_{f})}{4}\right) \right)This allows me to plot A as a function of p_f, using T = 298.15 K, N=N_A, and N_A\varepsilon_0=2.5 kJ/mol
As you can see, the free energy is a minimum when the probability of the folded state is about 0.4 or 0.407 to be more precise - precisely the value predicted by the Boltzmann distribution: p_f=1/q.
The total entropy is a maximum at equilibrium
This figure shows the internal energy and entropy contributions to the free energy as function on p_f.
When seeing this plot for the first time many people are surprised that the entropy is not a maximum (i.e. -TS is not a minimum) at equilibrium. Does this simple system violate the second law of thermodynamics? No! I am plotting the entropy of the system and not the total entropy that the second law refers to. The total entropy is indeed a maximum at equilibrium:dA_{\text{system}}=0\\dU_{\text{system}}-TdS_{\text{system}}=0\\-\frac{dU_{\text{system}}}{T}+dS_{\text{system}}=0\\ \frac{dq_{\text{surroundings}}}{T}+dS_{\text{system}}=0\\ dS_{\text{surroundings}}+dS_{\text{system}}=0\\dS_{\text{total}}=0 Increasing the number of molecules in the unfolded state requires energy (U increases at p_f decreases) and this energy has to come from somewhere. It is transferred to the system from the surroundings at heat: dq_{\text{surroundings}}=-dU_{\text{system}}and this lowers the entropy of the surroundings bydS_{\text{surroundings}}=\frac{dq_{\text{surroundings}}}{T}The entropy of the system is highest (-TS is lowest) when p_f=\frac{1}{5}, i.e. when all microstates are equally populated because this leads to the highest multiplicity (W_{\text{system}}):S_{\text{system}}=k\ln\left(\frac{N!}{N_f!N_{uf1}!N_{uf2}!N_{uf3}!N_{uf4}!}\right)
Some technical stuff you can skip if you want
Another reason you might think S_{\text{system}} should be a maximum at equilibrium is that the Boltzmann distribution is derived by maximizing the entropy:\frac{\partial S}{\partial N_i}=0\text{ for }i=1,2,3.,..However, what is actually maximized is a Lagrangian function: L=S+\alpha \left( N-\sum_i^{\begin{array}{ c }\text{micro} \\ \text{states} \\ \end{array}}N_i \right)-\beta \left( U-\sum_i^{\begin{array}{ c }\text{micro} \\ \text{states} \\ \end{array}}\varepsilon_i N_i \right) that conserves the number of particles (N) and the internal energy (U). Because of the latter requirement \frac{\partial L}{\partial N_i}=0\text{ for }i=1,2,3.,.. maximizes S_{\text{total}} rather than S_{\text{system}} because it is only the total internal energy that is conserved: dU_{\text{system}}+dU_{\text{surroundings}}=0

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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