Spontaneous Reactions

A change in the macrostate of the system is said to happen spontaneously if we do not need to apply a (thermodynamic) driving force to induce the change. It is then clear that, in large systems, spontaneous reactions cannot increase the entropy of the world, \(0\leq\Delta S_{\text{tot}},\) which is one formulation fo the second law.

The total entropy change can be decomposed into entropy changes \(\Delta S_S\) and \(\Delta S_R\) of system and reservoir, respectively. Assuming that the generalized coordinates of the reservoir don’t change, we can Taylor expand the reservoir entropy change \(\Delta S_R=\Delta E_R/T_R\) in terms of the energy change \(\Delta E_R\) of the reservoir (just as we did to derive the Boltzmann factor). In terms of the the heat \(Q\equiv-\Delta E_R\) added to the system, we thus have

\[0\leq\Delta S_{\text{tot}}=\Delta S_S+\Delta S_R=\Delta S_S-\frac{Q}{T_R}\]

and obtain the Clausius inequality

\[\boxed{\Delta S_S\geq\frac{Q}{T_R}}\;,\]

where the greater sign applies to spontaneous processes and the equal sign applies to reversible processes, which are the only ones that occur in equilibrium.

Next, we invoke the first law of thermodynamics, to express the change in internal energy of the system as

\[\Delta E_s=Q+W \]

in terms of the heat \(Q\) and the \(W\) is the work done on the system through some external force.

Also a footnote about \(U\): In the context of spontaneous processes, we define it as an ensemble average of the energy of the system. That ensemble average can have a time dependence if the system starts out of equilibrium and is allowed to relax. Example: Consider a bottle of initially cold water that we through into a hot tub. The temperature of the bottle will increase over time until the temperature of the bottle matches the temperature of the hot tub. The energy \(E_S(t)\) of the bottle will generally increase towards the peak energy (from above), but the time series will show fluctuations: If we repeat the entire experiment, \(E_S(t)\) will look slightly different because the fluctuations are different. But the ensemble average \(U(t)\) over (infinitely) many realizations of the non-equilibrium process will be a deterministic time-dependent function.

Now, we consider isothermal reactions in which the temperature is held constant, i.e. \(\Delta T=0\). Assuming no work done, \(W=0\), the Clausius inequality implies

\[\Delta E_S-T_R\Delta S_S\leq 0.\]

Note that the RHS is equal to the change \(\Delta F\) of the free energy (only in this isothermal case!). Thus, in a spontaneous process, the change in free energy is never positive,

\[\boxed{\Delta F\leq0\qquad\text{(isothermal processes)}.}\]

If the system does work, the Clausius inequality gives us a bound on how much work we can get out of a process that involves change \(\Delta F<0\) in free energy, namely

\[\boxed{W\geq\Delta F\qquad\text{(isothermal processes)}.}\]

Summary:

• Spontaneous reactions reduce the free energy \(F\).

• In equilibrium, the free energy is minimal.

• The work done on a system during a spontaneous process is typically larger than the change in free energy

• unless a process is reversible, in which case the Work just equals the change in free energy.

The above discussion also implies relationships between differentials for general reversible processes (not only isothermal ones):

\[ dF=dE-T dS -S dT=+dW -S dT\]

or

\[ dU=T dS+ dW\]

Example: ATP consumption Most cellular processes are powered by $\(ATP\rightarrow ADP + P_i\)$