0.16 Equilibrium and the second law of thermodynamics  (Page 7/11)

Free energy

How can the Second Law be applied to a process in a system that is not isolated? One way to view the lessons ofthe previous observations is as follows: in analyzing a process to understand why it is or is not spontaneous, we must consider boththe change in entropy of the system undergoing the process and the effect of the heat released or absorbed during the process on the entropy of the surroundings. Although wecannot prove it here, the entropy increase of a substance due to heat $q$ at temperature $T$ is given by $\Delta (S)=\frac{q}{T}$ . From another study , we can calculate the heat transfer for a process occurring under constant pressurefrom the enthalpy change, $\Delta (H)$ . By conservation of energy, the heat flow into the surroundings mustbe $-\Delta (H)$ . Therefore, the increase in the entropy of the surroundings due toheat transfer must be $\Delta (S_{\mathrm{surr}})=-\left(\frac{\Delta (H)}{T}\right)$ . Notice that, if the reaction is exothermic, $\Delta (H)< 0$ so $\Delta (S_{\mathrm{surr}})> 0$ .

According to our statement of the Second Law, a spontaneous process in an isolated system is always accompaniedby an increase in the entropy of the system. If we want to apply this statement to a non-isolated system, we must include thesurroundings in our entropy calculation. We can say then that, for a spontaneous process, $$\Delta (S_{\mathrm{total}})=\Delta (S_{\mathrm{sys}})+\Delta (S_{\mathrm{surr}})> 0$$ Since $\Delta (S_{\mathrm{surr}})=-\left(\frac{\Delta (H)}{T}\right)$ , then we can write that $\Delta (S)-\frac{\Delta (H)}{T}> 0$ . This is easily rewritten to state that, for a spontaneousprocess:

[link] is really just a different form of the Second Law of Thermodynamics.However, this form has the advantage that it takes into account the effects on both the system undergoing the process and thesurroundings. Thus, this new form can be applied to non-isolated systems.

[link] reveals why the temperature affects the spontaneity of processes. Recall that the condensation of water vapor occurs spontaneously attemperature below 100°C but not above. Condensation is an exothermic process; to see this, consider that the reverse process,evaporation, obviously requires heat input. Therefore $\Delta (H)< 0$ for condensation. However, condensation clearly results in a decrease in entropy, therefore $\Delta (S)< 0$ also. Examining [link] , we can conclude that $\Delta (H)-T\Delta (S)< 0$ will be less than zero for condensation only if the temperature is not too high. At high temperature, the term $-\Delta (S)$ , which is positive, becomes larger than $\Delta (H)$ , so $\Delta (H)-T\Delta (S)> 0$ for condensation at high temperature. Therefore, condensation only occurs at lower temperatures.

Because of the considerable practical utility of [link] in predicting the spontaneity of physical and chemical processes, it is desirable tosimplify the calculation of the quantity on the left side of the inequality. One way to do this is to define a new quantity $G=H-TS$ , called the free energy . If we calculate from this definition the change in the free energy which occurs during a process at constanttemperature, we get $$\Delta (G)=G_{\mathrm{final}}-G_{\mathrm{initial}}=H_{\mathrm{final}}-T{S}_{\mathrm{final}}-H_{\mathrm{initial}}-T{S}_{\mathrm{initial}}=\Delta (H)-T\Delta (S)$$ and therefore a simplified statement of the Second Law of Thermodynamics in [link] is that

for any spontaneous process. Thus, in any spontaneous process, the free energy of the system decreases. Notethat $G$ is a state function, since it is defined in terms of $H$ , $T$ , and $S$ , all of which are state functions. Since $G$ is a state function, then $\Delta (G)$ can be calculated along any convenient path. As such, the methods used to calculate $\Delta (H)$ in another study can be used just as well to calculate $\Delta (G)$ .