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Home Physical Chemistry Equilibrium and the Second Law of Thermodynamics Free Energy  
See also: Standard Enthalpies, Free Energies of Formation, Standard Entropies  
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Free EnergyAuthor: John Hutchinson
How can the Second Law be applied to a process in a system that is not isolated? One way to view the lessons of the previous observations is as follows: in analyzing a process to understand why it is or is not spontaneous, we must consider both the 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 we cannot prove it here, the entropy increase of a substance due to heat q at temperature T is given by ΔS = q/T. From another study, we can calculate the heat transfer for a process occurring under constant pressure from the enthalpy change, ΔH. By conservation of energy, the heat flow into the surroundings must be (ΔH). Therefore, the increase in the entropy of the surroundings due to heat transfer must be ΔS_{surr} = (ΔH/T). Notice that, if the reaction is exothermic, ΔH < 0 so ΔS_{surr} > 0. According to our statement of the Second Law, a spontaneous process in an isolated system is always accompanied by an increase in the entropy of the system. If we want to apply this statement to a nonisolated system, we must include the surroundings in our entropy calculation. We can say then that, for a spontaneous process, ΔS_{total} = ΔS_{sys} + ΔS_{surr} > 0 Since ΔS_{surr} = (ΔH/T), then we can write that ΔSΔH/T > 0. This is easily rewritten to state that, for a spontaneous process:
Equation 2 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 the surroundings. Thus, this new form can be applied to nonisolated systems. Equation 2 reveals why the temperature affects the spontaneity of processes. Recall that the condensation of water vapor occurs spontaneously at temperature 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 ΔH < 0 for condensation. However, condensation clearly results in a decrease in entropy, therefore ΔS < 0 also. Examining equation 2, we can conclude that ΔH  TΔS < 0 will be less than zero for condensation only if the temperature is not too high. At high temperature, the term (ΔS), which is positive, becomes larger than ΔH, so ΔHTΔS>0 for condensation at high temperature. Therefore, condensation only occurs at lower temperatures. Because of the considerable practical utility of equation 2 in predicting the spontaneity of physical and chemical processes, it is desirable to simplify the calculation of the quantity on the left side of the inequality. One way to do this is to define a new quantity G = HTS, called the free energy. If we calculate from this definition the change in the free energy which occurs during a process at constant temperature, we get ΔG = G_{final}G_{initial} = H_{final}TS_{final}(H_{initial}TS_{initial}) = ΔHTΔS and therefore a simplified statement of the Second Law of Thermodynamics in equation 2 is that
for any spontaneous process. Thus, in any spontaneous process, the free energy of the system decreases. Note that 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 ΔG can be calculated along any convenient path. As such, the methods used to calculate ΔH in another study can be used just as well to calculate ΔG.


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