edited by
RALPHK. BIRDWHISTELL University of West Florida Pansawla, FL 32504
textbook forum Reaction Thermodynamics: A Flawed Derivation Fred M. Hornack University of Nolth Carolina at Wilmington, Wilmington, NC 28403
In mathematics classrooms, the learning experience is sometimes enhanced by introducing simple proofs that lead to obvious contradictions. For example, a very popular one begins with the equations = t. The equation is multiplied by s, and t2is subtracted &om both sides. The resulting equation can be expressed as (s + t)(s - t) = t(s - t) and when the factor (s - t) is removed, there remains s + t = t. Since s = t, it follows that 2t = t or 2 = 1. In puzzling over this result students learn mathematics, and what they learn is apt to be etched in memory. It is in this same vein that the following problem is presented. The mathematics is quite simple so that students can participate in the derivations, search for the "flies in the ointment," and perhaps prepare a n essay summarizing their findings. In the process, they will learn something about mathematical logic and thermodynamic principles. It should become evident to them that even a convincing proof may have a fatal defect. The conventional derivation of the familiar van't Hoff equation is based on differential calculus.
The same relation can be obtained from the following equations using algebra only:
In this treatment, the temperature interval is assumed to be reasonably small, @and ASo are considered to be constant, and the free energy and entropy terms are systematically eliminated. Conversely, the enthalpy and free energy changes can be eliminated1 and one obtains the little known relation: R(T2In K2- TI in Kl) @= 72 ' - Ti Hence ASo,like m,can be calculated from equilibrium constants at two temperatures. 'A simple first step in both of these derivations is to eliminate the free energy changes and obtain AH - TI@ = -RT1 In Kl fl- T2ASo= -RT2 In Kz From these, A* can be conveniently eliminated and an expression forA 9 developed. If the equations aredivided by the respective temperatures T, and T2, ASocan be eliminated easily. 2Handbookof Chemistryand Physics, 66th ed.; Weast. R.C., Ed.; CRC Press: Boca Raton. FL, 1985-1986, pp D55-D59.
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Journal of Chemical Education
I t can now be shown that when AtT for a reaction is positive, the value for ASo must be positive also. According to the van't Hoff equation, if a reaction is endothermic the equilibrium constant increases when the temperature increases. If Tz is the higher temperature, then In K2 is larger than In Kl. Since Tz is larger than TI and In KZis larger than In Kl, i t follows that TzlnKz>TllnKl The numerator of the expression for AS' is therefore positive as is TZ- T1 in the denominator. Hence, ASomustbe positive for a n endothermic reaction and the following basic hypothesis emerges: If @for a reaction is positive, ASois also necessarily positive. There is concrete support for this-conclusion: (1)Gaseous dissoeiation reactions like Nz04(g)+ 2NOz(g) and PCIS(g)+ PCI3(g)+ C12(g)are endothermic and AS' calculated from standard tables is always found to be positive. This is reasonable from the "microscopic"or molecular point of view. In dissociation, bonds are broken and the system requires the absorption of "hand energy". At the same time, the number of moles of gas increases and a rise in entropy is expected because of greater disorder at the molecular level. (2) The "law of entropy" proves the hypothesis. In an endothermic process, heat is absorbed from the environment and A9 for the thermal swoundings is negative. If AS for the system is also negative, the overall change in entropy would he negative and the process would he forbidden. Hence, a system undergoing an endothermic change cannot s&er a decrease in entropy.
Although the foregoing arguments are convincing, a damaging piece of information is uncovered. Here as in mathematics. the existence of a counter example is enough to cause the &mplete collapse of a theory. he reaction C(graphite1+ 2Brz(g)-r CBr4(g) is endothermic with AtT = 16 kJ. As expected from the decreased number of moles, ASo is negative and is calculatedZ to be 4 0 Jldeg. The basic hypothesis is obviously invalid. What errors in reasoning have been made in its derivation and justification? Solution The mathematical error arises in the conclusion that if a>c and b>d then a b x d . This is always true when all of the values are positive. In the present case, if the equilibrium constants are less than unity the conclusion may or may not be true since the logarithms will be negative. A negative AS" requires that Tz In Kz < TI In Kl which is equivalent to
I(2
