150 Chemical thermodynamics
Thereisanimportant wordofwarninginthecalculation ofK(V,T).Thepartition
functions for the various components must all be calculated using the same energy
zero. (Thatiswhere alot ofthechemistry comesin. After all, you expect the reaction
togofrom ahighenergystate to alow one!) What this meansisthat the partition
functionstaketheform
Z(s)=V(
2 πM(s)kkkBT
h^2) 3 / 2
·ZZZint·exp(−W(s)/kkkBT)whereinthefinalterm a constantW(s)isincludedtoget the energy levelsofthat
component right, i.e. toget theground state of the translational motion (usually
taken as zero in a one-component situation) at the right energy. The result of this is
thatK(V,T)willcontainafactor oftheform exp(−W/kkkBT),whereinthis case
W=W(AB)−W(A)−W(B). The sign and magnitude ofWhas agreat influence
on the equilibrium condition of the reaction. IfWis positive, the forward reaction
will beareluctant starter particularlyatlow temperatures, since the exponentialfactor
inKwill be small and henceNNABwill also be small. On the other hand, a negativeW
will favour the forward equilibrium of the reaction, particularly at low temperatures.
The exponentialwill belargeandhence so will beits effect onKandonNNAB.This
is as expected. A negativeWmeans that the AB side of the equation has a lower
energy than theA+Bside.
Equation (13.24) also allows us to understandthevolumedependence ofthe reac-
tion. All thepartition functions haveVas a simple factor arising only from the
translationalpart. HenceK∝ 1 /V.Thus a small volumewill favour theforward
reaction. Againthisis reasonable, since theformation ofAB causes a reductionin
the total number of molecules, hence a reduction in the pressure (equation (13.19))
andhenceinthePVenergy ofthe system.
1 3.4.5 Reaction rates
Thelaw ofmass action (13.24) tellsusthe equilibrium concentrationsin a reaction.
That is a fine achievement, but it is not always enoughtodojustice to chemical reality.
For instance, when we consider the reaction A+B→AB, we have seen above that
whenWisnegative theyieldofAB willtendtoimprove atlower temperatures.
However, as anycook knows, reactionratesget smaller at lower temperatures, so
that one may have to wait an unrealistic time for the theoretical equilibrium to occur.
The reaction canbefrozeninto a metastable stateinwhichthechemicalpotentials
do not have time to equalize. This is a familiar thermodynamic situation. We have
discussed earlier (see Chapter 10) the existence of frozen-in disorder (i.e. non-zero
entropy)atlow temperatures.
In chemical reactions, the usual situation is that illustrated in Fig. 13.1. Consider the
transitionbetween separate A andB atoms towardstheformation ofan AB molecule,
forasysteminwhichWisnegative. Thefigure showsaschematicgraphshowing