An Elementary Discussion of Chemical Equilibrium Carl W. David University of Connecticut, Storrs. CT 06268 The crowning glory of thermodynamics is the understanding that we gain of chemical equilibrium. Without relying on kinetics, we obtain a sense of why chemical equilihrium is achieved by having the system follow "time's arrow". The standard derivation of the equation usually does not explicitly show how G is a minimum as function of .t. This is because the generality of the original derivation would be violated if a specific reaction were used. As a result the relationship between eq 1and any minimization is obscure.' In Hill2 an example is used that relates the minimization to the attainment of equilibrium, hut the example involves an equilibrium,
each of these compounds will differ from the starting values. During the process of equilibrating, at some instant of time, the number of moles of HZwill he nH,, the number of moles of NZ will he n ~ , and , the instantaneous number of moles of NH3 will be nNH,. If we were to define the extent of reaction (.t) OHz
then it takes some thought to realize that this definition (except for the sign of 0 is applicable to N2 and NH3 in addition to Hz. Thus
f=
I ~ N ,- ~ ' N J " ~ 1
which is unique and not representative of the complexities of reactions that do not have equal numbers of moles on both sides of the equation. In this discussion, a more difficult reaction is used as the prototype, and extra lines of calculus are included to make the discussion accessible to persons who can mechanically take partial derivatives. A simple reaction worth studying, prototypical of almost all aspects of this subject, is 3 H M + Ndg)
2NH,k)
(2)
The Gibbs free energy per mole of the Hz is
is valid, as is
f=
~ N H :, u
no~~d ~
~
I
The only problem we have is that if HZand N2 are disappearing, then NH3 is appearing, while if Hq and Nz are appearing, then NH3 is disappearing. Thus f defined on the right-hand side of the equilibrium equation (i.e., NH3) is opposite in sien to t defined on the left-hand side of the eauilihrium equation (i.e., HZand Nz). Basine our definition of o n the ripht-hand side of the equation (i.e., on NH3) means reversing the sign of zeta in the case of Hz and Nz, so that
while the Gibhs free energy per mole of the NZis
while, of course, the Gihbs free energy per mole of NH3 is
At equilihrium, we have a mixture of the three suhstances: HZ,N2, and NH3. The central question of chemical thermodynamics is how much of each substance will be present "at equilibrium". We want to phrase this in a way that is both correct and understandable, so we have to distinguish carefully between the number of moles of the various suhstances which we actually mix together, the number of moles of each substance that we measure to he present "at equilihrium", and the "extent of the reaction". This is because the "extent of reaction" is a theoretical concept, implying that we could artificially "shift the reaction". The stoichiometry numhers of the reaction we are studying, (3, 1, and 2) are found by writing
where U H , is 3, UN, is 1, and U N H ~is 2. If we start with noH2 original moles of HZ, no^, original moles of Nz, and ~ ' N H , original moles of NH3, then the actual numbers of moles of
' Castellan. G. W. Physical Chemistry, 3rd ed.; Addisan-Wesley:
and
We will solve now for nH,, nN,, and nNH,, the instantaneous mole numbers of the substances involved in the equilihrium. We have nH2= no,, -
We need these in order to obtain the Gibbs free energy, not for one mole of these substances, but for any number of moles of them. Thus, if there are nH, moles of Hz present a t some instant of time, then (assuming all three substances are ideal gases so that we can substitute P for the fugacity) we have (using eq 4): G H=~n ~ g ' n *+ ~ H P TP H ~
and similarly
Readina. MA. 1983:o 230. HII(T.~ . ' ~ e c t u & on Matter& Equilibrium: Benjamin: New York. 1966;p 250.
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Number 5
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What is the Gibhs free energy of the mixture of n ~moles , of Hz, n ~ of* N2 and ~ N H moles , of NH3? I t is just the sum of the Gihbs free energy of all three components (using eq 5 ) ,
+ GH, + G N=~ ~ N H $ ' N H ~ + n ~ & ' ~+,n ~ & ' ~ x
Grnkture= GNH,
+ nNH:RTin PNq + nHIRTIn PH1+ nN,RT lnPNx
We define AGO
= ~NH$'NH~ -
UHB'H;, -u ~ $ o ~ l
so that ,G ,,,
= n o ~ H & ++ ~R ~ T3 1nPNHJ+ n',(gDH2
+ noN3W,9+ RT l n P N J
+ R T I n PHJ
+ TAGo
+ XuNHl In PNH3 - uH2 I n PH1- uN2 I n PNJRT
(10)
We now take the derivative of G,iXtu, with respect to f and obtain Notice that GmiXt.,. is both an explicit and an implicit function of {. This is because you can see {(explicitly), and i t is also present in hidden form, hidden in the various partial pressures, that is, n;
(7)
Pi = *;P,,,.] = -P,l nto+A
for each of the three components, ni contains {through the expression ni = noi f fuj
where the i requires us to choose the appropriate sign. For nwt.l we have nml
= ~ ' N H , + ~ U N H : , f nSH2- fuH3 + n ' ~ l- @N1
Continuing, using eq 7, we have
which becomes = noNH,,
+ noHl+ noN2+ fAu
with A u being -2 in our particular case! Defining ~ " N H , n ' ~ , n ' ~ , we have
+
+
not.,~las
where we have held the total pressure constant during the differentiation process. Following the algebraic trail we have
n,,,.~ = nDb,1 + f A u
so that the mole fractions would be
Notice that laooears in both the numerator and the denominator of each mole fraction. The central thermodynamic point is that G tends to a minimum at constant T and P. That must mean that, if we take the derivative of G,iXt,,, with respect to { (at constant temperature and total pressure) and set it equal to zero, we will find what special value of (corresponds to an extremum in G ; if we could take the second derivative of G,iXt,,, with respect to f, we could ascertain whether in fact the extremum was a maximum or a minimum. Rewriting eq 6 we have Gmixture= ~ ' N H , , ~ N H :+, R T I ~ P N H )+ noH9(gOH2
+ RT I n pH.,)
+ noNr@ON, + R
T InPNJ
+ ~('JNH:~'NH:, - ~ H & ' ~ H ?- UNB'N,)
+ ~GJNH:,In PNH:, - UH* I n PH?- ON, In PN,)RT 408
Journal of Chemical Education
(9)
and similarly
and
so that
Wondrously, the last six terms in this expression cancel, and we obtain
which shows that the partial pressures adjust themselves is until G,iXt,,, finds its minimum value. When dG,iXt,,./a[ zero, all the partial pressures have attained their equilibrium values and we obtain A G O
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= - RT
In K,,
Number 5
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May 1988
409
