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Chemistry 75
The Degree of Advancement Variable
In discussions of reactions approaching an equilibrium, it is helpful to have a single
variable that measures the progress of all species as they move from their initial amounts
to their final, equilibrium amounts. This variable is called the degree of advancement or
extent of reaction variable, which we will symbolize .
Imagine a schematic reaction in which reactants A and B form products C and D
according to
A + B C + D
with , , , and the appropriate stoichiometric coefficients. We imagine mixing, at a
given temperature and pressure, arbitrary amounts of each species:
o = initial amount of A
nA
nBo = initial amount of B
nCo = initial amount of C
o = initial amount of D
nD
Reaction then starts, and at any later instant imagine finding nA, nB, etc., moles of A, B,
o , n o, etc., and to the stoichiometric coefficients
etc. These amounts are related to nA
B
through the equalities that define :
o – n
o
o
o
nA
A = nB – nB = nC – nC = nD – nD = .
Note that has units of moles. The differential degree of advancement, d, is
dn
dn
dnC
dnD
d = – A = – B =
=
.
(1)
(2)
For d > 0, dnA and dnB are negative, corresponding to the disappearance of reactants,
while dnC and dnD are positive, as more of the products appear. Thus, d > 0
corresponds to a reaction progressing from left to right as written.
The degree of advancement starts at zero, no matter what the initial composition,
o , etc., at the start. As reaction ensues, can either increase (become
since nA = nA
positive) or decrease (become negative), according to the spontaneous direction of the
reaction away from the initial composition and towards the equilibrium composition.
Note that cannot increase or decrease without limit, however; its range is governed by
the initial composition and the reaction stoichiometry.
Chemistry 75
Example 1 In preparation for studying the following reaction
3H2(g) + N2(g) 2NH3(g)
we mix 4 mol H2, 2 mol N2, and 1 mol NH3. What are the limits on for this
mixture?
Solution The initial composition is given by
o = 4 mol
o = 2 mol
o
nH
nN
nNH
= 1 mol
2
2
3
and = 3, = 1, = 2, and = 0. Imagine first all the ammonia reverting to
H2 and N2 so that nNH3 = 0. Then, using Eq. (1),
2 – nN2
4 – nH2
0 – 1
=
=
= – 1 mol = (nNH3 = 0).
3
1
2
2
The most negative can be is thus –1/2 mol. (Note, too, that nH2 = 11/2 mol
and nN2 = 5/2 mol when = –1/2 mol.)
If N2 and H2 react to form more NH3, then will be positive. The
maximum will be determined by the limiting reagent: does one deplete H2
first, or N2? Since every mole of N2 that reacts consumes 3 mol H2, we see
that when nH2 = 0, nN2 will be 2/3 mol, and thus H2 is the limiting reagent for
this particular reactant mixture. The largest positive value for will thus be
4 – 0
= 4 mol = (nH2 = 0).
3
3
and thus nNH3 = 11/3 mol is the maximum amount of NH3 possible from this
mixture because
o
nNH3 = 2 + nNH
= 2 4 + 1 mol = 11 mol.
3
3
3
Suppose the reaction had been written
6H2(g) + 2N2(g) 4NH3(g).
What would the limits on be for this reaction, given the same initial
mixture? Since is inversely proportional to stoichiometric coefficients,
multiplying a balanced reaction by a given factor divides the limits by the
same factor. Thus, –1/4 mol 2/3 mol for the same initial mixture
viewed in the context of this balanced reaction.
Example 2 For the reaction
2N2O5(g) 4NO2(g) + O2(g)
by how many moles must the amounts of each substance change to advance by 1 mol?
Solution By writing Eq. (1) in the form
no
n
= N2 O5 – N2 O5 = 1 mol
2
2
we can solve for no – n, yielding
o
nN
– nN2 O5 = 2 mol,
2 O5
implying 2 moles of N2O5 must disappear to advance by 1 mol. Similarly 4
mol NO2 and 1 mol O2 must appear as the 2 mol N2O5 disappear. Note that if
the initial mixture does not contain N2O5, then can not advance toward
positive values.