Page 158 - Modern physical chemistry

150 Relationships among Reactants

Whenever the correction is small, the error introduced by using the Berthelot equa-
tion is very small.

ExampleZ5

Calculate the entropy correction for gas imperfection of nitrogen at 77.32 K and 1.000
bar, if its critical temperature is 126 K and its critical pressure 33.9 bar.
Employ formula (7.31):

SO - S = 27 (8.3145 J K-l mOI-l) 1.000 [ 126)3 = 0.90 J K-l mol-l.
32 33.9 77.32

Thus, the entropy of 1.000 mol nitrogen in its standard state is 0.90 J Kl more than
its actual entropy at 1.000 bar and at 77.32 K. The gas imperfection imposes a slight
amount of order on the system.

Z8 Correcting the Calorimetric Entropy for Frozen-in Disorder
The order in a crystal involves not only where the molecules or ions go but also how
they are oriented. When two or more different orientations are almost equally stable, dis-
order is frozen in. This produces a residual entropy that must be added to the calori-
metric entropy to get the absolute entropy for the given substance.
Consider a linear molecule that is nearly as stable when placed in the crystal lattice
backward as forward. At the temperature of formation, the Boltzmann factor e-£/kT is nearly
the same for each configuration and the number put in each way is nearly the same.
Since each molecule may have two different configurations, N molecules in a crystal
possess 2!' configurations. In the approximation that each of these is equally likely, state
number W in formula (5.55) equals 2N and the residual entropy at 0 K is

[7.32J
Using statistical mechanics, one can calculate state number W from molecular para-
meters. These may be determined from spectroscopic data. One can then compare the
absolute entropies calculated from these data with the entropies that the third law pro-
vides. The discrepancies are attributed to the disorder frozen into the crystals. Some
results appear in table 7.2.
The CO and N 20 molecules are linear; so number (7.32) gives the frozen-in entropy
if there were complete disorder over the two orientations. The excess of this number
over the corresponding number in table 7.2 indicates that there is a small amount of
ordering.
In solid nitric oxide, the NO molecules dirnerize to produce a diamagnetic crystal. As
a consequence, the number in table 7.2 is for 112 mole of the solid. For 1.000 mol, the
observed value is 6.28 J Kl , somewhat higher than number (7.32).
In the interior of ice I, there are four close oxygen atoms to each given oxygen atom.
These are at the comers of a tetrahedron centered on the given atom. Now, the four neigh-
bors are linked to the central atom by hydrogen bonds. But an 0 - H-O bond is not sym-
metric; the H is 0.99 A from one 0 and 1.77 A from the other at eqUilibrium. Consequently,
each H in the crystal has the choice of two positions. In a crystal of N molecules there
are 2N hydrogen atoms, yielding a total of 22N possible configurations. However, many of
these correspond to ionized structures that are unstable. If all configurations occurred in
a given OH 4 unit, they would be 24 or 16 in number. But one of these is OH 4 ++, four are
OH3+, four are OH- , and one is 0=. By difference, we see that six correspond to OH 2 • So