Page 154 - Essentials of physical chemistry

116 Essentials of Physical Chemistry

sticky side. That is a fortuitous situation but often forensic analysis is faced with smudged partial
prints on rough surfaces like cloth or even porous ceramic surfaces. In those cases, any enhancement
of the fingerprint image can be helpful in solving a criminal case. The application of iodine vapor is a
very old technique where an investigator breaths out through a tube containing calcium carbonate
drying agent to remove breath moisture and then the dry breath air flows over iodine crystals as a
vapor on to the fingerprint surface. There it is preferentially absorbed into the fingerprint and shows as
a dark purple image. This is a very old technique, which fell into disfavor because the image tends to
fade away after a few hours. However, recently the use of digital photography has brought this
technique back to active use [3]. In particular, the fading aspect is actually an advantage because a
fresh iodine image on a rough surface like cloth may be only recognizable as a smudge. The trick is to
make a digital photograph of the original image, wait until the iodine fades, and then make another
digital photograph without moving the sample. Then simple digital subtraction of the background
image leaves only the iodine image. Thus, iodine treatment of fingerprints has undergone renewed
interest and use for prints on rough surfaces including cloth with a background pattern.
How does the iodine fingerprint technique work? The basic principle is that solid iodine is one of
the few substances that has a significant vapor pressure from the solid and actually sublimes directly
from the solid into the gas phase. Although it is common in organic laboratory procedures to purify
volatile compounds by sublimation in a vacuum pistol at very low pressure, carbon dioxide and
iodine are among only a very few substances that sublime easily at room temperature and pressure.
Thus, the technique of sweeping iodine vapor over a surface is a ‘‘dry’’ procedure without any liquid
and yet some data is available for the liquid [4].
Now we come to the most interesting type of equilibrium that occurs in the qualitative phase
diagrams of Figure 6.4, the place where all three state phase boundaries come together in what is
called the ‘‘triple point.’’ The triple point is the (P, T) point in the phase diagram where there are
three simultaneous equilibria: solid–liquid, solid–gas, and liquid–gas. While it looks like the triple
point is the melting point (it is close), it is slightly different. The triple point for water is 273.168K
but the melting point is 0.00988 lower, which leads to the standard value of 273.158K. Let us use the
0
data from Ref. [5], as shown in Table 6.3. The values of DH 0 fus and DH vap are given so we can add
0
them to obtain a value for the hypothetical DH sub value for the energy required for a whole mole to
flash from the solid directly to the vapor, a pure sublimation process without going through the
liquid phase. We have all probably seen ‘‘dry ice’’ sublime as solid carbon dioxide goes directly
from the solid to the gas and the process here might be called ‘‘dry iodine,’’ which is why it is so
useful in developing fingerprints without any liquid mess. We should note for future forensic
investigators that pure iodine is corrosive and poisonous so one should avoid inhaling it and the
use of a squeeze bulb is better than mouth-on-tubing when applying the vapor. We can add the
numbers using the ‘‘after-minus-before’’ principle of thermodynamics even though it might be
difficult to actually measure this DH 0 sub value. With the DH 0 sub value we can calculate the hypothet-
ical temperature at which the total sublimation would occur using the Clausius–Clapeyron equation.
A word of caution is needed here in that there are a number of 1=T values in the calculation and one
should not round off the calculation until the end because these reciprocal values are very small and
rounding them too soon can lead to large errors. Another problem with the data in Table 6.3 is that
the temperatures at which the pressures are very low are probably more uncertain due to the
difficulty in measuring such low pressures while the highest vapor pressure of the solid might be
contaminated experimentally with interference with some slight melting, so we choose the data
point at 68.78C at a pressure of 1 kPa. Then we can solve for the hypothetical T sub temperature.
101,325 Pa
(8:314 J=mol K) ln 1 1

1000 Pa ,so T sub ¼ 443:913 K, the

( 57, 090 J=mol) þ (68:7 þ 273:15) K ¼ T sub

hypothetical number at which the process of only sublimation would produce 1.01325 bar pressure
(1 atm). At the triple point P solid ¼ P liquid , so we can write