Page 149 - Essentials of physical chemistry

Gibbs’ Free Energy and Equilibria 111

DH vap 1 DH vap 1
þ ln (760) ,
ln(P) ¼ þ
R T(K) R 373:15
or

y ¼ mx þ b:

Table 6.2 tells us nothing about the vapor pressure of these five liquids below 08C but provides a
more detailed view of the vapor pressures in the easily accessible laboratory range between ice water
and boiling water. The data for water also give more detail relative to the vapor pressure of water
that might be needed for Dalton’s law when a gas is collected over water and also give us an
appreciation for the relationship between water vapor in the air and a dew point temperature.

DH vap
¼ 5202:9 and we find
A direct plot of ln (P vap ) versus (1=T(K)) yields a slope of
R
DH vap ¼ (5202:9)(8:314 J=mol) ¼ 43256:9 J=mol. The value given in the 90th Edn. of the CRC
Handbook is 43990 J=mol at 258C. This handbook clearly shows that DH vap is not constant but
varies slightly with temperature. Let us check that value using two values from Table 6.2

P 2 DH vap 1 1 þDH vap T 2 T 1 T 2 T 1 P 2
ln ¼ ¼ or R ln ¼þDH vap . Note
P 1 R T 2 T 1 R T 2 T 1 T 2 T 1 P 1
we have used the C temperature difference in the denominator to remind ourselves that it is a

difference of two temperatures and most importantly that the size of 18K is exactly the same as that
of 18C.
2
We see from the plot in Figure 6.3 that a least-squares line produces a R value of 0.9999, which
indicates a near perfect line. We need to comment on the slope that was found to be 5202.9 from
the best line fit. In the Clausius–Clapeyron equation, the log ratio of the pressures would cancel
whatever units the pressures are expressed in but the units get mixed up when we separate the
pressures and move the ln (760 mm) to the right side of the equation. This is a case where you
should cancel the units in the original equation before putting it into the linear form. In addition, you
could always use the ratio of the pressure in whatever units to the standard pressure in the same
units. The question of units occurs here because we used 760 mmHg=atm. The argument of a
logarithm has to be a unitless number.
This value is within the uncertainty of the other values because the pressures in Table 6.1 are
only given to three significant figures. It will be found that if you pick any two values for the vapor
pressure of water from Table 6.2, you may get slightly different numbers for each pair of vapor
pressures that is consistent with the temperature dependent DH vap value.

TABLE 6.3
The Vapor Pressure of Solid (Rhombic) I 2 at 8C Temperatures
for Known Pressure

P 1 Pa 10 Pa 100 Pa 1 kPa 10 kPa 100 kPa
T, 8C 12.8 (solid) 9.3 (solid) 35.9 (solid) 68.7 (solid) 108 (solid) 184.0 (liquid)
Source: Lide, D.R., CRC Handbook of Chemistry and Physics, 90th Edn., CRC Press, Boca Raton, FL,
2009–2010, pp. 6–72. With permission.
0
I 2(solid) mp ¼ 113:7 C ¼ 386:9 K, DH fus ¼ 15:52 kJ=mol ¼ 3:709 kcal=mol.

I 2(liquid) bp ¼ 184:4 ¼ 457:6 K, DH 0 vap ¼ 41:57 kJ=mol ¼ 9:936 kcal=mol.

0
0
DH 0 sub ¼ DH fus þ DH vap ¼ 15:52 kJ=mol þ 41:57 kJ=mol ¼ 57:09 kJ=mol ¼ 13:64 kcal=mol.