The thermodynamic description of mixtures
3.10 Ideal-dilute solutions
chemical potentials of A and B fall. Using eqn 3.15, the final Gibbs energy of the system is
GfnAAnBB
nA{A*RTlnxA}nB{B*RTlnxB} nAA*nARTlnxAnBB*nBRTlnxB
where the xJare the mole fractions of the two components in the mixture. The differenceGfGiis the change in Gibbs energy that accompanies mixing. The standard chemical potentials cancel, so
GRT{nAlnxAnBlnxB}
BecausexJnJ/n, we can substitute nAxAnandnBxBninto the expression above and obtain
GnRT{xAlnxAxBlnxB} which is eqn 3.16.
Equation 3.16 tells us the change in Gibbs energy when two components mix at constant temperature and pressure (Fig. 3.27). The crucial feature is that be- causexAandxBare both less than 1, the two logarithms are negative (ln x0 if x1), so G0 at all compositions. Therefore, mixing is spontaneous in all pro- portions. Furthermore, if we compare eqn 3.16 with G HTS, we can con- clude that:
1. Because eqn 3.16 does not have a term that is independent of temperature,
H0 (3.17a)
2. Because G0TSnRT{xAlnxAxBlnxB},
S nR{xAlnxAxBlnxB} (3.17b)
The value of Hindicates that although there are interactions between the molecules, the solute-solute, solvent-solvent, and solute-solvent interactions are all the same, so the solute slips into solution without a change in enthalpy. There is an increase in en- tropy, because the molecules are more dispersed in the mixture than in the unmixed component. The entropy of the surroundings is unchanged because the enthalpy of the system is constant, so no energy escapes as heat into the surroundings. It follows that the increase in entropy of the system is the “driving force” of the mixing.
Raoult’s law provides a good description of the vapor pressure of the solvent in a very dilute solution, when the solvent A is almost pure. However, we cannot in general expect it to be a good description of the vapor pressure of the solute B be- cause a solute in dilute solution is very far from being pure. In a dilute solution, each solute molecule is surrounded by nearly pure solvent, so its environment is quite unlike that in the pure solute, and except when solute and solvent are very similar (such as benzene and methylbenzene), it is very unlikely that the vapor pres- sure of the solute will be related in a simple manner to the vapor pressure of the pure solute. However, it is found experimentally that in dilute solutions, the vapor pressure of the solute is in fact proportional to its mole fraction, just as for the sol- vent. Unlike the solvent, though, the constant of proportionality is not in general the vapor pressure of the pure solute. This linear but different dependence was dis- covered by the English chemist William Henry (1774–1836) and is summarized as Henry’s law:
The vapor pressure of a volatile solute B is proportional to its mole fraction in a solution:
pBxBKB (3.18)
HereKB, which is called Henry’s law constant, is characteristic of the solute and chosen so that the straight line predicted by eqn 3.18 is tangent to the experimental curve at xB0 (Fig. 3.28).
Henry’s law is usually obeyed only at low concentrations of the solute (close toxB0). Solutions that are dilute enough for the solute to obey Henry’s law are calledideal-dilute solutions.
The Henry’s law constants of some gases are listed in Table 3.2. The values given there are for the law rewritten to show how the molar concentration depends on the partial pressure, rather than vice versa:
[ J]KHpJ (3.19)
Henry’s constant, KH, is commonly reported in moles per cubic meter per kilopas- cal (mol m3kPa1). This form of the law and these units make it very easy to calculate the molar concentration of the dissolved gas, simply by multiplying the partial pressure of the gas (in kilopascals) by the appropriate constant. Equation 3.19 is used, for instance, to estimate the concentration of O2in natural waters or the concentration of carbon dioxide in blood plasma.
COMMENT 3.3 The Web site contains links to online databases of Henry’s law constants.■
Ideal dilute solution(Henry's law)
Mole fraction of B, x2
0 1
0
KB
Pressure
pB*
Ideal solution (Raoult's law)
Fig. 3.28 When a component (the solvent) is almost pure, it behaves in accord with Raoult’s law and has a vapor pressure that is proportional to the mole fraction in the liquid mixture and a slope p*, the vapor pressure of the pure substance. When the same substance is the minor component (the solute), its vapor pressure is still proportional to its mole fraction, but the constant of proportionality is now KB.
Table 3.2
Henry’s law constants for gases dissolved in water at 25°CKH/(mol m3 kPa1) Carbon dioxide, CO2 3.39 101
Hydrogen, H2 7.78 103
Methane, CH4 1.48 102
Nitrogen, N2 6.48 103
Oxygen, O2 1.30 102
EXAMPLE 3.3 Determining whether a natural water can support aquatic life
The concentration of O2in water required to support aerobic aquatic life is about 4.0 mg L1. What is the minimum partial pressure of oxygen in the atmosphere that can achieve this concentration?
Strategy The strategy of the calculation is to determine the partial pressure of oxygen that, according to Henry’s law (written as eqn 3.19), corresponds to the concentration specified.
Solution Equation 3.19 becomes pO2
[ K O
H 2]
We note that the molar concentration of O2is
[O2]
mol m3
where we have used 1 L103m3. From Table 3.2, KHfor oxygen in water is 1.30 102mol m3kPa1; therefore the partial pressure needed to achieve the stated concentration is
pO2 9.6 kPa
The partial pressure of oxygen in air at sea level is 21 kPa (158 Torr), which is greater than 9.6 kPa (72 Torr), so the required concentration can be maintained under normal conditions.
A note on good practice: The number of significant figures in the result of a cal- culation should not exceed the number in the data.
SELF-TEST 3.9 What partial pressure of methane is needed to dissolve 21 mg of methane in 100 g of benzene at 25°C (KB5.69 104kPa, for Henry’s law in the form given in eqn 3.18)?
Answer: 57 kPa (4.3 102Torr)■
CASE STUDY 3.1 Gas solubility and breathing
We inhale about 500 cm3of air with each breath we take. The influx of air is a result of changes in volume of the lungs as the diaphragm is depressed and the chest expands, which results in a decrease in pressure of about 100 Pa relative to atmospheric pressure. Expiration occurs as the diaphragm rises and the chest contracts and gives rise to a differential pressure of about 100 Pa above atmo- spheric pressure. The total volume of air in the lungs is about 6 L, and the ad-
(4.0/32) mol m3 1.30 102mol m3kPa1
4.0 32
mol m3 4.0 103 32 103 mol
L 4.0 103 32 4.0 103g L1
32 g mol1
ditional volume of air that can be exhaled forcefully after normal expiration is about 1.5 L. Some air remains in the lungs at all times to prevent the collapse of the alveoli.
A knowledge of Henry’s law constants for gases in fats and lipids is impor- tant for the discussion of respiration. The effect of gas exchange between blood and air inside the alveoli of the lungs means that the composition of the air in the lungs changes throughout the breathing cycle. Alveolar gas is in fact a mix- ture of newly inhaled air and air about to be exhaled. The concentration of oxy- gen present in arterial blood is equivalent to a partial pressure of about 40 Torr (5.3 kPa), whereas the partial pressure of freshly inhaled air is about 104 Torr (13.9 kPa). Arterial blood remains in the capillary passing through the wall of an alveolus for about 0.75 s, but such is the steepness of the pressure gradient that it becomes fully saturated with oxygen in about 0.25 s. If the lungs collect fluids (as in pneumonia), then the respiratory membrane thickens, diffusion is greatly slowed, and body tissues begin to suffer from oxygen starvation. Carbon dioxide moves in the opposite direction across the respiratory tissue, but the partial pres- sure gradient is much less, corresponding to about 5 Torr (0.7 kPa) in blood and 40 Torr (5.3 kPa) in air at equilibrium. However, because carbon dioxide is much more soluble in the alveolar fluid than oxygen is, equal amounts of oxygen and carbon dioxide are exchanged in each breath.
A hyperbaric oxygen chamber, in which oxygen is at an elevated partial pres- sure, is used to treat certain types of disease. Carbon monoxide poisoning can be treated in this way, as can the consequences of shock. Diseases that are caused by anaerobic bacteria, such as gas gangrene and tetanus, can also be treated because the bacteria cannot thrive in high oxygen concentrations. ■
Henry’s law lets us write an expression for the chemical potential of a solute in a solution. By exactly the same reasoning as in Derivation3.5, but with the em- pirical constant KBused in place of the vapor pressure of the pure solute, pB*, the chemical potential of the solute when it is present at a mole fraction xBis
BB*RTlnxB (3.20)
This expression, which is illustrated in Fig. 3.29, applies when Henry’s law is valid, in very dilute solutions. The chemical potential of the solute has its pure value when it is present alone (xB1, ln 10) and a smaller value when dissolved (whenxB1, ln xB0).
We often express the composition of a solution in terms of the molar concen- tration of the solute, [B], rather than as a mole fraction. The mole fraction and the molar concentration are proportional to each other in dilute solutions, so we write xBconstant [B]. To avoid complications with units, we shall interpret [B] wher- ever it appears as the numerical value of the molar concentration in moles per liter.
Thus, if the molar concentration of B is 1.0 mol L1, then in this chapter we would write [B]1.0. Then eqn 3.20 becomes
BB*RTln(constant)RTln [B]
We can combine the first two terms into a single constant, which we denote B両, and write this relation as
BB両RTln [B] (3.21)
−∞
Chemical potential, B
0
Mole fraction of solvent, xB
1 B°
Fig. 3.29 The variation of the chemical potential of the solute with the composition of the solution expressed in terms of the mole fraction of solute. Note that the chemical potential of the solute is lower in the mixture than for the pure solute (for an ideal system). This behavior is likely to be shown by a dilute solution in which the solvent is almost pure and the solute obeys Henry’s law.
Figure 3.30 illustrates the variation of chemical potential with concentration pre- dicted by this equation. The chemical potential of the solute has its standard value when the molar concentration of the solute is 1 mol L1.