c12 JWBS043-Rogers September 13, 2010 11:27 Printer Name: Yet to Come
188 SOLUTION CHEMISTRYWritten in terms of Gibbs chemical potentials, this gives usμ 2 (g)=μ◦ 2 (g)+RTlnp 2
p◦=μ◦ 2 (g)+RTlnp 2whereμ◦ 2 (g) is the standard state free energy for an ideal gas at a partial pressurep 2
in a mixture of gases. The analogous expression for an ideal solute at mole fraction
X 2 in a mixture of solutes isμ 2 (sol)=μ◦ 2 (sol)+RTlnX 2(We shall usually treat the special case of only one solute dissolved in one solvent.)
Equality of the two Gibbs potentials at equilibrium givesμ 2 (sol)=μ 2 (g)
μ◦ 2 (g)+RTlnp 2 =μ◦ 2 (sol)+RTlnX 2orRTlnp 2
X 2=μ◦ 2 (sol)−μ◦ 2 (g)This leads to an expression with only constants on the right-hand side:lnp 2
X 2=
μ◦ 2 (sol)−μ◦ 2 (g)
RTorp 2
X 2=eμ◦ 2 (sol)−μ◦ 2 (g)
RT =econst=kwhich is Henry’s law,p 2 =kX 2. Notice that the difference between Raoult’s law
and Henry’s law is merely in the arbitrary selection of a standard state.12.7 BOILING POINT ELEVATION
The utility of Henry’s law becomes apparent when we consider a very important
class of binary solutions, the class of nonvolatile solutes in a volatile solvent—for
example, sugar (sucrose) in water. The nonvolatile solute has no measurable vapor
pressure, hence Raoult’s law does not apply because there is nop◦Bin Figs. 12.3 and
12.4. The total vapor pressure is due to the solvent:p=X 1 p 1 ◦