CHAPTER 11 REACTIONS AND OTHER CHEMICAL PROCESSES

11.4 ENTHALPIES OFSOLUTION ANDDILUTION 332

CLis equal to the limiting slope ofLversus
p
mB, ofÅHm(sol,mB) versus
p
mB,

and ofÅHm.dil,m^0 B!m^00 B/versus

p
m^0 B. The value given by Eq.11.4.29can be used for
extrapolation of measurements at 25 C and low molality to infinite dilution.

Equation11.4.28can be derived as follows. For simplicity, we assume the pressure

is the standard pressurep. At this pressureHB^1 is the same asHB, and Eq.11.4.17

becomesLBDHBHB. From Eqs.12.1.3and12.1.6in the next chapter, we can

write the relations

HBDT^2



@.B=T /

@T



p;fnig

HBDT^2

d.m;B=T /

dT

(11.4.30)

Subtracting the second of these relations from the first, we obtain

HBHBDT^2

@.

Bm;B/=T

@T



p;fnig

(11.4.31)

The solute activity on a molality basis,am;B, is defined byBm;BDRTlnam;B.

The activity of an electrolyte solute at the standard pressure, from Eq.10.3.10, is given

byam;BD.CC/

.mB=m/. Accordingly, the relative partial molar enthalpy

of the solute is related to the mean ionic activity coefficient by

LBDRT^2 



@ln (^) 
@T

p;fnig
(11.4.32)
We assume the solution is sufficiently dilute for the mean ionic activity coefficient
to be adequately described by the Debye–Huckel limiting law, Eq. ̈ 10.4.8: ln (^) D
ADHjzCzj
p
Im, whereADHis a temperature-dependent quantity defined on page
294. Then Eq.11.4.32becomes
LBDRT^2 jzCzj
p
Im

@ADH
@T

p;fnig
(11.4.33)
(very dilute solution)
Substitution of the expression given by Eq.10.4.9on page 297 forImin a solution of
a single completely-dissociated electrolyte converts Eq.11.4.33to
LBD
"
RT^2
p
2

@AADH
@T

p;fnig

jzCzj

3=2

p
mB
DCLB
p
mB (11.4.34)
(very dilute solution)
The coefficientCLB(the quantity in brackets) depends onT, the kind of solvent, and
the ion charges and number of ions per solute formula unit, but not on the solute
molality.
LetCLrepresent the limiting slope ofLversuspmB. In a very dilute solution
we haveLDCLpmB, and Eq.11.4.27becomes
LBDLC
pm
B
2
dL
dpmB
DCL
p
mBC
pm
B
2
CL (11.4.35)
By equating this expression forLBwith the one given by Eq.11.4.34and solving for
CL, we obtainCLD.2=3/CLBandLD.2=3/CLB
p
mB.