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172 REACTION SPONTANEITY AND THE DIRECTION OF THERMODYNAMIC CHANGE

the van’t Hoff isochore:

− H 1 1
O
K s(2) (cryst)
ln = − (4.78)
K s(1) R T 2 T 1
where R is the gas constant, H O is the change in enthalpy associated with
(cryst)
crystallization, and the two temperatures are expressed in kelvin, i.e. thermodynamic
temperatures.

Worked Example 4.14 The solubility s of potassium nitrate is 140 g per 100 g of water
◦
◦
at 70.9 C, which decreases to 63.6 g per 100 g of water at 39.9 C. Calculate the enthalpy
of crystallization, H O .
(cryst)
Strategy. For convenience, we will call the higher temperature T 2 and the lower temper-
ature T 1 . (1) The van’t Hoff isochore, Equation (4.78), is written in terms of a ratio,so
we do not need the absolute values. In other words, in this example, we can employ the
solubilities s without further manipulation. We can dispense with the units of s for the
same reason. (2) We convert the two temperatures to kelvin, for the van’t Hoff isochore
requires thermodynamic temperatures, so T 2 = 343.9K and T 1 = 312.0K. (3) We insert
values into the van’t Hoff isochore (Equation (4.78)):

140 − H (cryst) 1 1
O
Note how the two ln 63.6 =− 8.314 J K −1 mol −1 × 343.9K − 312.0K
minus signs on the
right will cancel. − H O
(cryst) −4 −1
ln(2.20) = −1 × (−2.973 × 10 K )
8.314 J K −1 mol
ln 2.20 = 0.7889
−1
We then divide both sides by ‘2.973 × 10 −4 K ’, so:
O
0.7889 H (cryst)
=
2.973 × 10 −4 K −1 8.314 J K −1 mol −1
3
The term on the left equals 2.654 × 10 K. Multiplying both sides by
Only when the differ- R then yields:
ence between T 2 and
T 1 is less than ca. H O = 2.654 × 10 K × 8.314 J K −1 mol −1
3
40 K can we assume (cryst)
the reaction enthalpy so
H O is independent of
O
temperature. We oth- H (cryst) = 22.1kJ mol −1
erwise correct for the
temperature depen-
dence of H O with
the Kirchhoff equation SAQ 4.12 The simple aldehyde ethanal (VII)reacts with
(Equation (3.19)). the di-alcohol ethylene glycol (VIII) to form a cyclic
acetal (IX):