(1)If we have sufficient heat capacity data (and the data on phase changes) we could write
(If there is a phase change between 0 K and T we would have to add the entropy of the phase change.) If Cp were constant near T = 0 we would have,
which is undefined. Fortunately, experimentally Cp → 0 as T → 0. For nonmetals Cp is proportional to T 3 at low temperatures. For metals Cp is proportional to T 3 at low temperatures but shifts over to being proportional to T at extremely low temperatures. (The latter happens when the atomic motion "freezes out" and the heat capacity is due to the motion of the conduction electrons in the metal.)
Equation 1 could be used to calculate absolute entropies for substances if we knew what the entropy is at absolute zero. Experimentally it appears that the entropy at absolute zero is the same for all substances. The third law of thermodynamics codifies this observation and sets
for all elements and compounds in their most stable and perfect crystalline state at absolute zero and one atmosphere pressure. (All except for helium, which is a liquid at the lowest observable temperatures at one atmosphere.)
The advantage of this law is that it allows us to use experimental data to compute the absolute entropy of a substance. For example, suppose we want to calculate the absolute entropy of liquid water at 25o C. We would need to know the Cp of ice from 0 K to 273.15 K and the Cp of liquid water from 273.15 K to 298.15 K. We also need the heat of fusion of water at its normal melting point. With all of this data, which can be obtained partly from theory and partly from experiment, we find
Some substances may undergo several phase changes.(2)
Entropy Changes in Chemical Reactions
We can use the third law entropies to calculate entropy changes for chemical reactions. For a typical reaction,
a A + b B → c C + d D. (3)the entropy change is
Notice two things:(4)
1. We did not define or use an entropy of formation, ΔfS o.As we have said before, ΔrS o and ΔrH o are independent of each other. They can not be calculated from each other. They must be calculated from Equation 4 and a comparable equation using heats of formation.
2. S oelement is not zero.
There is another way to calculate ΔrS o,
As we have seen before, the ΔrG o and ΔrH o can be calculated from free energies and heats of formation.(5)
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