6.5 Third-Law Entropy and Residual Entropy
339
S spec (nH 2 ; 0) =
3
4 S spec (oH 2 ; 0) +
1
4 S spec (pH 2 ; 0) .
The entropy at absolute zero for oH 2 will be given by S spec (oH 2 ; 0) = R ln j =1 =
R ln 3, while the entropy for pH 2 will be given by S spec (pH 2 ; 0) = R ln j =0 = 0.
This gives
S spec (nH 2 ; 0) =
3
4 R ln 3 0.824R ≈ 6.86 J mol
−1 K
−1 ,
which should be compared with S spec (nH 2 ) − S cal (nH 2 ) = 6.40 J mol
−1 K −1 . A
similar calculation for nD 2 gives an additional contribution of
S spec (nD 2 ; 0) =
1
3 R ln 3 3.054 J mol
−1 K
−1 ,
to be compared with S spec (nD 2 ) − S cal (nD 2 ) = 3.096 J mol
−1 K −1 . Note that
we need not consider the entropy of mixing, S mix , because it is the same for
all temperatures and, as S cal (T ) represents the difference between the entropy
measured at temperature T and the entropy at absolute zero, the entropy of mixing
cancels, and it is S spec (T ) without the entropy of mixing that must be compared with
S cal (T ). It is important to note that such a contribution to the spectroscopic entropy
for the hydrogen isotopologues gives different values of the residual entropy for H 2
and D 2 . This difference is observed.
The case of water ice, for which detailed calculations have been given, is an
especially interesting one. In the 1920s and early 1930s the residual entropy of ice
was known to be about 4.184 J mol
−1 K −1 . An initial explanation of this value was
given by Giauque and Ashley [41] along the lines of the successful explanation of
the residual entropy of H 2 , as being associated with the presence of the metastable
ground state of ortho-water. However, when improved calorimetric measurements
of Giauque and Stout [40] became available, the difference between S cal and S spec
was found to be only 3.431 J mol
−1 K −1 . Pauling [42] provided an explanation
based upon partial randomness in crystalline ice, in which the ordered arrangement
corresponding to zero entropy is not attained. From X-ray work it was known that
each oxygen atom in ice is tetrahedrally surrounded by four other oxygen atoms at
2.76 Å distance, bonded by hydrogen bonds. Each H atom is about 0.95 Å from one
oxygen atom (the same as in an isolated water molecule) and 1.81 Å from another:
moreover, they assume positions such that each O atom will have two H atoms
attached to it. Pauling made the following assumptions:
(i) In ice each O atom has two H atoms attached to it at distances of about 0.95 Å,
forming a water molecule, with the HOH angle being about 105 ◦ ;
(ii) each H 2 O molecule is oriented so that its two H atoms are directed approximately toward two of the four O atoms which surround it tetrahedrally, forming
hydrogen bonds;
(iii) the orientation of adjacent water molecules are such that only one H atom lies
approximately along each O–O-axis;
339
S spec (nH 2 ; 0) =
3
4 S spec (oH 2 ; 0) +
1
4 S spec (pH 2 ; 0) .
The entropy at absolute zero for oH 2 will be given by S spec (oH 2 ; 0) = R ln j =1 =
R ln 3, while the entropy for pH 2 will be given by S spec (pH 2 ; 0) = R ln j =0 = 0.
This gives
S spec (nH 2 ; 0) =
3
4 R ln 3 0.824R ≈ 6.86 J mol
−1 K
−1 ,
which should be compared with S spec (nH 2 ) − S cal (nH 2 ) = 6.40 J mol
−1 K −1 . A
similar calculation for nD 2 gives an additional contribution of
S spec (nD 2 ; 0) =
1
3 R ln 3 3.054 J mol
−1 K
−1 ,
to be compared with S spec (nD 2 ) − S cal (nD 2 ) = 3.096 J mol
−1 K −1 . Note that
we need not consider the entropy of mixing, S mix , because it is the same for
all temperatures and, as S cal (T ) represents the difference between the entropy
measured at temperature T and the entropy at absolute zero, the entropy of mixing
cancels, and it is S spec (T ) without the entropy of mixing that must be compared with
S cal (T ). It is important to note that such a contribution to the spectroscopic entropy
for the hydrogen isotopologues gives different values of the residual entropy for H 2
and D 2 . This difference is observed.
The case of water ice, for which detailed calculations have been given, is an
especially interesting one. In the 1920s and early 1930s the residual entropy of ice
was known to be about 4.184 J mol
−1 K −1 . An initial explanation of this value was
given by Giauque and Ashley [41] along the lines of the successful explanation of
the residual entropy of H 2 , as being associated with the presence of the metastable
ground state of ortho-water. However, when improved calorimetric measurements
of Giauque and Stout [40] became available, the difference between S cal and S spec
was found to be only 3.431 J mol
−1 K −1 . Pauling [42] provided an explanation
based upon partial randomness in crystalline ice, in which the ordered arrangement
corresponding to zero entropy is not attained. From X-ray work it was known that
each oxygen atom in ice is tetrahedrally surrounded by four other oxygen atoms at
2.76 Å distance, bonded by hydrogen bonds. Each H atom is about 0.95 Å from one
oxygen atom (the same as in an isolated water molecule) and 1.81 Å from another:
moreover, they assume positions such that each O atom will have two H atoms
attached to it. Pauling made the following assumptions:
(i) In ice each O atom has two H atoms attached to it at distances of about 0.95 Å,
forming a water molecule, with the HOH angle being about 105 ◦ ;
(ii) each H 2 O molecule is oriented so that its two H atoms are directed approximately toward two of the four O atoms which surround it tetrahedrally, forming
hydrogen bonds;
(iii) the orientation of adjacent water molecules are such that only one H atom lies
approximately along each O–O-axis;