Page 176 - Thermodynamics of Biochemical Reactions

10.3 Calculation of Standard Transformed Entropies of Biochemical Reactions 175

since

(1 0.3-6)

for a species.
Substituting (see Section 4.3)

A,Hio = AfHy - NH(j)AfHo(H+) - N,,(j)AfHo(Mg2+) (10.3-7)
and

A,GS" = A,Gy - NH(j)[AfGo(H+) - RTln(lO)pH]
-NMg(j)[A, G"(MgZ+) - RTln(lO)pMg] (10.3-8)

in equation 10.3-6 yields

Earlier the fundamental equation for G' (equation 4.1-22) was used to show
that S' = S - n,(H)S,(H+) when hydrogen ions are bound, and that can be
extended to

S' = S - n,(H)S,(H') - n,(Mg)Sm(Mg2+) (10.3-10)
This can be written

1 njSk(j) = 1 .,iSm(j) - n,(H)S,(H+) - n,(Mg)Sm(Mg2+) (10.3-11)
Substituting n,(H) = XN,(j)nj and n,(Mg) = CN,,(j)nj yields the expression for
the standard transformed molar entropy of a species:

S:(.j) = S:(j) - N,(j)[Si(H+) + Rln(lO)pH]
- NM,(j)[Si(Mg2') + Rln(1O)pMgl (10.3-12)
which can be compared with equation 10.3-9.
Equations 10.3-9 and 10.3-12 raise an issue about conventions for the
hydrogen ion in thermodynamic tables. Since it is not possible to connect the
standard thermodynamic properties of H+ to those of molecular hydrogen, the
convention is that AfGo(H+) = 0 and AfHo(H+) = 0 at each temperature. This
indicates that the standard entropy of formation of a hydrogen ion AfSo(H+)
should be taken as zero at each temperature, but, for historical reasons, the
convention adopted in current thermodynamic tables is Si(H+) = 0 at each
temperature. In principle, the value of SZ(H+) should be calculated from
ArS'(H+) for the formation reaction for H+. One way to write this reaction is

$H,(g) = H+(aq) + e- (10.3-13)
where c- is the formal electron, not a hydrated electron in water. Both
AfSo{H+(ao)) = 0 and Si(H+) = 0 can be satisfied if the formal electron is
treated as a reactant and assigned a standard molar entropy of
Si(e-) = (i)S;(H2, g) at each temperature. A different, but perhaps more logical
convention would be to assign AfGo(e-) = 0, AfHo(e-) = 0, and AfSo(e-) = 0 so
that equation 10.3-13 would lead to
A,SO[H+(~O)] = s~~(H+)+s;(~-) - (+)S"~(HJ = s:(H+) + o - (+)S;(H,,~) = o

( 10.3- 14)

In this case, S:(H+) = (+)Si(H2, g), rather than zero.