Page 333 - Thermodynamics of Biochemical Reactions

Systems of Biochemical Reactions 333

lnkra=Log[{6.05499*10A-8,1.9489*10A-311

{-16.6198, -6.24049)

The initial amount vector is unchanged, and so the equilibrium composition is given by

equcalcm t tnua, lnkra, noal
-'I
{1.70726 10 , 0.00354638, 0.00999983, 0.00354656, 0.00645362)

TableForm[{equcalcm[tnua,lnkra,noal 1 ,TableHeadings->{ {"c/M"), {"H", "Mg", "H2P04-
'I, "HP042- n, "MgHPO4 " 1 1 1
H Mg H2P04- HP04 2 - MgHPO4
-7
c/M 1.70726 10 0.00354638 0.00999983 0.00354656 0.00645362
6.7 When 0.01 M glucose 6-phosphate is hydrolyzed to glucose and phosphate at 298.15 K in the pH range 9-10, only four
species have to be considered.
GlcP2- + Hz 0 = Glc + HP04 '-
What is the equilibrium composition assuming the ionic strength is zero?

This calculation using equcalcc involves the problem that a, Go H2 0) is used in the calculation of the equilibrium constant
K, but the expression for the equilibrium constant does not involve the concentration of Hz 0. Thus in effect oxygen atoms
are not conserved, because in dilute aqueous solutions they are drawn for th essentially infinite reservoir of the solvent.
Therefore, the further transformed Gibbs energy G has to be used. The conservation matrix with C and P as components and
GlcP2-, Glc, and HPOd2-as species is given by

TableForm[abl
6 6 0
1 0 1
The H row and charge row are redundant, and so they are omitted. The transformed Gibbs energy has to be used because the
"concentration" of water is held constant. These transformed Gibbs energies of formation are given by

tgeGlcP=-1763.94-9*(-237.19)
370.77

tgeGlc=-915.9-6*(-237.19)
507.24

tgeHPO4=-1096.1-4*(-237.19)
-147.34

where the standard Gibbs energy of formaton of H20 is -237.19 kJ mol-l.