Page 261 - Thermodynamics of Biochemical Reactions

Chemical Equilibrium in One Phase Systems 261

calckrx[eq_,islist~l:=Module[{ener~,dG},(*Calculates the equilibrium constant K for a
chemical equation typed in the form atpHl+de==hydroion+atpHO.*)
energy=Solve [eq,del ;
dG=energy[[l,l,2l]/.is->islist;
Exp[-dG/(8.31451*.29815)11

atpkl=calckrx[at~Hl+de==hydro~on+atpHO,{0,.1,.25}1
-7 -7
I2.51302 1.81144 10 , 3.4275 10 }

TableForm[{at~kl,atpk2,adpkl,adpk2,~ikl,kref},TableHeadings-
> { {llatpkl", "atpk2". aadpklcl, "adpk2", "piklt', "kref 'I },iO, . 1, .25 1 11
0 0.1 0.25
-8 -7 -7
atpkl 2.51302 10 1.81144 10 3.4275 10
atpk2 0.0000210082 0.0000924186 0.000149099
-8 -7 -7
adpkl 6.64366 10 2.92266 10 4.71512 10
adpk2 0.0000437761 0.000117531 0.000161669
-8 -7 -7
pikl 6.05499 10 1.62565 10 2.23617 10
kref 0.291014 0.177606 0.151432
3.2 For the solution reaction A = B, assume that the standard Gibbs energy of formation of A is 20 kJ mol-' and of B is 18
kJ mol-' at 298.15 K. (a) For a reaction starting with a mole of A at a concentration of 1 M, plot the Gibbs energy of the
mixture versus extent of reaction from zero to unity and identify the approximate extent of reaction at equilibrium. (b)
Identify the equilibrium extent of reaction more precisely by plotting the derivative of the Gibbs energy of the mixture with
respect to extent of reaction. (c) Calculate the equilibrium constant and verify the equilibrium extent of reaction.

(a) The Gibbs energy of the reaction mixture is given by

Clear [XI

g= (1-X) *20+18*~+ (8.31451*10"-3) *298.15* ((I-X) *LOg[l-X] +X*LOSf[XI)
20 (1 - X) + 18 x + 2.47897 ((1 - X) LOg[l - XI + x Lo~[xI)

Plot[g, {X, 0, I}, Axesorigin-> {O, 15},
AxesLabel- {llg/molul, "G/kJ mol-l"}, PlotRange -f {15, 21}];