Page 271 - Thermodynamics of Biochemical Reactions
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Chemical Equilibrium  in One Phase Systems   271



                TableForm~calcdHTsp~atpsp,C283.15,298.15,313.15},01,T~1eHeadings-
                > C C "ATP-4 " , "HATP-3 'I,  "H2ATP-2" 1 , "2 83.15  K" , "2 98.15  K" , "3 13.15  K" 1 1 I
                                                  {
                           283.15 K    298.15 K    313.15 K
                ATP-4      -3619.21    -3619.21    -3619.21
                HATP-3     -3612.91  -3612.91      -3612.91
                H2ATP-2    -3627.91  -3627.91  -3627.91

         This calculation  is based on the assumption that the standard enthalpies of formation of the three species are independent of
         temperature, and so this is the expected result.

                TableForm[calcdHTsp[at~s~,{283.15,298.15,313.15},.l],TableHeadings-
                 > C C "ATP-4 'I,  "HATP-3 " , "H2ATP-2 " 1, { "2 83 -15  K" , "298.15  K" , "313.15  K" 1 1 I

                           283.15 K    298.15 K    313.15 K
                ATP-4      -3615.12  -3614.23      -3613.22
                HATP-3     -3610.61  -3610.11      -3609.54

                H2ATP-2    -3626.89    -3626.66    -3626.41
         Even though the standard enthalpies of formation are independent of temperature at zero ionic strength, they are not indepen-
         dent of temperature  at a finite ionic strength.  The largest effect is for the -4 ion.

                 TableForm[calcdHTsp[atpspr(283.15,298.15,313.15},.25l,TableHeadings-
                 > C I "ATP-4 'I,  " IiATP-3 'I,  "H2ATP-2 " 1,  C "2 83 .15 K" , "2 98.15  K" , "313 .15 K" 1 1 1
                           283.15 K    298.15 K    313.15 K
                 ATP-4     -3613.8     -3612.62  -3611.29

                 HATP-3    -3609.86    -3609.2     -3608.46
                 H2ATP-2    -3626.56  -3626.26  -3625.93
         3.9  There are two  ways to obtain values for the enthalpy  coefficient in equation 3.6-5 as a function  of temperature: (a)
         Calculate the derivative of the Gibbs energy coefficient divided by T.  (b) Fit the enthalpy coefficients of Clarke and Glew to
         AT2 + BT3. Use both of these methods and make plots to compare these functions with the values in Table 3.1

         (a)  According to Problem 3.5, RTa is given by

                 rtal~ha=9.20485*10A-3*t-l.28466*lOA-5*tA2+4.95l99*lOA-8*tA3
                                                               -8  3
                 0.00920485 t -  0.0000128466 t2 +  4.95199 10    t

          Using the Gibbs-Helmholtz equation

                 rt2Dalpha=Expand[tA2*D[rtalpha/t,tll

                                                      -8  3
                 0. t -  0.0000128466 t2 +  9.90398 10    t



          The following are the enthalpy coefficients given by Clarke and Glew.
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