Adv P Chem Final - Maggie Purvis
Rate Constants
In[]:=
k1f=249918.96;​​k1r=4044.1996;​​k2f=43301953;​​k2r=7767323.4;​​k3f=0.134715;​​k3r=0.1899605;​​k4=kp*7.34928*10^-14;​​k5=kp*7.34928*10^-14;​​k6=kp*1.41293*10^-20;
Reaction Rate Equations
In[]:=
c1f=k1f*Am[t];​​c1r=k1r*B1[t];​​c2f=k2f*B1[t];​​c2r=k2r*B2[t];​​c3f=k3f*Am[t];​​c3r=k3r*B3[t];​​p1f=kp*A[t]*KCO3[t];​​p1r=kp*Am[t]*KHCO3[t];​​{p2f,p3f,p4f}=kp*{B1[t],B2[t],B3[t]}*KHCO3[t];​​{p2r,p3r,p4r}={k4*Z5[t],k5*E5[t],k6*E6[t]}*KCO3[t];
Test Case 1 - Protonation/Deprotonation of A
In[]:=
eqs1={KHCO3'[t]==p1f-p1r,​​KCO3'[t]==-p1f+p1r,​​A'[t]==-p1f+p1r,​​Am'[t]==p1f-p1r,​​B1​​KHCO3[0]==0,​​KCO3[0]==0.5,​​A[0]==0.5,​​Am[0]==0};
In[]:=
vars1={KHCO3[t],KCO3[t],A[t],Am[t]};
In[]:=
s1=NDSolve[eqs1/.kp10^11,vars1,{t,0,1}];
In[]:=
Plot[Evaluate[A[t]/.s1],{t,0,.0000000002},PlotRange{0,0.5}]
Out[]=
0.0
5.0e-11
1.0e-10
1.5e-10
2.0e-10
0.1
0.2
0.3
0.4
0.5
​
​Test Case 2 - Include B1 and Z5
In[]:=
eqs2={KHCO3'[t]==p1f-p1r-p2f+p2r,​​KCO3'[t]==-p1f+p1r+p2f-p2r,​​A'[t]==-p1f+p1r,​​Am'[t]==p1f-p1r-c1f+c1r,​​B1'[t]==-p2f+p2r+c1f-c1r,​​Z5'[t]==p2f-p2r,​​KHCO3[0]==0,​​KCO3[0]==0.5,​​A[0]==0.5,​​Am[0]==0,​​B1[0]==0,​​Z5[0]==0};
In[]:=
vars2={KHCO3[t],KCO3[t],A[t],Am[t],B1[t],Z5[t]};
In[]:=
s2=NDSolve[eqs2/.kp10^11,vars2,{t,0,1}];
In[]:=
Plot[Evaluate[Z5[t]/.s2],{t,0,0.0001},PlotRange{0,0.5}]
Out[]=
0.00000
0.00002
0.00004
0.00006
0.00008
0.00010
0.1
0.2
0.3
0.4
0.5
Test Case 3 - Include E5 and B2
In[]:=
eqs3={KHCO3'[t]==p1f-p1r-p2f+p2r-p3f+p3r,​​KCO3'[t]==-p1f+p1r+p2f-p2r+p3f-p3r,​​A'[t]==-p1f+p1r,​​Am'[t]==p1f-p1r-c1f+c1r,​​B1'[t]==-p2f+p2r+c1f-c1r-c2f+c2r,​​Z5'[t]==p2f-p2r,​​E5'[t]==p3f-p3r,​​B2'[t]==-p3f+p3r+c2f-c2r,​​KHCO3[0]==0,​​KCO3[0]==0.5,​​A[0]==0.5,​​Am[0]==0,​​B1[0]==0,​​Z5[0]==0,​​E5[0]==0,​​B2[0]==0};
In[]:=
vars3={KHCO3[t],KCO3[t],A[t],Am[t],B1[t],Z5[t],B2[t],E5[t]};
In[]:=
s3=NDSolve[eqs3/.kp10^11,vars3,{t,0,1}];
In[]:=
Plot[Evaluate[Z5[t]/.s3],{t,0,0.0001},PlotRange{0,0.5}]
Out[]=
0.00000
0.00002
0.00004
0.00006
0.00008
0.00010
0.1
0.2
0.3
0.4
0.5
Final - Plot Z5 as a Function of Time for 2 Cases: kp=10^11 and kp=10^8
In[]:=
eqs={KHCO3'[t]==(-p1f+p2f+p3f+p4f)-(-p1r+p2r+p3r+p4r),​​KCO3'[t]==-(-p1f+p2f+p3f+p4f)+(-p1r+p2r+p3r+p4r),​​A'[t]==-p1f+p1r,​​Am'[t]==p1f-p1r-c1f+c1r-c3f+c3r,​​B1'[t]==-p2f+p2r+c1f-c1r-c2f+c2r,​​B2'[t]==-p3f+p3r+c2f-c2r,​​B3'[t]==-p4f+p4r+c3f-c3r,​​Z5'[t]==p2f-p2r,​​E5'[t]==p3f-p3r,​​E6'[t]==p4f-p4r,​​KHCO3[0]==0,​​KCO3[0]==0.5,​​A[0]==0.5,​​Am[0]==0,​​B1[0]==0,​​B2[0]==0,​​B3[0]==0,​​Z5[0]==0,​​E5[0]==0,​​E6[0]==0};
In[]:=
vars={KHCO3[t],KCO3[t],A[t],Am[t],B1[t],B2[t],B3[t],Z5[t],E5[t],E6[t]};
Trial 2 - kp = .001*10^11 = 10^8