Your program can now be modified easily to include the backreactions.
The reverse rate coefficients are denoted km1 and km2.
 | Clear[h,nsteps,k1,k2,km1,km2,a,b,c,data];
totaltime=100;
h=0.5;
nsteps = Round[totaltime/h];
k1=1.0 ;
k2=0.1 ;
km1=0.05;
km2=0.1;
a[0]=1;
b[0]=0;
c[0]=0;
Do[
{
a[n] = a[n-1] + h*( km1*b[n-1] - k1*a[n-1]),
b[n] = b[n-1] + h*( k1*a[n-1] + km2*c[n-1] - k2*b[n-1] - km1*b[n-1]),
c[n] = c[n-1] + h*( k2*b[n-1] - km2*c[n-1])
},
{n,1,nsteps,1}] //N;
data = Table[{a[n], b[n], c[n]}, {n, 0, nsteps}] |
to get a simplified rate law. Hint: You can use the expression for
d[B]/dt to isolate [B] and then eliminate [B] from the
expression for d[A]/dt. Q: What is the resulting rate law
and what is its solution [A](t)? Carry out computations using the
steady state approximation for differing values of the rate constants
and compare results with your numerical calculations for eh exact
reaction scheme by plotting the two on the same graph. Can you give a
simple set of rules indicating the range of validity of the steady
state approximation?