How to numerically solve a reaction network with fast equilibrium steps?

For example If I have to solve the following reaction numerically

\begin{align} *\ce{ + A(g) &<=>[K1] [A*]} \tag{1}\\ \ce{[A{*}] &<=>[k1][k-1] [B{*}]} \tag{2}\\ \ce{[B*] &<=>[K2] [B](g) +}* \tag{3} \end{align}

Where the first and third reactions are fast equilibrium steps.

[* ] represents vacnt site and [A*] and [B*] represent adsorbed species.

The reaction is carried out in a closed container, therefore the concentrations of A(g) and B(g) are not constant. Given initial A(g) and B(g) pressure and assuming zero coverage of reactants initially, how do I solve a system like this numerically.

• After you specify all the parameters, it's essentially a differential equation problem. I would suggest use matlab (ode45 or other solver). If it fails, use BZZmath or SUNDIALS solver for stiff problem like the one you mentioned. Sep 12 '17 at 4:09
• I understand that its a differential equation problem and have solved problems of this nature before. But the rapid equilibrium steps confuse me, I have solved problems which dont have rapid equilibration. I dont have a reverse and backward constant for the first and third reaction, only an equilibrium constant. Sep 12 '17 at 4:32

For A adsorption: $$r_1=0$$ because the reaction is in equilibrium.So, $$K_1=\frac{[A]}{P_A \times [*]}$$ From this equation you can relate $[A]$ to $[*]$.Similarly, using the last reaction you can relate $[B]$ to $[*]$. Your [A] and [B] are connected via reaction 2. $$\frac{d[A]}{dt}=-\frac{d[B]}{dt}=-k_1[A]+k_{-1}[B]$$ This is a first order differential equation, so you need only one boundary condition. You have three unknown and three boundary conditions $([A]_0=[B]_0=0$ and $[*]=1.00)$.