How to derive the steady-state solution for simple two-step reaction with differential equations? [closed]

Question

Consider this simple two step reaction, a variant of a Michaelis-Menten type of problem, where $\ce{A}$ and $\ce{B}$ reversibly bind to make $\ce{AB}$, and $\ce{AB}$ and $\ce{C}$ reversibly bind to make $\ce{ABC}$:

\begin{align} \ce{A + B &<=> AB}\\ \ce{AB + C &<=> ABC} \end{align}

Assume all the rate constants are equal to $k$ for simplicity, and that $[\ce{A}]_0$, $[\ce{B}]_0$, $[\ce{C}]_0$ are initial values for $\ce{A}$, $\ce{B}$, $\ce{C}$, and $[\ce{AB}]_0 = [\ce{ABC}]_0 = 0$.

What is the proper way to derive the steady-state concentration of $\ce{ABC}$ as a function of initial values? here is an attempt to get $\mathrm{d}[\ce{AB}]/\mathrm{d}t$: $$\frac{\mathrm{d}[\ce{AB}]}{\mathrm{d}t} = ([\ce{A}][\ce{B}]k + [\ce{ABC}]k) - ([\ce{AB}]k + [\ce{AB}][\ce{C}]k)$$

The first parenthesis of the sum are the two ways to make $\ce{AB}$ and second parenthesis are the two ways to lose $\ce{AB}$. We can set $\mathrm{d}[\ce{AB}]/\mathrm{d}t = 0$ and try to solve for $[\ce{ABC}]$, but I am not sure how this helps and where initial values come in.

Answer 1

Based on the discussion in comments, the OP is interested in the equilibrium concentration of the species ABC in this reaction scheme:

$$\ce{A + B + C <=>[k_1][k_2] AB + C <=>[k_3][k_4] ABC}$$

Equilibrium is reached when the rates of the forward and backward reactions of each step are equal. Thus, we have have two equations describing the equilibrium:

$$k_1\ce{[A][B]}=k_2\ce{[AB]}$$ $$k_3\ce{[AB][C]}=k_4\ce{[ABC]}$$

We also know from mass conservation that $$\ce{[A]_o}=\ce{[A] + [AB] + [ABC]}$$ $$\ce{[B]_o}=\ce{[B] + [AB] + [ABC]}$$ $$\ce{[C]_o}=\ce{[C] + [ABC]}$$

These five equations can be manipulated to find an expression for $\ce{[ABC]}$ in terms of $\ce{[A]_o}$, $\ce{[B]_o}$, and $\ce{[C]_o}$.