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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.

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    $\begingroup$ Does ABC form a product? If not, there isn’t a steady state. The system approaches equilibrium and then concentrations remain constant after equilibrium is reached $\endgroup$ – Andrew Feb 4 at 2:08
  • $\begingroup$ @Andrew: it doesn't form a product. why isn't there a steady state? i would like to analyze the relationship between initial values of A and B and the maximal amount of ABC they yield, i.e. when d[ABC]/dt = 0. $\endgroup$ – user64296 Feb 4 at 2:19
  • $\begingroup$ “Steady state” usually refers to a system in flux, where reactant and product concentrations are changing, but an intermediate concentration is constant. It sounds like you are talking about an equilibrium where no concentrations are changing? Or are you interested in the time evolution of the approach to equilibrium? $\endgroup$ – Andrew Feb 4 at 2:48
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    $\begingroup$ To elaborate - the reason for my confusion is that you asked about the "steady-state concentration of ABC". Here, ABC is the final product, and products don't have constant steady-state concentrations; only intermediates do. The products will only have a constant concentration when the system reaches equilibrium. Equilibrium concentrations can be determined using the equilibrium constants with no time factor involved. $\endgroup$ – Andrew Feb 4 at 13:30
  • $\begingroup$ If you are only interested in the final concentration of ABC, then remove the term "steady-state" completely. Your question is "what is the equilibrium concentration of ABC?". $\endgroup$ – Andrew Feb 4 at 14:25
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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 $\ce{[C]_o}$.

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  • $\begingroup$ can you say more about why steady state is wrong terminology? in equilibrium we will have d[ABC]/dt = 0 so mathematically why can't we solve for [ABC] as function of initial values that way? $\endgroup$ – user64296 Feb 4 at 14:44
  • $\begingroup$ Generally, we use "steady-state" to describe a system in which reactant and product concentrations are still changing, but an intermediate concentration remains essentially constant. Here, you are interested in a product. Its concentration is only constant at the end, when all of the concentrations are constant. Mathematically, you will note that I did use the fact that d[ABC]/dt=0 in setting up the two kinetic equations. It just isn't a scenario we would describe as a steady-state. $\endgroup$ – Andrew Feb 4 at 14:52
  • $\begingroup$ You must have made a mistake in your algebra. I started with C=Co-ABC. Then AB=(k4ABC)/(k3C). Substitute the expression for C into that, and you've got AB in terms of ABC. Then A and B are both done the same way. Eg A=Ao-AB-ABC and substitute the expression you got above for AB and you have A in terms of only ABC. Finally, subsitute into the first kinetic equation k1A.B=k2AB and isolate ABC. It's a big mess, ultimately quartic in ABC, but once you put in values for the initial concentrations and rate constants, easily solvable. $\endgroup$ – Andrew Feb 5 at 2:05

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