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I was helping some students today, quite successfully, until I came across this problem:

$$\begin{align} \ce{A &<=>[$k_+$][$k_-$] B + C} \\[5pt] \ce{C + D &->[$k_1$] E} \end{align}$$

The first equation is a fast process with forward and reverse rate constants $k_+$ and $k_-$, respectively. The second process is slow, with rate constant $k_1$.

The objective is to write down a rate law for the resulting equation:

$$\ce{A + D -> B + E}$$

The technique that had been working was to write down the rate for the slow process...

$$r = k_1[\ce{C}][\ce{D}]$$

Since $\ce{C}$ is a product in the fast reaction, which is an equilibrium equation, we also have:

$$k_+[\ce{A}]=k_-[\ce{B}][\ce{C}]$$

This is where things went awry, because the rate law that we seek is a function of $[\ce{A}]$ and $[\ce{D}]$ only. Considering the stoichiometry, it seemed appropriate to put $[\ce{B}]=[\ce{C}]$. This leads to the substitution $[\ce{C}]=(\frac{k_+}{k_-})^{1/2}[\ce{A}]^{1/2}$. The interface has declared the result to be invalid, so it must be an illegal maneuver. How, then, do I eliminate $[\ce{B}]$ and $[\ce{C}]$ and come up with the desired reaction rate law?

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You can't write $[\ce{B}] = [\ce{C}]$ because $\ce{C}$ is consumed in the second step but $\ce{B}$ isn't. However, you were pretty much on the right track.

Since $\ce{E}$ is produced from $\ce{D}$, you can write $[\ce{E}] = [\ce{D}]_i - [\ce{D}]$ where $[\ce{X}]$ denotes the concentration of $\ce{X}$ at a certain point in time and $[\ce{X}]_i$ denotes the initial concentration of $\ce{X}$. This is assuming that the initial concentration of $\ce{E}$ is $0$, of course.

On top of this, $\ce{C}$ is either produced from $\ce{A}$, or consumed in the production of $\ce{E}$. Therefore you can write $[\ce{C}] = [\ce{A}]_i - [\ce{A}] - [\ce{E}] = [\ce{A}]_i - [\ce{A}] - [\ce{D}]_i + [\ce{D}]$. (There is the reverse step in which $\ce{C}$ produces $\ce{A}$, but the equation still holds, since you can just consider this to be "negative" production of $\ce{C}$ from $\ce{A}$.)

Your rate law is therefore: $r = k_1[\ce{D}]\big([\ce{A}]_i - [\ce{A}] - [\ce{D}]_i + [\ce{D}]\big)$. The initial concentrations are merely constants, so you could view this as being of the form $r = k_a[\ce{D}] + k_b[\ce{A}][\ce{D}] + k_c[\ce{D}]^2$. I can't think of any way to simplify this (the steady-state approximation is not very valid here since $\ce{C}$ is produced quickly and consumed slowly, and in any case it does not yield a nicer expression). I imagine that one could use some form of an integrated rate law to find expressions for $[\ce{A}]$ and/or $[\ce{D}]$, but it would just make things even uglier.

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  • $\begingroup$ Very nice explanation of how to eliminate C. But B can also be eliminated by pretty much the same stoichiometric considerations. You're right that the math is messy though. $\endgroup$ – Curt F. Jul 20 '15 at 17:36
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The best first step in any kinetics problem is writing down materials balances on all relevant species.

$$\frac{dA}{dt}=-k_+ A + k_-B C$$ $$\frac{dB}{dt}=+k_+ A - k_-B C$$ $$\frac{dC}{dt}=+k_+ A - k_-B C - k_1 C D$$ $$\frac{dD}{dt}=- k_1 C D$$ $$\frac{dE}{dt}= k_1 C D$$

We want to solve for $\frac{dE}{dt}$ by eliminating $C$. The assumption that the first reaction is "fast" means we can make the equilibrium assumption on $A$ and $B$ to solve for $BC$

$$BC = \frac{k_+}{k_-}A$$

We can eliminate $C$ by stoichiometric considerations. The amount of $C$ in the system at any time is just equal to the net forward progress of reaction 1 minus the forward progress of reaction 2. So $C=A_0-A - (D_0-D)=(A_0-D_0)-(A-D)$. orthocresol's answer does a great job of explaining why. But we can also make a similar argument for B: the amount of B is just equal to the net forward progress of reaction 1, or $B=A_0-A$. So after application of stoichiometry, the equilibrium equation becomes:

$$BC = \frac{k_+}{k_-}A \implies \left[A_0-A \right] \left[(A_0-D_0)-(A-D)\right] = \frac{k_+}{k_-}A$$

That is a quadratic equation in $A$ (I won't show the math) that gives $A$ if you know the constants $A_0$, $D_0$, $k_+$, and $k_-$, and the variable $D$.

Substituting our expression for $C$ into the equation for $D$:

$$\frac{dD}{dt}= - k_1 C D = -k_1 D [(A_0-D_0)-(A-D)] $$

We could substitute the quadratic solution for $A$ into this equation and get a single differential equation that depends only on $D$ and constants. Since the ultimate rate we care about, $\frac{dE}{dt}$, is equal to $-\frac{dD}{dt}$, this means we have a rate law. The algebra is messy but that is the path forward.

Some problems are harder than they look!

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