Determine the rate equation (intermediate reaction) I was helping some students today, quite successfully, until I came across this problem:  

$$\begin{align}
\ce{A &amp;&lt;=&gt;[$k_+$][$k_-$] B + C} \\[5pt]
\ce{C + D &amp;-&gt;[$k_1$] E}
\end{align}$$

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:

$$\ce{A + D -&gt; B + E}$$

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?
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.