Is the following rate law possible for the given reaction? A researcher is investigating the following overall reaction:
$$\ce{2C + D -&gt; E}$$
The researcher claims the rate law for the reaction is written as follows: $\text{Rate} = k[\ce{C}][\ce{D}]$.

Is the rate law equation possible for the given reaction?
If so, suggest a mechanism that would match the rate law. If not, explain why not.


The answer is yes. I don't know how to arrive at this answer. From what I  have learned, the following must be true for a) to be true:

The rate law for the slow step agrees with the experimental rate law
The elementary steps of add up to the overall reaction.

I would assume the following elementary steps from the rate the researcher has confirmed:
$$
\begin{align}
\ce{C + D &amp;-&gt; CD}\\
\ce{C + CD &amp;-&gt; E}
\end{align}
$$
But since $\ce{CD}$ has a larger size than $\ce{D},$ wouldn't it be less likely to have the proper alignment in a collision to enter the active state? Thus, wouldn't it be a slower reaction than $\ce{C + D}?$
Please inform me where my logic is wrong and why the answer is yes.
You should use steady state approximation which leads to the mentioned answer:

$$\frac{\mathrm{d}[\ce{CD}]}{\mathrm{d}t} = 0$$

$$
\begin{align}
\ce{C + D &amp;-&gt;[$k_1$] CD}\\
\ce{C + CD &amp;-&gt;[$k_2$] E}
\end{align}
$$

$$\ce{-&gt;[QSSA]}\quad k_1[\ce{C}][\ce{D}] = k_2[\ce{C}][\ce{CD}] \quad\to\quad k_1[\ce{D}] = k_2[\ce{CD}]$$

$$\frac{\mathrm{d}[\ce{E}]}{\mathrm{d}t} = k_2[\ce{C}][\ce{CD}] = k_1[\ce{C}][\ce{D}]$$