Effect of Catalysis on Reaction Rate in elementary Reaction
If it is in equilibrium, you are already looking at it from a macroscopic scale. It can therefore not be an elementary reaction. For example, $$\ce{A + B -&gt; C + D}$$ is an elementary reaction, the reverse $$\ce{C + D -&gt; A + B}$$ is a different elementary reaction. The rates of both reactions 
\begin{align}
r&amp;=k_\mathrm{fwd}[\ce{A}][\ce{B}]\\
r&amp;=k_\mathrm{rev}[\ce{C}][\ce{D}]\\
\end{align}
would determine the location of the equilibrium 
\begin{align}
\ce{A + B &amp;&lt;=&gt; C + D}&amp; K&amp;=\frac{k_\mathrm{fwd}}{k_\mathrm{rev}}.
\end{align}


For a true termolecular reaction, i.e. three species react in one concerted transition state,
$$\ce{A + B + C -&gt; D + E + F},$$
the rates are always given as
\begin{align}
r&amp;=k_\mathrm{fwd}[\ce{A}][\ce{B}][\ce{C}]\\
r&amp;=k_\mathrm{rev}[\ce{D}][\ce{E}][\ce{F}].\\
\end{align}
You cannot simplify a rate law of an elementary reaction.$%edit$