Effect of Catalysis on Reaction Rate in elementary Reaction I understand that there are many different catalytic mechanisms that suggest different ways of accounting for the effect of a catalyst on reaction rate. My question is:

Given a termolecular elementary (it is known it is an elementary reaction) reaction of the form:
$$\ce{A + B + C &lt;=&gt; D + E + C}$$
Does the rate of reaction simplify down (according to collision theory) to:
$$r = [\ce{C}]\bigr(k_\mathrm{fwd}[\ce{A}][\ce{B}] - k_\mathrm{rev}[\ce{D}][\ce{E}]\bigr)$$

Edit: To clarify my question, I have a gas phase elementary reaction of the form:
$$\ce{A + B + C -&gt; D + E + C}$$
I have another of the form:
$$\ce{D + E + C -&gt; A + B + C}$$
Whether they are exactly the same mechanism is irrelevant. My question relates to how quantitatively does a collision partner's concentration in a gas mixture affect reaction rate. i.e. is forward rate defined by (assuming that I know the reaction acts as an elementary reaction in that there is no intermediate):
$$r_\mathrm{fwd} = k_\mathrm{fwd}[\ce{A}][\ce{B}][\ce{C}]$$
and reverse rate defined by (reverse is also elementary reaction - not concerned if same mechanism just also acts in the rate law as a single collision event):
$$r_\mathrm{rev} = k_\mathrm{rev}[\ce{D}][\ce{E}][\ce{C}]$$
Therefore, the total rate, not assuming they're the same reaction just two elementary reactions that go in opposite directions and we've called on forward and one reverse, the total rate forward, taking into account our other reaction, is:
$$r = r_\mathrm{fwd} - r_\mathrm{rev}$$
$$r = k_\mathrm{fwd}[\ce{A}][\ce{B}][\ce{C}] - k_\mathrm{rev}[\ce{D}][\ce{E}][\ce{C}]$$
$$r = [C]\left(k_\mathrm{fwd}[\ce{A}][\ce{B}] - k_\mathrm{rev}[\ce{D}][\ce{E}]\right)$$
Therefore quantitatively a collision partner does not change the overall rates direction but rather enhances it in both directions. Is this correct?
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.$%edit$