# If we know the rate constant, can't we find the reaction rate just by looking at the coefficients?

May 24, 2016

For any rate law, you have:

$\setminus m a t h b f \left(r \left(t\right) = k {\left[A\right]}^{m} {\left[B\right]}^{n} \cdot \cdot \cdot\right)$

for $N$ number of reactants.

Let's just say there are two for simplicity. Then...

• $r \left(t\right)$ is the reaction rate as a function of time, with units of $\text{M/s}$.
• $k$ is the rate constant, with units of $\frac{1}{\text{M"^(-1+ m+n)cdot"s}}$.
• $\left[\text{X}\right]$ is the concentration of reactant $\text{X}$.
• $m , n$ are the orders of reactants $\text{A}$ and $\text{B}$, respectively.

Even if you know $k$, only if you additionally know the order of each reactant (the contribution of the reactant concentration to the rate) will you know the reaction rate at that temperature.

ISSUES WITH KNOWING A REACTION MECHANISM

Any given reaction can undergo multiple different mechanisms and still yield the same products.

That's how catalysts are valid choices to speed up a reaction without changing the final product(s) made.

For a general reaction

$\text{A" + "B" stackrel(k_("obs")" ")(->) 2"C} ,$

you could have a single elementary step, written as

$\textcolor{b l u e}{\text{A" + "B" stackrel(k_(1)" ")(=>) 2"C}} .$

"$\implies$" indicates an elementary step, which means the reaction step can be taken as-written without any underlying implications (i.e. no hidden mechanism you don't know about). So, ${k}_{1} = {k}_{\text{obs}}$.

But, if we add a catalyst $\text{D}$, we might have, say,

$\text{A}$ $+$ $\cancel{\text{D}}$ $\stackrel{{k}_{1} \text{ }}{\implies}$ $\cancel{\text{E}}$ $+$ $\text{C}$
$\cancel{\text{E}}$ $+$ $\text{B}$ $\stackrel{{k}_{2} \text{ }}{\implies}$ $\text{C}$ $+$ $\cancel{\text{D}}$
$\text{-----------------------}$
$\textcolor{b l u e}{\text{A}}$ $\textcolor{b l u e}{+}$ $\textcolor{b l u e}{\text{B}}$ color(blue)(stackrel(k_("obs")" ")(->)) $\textcolor{b l u e}{2 \text{C}} ,$

where $\text{E}$ is an intermediate, which appears in the middle of a reaction and disappears by the time the reaction is over. Here, ${k}_{\text{obs}}$ is harder to figure out. It is some non-obvious combination of ${k}_{1}$ with ${k}_{2}$.

Notice that the mechanism is now different. However, we still get $2 \text{C}$, just faster.

CONCLUSIONS

• In the above catalyzed reaction, the order is harder to figure out because the reaction mechanism is actually multiple steps instead of just one and you need to work with very meticulous data.

• In the non-catalyzed reaction, ${k}_{1} = {k}_{\text{obs}}$, so it is easier to figure out the order based on experimental data.

That means you would need to know the details of the mechanism first before you can figure out the order.