# What are some common mistakes students make with rate law?

Jun 18, 2017

Here are two things I've seen happen, or anticipate happening.

First off, here's how you write a rate law for the reaction $A + B \to C$ in general:

$r \left(t\right) = k {\left[A\right]}^{m} {\left[B\right]}^{n}$,

where $r \left(t\right)$ is the initial rate, $k$ is the rate constant, $\left[A\right]$ is the molar concentration of reactant $A$ and $m$ is its order. These orders have no actual dependence on the reaction stoichiometry in many cases, especially for a complex reaction.

The following are two common errors.

## 1

When determining the order with respect to, say, reactant $A$, given a table of data, you are supposed to choose two trials $i$ and $j$ where reactant $B$'s concentration does NOT vary, and get rid of that in the rate law, since ${1}^{n} = 1$.

That would give, for trials 2 and 1 let's say:

$\frac{{r}_{2} \left(t\right)}{{r}_{1} \left(t\right)} = {\left(\frac{{\left[A\right]}_{2}}{{\left[A\right]}_{1}}\right)}^{m}$

The following three lines are what NOT to do:

$\frac{{r}_{2} \left(t\right)}{{r}_{1} \left(t\right)} = {\left(\frac{{\left[B\right]}_{2}}{{\left[B\right]}_{1}}\right)}^{n} = {\left(\frac{0.1}{0.1}\right)}^{n} = {1}^{n}$

$\implies \ln \left(\frac{4.0 \times {10}^{- 4}}{2.0 \times {10}^{- 4}}\right) = \ln {1}^{n} = n \ln 1$

$\textcolor{red}{\text{Therefore, }}$$\textcolor{red}{n = \frac{\ln 2}{\ln 1} = 0}$.

The error here is in choosing the wrong reactant terms to cancel out in the rate law; $A$ did vary, so it does not go to 1 for arbitrary orders.

Furthermore, I think you know that you can't divide by zero. Lastly, you cannot say that anything divided by zero is zero... I honestly don't know why this student (presumably) did not use his calculator to check his work, because he would have seen his calculator complain about dividing by zero.

But this has actually happened before. Who knew!

## 2

Some students have trouble with

• $\left(i\right)$ rationalizing where the coefficients in the rate of disappearance and appearance come from, and
• $\left(i i\right)$ how to reconcile that with the fact that the reactant orders have nothing to do with the stoichiometric coefficients in an overall reaction.

For example, say you have the overall (complex) reaction:

$\textcolor{b l u e}{2} A + \textcolor{b l u e}{3} B \to C$

You would write the rate law as:

$r \left(t\right) = k {\left[A\right]}^{m} {\left[B\right]}^{n}$

$= - \frac{1}{\textcolor{b l u e}{2}} \frac{\Delta \left[A\right]}{\Delta t} = - \frac{1}{\textcolor{b l u e}{3}} \frac{\Delta \left[B\right]}{\Delta t} = + \frac{1}{\textcolor{b l u e}{1}} \frac{\Delta \left[C\right]}{\Delta t}$

You in fact, without more information (e.g. the reaction mechanism or some tabled kinetics data), would not know that $m$ and $n$ are.

How you do use the coefficients is that the VALUES of each rate of disappearance or appearance were scaled down by the reciprocated coefficients in front.

You need to work $\boldsymbol{x}$ times as fast to consume a reactant or make a product with a stoichiometric coefficient of $\boldsymbol{x}$ in a set amount of time.