Rate-Limiting Step — Definition & Overview

Imagine a movie theater. It's the end of the movie. The credits are rolling, and everyone begins to leave. How quickly will the theater empty? That depends on how many people can fit through the doors. Typically, only a couple of people can exit the theater at a time. So things bottleneck. People end up standing in line to leave.

Chemical reactions are like movie theaters. The slowest part makes everything backup. As we've studied kinetics, we looked at reaction mechanisms. A mechanism gives the steps involved. Each step is an elementary reaction. In elementary reactions, the molecules collide in a single step. The molecularity is the number of molecules that come together. Some of the steps are faster than others. The slowest step is the rate-determining step. Sometimes it's called rate-limiting. The overall reaction's rate law equals the rate-determining step's rate law.

Let's look at a few examples.

Slow First Step Example

Tert-butyl bromide ($(CH_{3})_{3}CBr$) has a central carbon with four single bonds. Its VSEPR molecular geometry is tetrahedral. One bond is to the bromine. The other three are to methyl groups ($CH_{3}$). A weird thing can happen to tert-butyl bromide! Sometimes, the bromine falls off. The result is a tert-butyl cation and bromide anion. When it falls off, the carbon reacts with water. The product is tert-butyl alcohol and hydrobromic acid. It's a double replacement reaction. Let's determine the rate law of the reaction.

Here is the mechanism.

$Step\,1: (CH_{3})_{3}CBr \rightarrow CH_{3}C^{+} + Br^{-}$ (Slow)

$Step\,2: CH_{3}C^{+}+ Br^{-} + H_{2}O \rightarrow CH_{3}COH + HBr$ (Fast)

It's difficult for an atom to fall off. So the first step is slow. It's our rate-determining step. So, our rate law is $rate=k[CH_{3}CBr]$.

Slow Second Step Example

Hydrogen peroxide ($H_{2}O_{2}$) decomposes into water and oxygen gas. The halogen iodine catalyzes the reaction. Remember, catalysts lower the activation energy. So, they speed up the reaction rate. Let's write the overall equation and find the rate law.

Here is the mechanism.

$Step\,1: H_{2}O_{2(aq)} + I_{(aq)}^{-} \rightarrow IO_{(aq)} ^{-} + H_{2}O_{(l)}$ (Fast)

$Step\,2: H_{2}O_{2(aq)} + IO^{-}_{(aq)} \rightarrow H_{2}O_{(l)} + O_{2(g)} + I^{-}_{(aq)}$ (slow)

To find the balanced equation, treat the mechanism as an addition question. First, add the reactants' side. Next, add the products' side. Anything on both sides cancels.

$Step\,1: H_{2}O_{2(aq)} + I_{(aq)}^{-} \rightarrow IO_{(aq)} ^{-} + H_{2}O_{(l)}$ (Fast)

$Step\,2: H_{2}O_{2(aq)} + IO^{-}_{(aq)} \rightarrow H_{2}O_{(l)} + O_{2(g)}+ I^{-}_{(aq)}$ (slow)

$overall: H_{2}O_{2(aq)} + \cancel{I_{(aq)}^{-}} + H_{2}O_{2(aq)} + \cancel{IO_{(aq)} ^{-}} \rightarrow \cancel{IO_{(aq)} ^{-}} + H_{2}O_{(l)} + H_{2}O_{(l)} + O_{2(g)} + \cancel{I_{(aq)}^{-}}$

So the balanced equation is: $2H_{2}O_{2(aq)} \rightarrow H_{2}O_{(l)} + O_{2(g)}$. The iodine is not consumed. So, it must be a catalyst.

To find the rate law, look at the rate-determining step. The second step is slow. So, it's rate-determining. Its rate law is, $rate=k[H_{2}O_{2(aq)}][IO^{-} _{(aq)}]$.

But, there's a problem. The rate depends on the oxoanion intermediate, hypoiodite. So we need to solve for its concentration. How? Its formation depends on the first step. So we need step one's rate law.

When the second step is slow, we assume step one reaches equilibrium. The forward and reverse reactions' rates are equal. So we can write two rate laws. They will equal each other.

$rate= k_{f}[H_{2}O_{2(aq)}][I^{-} _{(aq)}]=k_{r}[IO^{-} _{(aq)}][H_{2}O_{(l)}]$

Now we can solve for $[IO^{-} _{(aq)}]$.

$k_{f}[H_{2}O_{2(aq)}][I^{-} _{(aq)}]=k_{r}[IO_{(aq)} ^{-}][H_{2}O_{(l)}] \rightarrow [IO^{-} _{(aq)}]= \frac{k_{f}}{k_{r}}\frac{[H_{2}O_{2(aq)}][I^{-} _{(aq)}]}{[H_{2}O_{(l)}]}$

So our overall rate law becomes:

$rate=k[H_{2}O_{2(aq)}]\times \frac{k_{f}}{k_{r}} \times \frac{[H_{2}O_{2(aq)}][I^{-} _{(aq)}]}{[H_{2}O_{(l)}]} \rightarrow K[H_{2}O_{2(aq)}][I^{-} _{(aq)}]$.

You'll notice that I did not include water in the final answer. In this case, water is our solvent. You can see that hydrogen peroxide and iodide are aqueous. So, water is in excess. Changing water's volume will not affect the rate. You also know that water has a fixed concentration. It's $55.5 \frac{mol}{L}$. We define the reaction rate as the change in concentration over time. Water's concentration is not changing. So, do not include it. That's true for any pure solid or liquid.

This problem creates another challenge. The rate law depends on the iodide concentration. But, iodide is a catalyst. Catalyst concentration only affects the rate to a certain point. At low concentrations, adding more catalyst will change the reaction's rate. At high catalyst concentrations, you remove it from the rate law.

Slow Third Step Example

Let's look at the most complicated example. We have a three-step reaction mechanism. Nitryl chloride ($NO_{2}Cl$) undergoes decomposition. The products are nitrogen dioxide and diatomic chlorine. Both are gases. Let's find the rate law for the overall reaction.

This is the mechanism.

$Step\,1: 2NO_{2}Cl_{(g)} \rightleftharpoons ClO_{2(g)} + N_{2}O_{(g)} + ClO_{(g)}$ (fast)

$Step\,2: ClO_{2(g)} + N_{2}O_{(g)} \rightleftharpoons NO_{2(g)} + NOCl_{(g)}$ (fast)

$step\,3: NOCl_{(g)} + ClO_{(g)} \rightarrow NO_{2(g)} + Cl_{2(g)}$ (slow)

The third step is slow. So, it's the rate-determining step. Its rate law is $rate=k[NOCl][ClO]$. We need to solve for both [NOCl] and [ClO].

Step one is in equilibrium. So, we write a rate law for both the forward and reverse reactions.

$rate=k_{f}[NO_{2}Cl]^{2}=k_{r}[ClO_{2}][N_{2}O][ClO]$

Remember, for elementary reactions, the exponent equals the stoichiometric coefficient.

We can solve for [ClO].

$[ClO]= \frac{k_{f}}{k_{r}}\frac{[NO_{2}Cl]^{2}}{[ClO_{2}][N_{2})]} \rightarrow k'\frac{[NO_{2}Cl]}{[ClO_{2}][N_{2})]}$

Now let's find the rate law for step 2. Again, it's at equilibrium.

$rate=k''_{f}[N_{2}O][ClO_{2}]=k''_{r}[NO_{2}][NOCl]$

We can solve for [NOCL].

$[NOCl]=\frac{k''_{f}}{k''_{r}}\frac{[N_{2}O][ClO_{2}]}{[NO_{2}]}\rightarrow k'\frac{[N_{2}O][ClO_{2}]}{[NO_{2}]}$

Now that we know both unknowns, we can generate the overall rate law.

$rate=k[NOCl][ClO]\rightarrow k \times k''\frac{[N_{2}O][ClO_{2}]}{[NO_{2}]}\times k'\frac{[NO_{2}Cl]^{2}}{[ClO_{2}][N_{2}O]}\rightarrow K\frac{[NO_{2}Cl]^{2}}{[NO_{2}]}$

This example is unique. Notice that the rate law has a denominator term. What does that mean? The formation of $NO_{2}$ is part of an equilibrium reaction. So, the reverse process slows the reaction rate.

(Video) The Rate-Limiting Step

by Bozeman Science

In this video, Mr. Andersen describes the rate-limiting step. He starts with an analogy about cars on the highway. Then he explains in terms of molecules and collisions. Finally, he demonstrates two example problems. This is an excellent video if you want a better understanding of rate-limiting steps!