# Rate Determining Step

The Rate Determining Step is the slowest step of a chemical reaction that determines the speed (rate) of the overall reaction. The rate determining step can be compared to the neck of a funnel. The rate at which water flows through a funnel is limited/ determined by the width of the neck of the funnel and not by the rate at which the water is poured into the funnel. Like the neck of the funnel, the slow step of a reaction determines the rate of a reaction. Not every reaction will have a rate determining step, only those with a step that significantly slower than the other steps in the reaction.

### Introduction

The rate determining step is important in deriving the rate equation of a chemical reaction. For example, consider a multi-step reaction,

$A + B \rightarrow C + D$

And assume the elementary steps for this reaction are:

Step 1: Slow          $$A + A \rightarrow C + E$$   (with a rate constant, $$k_1$$)

Step 2: Fast          $$E + B \rightarrow A + D$$   (with a rate constant, $$k_2$$)

where $$E$$ is an intermediate product in step 1 and an intermediate reactant in step 2 that will not show up in the overall reaction. This is because when you add steps 1 and 2, intermediate $$E$$ will cancel out along with the extra reactant $$A$$ seen in step 1. Note that intermediate reactions do not show up in the overall reaction.

In the case of this hypothetical reaction, if step 1 is the slow step and step 2 is the fast step of the reaction, then the overall reaction rate will depend on step 1; this is because the slow step in a reaction is always the rate-limiting step, which is also called the rate determining step.

Therefore the rate equation will be,

$\text{rate} = k_1 [A] [A] = k_1 [A]^2$

Example 1
Consider an example of a reaction:

$NO_{2 \; (g)} + CO_{(g)} \rightarrow NO_{(g)} + CO_{2\;(g)}$

which occurs in two elementary steps:

$NO_2 + NO_2 \rightarrow NO + NO_3 \;\;\; \text{(slow)}$

$NO_3 + CO \rightarrow NO_2 + CO_2 \;\;\; \text{(fast)}$

Here, we see that for the second step to occur, the first step must occur first, and since the first step is the slowest step, the overall reaction cannot be proceed any faster than the rate of the first elementary step. The first elementary step in this example is therefore the rate-determining step. The rate equation for this reaction is equal to the rate constant of step 1 multiplied by the reactants of that first step. If we denote the rate constant of step 1 to be $$k_1$$ then the rate of the the first step in the reaction (and the total reaction) will be

$\text{rate} = k_1 [NO_2][NO_2] = k_1[NO_2]^2$

Example 2

A way to further understand rate-determining steps is to try to apply it to real life events. For example, consider putting together a toy that you just purchased. Let's assume that building a toy purchased from a store requires 4 steps:

1. Purchase toy parts and load into car
2. Drive toy parts to the house (or where assembly will take place)
3. Unloading toy parts  from the car and take toy parts inside
4. Assemble toy.

In this example, how fast the toy is assembled will be determined by whichever of the four steps takes the longest time to complete. If there was a very long line in the toy store, then step (a) will determine the rate of assembly of  the toys If the person driving the car was stuck in traffic or was driving very slowly then step (b) will determine the rate of building the toy.

### Practice Problem

A reaction between $$NO$$ and $$H_2$$ occurs in the following three-step process:

$NO + NO \rightarrow N_2O_2 \;\;\; \text{(fast)}$

$N_2O_2+ H_2 \rightarrow N_2O + H_2O \;\;\; \text{(slow)}$

$N_2O + H_2 \rightarrow N_2 + H_2O \;\;\; \text{(fast)}$

1. What is the rate determining step?
2. Write the balanced equation for the overall reaction.
3. Are there any intermediates? If so, state what they are.

1. The rate determining step is the second step because it's the slow step.
2. $$2 NO + 2 H_2 \rightarrow N_2+ 2 H_2O$$. This balanced reaction can be found by simply adding all 3 reactions together. In adding these reactions, we will be eliminating the intermediates that show up on both the reactant and product side while adding the remaining reactants with the reactants and the products together with the other products.
3. The intermediates in this reaction are $$N_2O_2$$ and $$N_2O$$. Both of them show up on both the reactant and product side. Therefore in adding all three elementary reactions together, we will be able to cancel out $$N_2O_2$$ and $$N_2O$$ and as a result, they will not show up in the overall reaction equation.

### Sources

1. Chang, R. (2004). Physical Chemistry for the Biosciences. Sausalito,Ca: University Science Books.
2. Hahn, H.H. & Stumm,W.(1968). Kinetics of coagualtion with hydrolyzed Al (III): the rate determining step. Journal of Colloid and Interface Science. 28, 134-144.
3. Fong, J.C. & Horst, S.(1978). On the rate determining step of fatty acid oxidation in heart. The Journal of Biological Chemistry. 253, 6917-6922.
4. Structural Biochemistry/Enzyme/Rate-limiting step. Wikibooks. 15 Mar 2009

### Contributors:

• Galaxy Mudda, Pamela Chaha, and Florence-Damilola Odufalu (UCD)

00:16, 24 Jul 2014

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