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Pre-equilibrium Approximation

Pre-equilibrium Approximation

All reactions have their own speeds. Reactions like sodium in water are fast and almost explosive, while reactions like the rusting of iron are slow and steady. These reactions are elementary reactions, as in they are a single step. Figuring out the rate of these reactions is easy due to their simplicity. But what about reactions with multiple steps? The extra steps make determining the rate tricky, which is why there are methods we use to help us out.

In this article, we will be covering the pre-equilibrium approximation. This is a tool that chemists use to determine the rate of multistep reactions. Together, we will be walking through what the approximation is and how to use it. We will also have a brief refresher on kinetics to make sure you are up to speed.

  • This article covers the pre-equilibrium approximation
  • First, we will define the pre-equilibrium approximation
  • Next, we will do a brief overview of the rate law, multistep reactions, and equilibrium
  • Then, we will explain what assumptions we are making to use the approximation and how to set it up
  • After that, we will walk through some examples
  • Lastly, we will explain the main difference between the pre-equilibrium approximation and another method called steady-state approximation.

Pre-equilibrium Approximation Definition

Let's start by looking at the definition of pre-equilibrium approximation.

The pre-equilibrium approximation is a method to calculate the rate law of a consecutive reaction.

A consecutive (or multistep) reaction is a multistep reaction process, where the product of the first step is the reactant of the second step, and so on.

Consecutive reactions can be very complex, so methods like the pre-equilibrium approximation are handy tools to sort them out.

Basics of Rate Law, Multistep Reactions, and Equilibrium

Before we dive into the concept of the pre-equilibrium approximation, let's do a quick refresher on the kinetics relating to it. Firstly, let's recap the rate law.

The rate law is the expression that shows how the concentration of reactants affects the rate of reaction. For a general reaction: $$A+B\xrightarrow{k_1}C$$

The rate law is: \(\text{rate}=k_1[A][B]\)

Where k1 is the rate constant (proportionality constant relating rate of reaction at a given temperature to the concentration of reactants).

Now that's what a rate law looks like for single-step reactions. For multistep reactions, it's a bit different. The rate law is written for the reaction's slow step. This step is called the rate-determining step. Think of it like being stuck in traffic, you can only drive as fast as the car in front.

Here's an example of a rate law for a multistep reaction: $$A+B\underset{k_{-1}} {\stackrel{k_1}{\rightleftharpoons}}I\,\text{(fast)}$$ $$I+C\xrightarrow{k_2}D\,\text{(slow)}$$

$$\text{rate}=k_2[I][D]$$

In this reaction, I is an intermediate.

An intermediate is a species both produced and consumed in the reaction.

You wouldn't actually leave an intermediate in your rate law, but this is just to show you the basics. We exclude intermediates from the rate law since they are much more difficult to measure than reactants.

The last thing we need to recap is the equilibrium constant.

The equilibrium constant (K) shows whether the products or reactants are favored in an equilibrium reaction. For a general reaction $$A+B\underset{k_{-1}} {\stackrel{k_1}{\rightleftharpoons}}C+D$$

The formula for Keq is: $$K_{eq}=\frac{[C][D]}{[A][B]}$$

K can also be set equal to the rate constants, for this example it would be:

$$K_{eq}=\frac{k_1}{k_{-1}}$$

Now that we've covered the basics, let's move on to covering our approximation.

Pre-equilibrium Approximation Method of Equilibrium

The pre-equilibrium approximation can only be used when the first step is fast and the second step (or any later step) is slow. This is because of the assumption the approximation method makes about the equilibrium. We assume that the first reaction is going quickly enough that the intermediates produced are "waiting in line" to react in the slower, second reaction. However, instead of waiting, the intermediates will instead participate in the reverse reaction, leading back to reactants, until the rates of the forward reaction and reverse reaction, of the first step, are equal.

Think of it like waiting in line at a busy attraction at an amusement park. Some people will get tired of waiting, so they leave and walk "backward" in line. At some point, the rate of being fed up and leaving will be the same as people entering.

Pre-equilibrium Approximation Rule

Now that we know what assumptions we are making, we can learn how to use the approximation rule. Here's a general example of a reaction that we could use the approximation for: $$A+B\underset{k_{-1}} {\stackrel{k_1}{\rightleftharpoons}}C\,\text{(fast)}$$ $$C+D\xrightarrow{k_2}E\,\text{(slow)}$$

Our first step is to write out the rate law as usual, which in this case would be: $$\text{rate}=k_2[C][D]$$

As mentioned previously, we cannot have an intermediate in the rate law. Here the intermediate is C, since it is produced in step 1, but consumed in step 2.

Next, we set up our approximation. We are assuming that the forward and reverse reactions are proceeding at equal rates: $$\text{rate}=k_1[A][B]=k_{-1}[C]$$

Another way to write this expression would be: $$K_{eq}=\frac{k_1}{k_{-1}}=\frac{[C]}{[A][B]}$$

Now we can solve for the concentration of the intermediate: $$[C]=K_{eq}[A][B]$$

Lastly, we substitute this expression into the rate equation to get the final rate law. $$\text{rate}=k[A][B][D]\,\,\,\text{where}\,k=k_2K_{eq}$$

Pre-equilibrium Approximation Examples

Let's walk through some examples together:

Use the pre-equilibrium approximation to write the rate law for the following multistep reaction:

$$O_3\underset{k_{-1}} {\stackrel{k_1}{\rightleftharpoons}}O_2+O\,\text{(fast)}$$

$$O+O_3\xrightarrow{k_2}2O_2\,\text{(slow)}$$

First, we set up the rate equation: $$\text{rate}=k_2[O][O_3]$$

We need the pre-equilibrium approximation so we can replace the intermediate (O or oxygen). Now we set up our approximation using the assumption that the rate of the forward equilibrium reaction is equal to the rate of the reverse reaction, of the first step: $$k_1[O_3]=k_{-1}[O_2][O]$$ $$K_{eq}=\frac{k_1}{k_{-1}}=\frac{[O_2][O]}{[O_3]}$$

Next, we put our equation in terms of the intermediate (O). $$[O]=K_{eq}\frac{[O_3]}{[O_2]}$$

Lastly, we substitute this expression into the rate law $$\text{rate}=k_2K_{eq}\frac{[O_3]}{[O_2][O_3]}$$

Simplifying this, we get the final rate law $$\text{rate}=k\frac{[O_3]^2}{[O_2]}\,\,\,\text{where}\,k=k_2K$$

Let's work on one more problem

Using the pre-equilibrium equation to write the rate law for the following multistep reaction: $$H^++HNO_2\underset{k_{-1}} {\stackrel{k_1}{\rightleftharpoons}}H_2NO_2^+\,\text{(fast)}$$ $$H_2NO_2^++Br^-\xrightarrow{k_2}NOBr+H_2O\,\text{(slow)}$$ $$NOBr+C_6H_5NH_2\xrightarrow{k_3}C_6H_5N_2^++H_2O+Br^-\,\text{(fast)}$$

Even though this reaction has three steps instead of two, our method doesn't change. When we are writing our rate law, everything after our rate-determining step doesn't matter. Like before, we start by setting up the rate law equation $$\text{rate}=k_2[H_2NO_2^+][Br^-]$$

Here, \(H_2NO_2^+\) is our intermediate.

$$k_1[H^+][HNO_2]=k_{-1}[H_2NO_2^+]$$

$$K_{eq}=\frac{k_1}{k_{-1}}=\frac{[H_2NO_2^+]}{[H^+][HNO_2]}$$

$$[H_2NO_2^+]=K_{eq}[H^+][HNO_2]$$

Lastly, we substitute in our expression for the intermediate and simplify $$\text{rate}=k_2K_{eq}[H^+][HNO_2][Br^-]$$

$$k=[H^+][HNO_2][Br^-]\,\text{where}\,k=k_2K_{eq}$$

Pre-equilibrium Approximation vs. Steady-state

The pre-equilibrium approximation isn't the only tool used to help calculate the rate law for a multistep reaction. One other method is the steady-state approximation.

The steady-state approximation is a method used to calculate the rate law of a multistep reaction. It assumes that (for an intermediate) the rate of consumption is equal to the rate of creation. This means that the net change in the concentration of the intermediate would be zero.

The main difference between these approximations is when they are used. Here is a diagram to explain:

Pre-equilbrium Approximation Pre-equilbrium versus steady state StudySmarterDiagram showing the main difference between the two approximation methods. StudySmarter Original.

Each approximation is used based on where the rate-determining step is in the consecutive mechanism. If it is first, we use the steady-state approximation. If it isn't, then we use the pre-equilibrium approximation. As with the pre-equilibrium method, the steady-state approximation is used to solve for the rate law when there is an intermediate. We use the method to solve for the intermediate in terms of the reactants, then substitute that expression into the rate equation.

Pre-equilibrium Approximation - Key takeaways

  • The pre-equilibrium approximation is a method to calculate the rate law of a consecutive reaction
  • To use the pre-equilibrium approximation, the first step must be the fast step.
  • When using the approximation, we assume that the rate of the forward and reverse reactions are equal.
  • The pre-equilibrium approximation is different from the steady-state approximation, which is used when the first step in the mechanism is the slow step.

Frequently Asked Questions about Pre-equilibrium Approximation

The pre-equilibrium approximation is a method to calculate the rate law of a consecutive reaction

The pre-equilibrium approximation is used to find the rate law.

We treat the first step like it is at equilibrium. We set the rates of the forward and reverse reactions equal to each other and solve for the concentration of the intermediate. 


Lastly, you substitute this value in for the intermediate in the rate equation.

For a general reaction:

            k1

A + B <--> C (fast)

            k-1

          k2

C + D --> E (slow)


We would set up the approximation like this:


k1[A][B]=k-1[C]

[C]=(k1/k-1)[A][B]



Each approximation is used based on where the rate-determining step is in the consecutive mechanism. If it is first, we use the steady-state approximation. If it isn't, then we use the pre-equilibrium approximation.  

  1. Must be a multistep reaction
  2. Has a fast first step
  3. The first step is treated as an equilibrium reaction

Final Pre-equilibrium Approximation Quiz

Question

What is the pre-equilbrium approximation?

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Answer

The pre-equilibrium approximation is a method to calculate the rate law of a consecutive reaction 

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Question

In what case would you use the pre-equilibrium approximation?

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Answer

Multistep reaction, fast first step

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Question

What is an intermediate?

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Answer

An intermediate is species both produced and consumed in the reaction

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Question

What do we assume when we use the pre-equilibrium approximation?

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Answer

The rate of the forward and reverse equilibrium reaction are equal.

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Question

Which of the following is true about the pre-equilibrium approximation and the steady-state approximation?

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Answer

The steady state-approximation is used for slow first steps, while the pre-equilibrium approximation is used for fast first steps.

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Question

What is a consecutive/multistep reaction?

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Answer

A consecutive (or multistep) reaction is a multistep reaction process, where the product of the first step is the reactant of the second step and so on.

Show question

Question

What is the steady-state approximation?

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Answer

The steady-state approximation is a method to calculate the rate law of a multistep reaction. It assumes that the rate of consumption for an intermediate is equal to the rate of creation, therefore the change in the concentration of the intermediate would be zero.

Show question

Question

What is a rate law?

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Answer

The rate law is the expression that shows how the concentration of reactants affects the rate of reaction.

Show question

Question

Why do we assume that the forward and reverse reactions have equal rates?

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Answer

The second step is the slow step. Instead of "waiting in line" to react, the reverse reaction will proceed instead. Eventually, the forward and reverse reactions will have the same rate.

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Question

Why can't intermediates be left in the rate law?

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Answer

It is much more difficult to measure their concentration that that of the reactant(s)

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