# 12.6: The Equilibrium Approximation

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In many cases, the formation of a reactive intermediate (or even a longer lived intermediate) involves a reversible step. This is the case if the intermediate can decompose to reform reactants with a significant probability as well as moving on to form products. In many cases, this will lead to a pre-equilibrium condition in which the equilibrium approximation can be applied. An example of a reaction mechanism of this sort is

$A + B \xrightleftharpoons [k_1]{k_{-1}} AB \nonumber$

$AB \xrightarrow{k_2} C \nonumber$

Given this mechanism, the application of the steady state approximation is cumbersome. However, if the initial step is assumed to achieve equilibrium, an expression can be found for $$[AB]$$. In order to derive this expression, one assumes that the rate of the forward reaction is equal to the rate of the reverse reaction for the initial step in the mechanism.

$k_{1}[A][B] = k_{-1}[AB] \nonumber$

or

$\dfrac{ k_{1}[A][B]}{k_{-1}} = [AB] \nonumber$

This expression can be substituted into an expression for the rate of formation of the product $$C$$:

$\dfrac{d[C]}{dt} = k_2[AB] \nonumber$

or

$\dfrac{d[C]}{dt} = \dfrac{ k_2 k_{1}}{k_{-1}}[A][B] \nonumber$

Which predicts a reaction rate law that is first order in $$A$$, first order in $$B$$, and second order overall.

##### Example $$\PageIndex{1}$$:

Given the following mechanism, apply the equilibrium approximation to the first step to predict the rate law suggested by the mechanism.

$A + A \xrightleftharpoons [k_1]{k_{-1}} A_2 \nonumber$

$A_2+B \xrightarrow{k_2} C + A \nonumber$

###### Solution

If the equilibrium approximation is valid for the first step,

$k_{1}[A]^2 = k_{-1}[A_2] \nonumber$

or

$\dfrac{ k_{1}[A]^2}{k_{-1}} \approx [A_2] \nonumber$

Plugging this into the rate equation for the second step

$\dfrac{d[C]}{dt} = k_2[A_2][B] \nonumber$

yields

$\dfrac{d[C]}{dt} = \dfrac{ k_2k_{1}}{k_{-1}} [A]^2[B] \nonumber$

Thus, the rate law has the form

$\text{rate} = k' [A]^2[B] \nonumber$

which is second order in $$A$$, first order in $$B$$ and third order over all, and in which the effective rate constant ($$k'$$ is

$k' = \dfrac{k_2k_1}{k_{-1}}. \nonumber$

Sometimes, the equilibrium approximation can suggest rate laws that have negative orders with respect to certain species. For example, consider the following reaction

$A + 2B \rightarrow 2C \nonumber$

A proposed mechanism for which might be

$A + B \xrightleftharpoons [k_1]{k_{-1}} I + C \nonumber$

$I+ B \xrightarrow{k_2} C \nonumber$

in which $$I$$ is an intermediate. Applying the equilibrium approximation to the first step yields

$k_{1}[A][B] = k_{-1}[I][C] \nonumber$

or

$\dfrac{ k_{1}[A][B]}{k_{-1}[C]} \approx [I] \nonumber$

Substituting this into an expression for the rate of formation of $$C$$, one sees

$\dfrac{d[C]}{dt} = k_{2} [I] [B] \nonumber$

or

$\dfrac{d[C]}{dt} = \dfrac{ k_{1}[A][B]}{k_{-1}[C]} [B] = \dfrac{ k_{2} k_{1}[A][B]}{k_{-1}[C]} \nonumber$

The rate law is then of the form

$\text{rate} = k \dfrac{[A][B]^2}{[C]} \nonumber$

which is first order in $$A$$, second order in $$B$$, negative one order in $$C$$, and second order overall. Also,

$k'=\dfrac{k_2k_1}{k_{-1}}. \nonumber$

In this case, the negative order in $$C$$ means that a buildup of compound $$C$$ will cause the reaction to slow. These sort of rate laws are not uncommon for reactions with a reversible initial step that forms some of the eventual reaction product.

This page titled 12.6: The Equilibrium Approximation is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Patrick Fleming.