5.5: The Equilibrium Approximation
- Page ID
- 516733
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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
\[ A + B \xrightleftharpoons [k_{-1}]{k_1} AB \nonumber \]
\[ AB \xrightarrow{k_2} C \nonumber \]
If the initial equilibrium is fast compared to the second step, one makes the assumption that the ratio of [AB] to [A][B] is always that expected at equilibrium. This results in the expression:
\[ k_{1}[A][B] = k_{-1}[AB] \nonumber \]
or
\[\frac{ k_{1}[A][B]}{k_{-1}} = [AB] \nonumber \]
This expression can be substituted into an expression for the rate of formation of the product \(C\):
\[\frac{d[C]}{dt} = k_2[AB] \nonumber \]
or
\[\frac{d[C]}{dt} = \frac{ k_2 k_{1}}{k_{-1}}[A][B] \nonumber \]
Which predicts a reaction rate law that is first order in \(A\), first order in \(B\), and second order overall. This is equivalent to making the steady state approximation and assuming that k2 is negligible.
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
\[\frac{ k_{1}[A]^2}{k_{-1}} \approx [A_2] \nonumber \]
Plugging this into the rate equation for the second step
\[\frac{d[C]}{dt} = k_2[A_2][B] \nonumber \]
yields
\[\frac{d[C]}{dt} = \frac{ k_2k_{1}}{k_{-1}} [A]^2[B] \nonumber \]
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
\[ k' = \frac{k_2k_1}{k_{-1}}. \nonumber \]
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
\[\frac{ k_{1}[A][B]}{k_{-1}[C]} \approx [I] \nonumber \]
Substituting this into an expression for the rate of formation of \(C\), one sees
\[\frac{d[C]}{dt} = k_{2} [I] [B] \nonumber \]
or
\[\frac{d[C]}{dt} = \dfrac{ k_{1}[A][B]}{k_{-1}[C]} [B] = \frac{ k_{2} k_{1}[A][B]}{k_{-1}[C]} \nonumber \]
The rate law is then of the form
\[\text{rate} = k \frac{[A][B]^2}{[C]} \nonumber \]
which is first order in \(A\), second order in \(B\), negative one order in \(C\), and second order overall. Also,
\[ k'=\frac{k_2k_1}{k_{-1}}. \nonumber \]
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.