5.3: Representative Reaction Mechanisms

Introduction

A major goal in chemical kinetics is to determine the sequence of elementary reactions, or the reaction mechanism, that comprise complex reactions. For example, Sherwood Rowland and Mario Molina won the Nobel Prize in Chemistry in 1995 for proposing the elementary reactions involving chlorine radicals that contribute to the overall reaction of \(O_3 \rightarrow O_2\) in the troposphere. In the following sections, we will derive rate laws for complex reaction mechanisms, including reversible, parallel and consecutive reactions.

Parallel Reactions

Consider the reaction in which chemical species \(\text{A}\;\) undergoes one of two irreversible first order reactions to form either species \(\text{B}\;\) or species \(\text{C}\;\):

\[\begin{align} \text{A} &\overset{k_1}{\rightarrow} \text{B} \\ \text{A} &\overset{k_2}{\rightarrow} \text{C} \end{align}\]

The overall reaction rate for the consumption of \(\text{A}\) can be written as:

\[\dfrac{d \left[ \text{A} \right]}{dt} = -k_1 \left[ \text{A} \right] - k_2 \left[ \text{A} \right] = - \left( k_1 + k_2 \right) \left[ \text{A} \right] \label{21.1}\]

Integrating \(\left[ \text{A} \right]\) with respect to \(t\), we obtain the following equation:

\[\left[ \text{A} \right] = \left[ \text{A} \right]_0 e^{-\left( k_1 + k_2 \right) t} \label{21.2}\]

Plugging this expression into the equation for \(\dfrac{d \left[ \text{B} \right]}{dt}\), we obtain:

\[\dfrac{d \left[ \text{B} \right]}{dt} = k_1 \left[ \text{A} \right] = k_1 \left[ \text{A} \right]_0 e^{- \left( k_1 + k_2 \right) t} \label{21.3}\]

Integrating \(\left[ \text{B} \right]\) with respect to \(t\), we obtain:

\[\left[ \text{B} \right] = -\dfrac{k_1 \left[ \text{A} \right]_0}{k_1 + k_2} \left( e^{-\left( k_1 + k_2 \right) t} \right) + c_1 \label{21.4}\]

At \(t = 0\), \(\left[ \text{B} \right] = 0\). Therefore,

\[c_1 = \dfrac{k_1 \left[ \text{A} \right]_0}{k_1 + k_2} \label{21.5}\]

\[\left[ \text{B} \right] = \dfrac{k_1 \left[ \text{A} \right]_0}{k_1 + k_2} \left( 1 - e^{-\left( k_1 + k_2 \right) t} \right) \label{21.6}\]

Likewise,

\[\left[ \text{C} \right] = \dfrac{k_2 \left[ \text{A} \right]_0}{k_1 + k_2} \left( 1 - e^{-\left( k_1 + k_2 \right) t} \right) \label{21.7}\]

The ratio of \(\left[ \text{B} \right]\) to \(\left[ \text{C} \right]\) is simply:

\[\dfrac{\left[ \text{B} \right]}{\left[ \text{C} \right]} = \dfrac{k_1}{k_2} \label{21.8}\]

An important parallel reaction in industry occurs in the production of ethylene oxide, a reagent in many chemical processes and also a major component in explosives. Ethylene oxide is formed through the partial oxidation of ethylene:

\[2 \: C_2 H_4 + O_2 \overset{k_1}{\longrightarrow} 2 \: C_2 H_4 O\]

However, ethylene can also undergo a combustion reaction:

\[C_2 H_4 + 3 \: O_2 \overset{k_2}{\longrightarrow} 2 \: CO_2 + 2 \: H_2 O\]

To select for the first reaction, the oxidation of ethylene takes place in the presence of a silver catalyst, which significantly increases \(k_1\) compared to \(k_2\). Figure \(\PageIndex{1}\) displays the concentration profiles for species \(\text{A}\), \(\text{B}\), and \(\text{C}\) in a parallel reaction in which \(k_1 > k_2\).

Figure \(\PageIndex{1}\): Plots of \(\left[ \text{A} \right]\) (solid line), \(\left[ \text{B} \right]\) (dashed line) and \(\left[ \text{C} \right]\) (dotted line) over time for a parallel reaction.

Consecutive Reactions

Consider the following series of first-order irreversible reactions, where species \(\text{A}\) reacts to form an intermediate species, \(\text{I}\), which then reacts to form the product, \(\text{P}\):

\[\text{A} \overset{k_1}{\longrightarrow} \text{I} \overset{k_2}{\longrightarrow} \text{P}\]

We can write the reaction rates of species \(\text{A}\), \(\text{I}\) and \(\text{P}\) as follows:

\[\dfrac{d \left[ \text{A} \right]}{dt} = -k_1 \left[ \text{A} \right] \label{21.9}\]

\[\dfrac{d \left[ \text{I} \right]}{dt} = k_1 \left[ \text{A} \right] - k_2 \left[ \text{I} \right] \label{21.10}\]

\[\dfrac{d \left[ \text{P} \right]}{dt} = k_2 \left[ \text{I} \right] \label{21.11}\]

As before, integrating \(\left[ \text{A} \right]\) with respect to \(t\) leads to:

\[\left[ \text{A} \right] = \left[ \text{A} \right]_0 e^{-k_1 t} \label{21.12}\]

The concentration of species \(\text{I}\) can be written as

\[\left[ \text{I} \right] = \dfrac{k_1 \left[ \text{A} \right]_0}{k_2 - k_1} \left( e^{-k_1 t} - e^{-k_2 t} \right) \label{21.13}\]

Then, solving for \(\left[ \text{P} \right]\), we find that:

\[\left[ \text{P} \right] = \left[ \text{A} \right]_0 \left[ 1 + \dfrac{1}{k_1 - k_2} \left( k_2 e^{-k_1 t} - k_1 e^{-k_2 t} \right) \right] \label{21.14}\]

Figure \(\PageIndex{2}\) displays the concentration profiles for species \(\text{A}\), \(\text{I}\), and \(\text{P}\) in a consecutive reaction in which \(k_1 = k_2\). As can be seen from the figure, the concentration of species \(\text{I}\) reaches a maximum at some time, \(t_\text{max}\). Oftentimes, species \(\text{I}\) is the desired product. Returning to the oxidation of ethylene into ethylene oxide, it is important to note another reaction in which ethylene oxide can decompose into carbon dioxide and water through the following reaction

\[C_2 H_4 O + \dfrac{5}{2} \: O_2 \overset{k_3}{\longrightarrow} 2 \: CO_2 + 2 \: H_2 O\]

Thus, to maximize the concentration of ethylene oxide, the oxidation of ethylene is only allowed proceed to partial completion before the reaction is stopped.

Finally, in the limiting case when \(k_2 \gg k_1\), we can write the concentration of \(\text{P}\) as

\[\left[ \text{P} \right] \approx \left[ \text{A} \right]_0 \left\{ 1 + \dfrac{1}{-k_2} k_2 e^{-k_1 t} \right\} = \left[ \text{A} \right]_0 \left( 1 - e^{-k_1 t} \right) \label{21.15}\]

Thus, when \(k_2 \gg k_1\), the reaction can be approximated as \(\text{A} \rightarrow \text{P}\) and the apparent rate law follows \(1^{st}\) order kinetics.

Figure \(\PageIndex{2}\): Plots of \(\left[ \text{A} \right]\) (solid line), \(\left[ \text{I} \right]\) (dashed line) and \(\left[ \text{P} \right]\) (dotted line) over time for consecutive first order reactions.

Consecutive Reactions With an Equilibrium

Consider the reactions

\[\text{A} \overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \text{I} \overset{k_2}{\rightarrow} \text{P}\]

We can write the reaction rates as:

\[\dfrac{d \left[ \text{A} \right]}{dt} = -k_1 \left[ \text{A} \right] + k_{-1} \left[ \text{I} \right] \label{21.16}\]

\[\dfrac{d \left[ \text{I} \right]}{dt} = k_1 \left[ \text{A} \right] - k_{-1} \left[ \text{I} \right] - k_2 \left[ \text{I} \right] \label{21.17}\]

\[\dfrac{ d \left[ \text{P} \right]}{dt} = k_2 \left[ \text{I} \right] \label{21.18}\]

The exact solutions of these is straightforward, in principle, but rather involved, so we will just state the exact solutions, which are

\[\begin{align} \left[ \text{A} \right] \left( t \right) &= \dfrac{ \left[ \text{A} \right]_0}{2 \lambda} \left[ \left( \lambda - k_1 + K \right) e^{-\left( k_1 + K - \lambda \right) t/2} + \left( \lambda + k_1 - K \right) e^{-\left( k_1 + K + \lambda \right) t/2} \right] \\ \left[ \text{I} \right] \left( t \right) &= \dfrac{ k_1 \left[ \text{A} \right]_0}{\lambda} \left[ e^{- \left( k_1 + K - \lambda \right) t/2} - e^{-\left( k_1 + K + \lambda \right) t/2} \right] \\ \left[ \text{P} \right] \left( t \right) &= 2 k_1 k_2 \left[ \text{A} \right]_0 \left[ \dfrac{2}{\left( k_1 + K \right)^2 - \lambda^2} - \dfrac{1}{\lambda} \left( \dfrac{ e^{-\left( k_1 + K - \lambda \right) t/2}}{k_1 + K - \lambda} - \dfrac{e^{-\left( k_1 + K + \lambda \right) t/2}}{k_1 + K + \lambda} \right) \right] \end{align} \label{21.19}\]

where

\[\begin{align} K &= k_2 + k_{-1} \\ \lambda &= \sqrt{\left( k_1 - K \right)^2 - 4k_1 k_{-1}} \end{align} \label{21.20}\]

Notice that these solutions have similar functional forms to consecutive reactions without the equilibrium. The differences are essentially in the constant values in the expressions. We would still expect a build up and then decrease in the intermediate as the reactants are used up and final products are produced.