Lecture 5 Rate Laws for Complex Reactions

5.1 Introduction

Many chemical reactions proceed through multiple elementary steps rather than a single molecular collision or transformation. Understanding how these complex reaction mechanisms give rise to experimentally observed kinetic behavior requires careful analysis of each step and consideration of how these steps work together to control the overall reaction rate.

5.2 Consecutive Reactions

Consider the following reaction mechanism that consists of two elementary steps and proceeds through a single intermediate B:

\[\mathrm{A} \xrightarrow{k_1} \mathrm{B} \xrightarrow{k_2} \mathrm{C}\]

This type of reaction sequence, where one product becomes the reactant for a subsequent step, is called a consecutive reaction. To analyze the kinetics, we need to write differential rate equations for each species:

For reactant A: \[\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{d}t} = -k_1[\mathrm{A}]\]

For intermediate B: \[\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{d}t} = +k_1[\mathrm{A}] - k_2[\mathrm{B}]\]

For product C: \[\frac{\mathrm{d}[\mathrm{C}]}{\mathrm{d}t} = +k_2[\mathrm{B}]\]

# Plot showing [A], [B], and [C] versus time

5.2.1 Solving the Rate Equations

Starting with the initial conditions at \(t = 0\): - \([\mathrm{A}] = [\mathrm{A}]_0\) - \([\mathrm{B}] = 0\) - \([\mathrm{C}] = 0\)

We can solve these equations step by step:

  1. First, the equation for A is a simple first-order decay: \[\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{d}t} = -k_1[\mathrm{A}]\]

    Which integrates to: \[[\mathrm{A}]_t = [\mathrm{A}]_0e^{-k_1t}\]

  2. For B, we substitute this result: \[\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{d}t} + k_2[\mathrm{B}] = k_1[\mathrm{A}]_0e^{-k_1t}\]

    This is a linear first-order differential equation. Its solution is: \[[\mathrm{B}]_t = \frac{k_1}{k_2-k_1}[\mathrm{A}]_0(e^{-k_1t} - e^{-k_2t})\]

  3. For C, we can use conservation of mass: \[[\mathrm{C}]_t = [\mathrm{A}]_0 - [\mathrm{A}]_t - [\mathrm{B}]_t\]

    Substituting gives: \[[\mathrm{C}]_t = [\mathrm{A}]_0\left(1 + \frac{k_1e^{-k_2t} - k_2e^{-k_1t}}{k_2-k_1}\right)\]

5.2.2 Rate-Determining Step

In consecutive reactions, one step often proceeds much more slowly than the others, effectively controlling the overall reaction rate. This rate-determining step emerges naturally from the mathematical analysis of the reaction kinetics.

When analyzing concentration profiles for consecutive reactions, two limiting cases arise depending on the relative magnitudes of \(k_1\) and \(k_2\):

When \(k_1 \gg k_2\):

# Plot showing case where k1 >> k2
  • B builds up quickly then decays slowly
  • The slow second step is rate-determining
  • \([\mathrm{C}]_t \approx [\mathrm{A}]_0(1-e^{-k_2t})\)

When \(k_1 \ll k_2\):

# Plot showing case where k1 << k2
  • B remains at low concentration
  • The slow first step is rate-determining
  • \([\mathrm{C}]_t \approx [\mathrm{A}]_0(1-e^{-k_1t})\)

5.3 Equilibrium Reactions

Many reactions reach a dynamic equilibrium rather than proceeding to completion:

\[\mathrm{A} \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} \mathrm{B}\]

The rate equations are:

\[\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{d}t} = -k_1[\mathrm{A}] + k_{-1}[\mathrm{B}]\]

\[\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{d}t} = +k_1[\mathrm{A}] - k_{-1}[\mathrm{B}]\]

At equilibrium, concentrations no longer change, so the net rates become zero:

\[\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{d}t} = \frac{\mathrm{d}[\mathrm{B}]}{\mathrm{d}t} = 0\]

This condition gives: \[k_1[\mathrm{A}] = k_{-1}[\mathrm{B}]\]

Rearranging: \[\frac{k_1}{k_{-1}} = \frac{[\mathrm{B}]}{[\mathrm{A}]} = K_\mathrm{eq}\]

This important result connects kinetics (rate constants) with thermodynamics (equilibrium constant).

5.4 Pre-equilibrium Mechanisms

Some reactions establish a rapid equilibrium between reactants and an intermediate before slowly converting to products. A common example is acid-catalyzed ester hydrolysis:

\(\ce{CH3COOCH3 + H3O+ <=> CH3COOCH4+ + H2O}\) (fast equilibrium) \(\ce{CH3COOCH4+ + H2O -> CH3COOH + CH3OH + H3O+}\) (slow)

This behavior, called a pre-equilibrium mechanism, occurs when: - The forward and reverse rates of the first step are much faster than the second step - The first step reaches equilibrium before significant product formation - The overall rate depends on both the equilibrium constant and the rate of the second step

In general terms, a pre-equilibrium mechanism has the form:

\[\mathrm{A} + \mathrm{B} \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} \mathrm{C} \xrightarrow{k_2} \mathrm{D}\]

The rate equations are:

\[\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{d}t} = \frac{\mathrm{d}[\mathrm{B}]}{\mathrm{d}t} = -k_1[\mathrm{A}][\mathrm{B}] + k_{-1}[\mathrm{C}]\]

\[\frac{\mathrm{d}[\mathrm{C}]}{\mathrm{d}t} = k_1[\mathrm{A}][\mathrm{B}] - k_{-1}[\mathrm{C}] - k_2[\mathrm{C}]\]

\[\frac{\mathrm{d}[\mathrm{D}]}{\mathrm{d}t} = k_2[\mathrm{C}]\]

When \(k_{-1} \gg k_2\), the first step rapidly reaches equilibrium:

\[\frac{[\mathrm{C}]}{[\mathrm{A}][\mathrm{B}]} = \frac{k_1}{k_{-1}} = K\]

This gives: \[[\mathrm{C}] = K[\mathrm{A}][\mathrm{B}]\]

The overall rate becomes: \[\frac{\mathrm{d}[\mathrm{D}]}{\mathrm{d}t} = k_2[\mathrm{C}] = k_2K[\mathrm{A}][\mathrm{B}] = \frac{k_2k_1}{k_{-1}}[\mathrm{A}][\mathrm{B}]\]

# Plot showing pre-equilibrium kinetics

This second-order rate law emerges from a more complex mechanism through the pre-equilibrium approximation. The overall rate constant \(k_\mathrm{obs} = k_2k_1/k_{-1}\) combines the individual rate constants from each step in the mechanism.