Question

The pre-equilibrium and the steady-state approximations are two different approaches to deriving a rate law from a proposed mechanism. For the following mechanism, determine the rate law (a) by the pre-equilibrium approximation and (b) by the steady-state approximation. (c) Under what conditions do the two methods give the same answer? (d) What will the rate laws become at high concentrations of \(\mathrm{Br}^{-}\)? $$ \begin{aligned} &\mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}^{+} \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}_{2}^{+} \text {(fast equilibrium) } \\ &\mathrm{CH}_{3} \mathrm{OH}_{2}^{+}+\mathrm{Br}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{Br}+\mathrm{H}_{2} \mathrm{O} \text { (slow) } \end{aligned} $$

Short answer

By pre-equilibrium approximation: Rate = \(k_{slow}K_{eq}[\mathrm{CH_3OH}][\mathrm{H^+}][\mathrm{Br^-}]\). By steady-state approximation, a similar expression is derived where the steady-state concentration of \(\mathrm{CH_3OH_2^+}\) is found. Both methods yield the same rate law when the concentration of intermediate is in a pseudo-steady-state. At high \(\mathrm{Br^-}\) concentrations, the rate law can become independent of \(\mathrm{Br^-}\).

Step-by-step solution

Pre-equilibrium Approximation - Identify Fast and Slow Steps

Under the pre-equilibrium approximation, one assumes that the reaction steps up to the slowest step reach a quasi-equilibrium state very quickly. Thus, the rate is determined by the slowest step. In the given mechanism, the fast step is the equilibrium between \(\mathrm{CH_3OH} + \mathrm{H^+} \rightleftharpoons \mathrm{CH_3OH_2^+}\) and the slow step is the reaction of \(\mathrm{CH_3OH_2^+} + \mathrm{Br^-} \rightarrow \mathrm{CH_3Br} + \mathrm{H_2O}\).

Write the Pre-equilibrium Expression

The equilibrium constant expression for the fast step can be written as \(K_{eq} = \frac{[\mathrm{CH_3OH_2^+}]}{[\mathrm{CH_3OH}][\mathrm{H^+}]}\). Solve this equation for the concentration of the intermediate \(\mathrm{CH_3OH_2^+}\) under equilibrium conditions, which will be used in determining the rate law.

Derive Pre-equilibrium Rate Law

Since the slow step determines the rate, the rate law is the rate of the slow step, which is \(Rate = k_{slow}[\mathrm{CH_3OH_2^+}][\mathrm{Br^-}]\). Using the pre-equilibrium expression to substitute for \(\mathrm{CH_3OH_2^+}\), the rate law by the pre-equilibrium approximation is \(Rate = k_{slow}K_{eq}[\mathrm{CH_3OH}][\mathrm{H^+}][\mathrm{Br^-}]\).

Steady-State Approximation - Setting Up the Approximation

For the steady-state approximation, the concentration of the intermediate is considered constant (does not accumulate). The rate of formation equals the rate of consumption for \(\mathrm{CH_3OH_2^+}\). Set up the rate equations for the formation and consumption of \(\mathrm{CH_3OH_2^+}\).

Solve for Steady-State Concentration

Balancing the formation and consumption rates gives us the steady-state concentration of \(\mathrm{CH_3OH_2^+}\). This involves solving the rate equations derived in the previous step.

Derive Steady-State Rate Law

The steady-state rate law is now determined by using the rate of the slow step and substituting the expression for \(\mathrm{CH_3OH_2^+}\) obtained from the steady-state approximation.

Conditions for Identical Rate Laws

The pre-equilibrium and steady-state approximations provide the same rate law under the conditions where the pre-equilibrium approximation applies, meaning that the first step is fast and reaches a quasi-equilibrium, and the intermediate does not accumulate.

Rate Laws at High \(\mathrm{Br^-}\) Concentration

At high concentrations of \(\mathrm{Br^-}\), the second step is no longer the rate-limiting step, and the rate may be determined by how quickly \(\mathrm{CH_3OH}\) and \(\mathrm{H^+}\) can form the intermediate \(\mathrm{CH_3OH_2^+}\). The rate laws may become independent of \(\mathrm{Br^-}\) concentration.

Key concepts

Pre-Equilibrium Approximation

Understanding the pre-equilibrium approximation is crucial for students tackling kinetics in chemical reactions. This concept supposes that certain steps in a reaction mechanism can quickly reach an equilibrium before the overall reaction proceeds. Let's break it down with an example.

Consider a fast equilibrium between substances A and B to form an intermediate C, followed by a slow reaction where C reacts with substance D to form the end products. According to the pre-equilibrium approach, the intermediate C is in dynamic equilibrium with A and B, and its concentration can be expressed in terms of A and B using the equilibrium constant.

• First, write the equilibrium expression: \(K_{eq} = \frac{[C]}{[A][B]}\).
• Next, solve for [C]: \([C] = K_{eq} \times [A][B]\).
• Finally, the rate of the overall reaction is controlled by the slow step, so the rate law will include [C]. However, using the pre-equilibrium relationship, the rate law is expressed in terms of reactants A and B: \(Rate = k_{slow}[C][D] = k_{slow}K_{eq}[A][B][D]\).

Applying this to our textbook exercise

In the given example, the fast equilibrium is between \(\mathrm{CH_3OH}\) and \(\mathrm{H^+}\) to form \(\mathrm{CH_3OH_2^+}\), leading to the rate law \(Rate = k_{slow}K_{eq}[\mathrm{CH_3OH}][\mathrm{H^+}][\mathrm{Br^-}]\) by replacing [C] with \(\mathrm{CH_3OH_2^+}\). This approximation is particularly helpful when the intermediate is hard to measure or exists in a very low concentration.

Steady-State Approximation

Now, let's consider the steady-state approximation, which is another method to derive the rate law in complex reactions. Unlike the pre-equilibrium approximation, the steady-state assumption posits that the intermediate's concentration remains constant throughout the reaction—not because it's at equilibrium, but because its rate of formation equals its rate of consumption.

• The steady-state assumption is applied to intermediates: \(\frac{d[C]}{dt} = 0\).
• Different rate equations for the formation and consumption of C are set up and equated to maintain a steady concentration.
• These rate equations are solved algebraically to express [C] in terms of the concentrations of the reactants and constants.

Execution in the exercise

For the reaction involving \(\mathrm{CH_3OH_2^+}\), we equate the rate of its formation from \(\mathrm{CH_3OH}\) and \(\mathrm{H^+}\) to its consumption in the reaction with \(\mathrm{Br^-}\). After finding the steady-state concentration, it substitutes into the rate of the slow step to give the rate law under steady-state conditions. This approach is often favored when the pre-equilibrium cannot be assumed or when intermediates are present at higher concentrations.

Rate Law Derivation

Derivation of the rate law is a cornerstone in understanding chemical kinetics. The rate law links the rate of reaction to the concentration of reactants, often revealing details about the reaction mechanism. To derive it, we either follow the rate of disappearance of reactants or the rate of appearance of products, depending on the situation.

• Examine the rate-determining step, often the slowest step in a multi-step reaction.
• Use stoichiometry to relate reactant concentrations to the progress of the reaction.
• Apply an appropriate approximation, such as pre-equilibrium or steady-state, to relate the concentrations of intermediates to those of reactants.
• Express the rate in terms of reactant concentrations, constants, and orders of reaction.

Insight from our textbook problem

In our exercise example, the derivation process reveals different rate laws depending on the approximation used. However, both approaches ultimately function to connect the concentration of the reactants to the observable rate of the reaction. Additionally, the conditions under which both approximations yield the same rate law, or the behavior at high reactant concentrations, provide deeper insight into the limitations and applications of each method for specific reaction conditions.