Question

Consider the following mechanism, which results in the formation of product \(P:\) \\[ \begin{array}{l} \mathrm{A} \stackrel{k_{1}}{\rightleftharpoons_{k-1}} \mathrm{B} \frac{k_{2}}{\rightleftharpoons_{-2}} \mathrm{C} \\ \mathrm{B} \stackrel{k_{3}}{\rightarrow} \mathrm{P} \end{array} \\] If only the species \(A\) is present at \(t=0,\) what is the expression for the concentration of \(\mathrm{P}\) as a function of time? You can apply the preequilibrium approximation in deriving your answer.

Short answer

The concentration of product P as a function of time is given by the expression: \([P](t) = k_3\frac{k_1}{k_{-1}}([A]_0 - [A](t))\)

Step-by-step solution

Write the rate equations for each species

First, let's write the rate equations for each species involved in the mechanism. We can denote the concentrations of the species A, B, C, and P as [A], [B], [C], and [P], respectively: Rate of formation of B: \(R_B = k_1[A] - k_{-1}[B]\) Rate of formation of C: \(R_C = k_2[B]^2 - k_{-2}[C]\) Rate of formation of P: \(R_P = k_3[B]\)

Apply the preequilibrium approximation

Now, let's apply the preequilibrium approximation. This implies that the first step reaches equilibrium very quickly, so the rate of the formation, and consumption, of B is equal: \(k_1[A] = k_{-1}[B]\) From this equation, we can express the concentration of B in terms of A: \([B] = \frac{k_1}{k_{-1}}[A]\)

Solve for the concentration of product P as a function of time

To find the concentration of product P as a function of time, we need to integrate the rate equation for the formation of P: \(\frac{d[P]}{dt} = k_3 [B]\) Substitute the expression for [B] from step 2: \(\frac{d[P]}{dt} = k_3 \frac{k_1}{k_{-1}}[A]\) Now, integrate this equation with respect to time: \(\int_{0}^{[P]} d[P] = k_3 \frac{k_1}{k_{-1}} \int_{[A]_0}^{[A]} d[A]\) Here, \([A]_0\) is the initial concentration of A at time t=0. Now, integrate both sides: \([P] = k_3\frac{k_1}{k_{-1}}([A]_0 - [A])\) Finally, we have the expression for the concentration of product P as a function of time: \([P](t) = k_3\frac{k_1}{k_{-1}}([A]_0 - [A](t))\) This expression gives the concentration of product P as a function of the concentration of reactant A at any given time t.

Key concepts

Preequilibrium Approximation

When studying complex chemical reactions, the preequilibrium approximation is a useful concept that simplifies the analysis of reaction mechanisms. Essentially, it's an assumption that certain steps in a reaction mechanism reach a temporary state of equilibrium before the overall reaction proceeds to completion.

Let's consider the application of this approach with our example. The reaction begins with the reversible conversion of reactant A to an intermediate B and then on to another intermediate C, before finally producing the product P. By invoking the preequilibrium approximation, we are saying that the conversion of A to B and back to A happens much faster than the subsequent steps. As a result, the forward and reverse reactions between A and B are in dynamic equilibrium during the early stages of the reaction.

This assumption allows us to say that the rate at which A is converted to B is approximately equal to the rate at which B reverts to A, leading to a simplification in the form of a straightforward ratio of the rate constants and concentrations, as shown in the provided solution. It's important for students to recognize that this approximation greatly simplifies the task of determining the concentration of intermediates and products at any moment during the reaction.

Reaction Mechanisms

Understanding the Path from Reactants to Products

Reaction mechanisms provide a detailed, step-by-step description of how reactants are transformed into products during a chemical reaction. Each step involves elementary reactions that can either be unimolecular or bimolecular processes, involving one or two reactant molecules, respectively.

In our textbook example, we see a mechanism that involves multiple steps and intermediates. Such details are crucial for understanding the kinetics of the overall reaction. Subsequently, it also informs us about the speed at which product P is formed from reactant A. Each step has its unique rate constant, symbolized as k with a subscript to indicate the specific step.

Equilibration and Rate Determining Steps

Not all steps within a reaction mechanism occur at the same rate. Some may be so fast that they are at equilibrium (as assumed by the preequilibrium approximation), while others may be much slower, determining the overall rate of the reaction. Insight into these steps is invaluable as it helps chemists control and optimize reactions for desired outcomes.

Rate Equations

The rate equations or rate laws are mathematical expressions that describe the speed of a chemical reaction concerning the concentrations of the reactants. For each species involved in a reaction, we can write a unique rate equation that takes into account both its formation and degradation.

In the mechanism provided, we derived rate equations for intermediates B and C, and for the final product P. These equations link the rate constants and the concentrations of the reacting species. Notice that for the final product P, the rate is dependent solely on the concentration of intermediate B, represented in the form of the differential equation \(\frac{d[P]}{dt} = k_3 [B]\), where \(\frac{d[P]}{dt}\) represents the rate of formation of product P.

Integrating this equation over time, as shown in the steps, leads to an expression that gives us the concentration of the product P at any given time. This is essential for predicting product formation and can be applied to control industrial chemical processes as well as in laboratory synthesis.