Question

The mechanism of the decomposition: \(A \rightarrow B+C\) can be written as a sequence of the elementary steps: Step I: \(A \stackrel{k_{1}}{\longrightarrow} D\) Step II: \(D \stackrel{k_{2}}{\longrightarrow} B+C\) (a) What type of a species is D? (b) For step I alone, give an expression that shows the rate of appearance of \(D\) (c) How is your answer to altered if you take into account the sequence of steps I and II? (d) Use the steady state approximation to show that the rate of formation of \(\mathrm{B}\) is directly proportional to the concentration of A.

Short answer

(a) D is an intermediate. (b) Rate of appearance of D is \( k_1[A] \). (c) Net rate is \( k_1[A] - k_2[D] \). (d) Rate of formation of B is \( k_1[A] \).

Step-by-step solution

Identify Species Type D

Species D is an intermediate. In a reaction mechanism, an intermediate is formed in one step of the reaction and consumed in another. It is not present in the overall reaction equation.

Rate of Appearance of D in Step I

For elementary reaction Step I: \( A \xrightarrow{k_1} D \), the rate of appearance of \( D \) is given by the rate law for this step. The rate is expressed as \( \text{Rate} = k_1[A] \). This assumes that the rate of appearance of \( D \) is directly proportional to the concentration of \( A \).

Consideration of Steps I and II

When Steps I and II are considered together, \( D \) is both produced in Step I and consumed in Step II. This can affect the net rate of change in the concentration of \( D \) over time. Specifically, the rate of appearance of \( D \) is \( k_1[A] - k_2[D] \), where \( k_2[D] \) represents the rate at which \( D \) is used up in Step II.

Steady State Approximation for D

Using the steady state approximation, we assume that the concentration of intermediate \( D \) is constant over time. Setting \( \frac{d[D]}{dt} = 0 \) gives us \( k_1[A] = k_2[D] \). Solving for \( [D] \), we get \( [D] = \frac{k_1}{k_2}[A] \).

Rate of Formation of B

In Step II, the formation of \( B \) is given by \( k_2[D] \). By substituting \( [D] = \frac{k_1}{k_2}[A] \) from the steady state approximation, we get \( \text{Rate of formation of B} = k_2 \left( \frac{k_1}{k_2}[A] \right) = k_1 [A] \). This shows that the rate of formation of \( B \) is directly proportional to the concentration of \( A \).

Key concepts

Intermediate Species

In a chemical reaction mechanism, intermediate species play a crucial role. They are substances formed in one step of the mechanism and consumed in another. Intermediates are not present in the overall balanced equation of the reaction. Instead, they exist momentarily and often cannot be detected in the final reaction mixture.
In the given exercise, species D is identified as an intermediate. This means it is formed in the first elementary step, where A converts to D, and is used up in the next step to form products B and C.
Recognizing intermediates is vital because it helps us understand the progress of the reaction and how different steps interlink. Often, intermediates are key to figuring out the actual pathway and rate at which the final products appear.

Steady State Approximation

The steady state approximation is a simplifying assumption used in reaction kinetics. This approach assumes that the concentration of an intermediate remains constant throughout the reaction. It's particularly useful for complex mechanisms where intermediates are involved.
To apply the steady state approximation, we set the rate of formation and rate of consumption of the intermediate equal, meaning the net change in its concentration is zero. Mathematically, this is expressed as \( \frac{d[D]}{dt} = 0 \). For our exercise, it results in \( k_1[A] = k_2[D] \), providing a quick way to find the concentration of the intermediate D.
Using this approximation simplifies calculations and helps derive rate expressions for the appearance of products, streamlining the process of understanding how reactions proceed at the molecular level.

Rate Laws

Rate laws express the relationship between the concentration of reactants and the rate of a reaction. They are crucial for predicting how fast a reaction proceeds. For elementary steps, the rate law is directly based on the molecularity of the reaction - that is, the number of molecules involved.
In Step I of the mechanism, the rate law is given by \( \text{Rate} = k_1[A] \), which means the rate depends linearly on the concentration of A. This highlights that the step is a first-order reaction with respect to A.
When we consider both steps of the mechanism, the rate law changes. The net rate of formation of D takes into account both its formation and consumption: \( k_1[A] - k_2[D] \). This illustrates how additional steps or presence of intermediates impact how we express and calculate rates, essential for understanding reaction kinetics.

Elementary Steps

Elementary steps are the basic individual stages in a reaction mechanism. Each one describes a simple process, typically involving a small number of molecules. These steps, when combined, give the full pathway from reactants to products.
In the given decomposition mechanism, there are two elementary steps: the conversion of A to D, and then D to B plus C. Each step can be described by its own rate law and involves distinct reactive events.
Understanding elementary steps helps in detailing how complex reactions actually unfold at a molecular level. Each step's rate law, determined by its molecularity, provides insight into how quickly or slowly that step will proceed relative to others. This knowledge is key to managing and optimizing reactions in both laboratory and industrial processes.