Question

For the mechanism, \(\mathrm{A}+\mathrm{B} \underset{\mathrm{k}_{2}}{\stackrel{\mathrm{k}}_{1}}{\mathrm{~T}} \mathrm{C} ; \mathrm{C} \stackrel{\mathrm{k}_{3}}{\longrightarrow} \mathrm{D}\) The equilibrium step is fast. The reaction rate, \(\mathrm{d} / \mathrm{dt}[\mathrm{D}]\) is a. \(\mathrm{k}_{1} \mathrm{k}_{2} \mathrm{k}_{3}[\mathrm{~A}][\mathrm{B}]\) b. \(\frac{\mathrm{k}_{1} \mathrm{k}_{3}}{\mathrm{k}_{2}}[\mathrm{~A}][\mathrm{B}]\) c. \(\frac{\mathrm{k}_{1} \mathrm{k}_{3}[\mathrm{~A}][\mathrm{B}]}{\mathrm{k}_{2}+\mathrm{k}_{3}}\) d. \(\frac{\mathrm{k}_{2} \mathrm{k}_{3}}{\mathrm{k}_{1}}[\mathrm{~A}][\mathrm{B}]\)

Short answer

The rate of reaction is option b: \( \frac{k_1 k_3}{k_2}[A][B] \).

Step-by-step solution

Identify the Mechanism

The mechanism involves two main steps: \( A + B \) forms \( C \) through a fast equilibrium step, and \( C \) subsequently decomposes to \( D \). This is shown as \( A + B \underset{k_{2}}{\stackrel{k}{\rightleftharpoons}} C \) and \( C \stackrel{k_{3}}{\longrightarrow} D \).

Express the Equilibrium Condition

For the fast equilibrium, the forward rate \( k_1 [A][B] \) equals the reverse rate \( k_2[C] \). Thus, we can write: \[ k_1 [A][B] = k_2 [C] \] Solving for \( [C] \), we get: \[ [C] = \frac{k_1}{k_2} [A][B] \]

Determine the Rate of Formation of D

The rate of formation of \( D \) is determined by the rate at which \( C \) converts to \( D \). This is given by: \[ \frac{d[D]}{dt} = k_3[C] \] Substitute \( [C] \) from Step 2: \[ \frac{d[D]}{dt} = k_3 \left(\frac{k_1}{k_2} [A][B]\right) \] Simplifying, we find: \[ \frac{d[D]}{dt} = \frac{k_1 k_3}{k_2} [A][B] \]

Compare with Given Options

Compare our derived expression \( \frac{k_1 k_3}{k_2} [A][B] \) with the provided options. It matches exactly with option b.

Key concepts

Reaction Mechanism

A reaction mechanism provides a detailed step-by-step description of how reactants transform into products in a chemical reaction. Understanding reaction mechanisms is crucial because they reveal the sequence of events at the molecular level that leads to the overall reaction. In our exercise, the reaction mechanism consists of two main steps:

• The first step involves reactants \( A \) and \( B \) interacting to form an intermediate compound \( C \). This process is reversible and reaches a rapid equilibrium, characterized by the reaction rates \( k_1 \) and \( k_2 \).
• The second step is the decomposition of the intermediate \( C \), resulting in the formation of the product \( D \). This step is characterized by the rate constant \( k_3 \).

The overall reaction is considered as a combination of these elementary steps. By breaking down complex reactions into individual steps, reaction mechanisms help chemists understand how different conditions or catalysts affect the reaction rates and pathways.

Equilibrium Step

An equilibrium step in a reaction means that the forward and reverse reactions occur at the same rate, resulting in constant concentrations of reactants and products. In our context, the reaction between \( A \) and \( B \) to form \( C \) is an equilibrium step. It can be described by the reversible reaction \( A + B \rightleftharpoons C \).
Due to the rapid equilibrium, the concentrations of the reactants \( [A] \) and \( [B] \) and the intermediate \( [C] \) quickly reach a steady state. This allows us to use the equilibrium expression:
\[ k_1 [A][B] = k_2 [C] \]
Solving for \( [C] \), we get:
\[ [C] = \frac{k_1 [A][B]}{k_2}\]
This helps us determine how much of \( C \) is present at equilibrium, which is crucial for calculating the rate of product formation. Understanding equilibrium steps allows chemists to predict how altering conditions like temperature or concentration affects the equilibrium position, giving insights into reaction yield and efficiency.

Rate of Formation

The rate of formation in a chemical reaction refers to how quickly a product is generated from the reactants. For our mechanism, where the ultimate product is \( D \), the rate of formation is influenced by the intermediate \( C \) being converted into \( D \). This step occurs through the rate constant \( k_3 \):
\[ \frac{d[D]}{dt} = k_3 [C]\]
By substituting our expression for \( [C] \) from the equilibrium step:
\[ \frac{d[D]}{dt} = k_3 \left(\frac{k_1 [A][B]}{k_2}\right)\]
Simplifying, we derive the rate equation:
\[ \frac{d[D]}{dt} = \frac{k_1 k_3 [A][B]}{k_2}\]
This formula illustrates how the rate of formation of \( D \) depends on the concentrations of \( A \) and \( B \), the intermediate step equilibrium, and the subsequent reaction step rate constant. Understanding the rate of formation is fundamental in kinetics as it helps in designing reactions that maximize product yield efficiently, by adjusting concentrations and other conditions.