= 10.2. Suppose that a reaction A + 2B = 2Y + 2Z is believed to occur according to the mechanism A → 2X (very rapid equilibrium) k₂ X + B Y + Z (slow) Obtain an expression for the rate of formation of the product Y.

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### Reaction Mechanism Analysis

#### Problem 10.2:

Suppose that a reaction \( A + 2B = 2Y + 2Z \) is believed to occur according to the mechanism:

\[ A \xrightarrow{k_1} 2X \quad (\text{very rapid equilibrium}) \]

\[ X + B \xrightarrow{k_2} Y + Z \quad (\text{slow}) \]

**Task: Obtain an expression for the rate of formation of the product Y.**

### Detailed Diagram Description

- The diagram presents a two-step reaction mechanism.
- The first step shows a reactant \( A \) converting into \( 2X \) with a rate constant \( k_1 \). This step is characterized by very rapid equilibrium.
- The second step involves the intermediate \( X \) reacting with \( B \) to form products \( Y \) and \( Z \) with a rate constant \( k_2 \). This step is considered slow, indicating it is the rate-determining step of the overall reaction.

### Rate of Formation of Product Y

To find the expression for the rate of formation of \( Y \), we need to consider the slow step in the mechanism:
\[ X + B \rightarrow Y + Z \]

Since this is the rate-determining step, the rate of formation of \( Y \) depends on the concentration of \( X \) and \( B \):

\[ \text{Rate}_{Y} = k_2 [X][B] \]

However, since \( [X] \) is produced in the first step and is in rapid equilibrium with \( A \):

\[ A \xrightarrow{k_1} 2X \]

We can assume the concentration of \( X \) reaches a steady state quickly. Therefore, we can express \( [X] \) in terms of \( [A] \), and substitute this into the rate equation for \( Y \).

Finalizing the equations and solving to derive the exact expression for \( [Y] \) requires additional steps in kinetics and equilibrium assumptions, often involving \( k_1 \) being much larger compared to \( k_2 \). The detailed kinetic derivation is left as an exercise.
Transcribed Image Text:### Reaction Mechanism Analysis #### Problem 10.2: Suppose that a reaction \( A + 2B = 2Y + 2Z \) is believed to occur according to the mechanism: \[ A \xrightarrow{k_1} 2X \quad (\text{very rapid equilibrium}) \] \[ X + B \xrightarrow{k_2} Y + Z \quad (\text{slow}) \] **Task: Obtain an expression for the rate of formation of the product Y.** ### Detailed Diagram Description - The diagram presents a two-step reaction mechanism. - The first step shows a reactant \( A \) converting into \( 2X \) with a rate constant \( k_1 \). This step is characterized by very rapid equilibrium. - The second step involves the intermediate \( X \) reacting with \( B \) to form products \( Y \) and \( Z \) with a rate constant \( k_2 \). This step is considered slow, indicating it is the rate-determining step of the overall reaction. ### Rate of Formation of Product Y To find the expression for the rate of formation of \( Y \), we need to consider the slow step in the mechanism: \[ X + B \rightarrow Y + Z \] Since this is the rate-determining step, the rate of formation of \( Y \) depends on the concentration of \( X \) and \( B \): \[ \text{Rate}_{Y} = k_2 [X][B] \] However, since \( [X] \) is produced in the first step and is in rapid equilibrium with \( A \): \[ A \xrightarrow{k_1} 2X \] We can assume the concentration of \( X \) reaches a steady state quickly. Therefore, we can express \( [X] \) in terms of \( [A] \), and substitute this into the rate equation for \( Y \). Finalizing the equations and solving to derive the exact expression for \( [Y] \) requires additional steps in kinetics and equilibrium assumptions, often involving \( k_1 \) being much larger compared to \( k_2 \). The detailed kinetic derivation is left as an exercise.
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