Question

The proposed mechanism for a reaction is (1) \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{X}(g) \quad\) [fast \(]\) (2) \(\mathrm{X}(g)+\mathrm{C}(g) \longrightarrow \mathrm{Y}(g) \quad\) [slow] (3) \(\mathrm{Y}(g) \longrightarrow \mathrm{D}(g) \quad\) [fast (a) What is the overall equation? (b) Identify the intermediate(s), if any. (c) What are the molecularity and the rate law for each step? (d) Is the mechanism consistent with the actual rate law: Rate \(=k[\mathrm{~A}][\mathrm{B}][\mathrm{C}] ?\) (e) Is the following one-step mechanism equally valid? \(\mathrm{A}(g)+\mathrm{B}(g)+\mathrm{C}(g) \longrightarrow \mathrm{D}(g) ?\)

Short answer

The overall reaction is \( \text{A} (g) + \text{B} (g) + \text{C} (g) \rightarrow \text{D} (g) \). Intermediates are \( \text{X} \) and \( \text{Y} \. The original mechanism aligns with the given rate law, whereas a one-step tri-molecular reaction is implausible.

Step-by-step solution

Write the Overall Equation

Combine the reactants and products of all steps. Cancel out intermediates on both sides of the equation. The overall reaction is:ewline(1) \(\text{A} (g) + \text{B} (g) \rightleftharpoons \text{X} (g) \) [fast](2) \( \text{X} (g) + \text{C} (g) \rightarrow \text{Y} (g) \) [slow](3) \( \text{Y} (g) \rightarrow \text{D} (g) \) [fast] Intermediates \( \text{X} (g) \) and \( \text{Y} (g) \) cancel, yielding: ewline \[ \text{A} (g) + \text{B} (g) + \text{C} (g) \rightarrow \text{D} (g) \]

Identify Intermediates

Intermediates are species that are produced and consumed within the reaction mechanism. Here, \( \text{X} (g) \) and \( \text{Y} (g) \) are intermediates.

Determine Molecularity and Rate Law for Each Step

For each elementary step, determine the number of molecules involved (molecularity), and write the rate law: (1) \( \text{A} (g) + \text{B} (g) \rightleftharpoons \text{X} (g) \) [fast]• Molecularity: Bimolecular• Rate Law: \[ r_1 = k_1 [ \text{A} ] [ \text{B} ] \] (2) \(\text{X} (g) + \text{C} (g) \rightarrow \text{Y} (g) \) [slow] • Molecularity: Bimolecular• Rate Law: \[ r_2 = k_2 [ \text{X} ] [ \text{C} ] \] (3) \( \text{Y} (g) \rightarrow \text{D} (g) \) [fast]• Molecularity: Unimolecular• Rate Law: \[ r_3 = k_3 [ \text{Y} ] \]

Consistency with Given Rate Law

The slow step (rate-determining step) determines the overall rate. \( \text{X} (g) \) is an intermediate, so from Step (1), \[ \text{X} = K_1 [ \text{A} ][ \text{B} ] \] Substitute \([ \text{X} ] \) into slow step's rate law ewline \[ r_{overall} = k_2 [ \text{X} ] [ \text{C} ] = k_2 K_1 [ \text{A} ][ \text{B} ][ \text{C} ] \] The rate law \ \text{r} = \text{ k } [ \text{A} ][ \text{B} ][ \text{C} ] \ matches the provided rate law, confirming consistency.

Check Validity of One-Step Mechanism

Analyze the proposed single-step reaction \(\text{A} (g) + \text{B} (g) + \text{C} (g) \rightarrow \text{D} (g) \). A one-step tri-molecular collision is rare and less likely. Thus, the proposed mechanism is more valid than the one-step mechanism.

Key concepts

Overall Equation

In chemical reactions, the overall equation represents the complete, simplest, and balanced form of what's happening throughout the entire process. It sums up all the balanced chemical reactions taking place. Let’s look at the steps to write the overall equation for the given mechanism:

• Step 1: Identify the reactants and products in each step.
• Step 2: Combine the reactants and products.
• Step 3: Cancel out any intermediates that appear on both sides of the reaction.

For the given mechanism:
(1) \(\text{A}(g) + \text{B}(g) \rightleftharpoons \text{X}(g) \) [fast]
(2) \(\text{X}(g) + \text{C}(g) \rightarrow \text{Y}(g) \) [slow]
(3) \(\text{Y}(g) \rightarrow \text{D}(g) \) [fast]
By combining these steps and canceling intermediates like \(\text{X}(g) \) and \(\text{Y}(g) \), the overall reaction is:
\[\text{A}(g) + \text{B}(g) + \text{C}(g) \rightarrow \text{D}(g) \]
This equation reflects the complete transformation from reactants to the final product while omitting ephemeral intermediates.

Reaction Intermediates

Reaction intermediates are species that form in one step of the reaction and are consumed in subsequent steps. They are not seen in the overall reaction equation because they don’t appear in the complete transformation from reactants to products. For this mechanism, consider the steps again:

• (1) \(\text{A}(g) + \text{B}(g) \rightleftharpoons \text{X}(g) \) [fast]
• (2) \(\text{X}(g) + \text{C}(g) \rightarrow \text{Y}(g) \) [slow]
• (3) \(\text{Y}(g) \rightarrow \text{D}(g) \) [fast]

Here, \(\text{X}(g) \) and \(\text{Y}(g) \) are intermediates. They are formed in one step and consumed in another. Because they are short-lived and don't appear in the overall balanced equation, identifying intermediates helps us understand the sequence of steps in the reaction mechanism.

Intermediates provide a bridge between elementary steps and are crucial for studying reaction dynamics and rates.

Molecularity

Molecularity describes the number of reactant molecules involved in an elementary step. It helps us understand the mechanism rather than just the overall reaction.

For each step in the example mechanism, determine the molecularity and write the rate law:

• (1) \(\text{A}(g) + \text{B}(g) \rightleftharpoons \text{X}(g) \) [fast]
  Molecularity: Bimolecular
  Rate Law: \[r_1 = k_1[\text{A}][\text{B}] \]


• (2) \(\text{X}(g) + \text{C}(g) \rightarrow \text{Y}(g) \) [slow]
  Molecularity: Bimolecular
  Rate Law: \[r_2 = k_2[\text{X}][\text{C}] \]


• (3) \(\text{Y}(g) \rightarrow \text{D}(g) \) [fast]
  Molecularity: Unimolecular
  Rate Law: \[r_3 = k_3[\text{Y}] \]

Molecularity helps us propose how molecules collide and interact during the reaction. Bimolecular steps involve two molecules, while unimolecular steps involve only one. This knowledge assists in understanding the detailed mechanisms behind chemical kinetics.

Rate Law

The rate law expresses the rate of a reaction in terms of the concentration of reactants. It allows us to predict the speed of the reaction under different conditions. For elementary steps, the rate law is directly proportional to the reactants' concentrations.

Given the mechanism’s slow step (rate-determining step), the rate law for the overall reaction should reflect this critical step. For the given mechanism:
(2) \(\text{X}(g) + \text{C}(g) \rightarrow \text{Y}(g) \) [slow]
Rate-determining step’s rate law: \[r_{2} = k_{2}[\text{X}][\text{C}] \]
Since \(\text{X}(g) \) is an intermediate, we substitute its value from step (1): \(\text{X} = K_1[\text{A}][\text{B}] \)
Substitute into the rate law:
\[r_{overall} = k_{2}[\text{X}][\text{C}] = k_{2}K_1[\text{A}][\text{B}][\text{C}] \]
This matches the given rate law for the overall reaction:
\[Rate = k[\text{A}][\text{B}][\text{C}] \]
Validating this confirms that the proposed mechanism aligns with the observed rate law, ensuring that our understanding and assumptions about the reaction steps are accurate.