7.4.1 Potential Energy Diagram and Reaction Mechanism

This topic explains how potential energy diagrams are used to understand reaction mechanisms. It shows how to propose a reaction mechanism from a potential energy diagram and a given rate law.

Definition

Potential Energy Diagram: A graph showing the energy changes that occur during a chemical reaction, including the energies of reactants, products, transition states, and intermediates.

Activation Energy (Eₐ): The minimum energy that colliding molecules must possess for a reaction to occur.

Transition State: The highest energy state along the reaction pathway.

Main Idea

Potential energy diagrams provide information about the number of steps in a reaction mechanism, the rate-determining step, and the activation energy of each step. By analyzing the peaks and valleys in the diagram along with the experimental rate law, a reaction mechanism can be proposed.

Core Concept

• Number of peaks in the potential energy diagram = number of elementary steps.
• Highest peak = rate-determining step (slowest step).
• Valley between peaks = reaction intermediate.
• Rate law tells us the number of molecules involved in the rate-determining step.

Detailed Explanation

Example 7.8: Proposing a Mechanism from a Potential Energy Diagram

Potential energy diagram for the reaction between NO₂ and F₂ is shown.

The experimental rate law for this reaction is given below:

Reaction: 2NO_2(g) + F_2(g) → 2NO₂F_(g)

Rate: Rate = k[NO₂][F₂]

Propose reaction mechanism.

Solution:

Step 1: Determine the number of elementary steps. - As the potential energy diagram shows two peaks, the reaction mechanism must involve two elementary steps.

Step 2: Determine the rate-determining step. - Because the activation energy of step 1 is greater than the activation energy of step 2, step 1 is a slow and rate-determining step.

Step 3: Use the rate law to determine the number of molecules involved in the rate-determining step. - The given rate law indicates that one NO₂ and one F₂ molecule are involved in this step.

• In the overall reaction, two NO₂ molecules and one F₂ molecule react to form two NO₂F molecules.
• However, in the rate-determining step, only one NO₂ molecule and one F₂ molecule need to react to form one NO₂F and a reaction intermediate.
• In the second initial step, the reaction intermediate must react with another NO₂ molecule to form another NO₂F molecule.
• A species that is produced in one step of a chemical reaction and consumed in another step is called a reaction intermediate.

Thus, the proposed mechanism is:

NO₂ + F₂ → NO₂F + F (slow)

NO₂ + F → NO₂F (fast)

Step 5: Add the two steps to get the overall reaction which must be same as the reaction under consideration.

NO₂ + F₂ → NO₂F + F

NO₂ + F → NO₂F

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2NO₂ + F₂ → 2NO₂F

Since the sum of elementary steps gives the reaction under consideration, the proposed mechanism may be acceptable.

Concept Assessment Exercise 7.3

1. The following data was collected for the reaction between H₂ and NO at 700°C. 2H₂ + 2NO → 2H₂O + N₂

Suggest a plausible mechanism that is consistent with the rate law. (Hint: assume the oxygen atom is reaction intermediate.)

Potential energy diagram for this reaction is given below.

2. Following mechanism has been proposed for a reaction. H₂O_2(aq) + I^-_(aq) → H₂O_(l) + OI^-_(aq) OI^-_(aq) + H₂O_2(aq) → H₂O_(l) + O_2(g) + I^-_(aq)

3. The rate law for the following reaction: 2H₂ + 2NO → N₂ + 2H₂O is rate = k[H₂][NO]². Is the following mechanism consistent with the rate law? Argue.

H₂ + NO → H₂O + N (slow)

N + NO → N₂ + O (fast)

O + H₂ → H₂O (fast)

4. The rate law for the reaction: 2N₂O → 2N₂ + O₂ is rate = k[N₂O]. Reaction occurs in two elementary steps. Assume O atom as a reaction intermediate. Write mechanism for the reaction.

Intermediate Level

Steps to Propose a Mechanism

Example 7.8

Reaction: 2NO₂ + F₂ → 2NO₂F Rate: Rate = k[NO₂][F₂]

• Two peaks → two steps.
• Step 1 has higher Eₐ → slow, RDS.
• Rate law: one NO₂ and one F₂ in RDS.

• NO₂ + F₂ → NO₂F + F (slow)
• NO₂ + F → NO₂F (fast)

Intermediate: F

Concept Assessment Exercise 7.3

• Data given.
• Suggest mechanism.

2. H₂O₂ + I⁻ Reaction: - Mechanism proposed.

• Rate = k[H₂][NO]²
• Check mechanism.

• Rate = k[N₂O]
• Two steps, O intermediate.
• Write mechanism.

Important Points

• 1. Potential energy diagram: Shows energy changes.
• 2. Peaks = transition states.
• 3. Valleys = intermediates.
• 4. Highest peak = rate-determining step.
• 5. Rate law: Shows molecules in RDS.
• 6. Example 7.8: 2NO₂ + F₂ → 2NO₂F, Rate = k[NO₂][F₂], mechanism with F intermediate.
• 7. Concept Assessment 7.3: Four exercises.

Remember

• - Potential energy diagram: Energy changes
• - Peaks: Transition states
• - Valleys: Intermediates
• - Highest peak: Rate-determining step
• - Rate law: Shows RDS
• - Example 7.8: 2NO₂ + F₂ → 2NO₂F
• - Intermediate: F

Common Mistakes

• 1. Confusing peaks and valleys: Peaks = transition states; Valleys = intermediates.
• 2. Thinking the highest peak is the overall reaction: It is the rate-determining step.
• 3. Forgetting the rate law: It tells us the molecules in the RDS.
• 4. Mixing up steps: Add steps to get overall reaction.

Quick Revision

• Potential energy diagram: Shows energy changes.
• Peaks = transition states.
• Valleys = intermediates.
• Highest peak = rate-determining step.
• Rate law: Shows molecules in RDS.
• Example 7.8: 2NO₂ + F₂ → 2NO₂F, Rate = k[NO₂][F₂], mechanism with F intermediate.
• Concept Assessment 7.3: Four exercises.

60-Second Revision

• Potential energy diagram: Energy changes
• Peaks: Transition states
• Valleys: Intermediates
• Highest peak: Rate-determining step
• Rate law: Shows RDS
• Example 7.8: 2NO₂ + F₂ → 2NO₂F
• Intermediate: F

QUALITY VALIDATION

• Topic: 7.4.1 Potential Energy Diagram and Reaction Mechanism
• Chapter: 7 (Chemical Kinetics)

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