Sketch a potential energy diagram for the decomposition of nitrous oxide. N,0(g) → N2(9) + O(g) The activation energy for the forward reaction is 278 kJ; AH° is +167 kJ. What is the activation energy for the reverse reaction? Activation energy = kJ
Sketch a potential energy diagram for the decomposition of nitrous oxide. N,0(g) → N2(9) + O(g) The activation energy for the forward reaction is 278 kJ; AH° is +167 kJ. What is the activation energy for the reverse reaction? Activation energy = kJ
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Title: Potential Energy Diagram for the Decomposition of Nitrous Oxide**
**Reaction:**
\[ \text{N}_2\text{O}(g) \rightarrow \text{N}_2(g) + \text{O}(g) \]
**Introduction:**
In this exercise, we will sketch a potential energy diagram to understand the decomposition process of nitrous oxide (\(\text{N}_2\text{O}\)). The potential energy diagram illustrates the energy changes during a chemical reaction.
**Details:**
- The activation energy for the forward reaction is 278 kJ.
- The standard enthalpy change (\(\Delta H^\circ\)) is +167 kJ.
**Objective:**
Determine the activation energy for the reverse reaction.
**Explanation:**
To calculate the activation energy for the reverse reaction, we use the following relationship:
1. The activation energy of the forward reaction (Ea_forward) is given as 278 kJ.
2. The enthalpy change (\(\Delta H^\circ\)) is +167 kJ.
3. Using the formula:
\[
\text{Ea}_{\text{reverse}} = \text{Ea}_{\text{forward}} - \Delta H^\circ
\]
4. Substitute the values:
\[
\text{Ea}_{\text{reverse}} = 278 \, \text{kJ} - 167 \, \text{kJ} = 111 \, \text{kJ}
\]
The activation energy for the reverse reaction is thus 111 kJ.
**Illustration of the Diagram:**
The diagram should show the reactants (\(\text{N}_2\text{O}(g)\)) on the left side, progressing to the products (\(\text{N}_2(g) + \text{O}(g)\)) on the right side. The energy level of the products is higher than that of the reactants, reflecting the positive \(\Delta H^\circ\). The peak of the diagram represents the transition state, where the energy is at its maximum. The difference between the reactants and the transition state represents the forward activation energy, while the difference between the products and the transition state represents the reverse activation energy.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe19fa2ed-8566-4b3b-96d0-3591866fab62%2F73a7b2eb-daaa-45d6-b0dc-7e10651c189c%2Fhu6tldm_processed.png&w=3840&q=75)
Transcribed Image Text:**Title: Potential Energy Diagram for the Decomposition of Nitrous Oxide**
**Reaction:**
\[ \text{N}_2\text{O}(g) \rightarrow \text{N}_2(g) + \text{O}(g) \]
**Introduction:**
In this exercise, we will sketch a potential energy diagram to understand the decomposition process of nitrous oxide (\(\text{N}_2\text{O}\)). The potential energy diagram illustrates the energy changes during a chemical reaction.
**Details:**
- The activation energy for the forward reaction is 278 kJ.
- The standard enthalpy change (\(\Delta H^\circ\)) is +167 kJ.
**Objective:**
Determine the activation energy for the reverse reaction.
**Explanation:**
To calculate the activation energy for the reverse reaction, we use the following relationship:
1. The activation energy of the forward reaction (Ea_forward) is given as 278 kJ.
2. The enthalpy change (\(\Delta H^\circ\)) is +167 kJ.
3. Using the formula:
\[
\text{Ea}_{\text{reverse}} = \text{Ea}_{\text{forward}} - \Delta H^\circ
\]
4. Substitute the values:
\[
\text{Ea}_{\text{reverse}} = 278 \, \text{kJ} - 167 \, \text{kJ} = 111 \, \text{kJ}
\]
The activation energy for the reverse reaction is thus 111 kJ.
**Illustration of the Diagram:**
The diagram should show the reactants (\(\text{N}_2\text{O}(g)\)) on the left side, progressing to the products (\(\text{N}_2(g) + \text{O}(g)\)) on the right side. The energy level of the products is higher than that of the reactants, reflecting the positive \(\Delta H^\circ\). The peak of the diagram represents the transition state, where the energy is at its maximum. The difference between the reactants and the transition state represents the forward activation energy, while the difference between the products and the transition state represents the reverse activation energy.
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