Under what temperature conditions would the reaction given on the slide occur spontaneously? Do not perform any calculations. all temperatures high temperatures low temperatures
Under what temperature conditions would the reaction given on the slide occur spontaneously? Do not perform any calculations. all temperatures high temperatures low temperatures
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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![**Question:**
Under what temperature conditions would the reaction given on the slide occur spontaneously? Do not perform any calculations.
- [ ] all temperatures
- [ ] high temperatures
- [ ] low temperatures](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbd446b52-8c58-4d2e-95c3-f01363988023%2F69f9cd53-5fea-4437-8ecb-2ece6c61f5df%2Fgfpeije_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question:**
Under what temperature conditions would the reaction given on the slide occur spontaneously? Do not perform any calculations.
- [ ] all temperatures
- [ ] high temperatures
- [ ] low temperatures
![### Problem:
Use Enthalpy and Entropy changes to predict the direction of spontaneous change. Under what temperature conditions would the following reactions occur spontaneously?
\[ \text{2NH}_4\text{NO}_3\text{(s)} \rightarrow \text{4H}_2\text{O(g)} + 2\text{N}_2\text{(g)} + \text{O}_2\text{(g)} \]
\[ \Delta H^\circ = -236.0 \text{ kJ} \]
### Explanation:
To determine the spontaneity of the reaction, we need to consider both the enthalpy change (\(\Delta H\)) and the entropy change (\(\Delta S\)) of the system.
For a reaction to be spontaneous, the Gibbs free energy change (\(\Delta G\)) must be negative. The Gibbs free energy is given by the equation:
\[ \Delta G = \Delta H - T\Delta S \]
Where:
- \(\Delta G\) is the Gibbs free energy change
- \(\Delta H\) is the enthalpy change
- \(T\) is the temperature in Kelvin
- \(\Delta S\) is the entropy change
Evaluating the spontaneity of the reaction at different temperature conditions involves analyzing the sign and magnitude of \(\Delta H\) and \(\Delta S\). In this case, \(\Delta H^\circ\) is given as \(-236.0 \text{ kJ}\), indicating an exothermic reaction.
To complete the assessment, the entropy change \(\Delta S\) should be known or estimated. Depending on the signs of \(\Delta H\) and \(\Delta S\), different temperature conditions will favor spontaneity.
#### Key Points for Assessing Spontaneity:
1. If \(\Delta H < 0\) (exothermic) and \(\Delta S > 0\), the reaction is spontaneous at all temperatures.
2. If \(\Delta H > 0\) (endothermic) and \(\Delta S < 0\), the reaction is non-spontaneous at all temperatures.
3. If \(\Delta H < 0\) and \(\Delta S < 0\), the reaction is spontaneous at low temperatures.
4. If \(\Delta H > 0\) and \(\Delta](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbd446b52-8c58-4d2e-95c3-f01363988023%2F69f9cd53-5fea-4437-8ecb-2ece6c61f5df%2Fv14v5jd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem:
Use Enthalpy and Entropy changes to predict the direction of spontaneous change. Under what temperature conditions would the following reactions occur spontaneously?
\[ \text{2NH}_4\text{NO}_3\text{(s)} \rightarrow \text{4H}_2\text{O(g)} + 2\text{N}_2\text{(g)} + \text{O}_2\text{(g)} \]
\[ \Delta H^\circ = -236.0 \text{ kJ} \]
### Explanation:
To determine the spontaneity of the reaction, we need to consider both the enthalpy change (\(\Delta H\)) and the entropy change (\(\Delta S\)) of the system.
For a reaction to be spontaneous, the Gibbs free energy change (\(\Delta G\)) must be negative. The Gibbs free energy is given by the equation:
\[ \Delta G = \Delta H - T\Delta S \]
Where:
- \(\Delta G\) is the Gibbs free energy change
- \(\Delta H\) is the enthalpy change
- \(T\) is the temperature in Kelvin
- \(\Delta S\) is the entropy change
Evaluating the spontaneity of the reaction at different temperature conditions involves analyzing the sign and magnitude of \(\Delta H\) and \(\Delta S\). In this case, \(\Delta H^\circ\) is given as \(-236.0 \text{ kJ}\), indicating an exothermic reaction.
To complete the assessment, the entropy change \(\Delta S\) should be known or estimated. Depending on the signs of \(\Delta H\) and \(\Delta S\), different temperature conditions will favor spontaneity.
#### Key Points for Assessing Spontaneity:
1. If \(\Delta H < 0\) (exothermic) and \(\Delta S > 0\), the reaction is spontaneous at all temperatures.
2. If \(\Delta H > 0\) (endothermic) and \(\Delta S < 0\), the reaction is non-spontaneous at all temperatures.
3. If \(\Delta H < 0\) and \(\Delta S < 0\), the reaction is spontaneous at low temperatures.
4. If \(\Delta H > 0\) and \(\Delta
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