I Review I Constants I Periodic Table Consider the evaporation of methanol at 25.0 °C: Find AG° at 25.0°C. CH3 OH (1) → CH3OH (g). Express the free energy change in kilojoules to one decimal place. You may want to reference (Page) Section 18.9 while completing this problem. AG = kJ Submit Request Answer The free energy change in the previous part is calculated based on the assumption of standard conditions. For Parts B through D, find AG at 25.0° C under the given nonstandard conditions. Part B PCH,OH = 159.0mmHg Express the free energy change in kilojoules to one decimal place. να ΑΣφ AG = kJ Submit Request Answer Part C PCH,OH = 110.0mmHg
I Review I Constants I Periodic Table Consider the evaporation of methanol at 25.0 °C: Find AG° at 25.0°C. CH3 OH (1) → CH3OH (g). Express the free energy change in kilojoules to one decimal place. You may want to reference (Page) Section 18.9 while completing this problem. AG = kJ Submit Request Answer The free energy change in the previous part is calculated based on the assumption of standard conditions. For Parts B through D, find AG at 25.0° C under the given nonstandard conditions. Part B PCH,OH = 159.0mmHg Express the free energy change in kilojoules to one decimal place. να ΑΣφ AG = kJ Submit Request Answer Part C PCH,OH = 110.0mmHg
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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
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![**Exercise 18.70 - Thermodynamics**
**Topic: Evaporation of Methanol**
Consider the evaporation of methanol at 25.0°C:
\[ \text{CH}_3\text{OH (l)} \rightarrow \text{CH}_3\text{OH (g)} \]
You may want to reference Section 18.9 while completing this problem.
**Find ΔG° at 25.0°C.**
Express the free energy change in kilojoules to one decimal place.
\[ ΔG° = \_\_\_\_\_ \text{kJ} \]
<div style="background-color: #f0f0f0; padding: 10px; margin-top: 10px;">
The free energy change in the previous part is calculated based on the assumption of standard conditions. For Parts B through D, find ΔG at 25.0°C under the given nonstandard conditions.
</div>
**Part B**
\[ P_{\text{CH}_3\text{OH}} = 159.0 \text{mmHg} \]
Express the free energy change in kilojoules to one decimal place.
\[ ΔG° = \_\_\_\_\_ \text{kJ} \]
**Part C**
\[ P_{\text{CH}_3\text{OH}} = 110.0 \text{mmHg} \]
Express the free energy change in kilojoules to one decimal place.
\[ ΔG° = \_\_\_\_\_ \text{kJ} \]
**Instructions:**
- Click "Submit" to check your answers.
- Use the provided pressure values to compute the free energy changes for methanol under nonstandard conditions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcbdd570d-a0c9-44fd-80de-e8b53629889d%2F8b1594d4-c3cc-45f9-8b62-345f62113888%2Fthab01a_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Exercise 18.70 - Thermodynamics**
**Topic: Evaporation of Methanol**
Consider the evaporation of methanol at 25.0°C:
\[ \text{CH}_3\text{OH (l)} \rightarrow \text{CH}_3\text{OH (g)} \]
You may want to reference Section 18.9 while completing this problem.
**Find ΔG° at 25.0°C.**
Express the free energy change in kilojoules to one decimal place.
\[ ΔG° = \_\_\_\_\_ \text{kJ} \]
<div style="background-color: #f0f0f0; padding: 10px; margin-top: 10px;">
The free energy change in the previous part is calculated based on the assumption of standard conditions. For Parts B through D, find ΔG at 25.0°C under the given nonstandard conditions.
</div>
**Part B**
\[ P_{\text{CH}_3\text{OH}} = 159.0 \text{mmHg} \]
Express the free energy change in kilojoules to one decimal place.
\[ ΔG° = \_\_\_\_\_ \text{kJ} \]
**Part C**
\[ P_{\text{CH}_3\text{OH}} = 110.0 \text{mmHg} \]
Express the free energy change in kilojoules to one decimal place.
\[ ΔG° = \_\_\_\_\_ \text{kJ} \]
**Instructions:**
- Click "Submit" to check your answers.
- Use the provided pressure values to compute the free energy changes for methanol under nonstandard conditions.
![# Thermodynamics: Methanol Evaporation at 25.0 °C
## Exercise 18.70 - Enhanced with Feedback
### Context:
Consider the evaporation of methanol at 25.0 °C.
\[ \text{CH}_3\text{OH (l)} \rightarrow \text{CH}_3\text{OH (g)} \]
You may want to reference Section 18.9 while completing this problem.
### Part D:
Given:
\[ P_{\text{CH}_3\text{OH}} = 13.00 \text{ mmHg} \]
**Task:**
Express the free energy change (\( \Delta G^\circ \)) in kilojoules to one decimal place.
\[ \Delta G^\circ = \boxed{\phantom{0}} \text{ kJ} \]
*Submit* *Request Answer*
### Part E:
The vapor pressure of methanol is 143 mmHg. Identify the best reason why methanol spontaneously evaporates in open air at 25.0 °C and standard pressure (760 mmHg).
**Options:**
1. Under these nonstandard conditions, the partial pressure of methanol is lower than its vapor pressure.
2. Heat can be added under these nonstandard conditions to increase the vapor pressure of methanol.
3. The calculated standard free energy change corresponds to a spontaneous reaction.
4. Calculations for a closed container show a partial pressure of methanol lower than its vapor pressure.
*Submit* *Request Answer*
---
Note: This exercise is designed to help students understand the principles of thermodynamics related to phase changes and vapor pressure.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcbdd570d-a0c9-44fd-80de-e8b53629889d%2F8b1594d4-c3cc-45f9-8b62-345f62113888%2Fhbk7twu_processed.jpeg&w=3840&q=75)
Transcribed Image Text:# Thermodynamics: Methanol Evaporation at 25.0 °C
## Exercise 18.70 - Enhanced with Feedback
### Context:
Consider the evaporation of methanol at 25.0 °C.
\[ \text{CH}_3\text{OH (l)} \rightarrow \text{CH}_3\text{OH (g)} \]
You may want to reference Section 18.9 while completing this problem.
### Part D:
Given:
\[ P_{\text{CH}_3\text{OH}} = 13.00 \text{ mmHg} \]
**Task:**
Express the free energy change (\( \Delta G^\circ \)) in kilojoules to one decimal place.
\[ \Delta G^\circ = \boxed{\phantom{0}} \text{ kJ} \]
*Submit* *Request Answer*
### Part E:
The vapor pressure of methanol is 143 mmHg. Identify the best reason why methanol spontaneously evaporates in open air at 25.0 °C and standard pressure (760 mmHg).
**Options:**
1. Under these nonstandard conditions, the partial pressure of methanol is lower than its vapor pressure.
2. Heat can be added under these nonstandard conditions to increase the vapor pressure of methanol.
3. The calculated standard free energy change corresponds to a spontaneous reaction.
4. Calculations for a closed container show a partial pressure of methanol lower than its vapor pressure.
*Submit* *Request Answer*
---
Note: This exercise is designed to help students understand the principles of thermodynamics related to phase changes and vapor pressure.
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