The heat of vaporization, AH°vap, of acetone is 31.3 kJ/mol. If the vapor pressure of liquid acetone is 100 torr at 7.7°C, what is the boiling point of acetone. Use the Clausius-Clapeyron equation. (1.00 atm = 760 torr) Show your work. See pl for equation.
The heat of vaporization, AH°vap, of acetone is 31.3 kJ/mol. If the vapor pressure of liquid acetone is 100 torr at 7.7°C, what is the boiling point of acetone. Use the Clausius-Clapeyron equation. (1.00 atm = 760 torr) Show your work. See pl for equation.
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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![### Problem Statement
The heat of vaporization, \( \Delta H^\circ_{\text{vap}} \), of acetone is 31.3 kJ/mol. If the vapor pressure of liquid acetone is 100 torr at 7.7°C, what is the boiling point of acetone? Use the Clausius-Clapeyron equation to find the boiling point.
\[ \text{(1.00 atm = 760 torr)} \]
### Instructions:
- Show your work
- See p1 for equation
### Detailed Explanation:
This problem involves calculating the boiling point of acetone using the Clausius-Clapeyron equation. Here are the steps to solve the problem:
#### Given Data:
- Heat of vaporization (\( \Delta H^\circ_{\text{vap}} \)): 31.3 kJ/mol
- Vapor pressure at 7.7°C ( \( P_1 \)): 100 torr
- Temperature \( T_1 \): 7.7°C (convert to Kelvin)
- Standard atmospheric pressure (\( P_2 \)): 760 torr
#### Clausius-Clapeyron Equation:
The Clausius-Clapeyron equation is:
\[ \ln \left( \frac{P_2}{P_1} \right) = \frac{\Delta H^\circ_{\text{vap}}}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
Where:
- \( P_1 \) and \( P_2 \) are the vapor pressures at temperatures \( T_1 \) and \( T_2 \) respectively
- \( R \) is the universal gas constant (8.314 J/mol·K)
- \( T_1 \) and \( T_2 \) are the temperatures in Kelvin
#### Conversion of Temperatures
- \( T_1 = 7.7°C = 7.7 + 273.15 = 280.85 \ \text{K} \)
- \( P_1 = 100 \ \text{torr} \)
- \( P_2 = 760 \ \text{torr} \)
#### Calculation:
1. Convert the heat of vaporization into J/mol: \( 31.3 \ \text{kJ/mol} = 31300 \ \text{J/mol](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F48deb947-3d41-43de-ac43-bef03e01cc59%2F681c5dd8-0eba-496f-aebb-d2fe961f9f36%2Fzcqqf1.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
The heat of vaporization, \( \Delta H^\circ_{\text{vap}} \), of acetone is 31.3 kJ/mol. If the vapor pressure of liquid acetone is 100 torr at 7.7°C, what is the boiling point of acetone? Use the Clausius-Clapeyron equation to find the boiling point.
\[ \text{(1.00 atm = 760 torr)} \]
### Instructions:
- Show your work
- See p1 for equation
### Detailed Explanation:
This problem involves calculating the boiling point of acetone using the Clausius-Clapeyron equation. Here are the steps to solve the problem:
#### Given Data:
- Heat of vaporization (\( \Delta H^\circ_{\text{vap}} \)): 31.3 kJ/mol
- Vapor pressure at 7.7°C ( \( P_1 \)): 100 torr
- Temperature \( T_1 \): 7.7°C (convert to Kelvin)
- Standard atmospheric pressure (\( P_2 \)): 760 torr
#### Clausius-Clapeyron Equation:
The Clausius-Clapeyron equation is:
\[ \ln \left( \frac{P_2}{P_1} \right) = \frac{\Delta H^\circ_{\text{vap}}}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
Where:
- \( P_1 \) and \( P_2 \) are the vapor pressures at temperatures \( T_1 \) and \( T_2 \) respectively
- \( R \) is the universal gas constant (8.314 J/mol·K)
- \( T_1 \) and \( T_2 \) are the temperatures in Kelvin
#### Conversion of Temperatures
- \( T_1 = 7.7°C = 7.7 + 273.15 = 280.85 \ \text{K} \)
- \( P_1 = 100 \ \text{torr} \)
- \( P_2 = 760 \ \text{torr} \)
#### Calculation:
1. Convert the heat of vaporization into J/mol: \( 31.3 \ \text{kJ/mol} = 31300 \ \text{J/mol
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