DRAFT: This module has unpublished changes.

Data Analysis

 

Figure 1: Pressure vs Time

 

 

* on graph 1: ethanol is the green line and acetone is the blue line.

** on graph 2: ethanol is the bottom line and acetone is the top line.

 

Table 1: Heat of Vaporization Determined for Ethanol and Acetone

Substance Slope (K) Heat of Vaporization (J/mol)
Ethanol -3000  24,942
Acetone -2440  20,286

Calculations

 

Conclusion

      The purpose of this lab was to determine how intermolecular interactions affect physical properties. This was done by heating acetone and ethanol to certain temperatures and then obtaining the temperature as each cooled down to a specific temperature. The slope of the line obtained from the graphs was then used in the Clausius-Clapeyron equation to determine the heat of vaporization. The heat of vaporization for ethanol was higher than acetone. This is accurate because ethanol has a higher heat of vaporization due to the type of intermolecular bond that it has. Ethanol has a hydrogen bond and the strength of that bond is very strong so more energy is needed to break that bond. Acetone has a London dispersion force which is much weaker and requires less energy to break apart that bond.  The ethanol molecule is also much lighter and has a higher boiling point than acetone because of the hydrogen bonds between the ethanol molecules. Some sources of error in the experiment could come from boiling the ethanol and acetone too much so there is no more liquid in the container while obtaining the reading and also not having the rubber stopper inserted tightly on the Erlenmeyer flask allowing some gas molecules to escape.

 

DRAFT: This module has unpublished changes.