Caculating normal boiling point using an equation of a line

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SUMMARY

The discussion centers on calculating the normal boiling point and heat of vaporization of a volatile liquid using the best fit line equation y = -5307.5x + 19.55. The heat of vaporization was determined to be 44.127 kJ/mol using the formula Heat of vaporization = -(slope x R), where R is 8.314 J/mol*K. The participant attempted to calculate the boiling point using the Clausius-Clapeyron equation, arriving at 99°C, but later realized an error in unit conversion from kJ to joules, which affected the accuracy of their results.

PREREQUISITES
  • Understanding of the Clausius-Clapeyron equation
  • Familiarity with the concept of heat of vaporization
  • Knowledge of unit conversions between kJ and J
  • Basic skills in linear regression and interpreting best fit lines
NEXT STEPS
  • Review the Clausius-Clapeyron equation and its applications in phase transitions
  • Study linear regression techniques for deriving equations of best fit
  • Learn about unit conversions, specifically between kJ and J
  • Investigate the properties of common volatile liquids and their boiling points
USEFUL FOR

Chemistry students, researchers in thermodynamics, and anyone involved in the study of phase changes and vapor pressure calculations will benefit from this discussion.

ahhppull
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Homework Statement



The equation for the best fit line was y = - 5307.5x + 19.55. Using the equation of the best fit line, calculate the heat of vaporization and the normal boiling point (in C) of this volatile liquid. (Hint: What is the value of Pvap when liquid boils at its normal boiling point?

Homework Equations



Heat of vaporization = -(slope x R) R= 8.314 j/mol*k

The Attempt at a Solution




The heat of vaporization is 44.127 kJ/mol, but I'm unsure of how to calculate the boiling point.

I tried using the Clausius-Clapeyron equation to find the boiling point. I got 99 C, but I'm unsure if its right. If it helps, the next problem following this problem lists several compounds and I have to find out what compound it is using the Normal Boiling point and heat of vaporization

Ethanol 78.3 C 42.4kJ/mol
iso-Propanol 82.3 C 45.4 kJ/mol
Cyclohexane 80.7 C 33.1 kJ/mol

I know that its not cyclohexane because its heat of vaporization differs from mines.
 
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Actually, the clausius equation doesn't work because I forgot to convert kJ to joules while doing the equation.
 

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