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posted by  KAthy07 on 11/18/2008 6:53:37 PM  |  status: Live  

Energy

Course Textbook Chapter Problem
General Chemistry N/A N/A N/A
Question Details:
Chloroform has \Delta H_{\rm vaporization} = 29.2{\rm kJ/mol} and boils at 61.2 ^\circ {\rm C}.
What is the value of \Delta S_{\rm vaporization} for chloroform?
  \Delta S_{\rm vaporization} =
 {\rm J/(mol \cdot K)}

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posted by Werner on 11/19/2008 8:01:47 AM  |  status: Live
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KAthy07's comment:
"Thanks"
Response Details:
  Given data
  \Delta H_{\rm vaporization} = 29.2{\rm kJ/mol} 
  Boiling point, Tb = 61.2 0C 
                            = 334.2 K
  \Delta S_{\rm vaporization} = \Delta H_{\rm vaporization} / Tb
                      = ( 29.2{\rm kJ/mol} ) / 334.2 K
                      = 0.08737 kJ /mol.K
                      = 87.37 J/mol.K
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