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Clausius-Clapeyron Equation

Simply put, this equation is used to determine Heat of Vaporization of a liquid given 2 vapor pressures and 2 temperatures.

Here is the equation

which useful for the C.C. Lab

 

for C.C. problems we use    

There are a few things you must watch out for. 

DH is in joules/mol, you will most likely get it in kJ/mol so convert it.

R= 8.314 J/mol K  (notice the J)

T must be in Kelvin

The 2 vapor pressures must have the same unit.

Most problems give you a "NORMAL" boiling point. You have to be clever enough to get the hint.

normal boiling points occur at normal pressures....760 torr (the question hide the pressure...very sneaky)

This is the piece most students miss. Also make sure you change kJ/mol to J/mol.

Sample Problems-

1.The heat of vaporization of water is 44.01kJ.mol and the normal boiling point of water is 100.oC. Calculate the atmospheric pressure in Denver where the boiling point is 97.10oC.

Covert 44.01kJ/mol to 44,010J/mol

T1= 100 +273=373K   and  since it is normal boiling point  P1 is 760 torr

T2=97.10 + 273=370.10  and we are solving for P2

plug and chug

ln(

   P2    

)= 44,010J/mol (

  1   

-

  1  

)
760. torr 8.314 J/mol K 373K 370.10K

ln(

   P2    

)=-0.111
760. torr

e up

   P2    

=0.895
760 torr

   P2    

=680. torr

 

2. The Vapor Pressure of Bromine (Br2) is 100.0 torr at 9.30oC and the enthalpy of vaporization is 30.91kJ/mol. Calculate the normal boiling point of Br2.

Convert 30.91kj/mol to 30,910J/mol

T1= 9.30+273= 282.30  and P1 is 100.0torr

T2 is the normal boiling point therefore P2 is normal pressure, so 760. torr.

Plug and chug

 

ln(

 760.torr    

)= 30910 J/mol (

  1   

-

  1  

)

100. torr

8.314 J/mol K 280.30K

T1

2.03=

3718

(

  1   

-

  1  

)
280.30K

T1

0.000546=

( 0.003567 -

 1  

)

T1

-0.00302=

- 1  

T1

T1=

- 1  

-0.00302

T1=331K  or  58oC

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