Question

The ΔH vap of a certain compound is 45.92 kJ/mol and its ΔS is 64.62 J/mol*K. What is the boiling point of this compound?

Any help?

Answer 1

To find the boiling point of this compound, we can do= \[B.P. =\]\[\frac{ \Delta Hvap }{ \Delta S }\] Make sure you multiply 64.62 J/mol*K by 1000 to get kJ/mok*K so that all of the units work out (Since the deltaH is in kJ/mol). If you do a unit analysis, you can see that this will work out: \[B.P ( \in Kelvin) = (\frac{ \frac{ kJ }{ mol } }{ \frac{ kJ }{ mol*K } })=(\frac{ kJ }{ mol})(\frac{ mol*K }{ kJ })=K(Kelvin) \] All of the units except K, or temperature, cancel out, leaving you with the boiling point.

Answer 2

thanks for the help!