Heat fusion, vaporization and entropy

AI Thread Summary
The discussion focuses on calculating the entropy changes for benzene transitioning from solid to liquid and from liquid to vapor, using its molar heat fusion (10.9 kJ/mol) and vaporization (31.0 kJ/mol) values. It highlights that the change in entropy (ΔS) for these two phase changes is expected to differ due to the distinct energy requirements for melting and vaporization. The physical significance of this difference is that it reflects the change in molecular characteristics as substances transition between solid, liquid, and gas states. Additionally, the heat of vaporization is typically greater than the heat of fusion because vaporization requires more energy to overcome intermolecular forces. Understanding these concepts is crucial for thermodynamic calculations involving phase transitions.
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Homework Statement


The molar heat fusion and vaporization of benzene are 10.9kJ/mol and 31.0 kJ/mol, respectively. The melting temperature of benzene is 5.5C and it boils at 80.1C.
a) calculate the entropy changes for solid to liquid, and liquid to vapor of benzene
b) would you expect the change in S FOR THESE TWO CHANGES TO BE ABOUT THE SAME?
c) comment on the physical significance of the difference in these two values.
d) why are the values for heat of vaporization usually so much greater than the heats of fusion?


Homework Equations


OK I DON'T REALLY KNOW HOW TO a) OR EQUATION TO DO IT: delta G system(free energy change)= delta H ( change in enthalpy) - T(temp.) System (represent the change in entropy
- q= c x m x deltaT

The Attempt at a Solution


b) I would expect the change in S for these two changes to not be the same because if one change so does the other right?
c)the physical significance of the difference on these two values they change their characteristics from solid to liquid and liquid to vapor (solid to liquid to gas)
d) the values for the heat vaporization usually is greater than the heat of fusion because heat cause water to evaporate in the air and if combined with another it may take longer

PLEASE CORRECT ME ON MY MISTAKES!
 
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yuuri14 said:

Homework Statement


The molar heat fusion and vaporization of benzene are 10.9kJ/mol and 31.0 kJ/mol, respectively. The melting temperature of benzene is 5.5C and it boils at 80.1C.
a) calculate the entropy changes for solid to liquid, and liquid to vapor of benzene
b) would you expect the change in S FOR THESE TWO CHANGES TO BE ABOUT THE SAME?
c) comment on the physical significance of the difference in these two values.
d) why are the values for heat of vaporization usually so much greater than the heats of fusion?


Homework Equations


OK I DON'T REALLY KNOW HOW TO a) OR EQUATION TO DO IT: delta G system(free energy change)= delta H ( change in enthalpy) - T(temp.) System (represent the change in entropy
- q= c x m x deltaT

I believe that you could use Hess's law to calculate it. You will need to find a table of standard entropies for benzene as a solid, liquid and vapor to do it.
 
The entropy changes can be calculated with the given information without checking tables. You need the relationship between \Delta S, \Delta H, and \Delta G, and you need to know the value of \Delta G for any phase transition at constant temperature and pressure.
 
If the two states are in equilibrium then \Delta G=0.
The problem gives all of the needed information.
 
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