Heat fusion, vaporization and entropy

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SUMMARY

The discussion focuses on calculating the entropy changes for benzene during phase transitions, specifically from solid to liquid and liquid to vapor. The molar heat fusion of benzene is 10.9 kJ/mol, and the molar heat vaporization is 31.0 kJ/mol. The melting point is 5.5°C, and the boiling point is 80.1°C. The participants conclude that the entropy changes for these transitions differ significantly due to the greater energy required for vaporization compared to fusion.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically entropy and enthalpy.
  • Familiarity with phase transitions and their associated heat values.
  • Knowledge of Hess's Law and its application in thermodynamics.
  • Ability to interpret standard entropy tables for substances.
NEXT STEPS
  • Calculate entropy changes using the formula ΔS = ΔH/T for phase transitions.
  • Research standard entropies for benzene in solid, liquid, and vapor states.
  • Explore the relationship between ΔG, ΔH, and ΔS in thermodynamic processes.
  • Examine the implications of phase transition energy requirements in real-world applications.
USEFUL FOR

Chemistry students, thermodynamics enthusiasts, and professionals involved in chemical engineering or materials science will benefit from this discussion.

yuuri14
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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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