Heat fusion, vaporization and entropy

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Discussion Overview

The discussion revolves around calculating the entropy changes associated with the phase transitions of benzene, specifically from solid to liquid and from liquid to vapor. Participants explore the implications of these changes and the differences in heat of fusion and heat of vaporization.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Some participants suggest calculating the entropy changes using the provided heat values and the relationship between enthalpy and entropy.
  • One participant expresses uncertainty about how to approach the calculations and asks for corrections on their understanding.
  • Another participant mentions the potential use of Hess's law and the need for standard entropy tables for benzene in different states.
  • It is noted that the entropy changes can be calculated directly from the given information without needing additional tables.
  • A participant states that for phase transitions at constant temperature and pressure, the Gibbs free energy change (\Delta G) is zero.

Areas of Agreement / Disagreement

Participants generally agree that the necessary information for calculations is provided, but there is no consensus on the best approach to solve the problem or the expected relationship between the entropy changes for the two phase transitions.

Contextual Notes

Some participants express uncertainty about the calculations and the relationships between the thermodynamic quantities involved, indicating potential limitations in their understanding of the concepts.

Who May Find This Useful

Students studying thermodynamics, particularly those interested in phase transitions and entropy calculations.

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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