# Heat fusion, vaporization and entropy

1. Sep 14, 2009

### yuuri14

1. The problem statement, all variables and given/known data
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?

2. Relevant 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

3. 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!

2. Sep 18, 2009

### chemisttree

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.

3. Sep 18, 2009

### Mapes

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.

4. Sep 18, 2009

### PhaseShifter

If the two states are in equilibrium then $$\Delta G=0$$.
The problem gives all of the needed information.