Vaporization latent heat vs boiling point

  • Thread starter yecko
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  • #1
yecko
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Homework Statement:
If the vaporization latent heat of Ethlyl alcohol under the standard conditions is 38.56 kJ/mol, estimate the boiling point of Ethlyl alcohol (in °C) according to the A26.
Relevant Equations:
vaporization latent heat = Hf-Hg = Hfg
delta G = delta H - T delta S
1621173385541.png


answer is 78oC


delta G = delta H - T delta S
-235,310- (-277,690) = -38.56e3- T (282.59-160.70)
T = -664K

I am not sure if my concept is correct. May anyone help a little bit on that please?
thank you
 
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  • #2
haruspex
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Seems to me you used the g and h columns in reverse. Your delta G is a difference of two h values.
 
  • #3
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The heat of formation value for the ethanol gas in the table is for a hypothetical ideal gas state of ethanol at 1 bar and 25 C. This is not an actual thermodynamic equilibrium state of ethanol, since, at 25 C and 1 bar, ethanol is actually a liquid. When they say that the heat of vaporization under standard conditions is 38.56 kJ/mol, what they mean is that the heat of vaporization at the boiling temperature and 1 bar is 38.56 kJ/mol. The hypothetical heat of vaporization from the liquid to the vapor at 25 C and 1 bar is, according to the table, 42.37 kJ/mol. You need to use Hess's law to express the heat of vaporization at the boiling temperature T and 1 bar in terms of the "heat of vaporization" at 25 C and 1 bar. This involves using the heat capacities of the liquid and vapor.
 

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