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3. Find the boiling point of ethanol based on the change in enthalpy of vaporization and change of entropy of vaporization.

\[
\Delta H_v = 39.3 \, \text{kJ/mole}
\]
\[
\Delta S_v = 112 \, \text{J/K mole}
\]

Answer :

The boiling point of ethanol can be determined based on the change in enthalpy of vaporization (ΔHv) and the change in entropy of vaporization (ΔSv). the boiling point of ethanol, based on the given values of ΔHvap and ΔSvap, is 351 Kelvin.


To find the boiling point, we need to use the equation:
ΔGvap = ΔHvap - TΔSvap
Where:
ΔGvap is the change in Gibbs free energy of vaporization
ΔHvap is the change in enthalpy of vaporization
T is the temperature (in Kelvin)
ΔSvap is the change in entropy of vaporization


At the boiling point, the change in Gibbs free energy of vaporization is zero, so we can set ΔGvap to zero in the equation:
0 = ΔHvap - TbpΔSvap
Rearranging the equation, we can solve for the boiling point (Tbp):
Tbp = ΔHvap / ΔSvap
Given that ΔHvap = 39.3 kJ/mol and ΔSvap = 112 J/K mol, we need to convert the units to be consistent:
ΔHvap = 39.3 * 10^3 J/mol
ΔSvap = 112 J/K mol
Now, we can substitute these values into the equation to find the boiling point:
Tbp = (39.3 * 10^3 J/mol) / (112 J/K mol)
Tbp = 351 K

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