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Determine the normal boiling point of ethanol ([tex]CH_3CH_2OH[/tex]), given that its enthalpy of vaporization is [tex]38.6 \, \text{kJ/mol}[/tex] and its entropy of vaporization is [tex]110 \, \text{J/mol·K}[/tex].

Answer :

Final answer:

The normal boiling point of ethanol is determined by setting the Gibbs free energy change at the boiling point to zero and using the enthalpy and entropy of vaporization to calculate the temperature, which is found to be approximately 77.85 °C.

Explanation:

To determine the normal boiling point of ethanol (CH3CH2OH) using its given enthalpy of vaporization (38.6 kJ/mol) and its entropy of vaporization (110 J/mol·K), we can apply the Clausius-Clapeyron equation and the concept of Gibbs free energy change (ΔG) at the boiling point.

At the normal boiling point, ΔG equals zero, thus:

ΔG = ΔH - TΔS = 0

Where ΔH is the enthalpy of vaporization, and ΔS is the entropy of vaporization. By rearranging the equation to solve for the boiling temperature (T), we get:

T = ΔH / ΔS

Converting the units of entropy from J to kJ to match the units of enthalpy:

110 J/mol·K = 0.110 kJ/mol·K

Now we can plug in the values:

T = 38.6 kJ/mol / 0.110 kJ/mol·K = 351 K

To convert from Kelvin to degrees Celsius, we subtract 273.15:

351 K - 273.15 = 77.85 °C

Therefore, the normal boiling point of ethanol is approximately 77.85 °C.

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