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Answer :
Final answer:
To calculate the normal boiling point of ethanol, the enthalpy of vaporization is divided by the entropy of vaporization, resulting in a temperature of 350.91 K or 77.76 °C.
Explanation:
The question is asking to calculate the normal boiling point of ethanol, given its enthalpy of vaporization and its entropy of vaporization. According to the Clausius-Clapeyron relation, the normal boiling point can be found where the vapor pressure equals the external atmospheric pressure (usually 1 atm or 101.3 kPa). This relation links the pressure, enthalpy of vaporization, entropy of vaporization, and temperature.
To find the normal boiling point, we can use the equation derived from the Clausius-Clapeyron relation:
ΔG = ΔH - TΔS = 0
At the boiling point, the Gibbs free energy of vaporization (ΔG) is zero. Therefore, we can set the enthalpy of vaporization equal to the temperature times the entropy of vaporization, then solve for the temperature (T).
ΔH = TΔS
T = ΔH / ΔS
T = (38.6 kJ/mol) / (110 J/(K⋅mol))
T = 38600 J/mol / 110 J/(K⋅mol)
T = 350.91 K
Converting kelvins to degrees Celsius by subtracting 273.15 gives us:
T(°C) = 350.91 K - 273.15
T(°C) = 77.76 °C
Therefore, the normal boiling point of ethanol is 77.76 °C.
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