Final answer:
To estimate the heat of vaporization of ethanol, the Clausius-Clapeyron equation is used with given vapor pressures at different temperatures to solve for the enthalpy of vaporization, resulting in a lower value than for water due to weaker hydrogen bonding.
Step-by-step explanation:
The theoretical value of the heat of vaporization of ethanol can be estimated using the Clausius-Clapeyron equation, which relates the change in vapor pressure with temperature to the enthalpy of vaporization (ΔHvap). The given vapor pressures at 20.0 °C (5.95 kPa) and at 63.5 °C (53.3 kPa) allow us to calculate ΔHvap for ethanol by establishing a linear relationship between the natural logarithm of vapor pressure and the inverse of temperature.
To determine the enthalpy change, we can use the equation ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1), where P1 and P2 are the vapor pressures at temperatures T1 and T2 respectively, and R is the ideal gas constant (8.314 J/(mol*K)). By rearranging this equation, we can solve for the enthalpy of vaporization of ethanol. With ethanol's weaker hydrogen bonding compared to water, its heat of vaporization is expected to be less than water's (~2,250 J/g). The energy required for the phase change, however, still offsets the increase in entropy to maintain equilibrium at the boiling point.