Final answer:
The theoretical value of heat vaporization of ethanol can be estimated using the Clausius-Clapeyron equation. Plugging in the given temperatures and vapor pressures, we find that the enthalpy of vaporization for ethanol is approximately 40.79 kJ/mol.
Step-by-step explanation:
The theoretical value of heat vaporization of ethanol can be estimated using the Clausius-Clapeyron equation. The equation is given by: ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1) where P1 and P2 are the vapor pressures at temperatures T1 and T2 respectively, ΔHvap is the enthalpy of vaporization, and R is the ideal gas constant. We are given the vapor pressures of ethanol at 20.0 °C (5.95 kPa) and 63.5 °C (53.3 kPa). Converting these temperatures to Kelvin (293.15 K and 336.65 K), we can substitute the values into the equation solve for ΔHvap.
Using the given temperatures and vapor pressures:
- T1 = 293.15 K, P1 = 5.95 kPa
- T2 = 336.65 K, P2 = 53.3 kPa
Plugging these values into the Clausius-Clapeyron equation:
ln(53.3 kPa/5.95 kPa) = -ΔHvap/(8.314 J/mol·K) * (1/336.65 K - 1/293.15 K)
Solving for ΔHvap, we find that the enthalpy of vaporization for ethanol is approximately 40.79 kJ/mol.