Answer :
Final answer:
To estimate the enthalpy of vaporization for ethanol, the Clausius-Clapeyron equation can be used with the provided vapor pressures at two different temperatures. Upon solving, this yields the estimated enthalpy of vaporization in kJ/mol.
Explanation:
To estimate the enthalpy of vaporization for ethanol using the given vapor pressures at two different temperatures, we can apply the Clausius-Clapeyron equation, which is a way to quantify how the vapor pressure of a liquid changes with temperature:
\[\ln\left(\frac{P2}{P1}\right) = \frac{-\Delta H_{vap}}{R}\left(\frac{1}{T2} - \frac{1}{T1}\right)\]
Where:
- P1 and P2 are the vapor pressures at temperatures T1 and T2 (in K).
- \(\Delta H_{vap}\) is the enthalpy of vaporization (in J/mol).
- R is the ideal gas constant (8.314 J/(mol·K)).
Given that at 20.0 °C (293.15 K) the vapor pressure is 5.95 kPa and at 63.5 °C (336.65 K) the vapor pressure is 53.3 kPa, we can rearrange the equation to solve for \(\Delta H_{vap}\):
\[\Delta H_{vap} = \frac{-R\ln\left(\frac{P2}{P1}\right)}{(1/T2) - (1/T1)}\]
By plugging in the values:
\[\Delta H_{vap} = \frac{-8.314 J/(mol\cdot K)\ln\left(\frac{53.3}{5.95}\right)}{(1/336.65) - (1/293.15)}\]
After calculating the above expression, you will arrive at the estimated enthalpy of vaporization for ethanol, expressed in kJ/mol.