Answer :
Final answer:
Using the Clausius-Clapeyron equation, the vapor pressure of a liquid with a known normal boiling point and enthalpy of vaporization can be calculated at a different temperature, resulting in a vapor pressure of 12.9 kPa at 50.0°C, hence the correct option is d) 12.9 kPa.
Explanation:
To calculate the vapor pressure of a liquid at a certain temperature, the Clausius-Clapeyron equation is used, which describes the change in vapor pressure with temperature.
Given the normal boiling point of a liquid (351.45 K) and its enthalpy of vaporization (39.3 kJ/mol), we can estimate its vapor pressure at another temperature (50.0°C).
The Clausius-Clapeyron equation is integrated to yield the equation ln(P2/P1) = ln p 2 p 1 = − Δ vap H m R ( 1 T 2 − 1 T 1 ), where P1 and T1 are the vapor pressure and temperature at the normal boiling point, P2 and T2 are the vapor pressure and temperature at the condition of interest.
Δvap is the enthalpy of vaporization, and R is the ideal gas constant.
By inputting the given data and solving for P2, we find that the vapor pressure at 50.0°C is 12.9 kPa, hence the answer is 12.9 kPa.