Question

For a certain liquid, the vapor pressure is 92.0 Torr at 23.0 °C and 230.0 Torr at 45.0 °C. What is the value of Δ∘vap for this liquid?

a) 131 kJ/mol
b) 95 kJ/mol
c) 36 kJ/mol
d) 22 kJ/mol

Answer 1

To find the enthalpy of vaporization (ΔHvap) for a liquid, the Clausius-Clapeyron equation is used with the provided vapor pressures and temperatures, giving an approximation of ΔHvap as 36 kJ/mol.

Step-by-step explanation:

The question involves finding the value of ΔHvap for a liquid, given two vapor pressure measurements at different temperatures. This can be done using the Clausius-Clapeyron equation, which is a way to express how the vapor pressure of a substance changes with temperature, in accordance with its enthalpy of vaporization (ΔHvap). The equation is ln(P2/P1) = (ΔHvap/R) (1/T1 - 1/T2), where P1 and P2 are vapor pressures at temperatures T1 and T2 respectively, ΔHvap is the enthalpy change of vaporization, and R is the ideal gas constant (8.314 J/mol·K).

Firstly, we must convert the temperatures given in °C to Kelvins (K): T1 = 23.0 + 273.15 = 296.15 K and T2 = 45.0 + 273.15 = 318.15 K. Substituting the given vapor pressures and the temperatures into the Clausius-Clapeyron equation allows us to solve for ΔHvap, which comes out to approximately 36 kJ/mol, when using the given value of R.