Respuesta :
Answer:
28.7664 kJ /mol
Explanation:
The expression for Clausius-Clapeyron Equation is shown below as:
[tex]\ln P = \dfrac{-\Delta{H_{vap}}}{RT} + c [/tex]
Where,
P is the vapor pressure
ΔHvap is the Enthalpy of Vaporization
R is the gas constant (8.314×10⁻³ kJ /mol K)
c is the constant.
The graph of ln P and 1/T gives a slope of - ΔHvap/ R and intercept of c.
Given :
Slope = -3.46×10³ K
So,
- ΔHvap/ R = -3.46×10³ K
ΔHvap = 3.46×10³ K × 8.314×10⁻³ kJ /mol K = 28.7664 kJ /mol
Based on the data provided, the enthalpy of vaporization of the substance is 28.7664 kJ /mol.
What enthalpy of vaporization?
The enthalpy of vaporization is the heat required to convert a liquid at its boiling point to vapor.
The Clausius-Clapeyron equation relates the enthalpy of vaporization and vapor pressure.
The expression for Clausius-Clapeyron Equation is shown below as:
- lnP = ΔHvap/RT + c
Where,
- P is the vapor pressure
- ΔHvap is the Enthalpy of Vaporization
- R is molar gas constant (8.314×10⁻³ kJ /mol K)
- c is the constant.
Slope of the graph of ln P and 1/T = - ΔHvap/ R
Given that the Slope = -3.46×10³ K
Then,
- ΔHvap/ R = -3.46×10³ K
ΔHvap = 3.46×10³ K × 8.314×10⁻³ kJ /mol K
ΔHvap = 28.7664 kJ /mol
Therefore, the enthalpy of vaporization of the substance.is 28.7664 kJ /mol.
Learn more about enthalpy of vaporization at: https://brainly.com/question/21504082
