Question

What is the enthalpy of vaporization of a compound that has a vapor pressure of 151 mmHg at 247 K and 1.52 mmHg * ((at 187 K? (R = 8.31 J/(K: mol)

O 636 kJ/mol
O 2 kJ/mol
O 4 kJ/mol
O 29 kJ/mol
O. 290kJ/mol

Answer 1

Using the Clausius-Clapeyron equation, convert the given vapor pressures to atm, substitute the values into the equation along with the provided temperatures in Kelvin, solve for the enthalpy of vaporization ΔHvap, and express it in kJ/mol.

Step-by-step explanation:

To calculate the enthalpy of vaporization (ΔHvap) of a compound, we will make use of the Clausius-Clapeyron equation, which relates ΔHvap to the change in vapor pressure with temperature. The equation is ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1), where P1 and P2 are the vapor pressures at temperatures T1 and T2 respectively, and R is the ideal gas constant (8.314 J/mol·K).

First, we convert the vapor pressures from mmHg to atmospheres by dividing by 760 mmHg/atm: P1 = 151 mmHg / 760 mmHg/atm = 0.1987 atm, P2 = 1.52 mmHg / 760 mmHg/atm = 0.002 atm.

Substitute the given values into the Clausius-Clapeyron equation:

ln(0.1987/0.002) = -ΔHvap / (8.314 J/mol·K) * (1/247 K - 1/187 K), solve for ΔHvap to obtain the heat of vaporization in joules per mole, and then convert this value to kilojoules per mole by dividing by 1000 J/kJ.