If the enthalpy of vaporization of chloromethane, CH₃Cl, is 21.5 kJ/mol at the normal boiling point, 249 K, calculate ΔS°vap. A) 86.3 J mol^-1 K^-1 B) 5.35 J mol^-1 K^-1 C) 11.6 J mol^-1 K^-1 D) 896 J mol^-1 K^-1

Question

Quizplus · Accepted Answer

We can use the Clausius-Clapeyron equation:

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 gas constant. At the boiling point, the vapor pressure is equal to the atmospheric pressure, so we can use the normal boiling point temperature and pressure (1 atm) as our reference values:

ln(1/1) = - ΔHvap/R * (1/249 K - 1/273 K)

Simplifying and solving for ΔS°vap:

ΔS°vap = -R * ln(1/1) / (-ΔHvap/R * (1/249 K - 1/273 K))

ΔS°vap = ΔHvap / T

ΔS°vap = 21.5 kJ/mol / 249 K = 86.3 J/(mol*K)

If the Enthalpy of Vaporization of Chloromethane, CH3Cl, Is 21

If the Enthalpy of Vaporization of Chloromethane, CH₃Cl, Is 21