ΔH_vap for water is 40.7 kJ/mol. Calculate the boiling point of water at 0.500 atm. A) 50.0°C B) 81.3°C C) 100.0°C D) 98.6°C E) none of these

Question

Quizplus · Accepted Answer

To solve this problem, we need to use the Clausius-Clapeyron equation:
ln(P2/P1) = -(ΔHvap/R)((1/T2)-(1/T1))
where
P1 = 1 atm
P2 = 0.500 atm
ΔHvap = 40.7 kJ/mol
R = 8.314 J/mol*K
T1 = 373.15 K (the boiling point of water at 1 atm)
T2 = ? (the boiling point of water at 0.500 atm)
We can solve for T2 by rearranging the equation:
(1/T2) = -(ΔHvap/R)(ln(P2/P1)) + (1/T1)
(1/T2) = -(40.7 kJ/mol / 8.314 J/mol*K)(ln(0.500/1)) + (1/373.15 K)
T2 = 352.8 K = 79.7°C
Therefore, the boiling point of water at 0.500 atm is 79.7°C, which is closest to choice B (81.3°C).

ΔHvap for Water Is 40

ΔH_vap for Water Is 40