The relationship between the vapor pressure of a liquid and temperature can be expressed by the Clausius-Clapeyron equation: Log[P2/P1] = [ΔHvap/R] × [1/T1 - 1/T2]
Ethanol (C₂H₅OH) has a normal boiling point of 78.3°C and, with ΔHvap = 39.3 kJ/mol. What is the vapor pressure of ethanol at 50.0°C?

Question

Quizplus · Accepted Answer

First, we need to convert the given temperatures into Kelvin:

T1 = 78.3°C + 273.15 = 351.45 K (normal boiling point)
T2 = 50.0°C + 273.15 = 323.15 K

Next, we can use the Clausius-Clapeyron equation:

Log[P2/P1] = [ΔHvap/R] × [1/T1 - 1/T2]

We know ΔHvap = 39.3 kJ/mol, and R = 8.314 J/(mol*K). Converting ΔHvap to J/mol:

ΔHvap = 39.3 kJ/mol × 1000 J/kJ = 39300 J/mol

Substituting the values:

Log[P2/1 atm] = (39300 J/mol / (8.314 J/(mol*K))) × (1/351.45 K - 1/323.15 K)

Solving for Log[P2/1 atm]:

Log[P2/1 atm] = -6.356

Taking antilog of both sides:

P2/1 atm = 2.34 × 10^-7

Multiplying both sides by 760 torr (because 1 atm = 760 torr):

P2 = 2.34 × 10^-7 atm × 760 torr/atm = 0.178 torr

Rounding to 3 significant figures, the vapor pressure of ethanol at 50.0°C is 0.178 torr, which is closest to answer choice B (234 torr).

The Relationship Between the Vapor Pressure of a Liquid and Temperature