Question

A liquid has an enthalpy of vaporization of 30.8 kJ/mol. At 273
K it has a vapor pressure of 102 mmHg. What is the normal boiling
point of this liquid? (*R* = 8.31 J/(K· mol))

Answer #1

**Solution :-**

Using the Clausius Clapeyron equation we can calculate the normal boiling point

Normal boling point means when the vapor pressure equals to the atmospheric pressure that is 760 mmHg

P1=102 mmHg , P2 = 760 mmHg , T1 = 273 K , T2 = ?

Delta H vap = 30.8 kJ/ mol *1000 J / 1 kJ = 30800 J/mol

ln(P2/P1) = DeltaH vap / R [(1/T1)- (1/T2)]

lets put the values in the formula

ln(760/102) = 30800 J per mol / 8.31 J per mol K [(1/T2)- (1/273)]

2.008 = 3706.4 [(1/T2)- 0.003663]

2.008/3706.4 = (1/T2) - 0.003663

0.0005418 = (1/T2 )- 0.003663

0.003663 – 0.0005418 = 1/T2

0.0031212 = 1/T2

1/0.0031212 = T2

320.4 K = T2

So the normal boiling point is 320.4 K

Or we can also write it as 320.4 K -273 = 47.4 ^{o}C

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