chemistry problem help please!?

Question

a compound has a vapor pressure of 2.23 torr at 45 Celsius. what is the normal boiling point of the compound. delta H vaporization is 44.09 kj/mole.

R =8.314

answer is: 216 Celsius

can someone work out the problem step by step so I understand?

Answer 1

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Clausius Clapeyron equation

ln(p2/p1) = heat of vaporization/R(1/T1 – 1/T2)

Normal boiling point is when vapor pressure = 760 torr

R = 8.315 Joules/moleK

T1= 45 Celsius = 273 + 45 = 318 Kelvin

Change heat of vaporization to Joules = 44.09 kJ x 1000 Joules/kJ = 44,090

ln(760/2.23 torr) = 44,090 kJ/8.3145 ( 1/318 – 1/T2)

5.83 = 5302(0.00314– 1/T2}

5.83 = 16.67 – 5302/T2

-10.84 = -5302/T2

T2 = -5302/10.84 = 489 Kelvin

Celsius = 489 -273 = 216 C