Question

A certain liquid has a vapor pressure of 92.0 Torr at 23.0 °C and 221.0 Torr at 45.0 °C. Calculate the value of AH vap for this liquid. AHMap = 1 41.69 kJ/mol Calculate the normal boiling point of this liquid. boiling point: 64.85

Answer 1

1)

T1 = 23 oC

=(23 + 273)K

= 296 K

T2 = 45 oC

=(45 + 273)K

= 318 K

P1 = 92 Torr

P2 = 221 Torr

use:

ln(P2/P1) = (ΔH/R)*(1/T1 - 1/T2)

ln(221/92) = ( ΔH/8.314)*(1/296.0 - 1/318.0)

0.8764 = (ΔH/8.314)*(2.337*10^-4)

ΔH = 31174 J/mol

ΔH = 31.2 KJ/mol

Answer: 31.2 KJ/mol

2)

T1 = 23 oC

=(23 + 273)K

= 296 K

P1 = 92 Torr

P2 = 760 Torr [At normal boiling point, pressure is 760 torr]

ΔH = 31.174 KJ/mol

= 31174 J/mol

use:

ln(P2/P1) = (ΔH/R)*(1/T1 - 1/T2)

ln(7.6*10^2/92) = (31174.0/8.314)*(1/296.0 - 1/T2)

2.1115 = 3749.579*(1/296.0 - 1/T2)

T2 = 355 K

= (355-273) oC

= 82 oC

Answer: 82 oC