Question

1. The following are the vapor pressures of some relatively common chemicals measured at 20 °C:

Benzene, C6H6	80 torr
Acetic acid, HC2H3O2	11.7 torr
Acetone, C3H6O	184.8 torr
Water, H2O	17.5 torr
Diethyl ether, C4H10O	442.2 torr

Arrange these substances in order of increasing boiling point.

2. At 0.00 °C, hexane, C₆H₁₄, has a vapor pressure of 45.37 mm Hg. Its ΔH_vap is 30.1 kJ mol^-1. What is the vapor pressure of hexane at 12.9 °C?
The vapor pressure of hexane at 12.9 °C is ? mm Hg.

3. The boiling point of 2,3,4-trimethylpentane, C8H18, at 760 torr is 113.47°C, and its molar enthalpy of vaporization is 37,600 J mol-1. What is its vapor pressure at 105.5°C?

Answer 1

1) the higher the vapour pressure faster the solvent evaporates and therefore lower is the boiling point

therefore increasing order of boiling point = diethyl ethet < acetone <benzene < water < acetic acid

2) ln (P2/P1) = (ΔHvap / R)(1/T1 - 1/T2) . . .let T1 be the lower temperature (0.00 C, 273.15 K) so P1 = 45.37 mmHg. T2 = 14.2 C = 287.35 K.

ln (P2/P1) = (30,100 J/mol / 8.31 J/mol K)(1/273.15 - 1/287.35)
ln (P2/P1) = 0.655
P2/P1 = e^0.655 = 1.926
P2 = (1.926)(P1) = (1.926)(45.37 mmHg) = 87.4 mmHg.

3) at boiling point, vapour pressure of the liquid is = atmpospheric pressure

So at 113.47 degree C, vapour pressure of 2,3,4-trimethylpentane = 760 torr

using the same formula as in question 2

ln (P2/760) = 37600/8.314 [1/(113.47 + 273.15) - 1/(105.5 + 273.15) = -0.246

P2/760 = e^-0.246 = 0.782

P2 = 0.782 * 760 = 594 torr