1. The following are the vapor pressures of some relatively common chemicals mea
ID: 973112 • Letter: 1
Question
1. The following are the vapor pressures of some relatively common chemicals measured at 20 °C:
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 Hvap 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?
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 torrExplanation / Answer
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) = (Hvap / 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
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