A liquid-vapor mixture has two components: n-hexane and n-octane. the compositio

Question

A liquid-vapor mixture has two components: n-hexane and n-octane. the composition of the liquid phase is 50.0 mole % n-hexane, and 50 mole % n-octane at 3933 K. the vapor pressures of n-hexane and n-octane at 393.3 K were previously estimated using the Antoine equation as 3003 mm Hg and 650.4 mm Hg respectively. Estimate: the partial pressures (mm Hg) of n-hexane and n-octane in the vapor phase. the mole % of n-hexane and mole % of n-octane in the vapor phase. the total pressure (mm Hg) in the vapor phase. the total pressure (mm Hg) in the liquid phase.

Explanation / Answer

MOLE FRACTION =MOLE PERCENT /100

MOLE FRACTION OF n-HEXANE = XA=50%/100 =0.5

MOLE FRACTION OF n-OCTANE = XB =50%/100 = 0.5

VAPOUR PRESSURE OF n-HEXANE GIVEN IS = PA0 =3003mm of Hg

VAPOUR PRESSURE OF n-OCTANE GIVEN IS =PB0 =650.4mm of Hg

(d)TOTAL PRESSURE IN LIQUID STATE =PT =PA0XA+PB0XB =3003X0.5+650.4X0.5 =1826.7mm of Hg.

(b)MOLE FRACTION OF n-HEXANE IN VAPOUR PHASE =PA0XA/PT =YA =3003X0.5/1826.7 =0.8219

MOLE % =MOLE FRACTION X100 =0.8219X100=82.19%

MOLE FRACTION OF n-OCTANE IN VAPOUR PHASE =PB0XB/PT = YB =650.4X0.5/1826.7 =0.17802

MOLE % = MOLE FRACTION X100 =0.17802X100 =17.802%

(c) TOTAL PRESSURE IN VAPOURE PHASE =PATIAL PRESSURE X MOLE FRACTION

PT IN VAPOUR PHASE =3003X0.8219+650.4X0.17802=2583.94991mm    (a) PARTIAL PRESSURE OF n-HEXANE = PTYA =2583.95X0.8219 = 2123.74843 mm of Hg

PARTIAL PRESSURE OF n- OCTANE = 2583.95X0.17802 =459.994763mm of Hg.

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