An initially unknown mixture of n-heptane and n-hexane have been removed from a

Question

An initially unknown mixture of n-heptane and n-hexane have been removed from a distillation column and placed into a sealed container. The container is currently at a temperature of 30 degree C, use the Antoine equation Jo estimate the pure component vapor pressure of each species. If the liquid composition were to be 40 mole% n-heptane and 60 mole% n- hexane, What would be the resulting vapor pressure at this temperature? Comment on What would happen if the pressure on the vessel was above or below this pressure.

Explanation / Answer

Antoine equation of vopour pressure

P = 10 A - (B / C + T)

A,B and C are antoine equation parameter for n-hexane

A = 7.01051 , B = 1246.33 , C = 232.988, P in mmHg T, in C

P = 186.808 mmHg for n-hexane

for n-heptane

A = 4.02832 , B = 1268.636 , C = -56.199

P = 58.3735 mmHg

P total = 245.1815 mmHg

b) P = xi Ptotal { xi = mole fraction}

n-hexane = 0.4 and n-heptane = 0.6

P n-hexane = 0.4 x 245.1815 = 98.0726 mmHg

Pn-heptane = 147.1089 mmHg

c) if pressure above or below from current pressure then vapour pressure of both n-hexane and n-heptane will change and boiling temperature will also change

at high pressureand boiling temp will increase

and at low pressure boiling point decrease.

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