An initially unknown mixture of n-heptane and n-hexane have been removed from a
ID: 1060811 • Letter: 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.