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A mixture of benzene and toluene is in equilibrium with its vapor at 50oC. What mole...

A mixture of benzene and toluene is in equilibrium with its vapor at 50oC. What mole fraction of benzene is in the vapor phase? You may assume that this system follows Raoult's Law. Note: when solving Antoine's equation it is best to keep all sig figs through calculation and then truncate them at the end.

Solutions

Expert Solution

Apply Raoults Equations in equilibrium

x1*P°1 = y1*PT

x2*P°2 = y2*PT

T = temperature in °C

P°1 = vapor pressure for benzene; from antoine equations (bar)

log10(P) = A − (B / (T + C)) A,B,C = 4.72583 1660.652 -1.46

P°2 = vapor pressure for toluene ; from antoine equations

log10(P) = A − (B / (T + C)) A,B,C = 4.23679 1426.448 -45.957

also

x2 = 1-x1 and y2 = 1-y1; Pt total pressure

substitute

x1*10 ^(4.72583 - 1660.652/ (T-1.46) ) = y1*PT

(1-x1)*10 ^(4.23679 - 1426.448 / (T-45.957) ) = (1-y1)*PT

T = 50°C

x1*10 ^(4.72583 - 1660.652/ (273+50-1.46) )  = y1*PT

(1-x1)*10 ^(4.23679 - 1426.448 / (273+50-45.957) ) = (1-y1)*PT

simplify

0.36403x1 = y1*PT

0.12244(1-x1) = (1-y1)*PT

Assume Pressure is 1 atm, so P = 1 bar approx

0.36403x1 = y1*1

0.12244(1-x1) = (1-y1)*1

simplify

y1 = 0.36403*x1

0.12244-0.12244*x1 = 1-y1

0.12244-0.12244*x1 = 0.36403*x1

(0.36403 + 0.12244)*x1 =0.12244

x1 = 0.12244 / ((0.36403 + 0.12244)) = 0.2516

so

y1 = 0.36403*x1 = 0.36403*0.2516 = 0.09158

queen_honey_blossom answered 2 years ago
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