Question

The vapor pressure of pure ethylene dibromide is 172 Torr and that of pure propylene dibromide is 128 Torr both at 358 K and 1 atm pressure. If these two components follow Raoult's law, estimate the total pressure in kPa and the vapor composition in equilibrium with a solution that is 0.600 mol fraction propylene dibromide.

Answer 1

A) PT (Total Pressure) = 19.41 Kpa B) composition in equilibrium: X ED = 0.47 (ED: ethylene dibromide) , X PD = 0.53 (PD: propylene dibromide)

Step-by-step explanation:

Part A) Data: P ED = 172 Torr = 22931.04 Pa , P PD = 128 Torr = 17064.96 Pa , X PD = 0.600

Where: P ED = vapor pressure of pure ethylene dibromide

P PD = vapor pressure of pure propylene dibromide

We can use then Raoult´s Law:

PT = P ED*X ED + P PD*X PD

PT = total pressure

X ED = mol fraction of ethylene dibromide in solution

X PD = mol fraction of propylene dibromide in solution

Also, we can considere that: 1 = X ED + X PD , then X ED = 1 - 0.600 = 0.400

PT = 22931.04Pa*(1 - 0.600) + 17064.96Pa*0.600

PT = 19411.39 Pa = 19.41 KPa

Part B) We cna use the following formula to determine the vapor composition (mol fraction) of each component

X i = (Y i*PT)/Pi

Where:

X i = mol fraction of component i in solution

Y i = mol fraction of component i in vapor phase

PT = total pressure

P i = Vapor pressure of component i

Y i = (X i *Pi)/PT

Y DE = [(1 - 0.600)*22931.04Pa]/19411.30 Pa

Y DE = 0.47 mol fraction of ethylene dibromide in vapor phase

Y PD = (0.600*17064.96 Pa)/19411.39 Pa

Y PD = 0.53 mol fraction of propylene dibromide in vapor phase