Question

Assume dibromoethene (dbe) and dibromopropene (dbp) form an ideal solution. At 358 K, the vapor pressures of pure dbe and dbp are 22.9 kPa and 17.1 kPa, respectively. Find the total vapor pressure in kPa of a mixture of dbe and dbp where the mole fraction of dbe is 0.2?

Answer 1

The total vapor pressure of a mixture of dibromoethene and dibromopropene with a mole fraction of dibromoethene as 0.2 is calculated using Raoult's Law, leading to a total vapor pressure of 18.26 kPa.

Step-by-step explanation:

To calculate the total vapor pressure of a mixture of dibromoethene (dbe) and dibromopropene (dbp) where the mole fraction of dbe is 0.2, we apply Raoult's Law, which states that the partial vapor pressure of each component in an ideal solution is equal to the product of the mole fraction of the component and the vapor pressure of the pure component. Therefore, the total vapor pressure (Ptotal) of the solution is the sum of the partial pressures of dbe and dbp.

Ptotal = (Xdbe * Pdbe) + (Xdbp * Pdbp)

Given that Xdbe = 0.2, Pdbe (vapor pressure of pure dbe) = 22.9 kPa, and since we have a binary mixture, Xdbp = 1 - Xdbe = 0.8, Pdbp (vapor pressure of pure dbp) = 17.1 kPa, the calculation would be:

Ptotal = (0.2 * 22.9 kPa) + (0.8 * 17.1 kPa) = 4.58 kPa + 13.68 kPa = 18.26 kPa