Answer:
Partial pressures:
PCl₅ = 0.558 atm
PCl₃ = 0.22 atm
Cl₂ = 0.22 atm
Explanation:
From the given information:
The number of moles of PCl₅ associated with the evaporation is:
Temperature of the gas = 250° C = (250 + 273.15) K
= 523.15 K
Using the Ideal gas equation to determine the pressure exerted by the completely vaporized PCl₅
PV = nRT
P = 0.558 atm
Thus, at 250° C, decomposition of PCl₅ occurs.
In the container, PCl₅ decomposes to PCl₃ and Cl₂.
i.e.
Using Dalton's Law:
where;
X = mole fraction
Then, the total no. of moles in the container is:
n = 0.023 mol
Now, the container contains a total amount of 0.023 mol where initially 0.013 mol are that of PCl₅ and remaining 0.005 mol of PCl₃ and 0.005 mol of Cl₂.
Thus, the partial pressure of PCl₃ is:
Thus, since the no of moles of PCl₃ and Cl₂ are the same, then the partial pressure for Cl₂ is = 0.22 atm