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:
[tex]n_{PCl_5}= \dfrac {weight \ of \ PCl_5} {M.Wt. \ of \ PCl_5}[/tex]
[tex]n_{PCl_5}= \dfrac {2.69 \ g} {208.5 \ g/mol}[/tex]
[tex]n_{PCl_5}= 0.013 \ mol[/tex]
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
[tex]P = \dfrac{nRT}{V}[/tex]
[tex]P = \dfrac{0.0013 \ mol \times 0.082 \ Latm^0 K^{-1} . mol ^{-1} \times 523.15 \ K}{1.0 \ L}[/tex]
P = 0.558 atm
Thus, at 250° C, decomposition of PCl₅ occurs.
In the container, PCl₅ decomposes to PCl₃ and Cl₂.
i.e.
[tex]PCl_{5(g)} \to PCl_{3(g)}+ Cl_{2(g)}[/tex]
Using Dalton's Law:
[tex]P_{total } =P_1 + P_2+P_3 +...[/tex]
[tex]P_1 = P_{Total} \times X_1[/tex]
where;
X = mole fraction
Then, the total no. of moles in the container is:
[tex]n = \dfrac{PV} {RT}[/tex]
[tex]n = \dfrac{1\ atm \times 1.0\ L}{0.0821 \ L \ atm \ K^{-1}.mol \times 523.15\ K}[/tex]
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:
[tex]P__{PCL_3} }= P_{total} \times \dfrac{no. \ of \ moles \ of PCl_5}{total \ no. \ of \ moles}[/tex]
[tex]P__{PCL_3}} = 1 \ atm \times \dfrac{0.005}{0.023}[/tex]
[tex]P__{PCL_3}} = 0.22 \ atm[/tex]
Thus, since the no of moles of PCl₃ and Cl₂ are the same, then the partial pressure for Cl₂ is = 0.22 atm
