A sample of PCl5 weighting 2.69 gram was placed in 1.00 Litter container and completely vaporized at 250C. The pressure observed at that temperature was 1.00 atm. The possibility exists that some of the PCl5 dissociated according to PCl5 (g) ! PCl3 (g) Cl2 (g) . What must be the partial pressures of PCl5 PCl3 and Cl2 under these experimental conditions

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

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