3. The equilibrium constant KP for the decomposition of phosphorus pentachloride to phosphorus trichloride and molecular chlorine according to the chemical equation given below is found to be 1.05 at 250 °C. If the equilibrium partial pressures of PCl5 and PCl3 are 0.875 atm and 0.463 atm, respectively, what is the equilibrium partial pressure of Cl2 at 250 °C? PCl5(g)  PCl3(g) + Cl2(g)

Explanation:

Hello,

In this case, given the reaction:

Thus, the law of mass action in terms of partial pressures is:

Hence, given the partial pressures of phosphorous penta and trichloride and the Kpa, one computes the equilibrium partial pressure of chlorine as shown below:

Best regards.

1.98 atm

Explanation:

To do this, let's write the equilibrium reaction of decomposition of PCl₅:

PCl₅ > PCl₃ + Cl₂      Kp = 1.05

Now, the Kp expression is the following:

Kp = PpCl₂ * PpPCl₃ / PpCl₅

Solving for Cl₂:

PpCl₂ = Kp * PpCl₅ / PpPCl₃

We already have the partial pressures of the reagents, so, replacing in the above expression we have:

PpCl₂ = 1.05 * 0.875 / 0.463

PpCl₂ = 1.98 atm

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