Question

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)

Answer 1

1.98 atm

Step-by-step 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

Answer 2

`p_(Cl_2)=1.97atm`

Step-by-step explanation:

Hello,

In this case, given the reaction:

`PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)`

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

`Kp=(p_(PCl_3)p_(Cl_2))/(p_(PCl_5))`

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

`p_(Cl_2)=(Kp*p_(PCl_5))/(p_(PCl_3))=(1.05*0.875atm)/(0.463atm) \\\\p_(Cl_2)=1.97atm`

Best regards.