Question

Calculate the volume % of Chlorine at equilibrium in PCL 5 under a total pressure in 1.5 ATM (Kp= 0.202)

Answer 1

we have to calculate the volume % of chlorine at equilibrium in PCl₅ under total pressure in 1.5 atm (Kp = 0.202)

solution : dissociation chemical reaction of PCl₅ is given as,

                                      PCl₅    ⇄  PCl₃   +   Cl₂

at t = 0                              1              0            0

at equilibrium              1 - α            α              α

so total no of moles = 1 - α + α + α = 1 + α

so the partial pressure of PCl₅ = (1 - α)/(1 - α) P

partial pressure of PCl₃ = α/(1 + α) P

partial pressure of Cl₂ = α/(1 + α) P

now Kp = $\frac{P_{Cl_2}\times P_{PCl_3}}{P_{PCl_5}}$ = α²P/(1 - α²)

⇒ 0.202 = α²(1.5)/(1 - α²) = 0.344

now the percentage volume of chlorine = mole fraction of chlorine × 100

      = α/(1 + α) × 100 = 0.344/1.344 × 100 = 25.6 %

therefore the volume % of chlorine at equilibrium is 25.6%