Question

At equilibrium, a sample of gas from the system is collected into a 4.00 L flask at 600 K. The flask is found to contain 3.86 g of PCl₅, 12.7 g of PCl₃, and 13.3 g of Cl₂. What are the values of Kc and Kp for this reaction?

Answer 1

Step-by-step explanation:

PCl₅ ⇄ PCl₃ + Cl₂

1 mole 1 mole 1 mole

molecular weight of PCl₅ = 208.5

molecular weight of PCl₃ = 137.5

molecular weight of Cl₂ = 71

moles of PCl₅ = .0185

moles of PCl₃ = .0924

moles of Cl₂ = .1873

Total moles = .2982 moles

mole fraction of PCl₅ = .062

mole fraction of PCl₃ = .31

mole fraction of Cl₂ = .628

If total pressure be P

partial pressure of PCl₅ = .062 P

partial pressure of PCl₃ = .31 P

partial pressure of Cl₂ = .628 P

Kp = .31 P x .628 P / .062 P

= 3.14 P

To calculate Total pressure P

PV = nRT

P x 4 x 10⁻³ = .2982 x 8.31 x 600

P = 371.7 x 10³

= 3.717 x 10⁵ Pa

Kp = 3.14 P = 3.14 x 3.717 x 10⁵ Pa

= 11.67 x 10⁵ Pa

Kp = Kc x
`( RT )^(\triangle n)`

`\triangle n` = 1

11.67 x 10⁵ = Kc x
`( 8.31* 600 )^(1)`

Kc = 234