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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?

Respuesta :

Answer:

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 10Âł

= 3.717 x 10⁔ Pa

Kp = 3.14 P = 3.14 x  3.717 x 10⁔ Pa

= 11.67 x 10⁔ Pa

Kp  = Kc x [tex]( RT )^{\triangle n}[/tex]

[tex]\triangle n[/tex] = 1

11.67 x 10⁔ = Kc x [tex]( 8.31\times 600 )^{1}[/tex]

Kc = 234

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