Question

PCl5 (g) <--> PCl3 (g) + Cl2 (g) K_p=0.0497 at 500 degrees Celsius. Determine the equilibrium concentration(M) of all species for a sealed gas cylinder charged with 1.53 atm of PCl5? Please show all work.

Answer 1

T = 500 + 273 = 773 K

Dn = product moles - reactant moles = 2-1 = 1

Kp = Kc (RT)^Dn

0.0497 = Kc (0.0821 x 773) ^1

Kc = 7.83 x 10^-4

Pressure of PCl5 = 1.53 atm

P V = n R T

P / R T = n / V

1.53 / 0.0821 x 773 = n / V = molarity

molarity = 0.024 M

PCl5 < ----------------------------> PCl3 + Cl2

0.024                                         0         0 -------------------> initial

0.024 -x                                    x           x   -------------------> equilibrium

Kc = [PCl3][Cl2]/[PCl5]

7.83 x 10^-4   = x^2 / 0.024 -x

x^2 + 7.83 x 10^-4 x - 1.879 x 10^-5 = 0

x = 0.00396

equilbrium concentration:

[PCl5] = 0.024 - x = 0.024 - 0.00396 = 0.02 M

[PCl3] = x = 0.00396 M

[Cl2] = x = 0.00396 M