For the reaction: PCl5(g) ⇌ PCl3(g) + Cl2(g) at 600.0 K, the equilibrium constant is 11.5. Suppose that 1.500 g of PCl5 (MW=208.22 g/mol) is placed in an evacuated 500.0 mL bulb, which is then heated to 600.0 K. What is the total pressure (in atm) in the bulb at equilibrium?

1.418688 atm

Explanation:

(a)  Moles of PCl5 = mass / molar mass

                      =1.5 g / 208.22 g/mol

                     = 0.0072 moles

Also given,

T = 600 K

V = 0.500 L

Pressure of PCl5, P = nRT / V

                             = 0.0072 mol×0.0821 L-atm / (mol.K)×600 K / 0.500 L

                             = 0.709344 atm

(b)  PCl5(g) ⇄ PCl3(g) + Cl2(g)

Initial               0.965         0               0

Change               -x            +x             +x

Equilibrium (0.709344 -x)         x               x

K_p = 11.5 = x×x / (0.965 -x)

solving, we get x = 0.67027

So partial pressure of PCl5 at equilibrium = 0.709344 - 0.67027 = 0.039074 atm

(c)   Partial pressure of PCl3 = Cl2 = 0.709344 atm

So total pressure = 0.709344+0.039074+ 0.67027= 1.418688 atm