Question

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?

Answer 1

1.418688 atm

Step-by-step 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