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?

Question

Charity Woods

Chemistry

24 November, 05:44

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?

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Kennedi Potts · Answer 1

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