75.0 g of PCl5 (g) is introduced into an evacuated 3.00-L vessel and allowed to reach equilibrium at 250ºC. PCl5 (g) ↔ PCl3 (g) + Cl2 (g) If Kp = 1.80 for this reaction, what is the total pressure inside the vessel at equilibrium (R = 0.0821 L·atm·K-1·mol-1) ?

total pressure inside the vessel is 7.42 atm

Explanation:

given data

PCl5 weight = 75 g

volume V = 3L

temperature T = 250ºC = 250 + 273 = 523 K

PCl5 (g) ↔ PCl3 (g) + Cl2 (g)

Kp = 1.80

R = 0.0821 L·atm·K-1·mol-1

to find out

the total pressure inside the vessel

solution

we know that molar mass of PCl5 = 208.24 g/mol

so moles of PCi5 (n) = 75 / 208.24 = 0.36 mole

and

initial pressure PCl5 = nRT / V

put all value

initial pressure PCl5 = 0.36 (0.0821) 523 / 3

initial pressure PCl5 = 5.149 atm

so

Kp = [PCl3 (g) ] * [Cl2 (g) ] / (PCl5)

and we know

at equilibrium x atm of product

and the 5.149 - x atm of reactant here

so we can say

1.8 = x² / (5.149 - x)

so x² + 1.8 x - 9.267 = 0

x = 2.274429

here 2.274 atm is equal to Cl2 + PCl3 pressure

and we know pressure by PCl 5 is 5.149 - 2.274 = 2.875 atm.

so

total pressure is = 2.87 + 2.27 + 2.27

total pressure inside the vessel is 7.42 atm