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) + C

Answer:

The cumulative pressure inside the container amounts to 7.42 atm

Explanation:

given information

Weight of PCl5 = 75 g

Volume V = 3L

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

The reaction can be represented as: PCl5(g) ↔ PCl3(g) + Cl2(g)

Kp = 1.80

R = 0.0821 L·atm·K-1·mol-1

to determine

the overall pressure within the container

solution

it is known that the molar mass of PCl5 = 208.24 g/mol

thus, moles of PCl5 ( n )= 75 / 208.24 = 0.36 moles

and

the initial pressure of PCl5 = nRT /V

substituting the values

initial pressure of PCl5 = 0.36 (0.0821 )523 / 3

initial pressure of PCl5 = 5.149 atm

thus,[Kp = [PCl3(g)] × [Cl2(g)] / ( PCl5 )

and given

at equilibrium x atm of product

and the 5.149 - x atm of reactant here

we can formulate

1.8 = x² / ( 5.149 - x )

thus, x² +1.8 x - 9.267 = 0

x = 2.274429

here 2.274 atm stands for Cl2 + PCl3 pressure

therefore, the pressure from PCl5 becomes 5.149 - 2.274 = 2.875 atm.

so finally

total pressure equates to 2.87 + 2.27 + 2.27

the total pressure inside the container is found to be 7.42 atm