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

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