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1.0 mole of nitrogen and 3.0 moles of PCl 5 are placed in 100 L vessel heated to 227 o C. The equilibrium pressure is 2.05 atm. Assuming ideal behaviour, calculate the degree of dissociation for PCl 5 and K p for the reaction, PCl 5 (g)⇌PCl 3 (g)+Cl 2 (g).
Asked by vishalrolaniya2005 | 20 Sep, 2023, 08:58: PM
Dear Student,

PCl5 (g) ⇌ PCl3 (g) + Cl2 (g)

Initial             3                 0                 0

At equi:   3(1 – x )          3x               3x

(where x is degree of dissociation)

Total moles  = 3 (1 − x) + 3x + 3x = 3(1 + x)

1 mole of nitrogen is present, so actual total number of moles at equilibrium

= 3 (1 + x) + 1
We know that, according to Ideal gas equation,
PV=nRT
Given,

P = 2.05 atm, V = 100 litres, R = 0.082 atm × L/(mol×K)
T = (273 + 227) = 500 K
So, n = 0.082 × 5002.05 × 100 ​= 5
or 3(1 + x) + 1 = 5
3x = 1 or x = 0.333
Degree of dissociation is 0.333
At equilibrium,

pPCl5 = [3(1 – x) / (3x+4) ] ⨯ 2.05 atm

pPCl3 = pCl2 = [3x / (3x +4) ] ⨯ 2.05 atm

Kp = pPCl3 ⨯ pPCl2 / pPCl5

= {[3x / (3x +4) ] ⨯ 2.05 atm}2 /  [3(1 – x) / (3x+4) ] ⨯ 2.05 atm

Kp = 0.204 atm

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