A 2 (g) and B 2 (g) having partial pressures 60 mm of Hg & 42 mm of Hg respectively, are present in a closed vessel. At equilibrium, partial pressure of AB(g) is 28 mm of Hg. If all measurements are made under similar condition, then calculate percentage of dissociation of AB (g). (Round of answers to nearest integer).
Text Solution
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A 2 (g) + B 2 (g) 
2AB(g)
at t = 0 60 42 0
at eq. (60–x) (42–x) (2x)
(Partial pressures)
⇒ 2x = 28 ⇒ x = 14
K p = 
= 
= 
For Δ n g = 0 ; K p = 
= 
Let degree of dissociation for AB is ‘x’ , then
2AB(g) 
A 2 (g) + B 2 (g) 
1 0 0
at t = 0 (1 – x) x x
= 
→ α =0.719
Hence percentage of dissociatin = 0.719 × 100 = 72 %
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