Question

`AB_(2)` dissociates as
`AB_(2)(g) hArr AB(g)+B(g)`. If the initial pressure is `500` mm of Hg and the total pressure at equilibrium is `700` mm of Hg. Calculate `K_(p)` for the reaction.

Answer 1

Let us suppose that after dissociation , the decrease in vapour pressure of `AB_(2)` is p mm.
`{:(,AB_(2)(g) ,hArr, AB(g) ,+, B(g)),("Initial pressure",500 mm,,0,,0),("Eqm. pressure"," (500-p)mm ",," pmm ",," Pmm "):}`
Total pressure at equilibrium =500 -p + p +p = (500 + P) mm
(500 + p ) = 700 or p = 700 -500 = 200 mm
`:. " At equilibrium point " , P_(AB_2) = (500 -200) = 300 mm, P_(AB) = 200 , P_(B) = 200 mm`
Applying Law of chemical equilibrium :
`K_(c) =(P_(AB)xxP_(B))/(P_(AB_2)) = ((200 mm)xx (200 mm))/((300mm))= 1.33 mm`