Question

`AB_(2) " dissociates as " : AB_(2) (g) hArr AB (g) + B (g).`
If the intial pressure is 500 mm of Hg and the total pressure at equilibrium is 700 mm 0f Hg, calculate `K_(p)` for the reaction.

Answer 1

After dissociation , suppose the decrease in the pressure of `AB_(2)` at equilibrium is p mm. Then
` {:(,AB_(2) (g),hArr,AB (g),+,B (g)),(" Intial pressure",500 mm,,0,,0),(" Pressures at eqm.",(500-)p " mm", ,p" mm",,p " mm"):} `
`:. " Total pressure at equilibrium " = 500 - p + p+ p = 500 + p" mm"`
`500 + p = 700" (Given) " or p= 200 " mm"`
Hence , at equilibrium `p _(AB_(2)) = 500 - 200 = 300 " mm" , 200 " mm " , p_(B) = 200 mm `
`:. K_(p) = (p_(AB) xx p_(B))/p_(AB_(2)) = (200 xx 200)/300 = 133*3mm`
Note. With respect to standard state pressure of 1 bar , i.e., 0*987 atm , i.e., 750 mm, `K_(p) = (133*3)/750=0*178.`