Question

The values of `K_(p1)` and `K_(p2)` for the two equilibrium reactions
`X hArr +Z and A hArr 2B`
are in the ratio 9, 1,. If degree of dissociation of X and A be equal , calculate the ratio of the total pressure of the equilibrium mixture in the two cases.

Answer 1

Let the degree of disscoiation in the two cases be `alpha`
(i) Calculattion of `K_(p1)`
`{:("For equilibrium", X,hArr, Y, +,Z),("initial moles" : ,1,,0,,0),("moles at eqm. point",(1:alpha),,alpha,,alpha):}`
Total no. of ,moles `=1 -alpha +alpha+alpha=(1+alpha)`
`K_(p1) = (P_(y)xxP_(z))/(P_(x)) = (((alpha)/(1+alpha)P_(1))xx((alpha)/(1+alpha)P_(1)))/(((1-alpha)/(1+alpha)P_(1)))`
`=(P_(1)^(2)alpha^(2))/((1+alpha)^(2))xx((1+alpha)/((1-alpha)P_(1)))=(P_(1)alpha)^(2)/((1+alpha)(1-alpha))`
(II) Calculation of `K_(p2)`.
`{:("For equilibrium",A, hArr, 2B),("initial moles",1,,0),("Moles at eqm. point",(1-alpha),,2alpha):}`
Total no. of moes `=1 alpha +2alpha =(1+alpha)`
`K_(p2) = (pB)^(2)/(pA)=((2alpha)/(1+alpha)P_2)^(2)/(((1-alpha)/(1+alpha)P_(2)))`
`P_(2)^(2) xx (4alpha^(2))/(1+alpha)^(2) xx ((1+alpha))/((1-alpha)P_(2))= (P_(2)xx4alpha^(2))/((1+alpha)(1-alpha))`
Dividing eqn. (i) with eqn. (ii),
`(K_p1)/(K_p2) = ((P_(1)alpha)^(2))/((1-alpha^(2)))xx ((1-alpha^(2)))/(P_(2)xx4alpha^(2)) = P_(1)/(4_(p_(2))`
`" Now " " "(K_(p1))/(K_(p2))=9/1 ("given") :. P_(1)/(4P_(2))=9/1 or P_(1)/P_(2) = (36)/(1)`