The values of `K_(p)` and `Kp_(2)` fot the reactions `XhArrY+Z` , (a)
and `A hArr 2B` , (b)
are in the ration of `9:1`. If the degree of dissociation of X and A is equal, then the total pressure at equilibriums (a) and (b) is in the ration

Options

A. `3:1`
B. `1:9`
C. `36:1`
D. `1:1`

Answer

C

Explanation

Let us assume that `P_(1)` and `P_(2)` are the total pressures at equilibrium for reactions (a) and (b), respectively.
First consider reaction (a):
`{:(,X hArr,Y+,Z),("Initial moles",1,0,0),("Equilibrium moles",1-alpha,alpha,alpha):}`
where `alpha` is the degree of dissociation of X.
`K_(p_(1))= (P_(y)P_(z))/(P_(x))`
Partial pressure = (Mole fraction) x (Total pressure)
`:. P_(x)= (1-alpha)/(1+alpha)P_(1)`, `P_(Y)=(alpha)/(1+alpha)P_(1)`, `P_(Z)=(alpha)/(1+alpha)P_(1)`
`:. K_(P_(1))=(((alpha)/(1+alpha)P_(1))^(2))/((1-alpha)/(1+alpha)P_(1)) = (alpha^(2))/((1+alpha)(1-alpha))P_(1)`
`= (alpha^(2))/(1-alpha^(2))P_(1)`
Now consider reaction (b):
`{:(,A hArr,2B),("Initial moles",1,0),("Equilibrium moles",1-alpha,2alpha):}`
where `alpha` is the degree of dissociation of A.
`P_(A) = (1-alpha)/(1+alpha)P_(2)`, `P_(B)= (2alpha)/(1+alpha)P_(2)`
`K_(P_(2))=(P_(B)^(2))/(P_(A))=(((2alpha)/(1+alpha)P_(2))^(2))/(((1-alpha)/(1+alpha)P_(2)))`
`= (4alpha^(2))/(1-alpha^(2))P_(2)`
Let us take the ratio of `K_(P_(1))` and `K_(P_(2))` :
`(K_(P_(1)))/(K_(P_(2)))=((alpha^(2))/(1-alpha^(2))P_(1))/((4alpha^(2))/(1-alpha^(2))P_(2))`
`= (P_(1))/(4P_(2))` (as `alpha` of X and A are equal)
Since `(K_(P_(1)))/(K_(P_(2)))=(9)/(1)`, we have
`(P_(1))/(4P_(2))=(9)/(1)`
or `(P_(1))/(P_(2))=(36)/(1)=36 : 1`