The values of `K_(p_(1))` and `K_(p_(2))` for the reactions
`X hArr Y+Z` ….(i)
and `A hArr 2B` …(ii)
are in ratio of 9 : 1. If degree of dissociation of X and A be equal, then total presure at equilibrium (i) and (ii) are in the ratio.

Options

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

Answer

C

Explanation

Suppose the total pressure at equilibrium for the reactions (1) and (2) are `P_(1)` and `P_(2)` respectively. Then
`{:(,X,hArr,Y,+,Z),("Initial",1 mol,,0,,0),("At eqm.",1-alpha,,alpha,,alpha):}`
Total no. of moles at eqm. `=1+alpha`
`P_(X)=(1-alpha)/(1+alpha)P_(1),P_(Y)=(alpha)/(1+alpha)P_(1),P_(Z)=(alpha)/(1+alpha)P_(1)`
`K_(P)=(((a)/(1+a)P_(1))^(2))/(((1-a)/(1+a))P_(1))=(alpha^(2)P_(1))/(1-alpha^(2))=alpha^(2)P_(1)`
`{:(,A,hArr,2B,),("At eqm.",1-alpha,,2alpha,):}`
Total no. of moles at eqm. `=1+alpha`
`P_(A)=(1-alpha)/(1+alpha)P_(2),P_(B)=(2alpha)/(1+alpha)P_(2)`
`K_(p_(2))=(((2alpha)/(1+alpha)P_(2))^(2))/(((1-alpha)/(1+alpha))P_(2))=(4alpha^(2))/(1-alpha^(2))P_(2)~~P_(2)`
`(K_(p_(1)))/(K_(p_(2)))=(alpha^(2)P_(1))/(4alpha^(2)P_(2))=(P_(1))/(4P_(2))=(9)/(1)`Given
or `(P_(1))/(P_(2))=(36)/(1)` or ` P_(1) : P_(2) =36 :1`