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 total pressure at equilibrium for reactions (1) and (2) are `P_(1) and P_(2)` respectively . Then
`{: (,X,hArr,Y,+,Z), (" Intial",1 "mole",,0,,0), ("At eqm",1-alpha,,alpha,,alpha"Total"=1+alpha), (,p_(X)=(1-alpha)/(1+alpa)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) P_(1))/(1-alpha^(2))cong alpha^(2)P_(1)`
`{: ( ,A,hArr,2B,) , (" Intial" ,1 "mole",,0,) ,(" At eqm" ,1-alpha,,2 alpha,"Total" = 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)))/((1-alpha)/(1+alpha)P_(2)) = (4alpha^(2))/(1-alpha^(2))P_(2)= 4 alpha^(2)P_(2)`
` K_(p_(1))/K_(p_(2)) = (alpha_(2)P_(1))/(4 alpha^(2)P_(2))=P_(1)/(4 P_(2))=9/1 ` (Given )
or ` (P_(1))/(P_(2))=36/1 = 36*1`