The equilibrium constant `K_(p1)` and `K_(p2)` for the reactions `X hArr 2Y` and `Z hArr P+Q`, respectively, are in the ratio of `1:9`. If the degrees of dissociation of `X` and `Z` are equal, then the ratio of total pressure at equilibria is

Options

A. `1:36`
B. `1:9`
C. `1:6`
D. `1:4`

Explanation

`{:("for",A,hArr,2B),("Initial",1,,0),("At equilibrium",1-alpha,,2alpha),("Total moles"=1-alpha+2alpha=1+alpha,,,):}`
`:. P_(A)=("mole fraction of A")xx "Initial pressure"=((1-alpha)/(1+alpha))P_(1)`
Similarly `P_(B)=((2alpha)/(1+alpha))P`
`:. K_(P_(1))=((P_(B))^(2))/((P_(A)))=([((2alpha)/(1+alpha))P_(1)]^(2))/(((1-alpha)/(1+alpha))P_(1))=(4alpha^(2)P_(1))/((1-alpha^(2)))`
`{:("for",P,hArr,Q,+,R),("Initial",1,,0,,0),("At equilibrium",1-alpha,,alpha,,alpha),("Total "moles""=1-alpha+alpha+alpha=1+alpha,,,,,):}`
`:. P_(p)=((1-alpha)/(1+alpha))P_(2), P_(Q)=(alpha/(1+alpha))P_(2),P_(R)=(alpha/(1+alpha))P_(2)`
`K_(p_(2))=(P_(R)xxP_(Q))/P_(P)=((alpha/(1+alpha))P_(2)xx(alpha/(1+alpha))P_(2))/(((1-alpha)/(1+alpha))P_(2))=(alpha^(2)P_(2))/((1-alpha^(2)))`
`:. K_(p_(1))/K_(p_(2))=2/3`(given) `=(4cancel(alpha^(2))P_(1))/cancel((1-alpha^(2)))xx(cancel(1-a^(2))/(cancel(alpha^(2))P_(2)))=(4P_(1))/P_(2)`
`:. P_(1)/P_(2)=1/6`