The equilibrium constants `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 degree of dissures at these equilibria is:

Options

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

Answer

C

Explanation

`{:("In equation",X,hArr,Y,+,Z),("Initial moles",1,,0,,0),("At equilibrium",(1-alpha),,alpha,,alpha):}`
Where `alpha=` degree of dissociation
Total number of moles `=1-alpha+alpha+alpha=(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_(P1)=([P_(y)][P_(z)])/([P_(x)])=(((alpha)/(1+alpha))p_(1)xx((alpha)/(1+alpha))p_(1))/(((1-alpha)/(1+alpha))p_(1))`
`=(((alpha)/(1+alpha))p_(1))/(((1-alpha)/(1+alpha))) (`i`)`
`{:("For equation",A,hArr,2B),("Initial moles",1,,0),(At equilibrium",(1-alpha),,2alpha):}`
Total number of moles at equilibrium `=(1+alpha)`
`P_(B)=((2alpha)/(1+alpha))p_(2)`
`P_(A)=((1-alpha)/(1+alpha))p_(2)`
`K_(p_(2))=([P_(B)]^(2))/([P_(A)])=([((2alpha)/(1+alpha))p_(2)]^(2))/(((1-alpha)/(1+alpha))p_(2))`
` =(((2alpha)/(1+alpha))p_(2))/(((1-alpha)/(1+alpha)))` (`ii`)
Eq. (`i`)(`ii`)
`(K_(p_(1)))/(K_(p_(2)))=(a^(2)xxp_(1))/(4alpha^(2)xxp_(2))`
`(9)/(1)=(p_(1))/(4p_(2))`
`(p_(1))/(p_(2))=(36)/(1)=36 : 1`