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

`{:("In equation,",X,hArr,Y,+,Z),("Initial moles",1,,0,,0),("At equil",(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_(p_(1))=([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))^(2)p_(1))/(((1-alpha)/(1+alpha)))`
`{:("For equation,",A,hArr,2B),("Initial moles",1,,0),("At equil.",(1-alpha),,2alpha):}`
Total number of moles at equilibrium `=(1+alpha)`
`p_(B)=((2alpha)/(1+alpha))p_(2)`
`p_(A)=((1-alpha)/(alpha))p_(2)`
`K_(p_(2))=([p_(B)]^(2))/([p_(A)])=([((2alpha)/(1+alpha))p_(2)]^(2))/(((1-alpha)/(1+alpha))p_(2))`
`K_(p_(2))=(((2alpha)/(1+alpha))^(2)p_(2))/(((1-alpha)/(1+alpha)))` ....(ii)
Eq. (i) divide by Eq. (ii)
`(K_(p_(1)))/(K_(p_(2)))=(alpha^(2)xxp_(1))/(4alpha^(2)xxp_(2))`
`(9)/(1)=(p_(1))/(4p_(2))`
`(p_(1))/(p_(2))=(36)/(1)= 36 : 1`