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

(3) In equation `{:("In equation", H, hArr, Y+Z),("Initial mole",1,0,0),("At equilibrium",(1-alpha),alpha,alpha):}`
where `alpha`= degree dissociation
Total number of mole `=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_(Pl)=([P_(y)][P_(z)])/([P_(x)])=(((alpha)/(1+alpha))p_(1)xx((alpha)/(1+alpha))p_(1))/(((1-alpha)/(1+alpha))p_(1))=(((1-alpha)/(1+alpha))^(2))/(((1-alpha)/(1+alpha)))`
`{:("For equation," ,AhArr,2B),("Initail 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)=((alpha-1)/(1+alpha))p_(2)`
`K_(sp)=([p_(B)]^(2))/([p_(A)])=([((2alpha)/(1+alpha))p_(2)]^(2))/(((alpha-1)/(1+alpha))p_(2))=(((2alpha)/(1+alpha))^(2)p_(2))/(((alpha-1)/(1+alpha)))`
Eq. (i)/ (ii)
`(K_(sp))/(K_(p2))=(alpha^(2)+xp_(1))/(4alpha^(2)xxp_(2)),(9)/(1)=(p_(1))/(4p_(2)),(p_(1))/(p_(2))=(36)/(1)=36:1`