The equilibrium constant `K_(p_(1))` and `K_(p_(2))` for the reactions `XhArr2Y` and `ZhArrP+Q`, respectively are in the ratio of `1:9`. If the degree of dissociation of `X` and `Z` be equal, then the ratio of total pressure at these equilibrium is:

Options

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

Answer

A

Explanation

`x Leftrightarrow 2y" "z Leftrightarrow p+Q`
`{:(,1,0),(,(1-x),2x):}" " {:(,1,0,0),(,(1-x),x,x):}`
`k_(p)=((2x)^(2))/((1-x)) ((p_(1))/(1+x))" "k_(p_(2))=(x^(2))/((1-x)) ((p_(2))/(1+x))`
Given that `k_(p_(1))/(k_(p_(2)))=1/9 rArr (4 xx p_(1))/(p_(2))=1/9`
`p_(1)/p_(2)=(1)/(36)`