The equilibrium constant `K_(p_(2))` and `K_(p_(2))` for the reactions `A hArr 2B` and `P hArr Q+R`, respectively, are in the ratio of `2:3`. If the degree of dissociation of A and P are equal, the ratio of the total pressure at equilibrium is,

Options

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

Answer

A, C

Explanation

`{:("Soln. for",X,hArr,2Y),("Initial moles",1,,0),("Eq. mol",1-x,,2x),("Total mol"=1-x+2x=1+x,,,):}`
`p_(x)=((1-x)/(1+x))P_(1)" "p_(y)=((2x)/(1+x))P_(1).`
`K_(P1)=((p_(y))^(2))/p_(x)=([((2x)/(1+x))P_(1)]^(2))/(((1-x)/(1+x))P_(1))`
`=(4x^(2)P_(1))/((1-x^(2)))`
`{:("Similary for",Z,hArr,P,+,Q),("Initial mol",1,,0,,0),("Equilibrium mol",1-x,,x,,x),("Total mol":1-x+x+x=1+x,,,,,):}`
`p_(z)=((1-x)/(1+x))P_(2), p_(Q)(x/(1+x))P_(2), p_(p)=(x/(1+x))P_(2)`
`k_(p_(2))=(p_(p)xxp_(Q))/p_(z)=((x/(1+x))P_(2)xx(x/(1+x))P_(2))/(((1-x)/(1+x))P_(2))`
`=(x^(2)P_(2))/((1-x^(2)))`
`K_(p_(1))/K_(p_(2))=(4cancel(x^(2))P_(1))/((1-cancel(x^(2))))xx((1-cancel(x^(2))))/(cancel(x^(2)) P_(2))=(4P_(1))/P_(2)=1/9` (Given)
`P_(1)/P_(2)=1/36`