The values of `K_(p)` and `Kp_(2)` for the reactions `XhArrY+Z` , (a)
and `A hArr 2B` , (b)
are in the ration of `9:1`. If the degree of dissociation of X and A is equal, then the total pressure at equilibriums (a) and (b) is in the ratio

Explanation

Suppose total pressure at equilibrium for reactions (1) and (2) are `P_(1)and P_(2)` respectively . Then
` {:(,X,hArr,Y,+,Z),("Intial.",1"mole",,0,,0),("At eqm.",1-alpha,,alpha,alpha,"Total"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)) = ((apha)/(1+alpha) P_(1))^(2)/((1-alpha)/(1+alpha)P_(1))=(alpha^(2)P_(1))/(1- alpha^(2) )cong alpha^(2) P_(1) `
`{:(,A,hArr,2B,),("Intial",1"mole",,0,),("At eqm.",1-alpha,,2 alpha,"Total"=1=alpha):}`
` p_(A) = (1-alpha)/(1 +alpha)p^(2),p^(B) = (2alpha)/(1 + alpha) P^(2)`
` K_(p_(2)) = ((2alpha)/(1+alpha)P_(2))^(2)/((1-alpha)/(1+alpha)P^(2))= (4 alpha^(2))/(1-alpha^(2)) P_(2) = 4 alpha ^(2) P_(2) `
` (K_(p_(1)))/(K_(p_(2)))= (alpha^(2) P_(1))/(4alpha^(2)P_(2)) = (P_(1))/(4 P_(2))= 9/1 ("Given") or (P_(1))/(P_(2))= 36 /1 = 36 :1 `