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For the reaction (1)and(2) A(g)hArrB(g...

For the reaction (1)and(2)
`A(g)hArrB(g)+C(g)`
` X(g) hArr2Y(g)`
Given , `K_(p1):K_(p2)=9:1`
If the degree of dissociation of A(g) and X(g) be same then the total pressure at equilibrium
(1) and (2) are in the ratio:

A

`3 : 1`

B

`1 : 9`

C

`36 : 1`

D

`1 : 1`

Text Solution

Verified by Experts

The correct Answer is:
C
`{:("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`
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