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The dissociation equilibrium of a gas AB...

The dissociation equilibrium of a gas `AB_2` can be represented as, `2AB_2(g) hArr 2AB (g) +B_2(g)`. The degree of disssociation is 'x' and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant `k_p` and total pressure P is

A

`(2K_(p)//P)^(1//2)`

B

`(K_(p)//P)`

C

`(2K_(p)//P)`

D

`(2K_(p)//P)^(1//3)`

Text Solution

Verified by Experts

The correct Answer is:
D
`2AB_(2)hArr2AB+B_(2)`
`{:(1,0,0,),("At eqm.",1-x,x,x//2):}`
`K_(p)=((P_(AB))^(2)(P_(B_(2))))/((P_(AB_(2)))^(2))impliesK_(p)=([(x)/((1+x//2))xxP]^(2)[(x//2)/(1+x//2)xxP])/([(1-x)/(1+x//2)xxP]^(2))`
`impliesK_(p)=(x^(3)P)/(2(1+x//2)(1-x)^(2))`
`1-x-=1" "1+x//2-=1`
`K_(p)=(x^(3)P)/(2)impliesx=((2K_(p))/(P))^(1//3)`
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