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At certain temperature compound AB(2)(g)...

At certain temperature compound `AB_(2)(g)` dissociates according to the reaction
`2AB_(2)(g) hArr2AB (g)+B_(2)(g)`
With degree of dissociation `alpha` Which is small compared with unity, the expression of `K_(p)` in terms of `alpha` and initial pressure P is :

Text Solution

Verified by Experts

For the given equilibrium, the equilibrium concentration are
`2AB_(2)(g)hArr2AB(g)+B_(2)(g)`
`{:("Equilibrium",c(1-alpha),,calpha,(calpha)/2),("concentration",,,,):}`
`:. K_(p)=((P_(B_(2)))(P_(AB))^(2))/(P_(AB_(2)))^(2)=`
`K_(p)=(alpha^(3)xxP_(T))/(2(1-alpha)^(2)(1+alpha/2))`
Since alpha is small compared to unity, so `1-alpha ~= 1` and `1+alpha/2 ~=1`.
`:. K_(p)=(alpha^(3)xxP_(T))/2`
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