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For the reaction 2AB(g)hArrA(2)(g)+B(2...

For the reaction
`2AB(g)hArrA_(2)(g)+B_(2)(g)`
The degree of dissociation `(alpha)` of `AB(g)` is related to equilibrium constant `K_(p)` by the expression

A

`(1+2sqrt(K_(p)))/(2sqrt(K_(p)))`

B

`sqrt((1+2K_(p))/2)`

C

`sqrt((2K_(p))/(1+2K_(p)))`

D

`(2sqrt(K_(p)))/(1+2sqrt(K_(p)))`

Text Solution

Verified by Experts

The correct Answer is:
D
`2AB(g)hArrA_(2)(g)+B_(2)(g)`
`{:(,1-apha,,alpha/2,alpha/2),(,,,,):}`
`K_(p)=((alpha/2xxP_(T))^(2))/((1-alpha)^(2)P_(T)^(2))`
`alpha/(1-alpha)=2sqrt(K_(p))`
`alpha=(2sqrt(K_(p)))/(1+2sqrt(K_(p)))`
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