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In a 0.25 L tube dissociation of 4 mol o...

In a 0.25 L tube dissociation of 4 mol of NO is taking place. If its degree of dissociation is 10%. The value of K​P​ for the reaction `2NO_((g)) to N_(2(g)) + O_(2(g)) ` is :-

A

`(1)/((18)^(2))`

B

`(1)/((8)^(2))`

C

`1/(16)`

D

`1/(32)`

Text Solution

Verified by Experts

The correct Answer is:
A
`NO_((g)) hArr N_(2)(g) + O_(2)(g)`
`T=0, 4" mol"" "0" "0`
At Eq. 4(1-0.1) mol `(4xx0.1)/(2)` mol `(4xx0.1)/(2)` mol
or `(4xx0.9)/(0.25) ` M or `(4xx0.1)/(0.25xx2)M, (4xx0.1)/(0.25xx2)M`
`K_(p)=([N_(2)]xx[O_(2)])/([NO]^(2))=((4xx0.1)/(2xx.025)xx(4xx0.1)/(2xx0.25))/((4xx0.9)/(0.25))`
`=(0.1xx0.1)/(2xx2xx9xx0.9)`
`=(1)/((18)^(2))`
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