1.

2 mol of N_(2) is mixed with 6 mol of H_(2) in a closed vessel of one litre capacity. If 50% of N_(2) is converted into NH_(3) at equlibrium, the value of K_(c) for the reaction N_(2(g))+3H_(2(g))hArr2NH_(3(g)) is

Answer»

`4//27`
`27//4`
`1//27`
`24`

Solution :`underset((2-x))(N_(2))+underset((6-3x))(3H_(2))hArrunderset((2x))(2NH_(3))`
`50%` Dissociation of `N_(2)` TAKE PLACE so,
At equlibrium `(2xx10)/(100)=1,` value of x=1
`K_(C)=([2]^(2))/([1][3]^(3))=4/27so, K_(c)=4/27`


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