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2 mol of N(2) is mixed with 6 mol of H(2...

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

A

`4//27`

B

`27//4`

C

`2//27`

D

`20`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
`K=([NH_(3)]^(2))/([H_(2)]^(3)[N_(2)])` for `50% N_(2)`
`K=4/27`
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