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`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` |
Answer» Correct Answer - A::B::C `K=([NH_(3)]^(2))/([H_(2)]^(3)[N_(2)])` for `50% N_(2)` `K=4/27` |
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