`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. `1//27`D. `24`

`2` mol of `N_(2)` is mixed with `6` mol of `H_(2)` in a closed vessel

1.	`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. `1//27`D. `24`
Answer» Correct Answer - A `underset((a-x))oversetaN_(2)+underset((b-3x))overset3H_(2)harrunderset((2x))overset02NH_(3)` `50%` Dissociation of `N_(2)` take palce so, At equilibrium `(2xx50)/(100) = 1` , value of x = 1 `K_(c) = ([2]^(2))/([1][3]^(3)) = (4)/(27)` So `K_(c) = (4)/(27)`c

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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. `1//27`D. `24`

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