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Equilibrium constant K(c) for the reacti...

Equilibrium constant `K_(c)` for the reaction , `N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g)` at 500 K is `0.061`, At particular time ,the analysis shows that the composition of the reaction mixture is 3.0 mol `L^(-1)` of `H_(2), 0.50` mol `L^(-1)` of `NH_(3)` . Is the reaction at equilibrium ?

Text Solution

Verified by Experts

The given reaction is `: N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g)`
According to available data
`N_(2) = [3.0] , H_(2) = [2,0] , NH_(3) = [0.50]`
`Q_(C) = ([NH_(3) (g)]^(2))/([N_(2) (g)] [H_(2) (g)]^(3)) = ([0.50]^(2))/([3.0][2.0]) = (0.25)/(24) = 0.0104`
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