Equilibrium constant, `K_(C )` for the reaction
`N_(2)(g) + 3H_(2)(g) hArr 2NH_(3)(g)` at 500 K is 0.061
At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol `L^(-1)N_(2), 2.0 "mol L"^(-1) H_(2)`, 0.5 mol `L^(-1) NH_(3)`. Is the reaction at equilibrium ? If not, in which direction does the reaction tends to proceed to reach equilibirum ?

Explanation

As we know that
`Q_(C ) = ([NH_(3)(g)]^(2))/([N_(2)(g)][H_(2)(g)]^(3)) = ([0.500]^(2))/([3.00][2.00]^(3)) = (0.25)/(24) = 0.0104`
Now `Q_(C ) lt K_(C)`
So the reaction will not be in equilibirum and will proceed in forward direction.