Question

A mixture of `1.57 mol` of `N_(2), 1.92 mol` of `H_(2)` and `8.13 mol` of `NH_(3)` is introduced into a `20 L` reaction vessel at `500 K`. At this temperature, the equilibrium constant `K_(c )` for the reaction `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)` is `1.7xx10^(2)`. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Answer 1

The reaction is: `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)`
`Q_(c )=([NH_(3)]^(2))/([N_(2)][H_(2)])`
`=((8.13//20 mol L^(-1)))/((1.57//20 mol L^(-1))(1.92//20 mol L^(-1)))=2.38xx10^(3)`
As `Q_(c )!=K_(c )`, the reaction mixture is not in equilibrium.
As `Q_(c )gt K_(c )`, the reaction will be in the backward direction.