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

`Q_(c) = 2.379 xx 10^(3)`, no. reaction is not at equilibrium.

Explanation

The given reaction is :
`N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g))`
The given concentration of various species is
`{:([N_(2)] = (1.57)/(20)"mol" L^(-1),[H_(2)]=(1.92)/(20)"mol"L^(-1)),([NH_(3)]= (8.13)/(20)"mol"L^(-1),):}`
Now, reaction quotient `Q_(c) ` is :
`Q_(c) = ([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))`
`= (((8.13))/(20))^(2)/((1.57/20)((1.92)/(20))^(2))`
`= 2.4 xx 10^(3)`
Since `Q_(c) ne K_(c)`, the reaction mixture is not at equilibrium.
Again `K_(c) gt K_(c)`. Hence, the reaction will proceed in the reverse direction.