The synthesis of ammonia is given as:
`N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g), DeltaH^(ɵ)=-92.6 kJ mol^(-1)` given `K_(c)=1.2` and temperature `(T)=375^(@)C`
Starting with `2` mol of each `(N_(2), H_(2) "and" NH_(3))` in `5.0 L` reaction vessel at `375^(@)C`, predict what is true for the reaction?

Options

A. The reaction is at equilibrium
B. The reaction proceed in forward direction.
C. The reaction proceed in backward direction
D. `Q_(c)` for the reaction is less then `K_(c)`

Answer

C

Explanation

`[N_(2)]=2/5M, [H_(2)]=2/5 M, [NH_(3)]=2/5 M`
`Q_(c)=([NH_(3)])/([N_(2)][H_(2)]^(3))=((2//5)^(2))/((2//5)(2//5)^(3))=(2//5)^(-2)=6.25`
Since `Q_(c)` is greater than `K_(c) (1.2)` the reaction takes place in backward direction.