Question

A reaction mixture containing `H_(2), N_(2)` and `NH_(3)` has partial pressures `2` atm, `1` atm, and `3` at,. Respectively, at `725 K`. If the value of `K_(p)` for the reaction, `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)` is `4.28xx10^(-5) atm^(-2)` at `725 K`, in which direction the net reaction will go?
A. Forward
B. Backward
C. No net reaction
D. Direction of reaction cannot be predicted.

Answer 1

`Q_(p)=((p_(NH_(3)))^(3))/(p_(N_(2))xx(P_(H_(2)))^(3))=((3)^(2))/((1)(2)^(3))=9/8 "atm"^(-2)=1.125 "atm"^(-2)`
Since the value of `Q_(p)` is larger then `K_(p) (4.28xx10^(-5) "atm"^(-2))`, it indicates net reaction will proceed in the backward direction.