A mixture of 1.57 mol of `N_(2),1.92` mol of `H_(2)` & 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 2NO_(3)(g)` is `1.7xx10^(2)`. is the reaction mixture at equilibrium ? if not, what is the direction of the net reaction?

Explanation

The reaction quotient for the given reaction:
`N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)` is `Q_(c)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))`
Molar concentrations of the substances in the initial mixture:
`[NH_(3)]=(8.13)/(20)=0.4065mol*L^(-1)`
`[N_(2)]=(1.57)/(20)=0.0785mol*L^(-1)`
and `[H_(2)]=(1.92)/(20)=0.096mol*L^(-1)`
`therefore Q_(c)=((0.4065)^(2))/(0.0785xx(0.096)^(3))=2.38xx10^(3)`
So, `Q_(c) gt K_(c)`. Hence, the reaction is not in equilibrium. as `Q_(c) gtK_(c)`, the net reaction will occur in the reverse direction.