A mixture of 1.57 mol of \mathrm{N}_{2}, 1.92 mol of \mathrm{H}_{2} and 8.13 \mathrm{~mol} of \mathrm{NH}_{3} is introduced into a 20 \mathrm{~L} reaction vessel at 500 \mathrm{~K}. At this temperature, the equilibrium constant, \mathrm{K}_{\mathrm{c}} for the reaction N _{2}( g )+3 H _{2}( g ) \rightleftharpoons 2 NH _{3}( g ) is 1.7 \times 10^{2} Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?
A mixture of 1.57 mol of \mathrm{N}_{2}, 1.92 mol of \mathrm{H}_{2} and 8.13 \mathrm{~mol} of \mathrm{NH}_{3} is introduced into a 20 \mathrm{~L} reaction vessel at 500 \mathrm{~K}. At this temperature, the equilibrium constant, \mathrm{K}_{\mathrm{c}} for the reaction N _{2}( g )+3 H _{2}( g ) \rightleftharpoons 2 NH _{3}( g ) is 1.7 \times 10^{2} Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Answer:

The given reaction is:

\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})

The given concentration of various species is

\left[\mathrm{N}_{2}\right]=\frac{1.57}{20} molL^{-1} ,\left[\mathrm{H}_{2}\right]=\frac{1.92}{20} \operatorname{mol} L^{-1}, \left[\mathrm{NH}_{3}\right]=\frac{8.31}{20} \mathrm{~mol} L^{-1}

At any given moment in time during a reaction, the reaction quotient (Q) quantifies the relative quantities of products and reactants present in the reaction mixture at that time.
Now, reaction quotient \mathrm{Q}_{\mathrm{c}} is:

Q=\frac{\left[N H_{3}\right]^{2}}{\left[N_{2}]\left[H_{2}\right]^{3}\right.}

=\frac{\left(\frac{(8.13)}{20}\right)^{2}}{\left(\frac{1.55}{20}\right)\left(\frac{1.92}{20}\right)^{3}}

=2.4 \times 10^{3}

Since, \mathrm{Q}_{\mathrm{c}} \neq \mathrm{K}_{\mathrm{c}}, the reaction mixture is not at equilibrium.

As, Q_{c}>K_{c}. Hence, the reaction will move in the reverse direction.