Question

The reaction quotient Q for:$ {N_2}(g) + 3{H_2}(g) \rightleftharpoons 2N{H_3}(g) $ is given by $ Q = \dfrac{{{{\left[ {N{H_3}} \right]}^2}}}{{\left[ {{N_2}} \right]{{\left[ {{H_2}} \right]}^3}}}. $ The reaction will proceed in a backward direction, when:
(A) $ Q = {K_C} $
(B) $ Q < {K_C} $
(C) $ Q > {K_C} $
(D) $ Q = 0 $

Answer 1

Hint: In the reversible reaction, the direction of the reaction can be forward or backward. So, while performing these types of reactions in the laboratories, it is important to check the direction of the reaction at a particular point of time. The direction of the reaction in which it is proceeding can be determined with the help of the reaction quotient and equilibrium constant.

Complete answer:
We know that when a reaction is reversible in nature, it can proceed in forward direction as well as in backward direction according to the conditions in which the reaction is taking place. So, it is important to check the direction of the reaction at a particular point of time.
The given reaction is:
 $ {N_2}(g) + 3{H_2}(g) \rightleftharpoons 2N{H_3}(g) $
For the given reaction, we can write the reaction quotient as:
 $ Q = \dfrac{{{{\left[ {N{H_3}} \right]}^2}}}{{\left[ {{N_2}} \right]{{\left[ {{H_2}} \right]}^3}}} $
So, if the concentration of product increases, then to attain equilibrium, the reaction will proceed in the backward direction. Hence we can say that when the value of reaction quotient (Q) is greater than the value of equilibrium constant $ ({K_C}) $ , then the reaction will move in the backward direction.
Hence, the correct option is (C) $ Q > {K_C} $ .

Note:
We should note that there are two types of equilibrium constants such as one which is constant at constant pressure $ ({K_P}) $ and the other one is, which is constant at a constant concentration of reactants and products. The equilibrium constant varies with the change in temperature and the equation of the chemical reaction.