Question

The equilibrium constants of the following are
${{\text{N}}_2} + 3{{\text{H}}_2} \rightleftharpoons 2{\text{N}}{{\text{H}}_3};{{\text{K}}_1}$
${{\text{N}}_2} + {{\text{O}}_2} \rightleftharpoons 2{\text{NO}};{{\text{K}}_2}$
${{\text{H}}_2} + \dfrac{1}{2}{{\text{O}}_2} \to {{\text{H}}_2}{\text{O;}}{{\text{K}}_3}$
The equilibrium constant $\left( K \right)$ of the reaction, $2{\text{N}}{{\text{H}}_3} + \dfrac{5}{2}{{\text{O}}_2} \rightleftharpoons 2{\text{NO}} + 3{{\text{H}}_2}{\text{O}}$ will be
A.${K_1}K_3^3/{K_2}$
B.${K_2}K_3^3/{K_1}$
C.${K_2}{K_3}/{K_1}$
D.$K_2^3{K_3}/{K_1}$

Answer 1

The equilibrium constant of a reaction is determined by the concentration of products at equilibrium divided by the concentration of reactants at equilibrium raised to the power of their respective stoichiometric coefficients in the balanced chemical equation. To solve this question, you must recall the formula for equilibrium constant of a reaction, and the combination of the equilibrium constants of various reactions to give the equilibrium constant of a single resultant reaction

Complete step by step answer:
The equilibrium constants of t two reactions are multiplied, when the two reactions are added, and the equilibrium constants of two reactions are divided when the two reactions are subtracted. If the reaction is multiplied by a scalar quantity, then the equilibrium constant of the reaction will be raised to the power of that number.
The reaction whose equilibrium constant is to be determined is given in the question as
$2{\text{N}}{{\text{H}}_3} + \dfrac{5}{2}{{\text{O}}_2} \rightleftharpoons 2{\text{NO}} + 3{{\text{H}}_2}{\text{O}}$
The equilibrium constant for the given reaction in terms of the concentration of its constituents can be written as, $K = \dfrac{{{{\left[ {NO} \right]}^2}{{\left[ {{H_2}O} \right]}^3}}}{{{{\left[ {N{H_3}} \right]}^2}{{\left[ {{O_2}} \right]}^{5/2}}}}$
The equilibrium constants already given in the question can be written in terms of the concentration of their respective constituents as,
${K_1} = \dfrac{{{{\left[ {N{H_3}} \right]}^2}}}{{\left[ {{N_2}} \right]{{\left[ {{H_2}} \right]}^3}}}$,
${K_2} = \dfrac{{{{\left[ {NO} \right]}^2}}}{{\left[ {{N_2}} \right]\left[ {{O_2}} \right]}}$ and
${K_3} = \dfrac{{\left[ {{H_2}O} \right]}}{{\left[ {{H_2}} \right]{{\left[ {{O_2}} \right]}^{1/2}}}}$
So we can write, $K = \dfrac{{{{\left[ {NO} \right]}^2}{{\left[ {{H_2}O} \right]}^3}}}{{{{\left[ {N{H_3}} \right]}^2}{{\left[ {{O_2}} \right]}^{5/2}}}} = \dfrac{{\left( {\dfrac{{{{\left[ {NO} \right]}^2}}}{{\left[ {{N_2}} \right]\left[ {{O_2}} \right]}}} \right){{\left( {\dfrac{{\left[ {{H_2}O} \right]}}{{\left[ {{H_2}} \right]{{\left[ {{O_2}} \right]}^{1/2}}}}} \right)}^3}}}{{\dfrac{{{{\left[ {N{H_3}} \right]}^2}}}{{\left[ {{N_2}} \right]{{\left[ {{H_2}} \right]}^3}}}}}$
Therefore, $K = \dfrac{{{K_2}K_3^3}}{{{K_1}}}$.

Thus, the correct answer is B.

Note:
The equilibrium constant is in general numerically calculated by allowing the reaction under consideration to proceed to equilibrium and then measuring the concentrations of each constituent present in the reaction mixture. Since the concentrations of the constituents of the reaction mixture are measured at equilibrium, the equilibrium constant always has the same value for a given reaction irrespective of the initial amount taken of the reactants.