Question

The following equilibrium constants are given:
${N_2} + 3{H_2} \rightleftharpoons 2N{H_3};{K_1}$
${N_2} + {O_2} \rightleftharpoons 2NO;{K_2}$
${H_2} + 1/2{O_2} \rightleftharpoons {H_2}O;{K_3}$
The equilibrium constant for the oxidation of the $N{H_3}$ by oxygen to give $NO$ is:
A.$\dfrac{{{K_1}{K_2}}}{{{K_3}}}$
B.$\dfrac{{{K_2}K_3^3}}{{{K_1}}}$
C.$\dfrac{{{K_2}K_3^2}}{{{K_1}}}$
D.$\dfrac{{K_2^2{K_3}}}{{{K_1}}}$

Answer 1

Hint:The equilibrium constant is given by the concentration at equilibrium of products divided by the concentration at equilibrium of reactants raised to the power of their respective stoichiometric coefficients. To answer this question, you must recall the formula of equilibrium constant of a reaction, that is $K$ and their combination to give the equilibrium constant of a resultant reaction.

Complete step by step answer:
When two reactions are added, the equilibrium constants of the two reactions are multiplied and when two reactions are subtracted, the equilibrium constants of the reactions are divided. If we multiply the reaction with a scalar quantity, then the equilibrium constant will be raised to the power of that number.
The reaction of oxidation of $N{H_3}$by oxygen to $NO$ can be written as:
$4N{H_3} + 5{O_2}\xrightarrow[{{{800}^{\text{o}}}C}]{{Pt(gauge)}}4NO + 6{H_2}O$
We can write the given equilibrium constant for the reaction as
$K = \dfrac{{{{\left[ {NO} \right]}^4}{{\left[ {{H_2}O} \right]}^6}}}{{{{\left[ {N{H_3}} \right]}^4}{{\left[ {{O_2}} \right]}^5}}}$
The equilibrium constants given in the question are,
${K_1} = \dfrac{{{{\left[ {N{H_3}} \right]}^2}}}{{\left[ {{N_2}} \right]{{\left[ {{H_2}} \right]}^3}}}$, ${K_2} = \dfrac{{{{\left[ {NO} \right]}^2}}}{{\left[ {{N_2}} \right]\left[ {{O_2}} \right]}}$and ${K_3} = \dfrac{{\left[ {{H_2}O} \right]}}{{\left[ {{H_2}} \right]{{\left[ {{O_2}} \right]}^{1/2}}}}$
So we can write, $K = \dfrac{{{{\left[ {NO} \right]}^4}{{\left[ {{H_2}O} \right]}^6}}}{{{{\left[ {N{H_3}} \right]}^4}{{\left[ {{O_2}} \right]}^5}}} = \dfrac{{\left( {\dfrac{{{{\left[ {NO} \right]}^2}}}{{\left[ {{N_2}} \right]\left[ {{O_2}} \right]}}} \right){{\left( {\dfrac{{\left[ {{H_2}O} \right]}}{{\left[ {{H_2}} \right]{{\left[ {{O_2}} \right]}^{1/2}}}}} \right)}^3}}}{{\dfrac{{{{\left[ {N{H_3}} \right]}^2}}}{{\left[ {{N_2}} \right]{{\left[ {{H_2}} \right]}^3}}}}}$
Therefore, $K = \dfrac{{{K_2}K_3^3}}{{{K_1}}}$.

Thus, the correct answer is B.
Note:
The numerical value of an equilibrium constant is generally obtained by allowing the reaction under consideration to proceed to equilibrium and then measuring the concentrations of each substance present in the reaction mixture. Since the concentrations of the reagents are measured at equilibrium, the equilibrium constant always remains the same for a given reaction irrespective of the initial concentrations of the reactants taken. Using this knowledge we are able to derive a standard expression that serves as a standard for all reactions. This basic model form of the equilibrium constant is given as
${K_{eq}} = \dfrac{{{{\left( {{a_C}} \right)}^c}}}{{{{\left( {{a_A}} \right)}^a}{{\left( {{a_B}} \right)}^b}}}$ for a reaction: $aA + bB \rightleftharpoons cC$