Question

The equilibrium constants of the following are:
${N_2} + 3{H_2} \rightleftharpoons 2N{H_3};{K_1}$
${N_2} + {O_2} \rightleftharpoons 2NO;{K_2}$
${H_2} + \dfrac{1}{2}{O_2} \rightleftharpoons {H_2}O;{K_3}$
The equilibrium constant (K) of the reaction:
$2N{H_3} + \dfrac{5}{2}{O_2}\xrightarrow{k}2NO + 3{H_2}O$, will be:
A.$\dfrac{{{K_2}^3{K_3}}}{{{K_1}}}$
B.$\dfrac{{{K_1}{K_3}^3}}{{{K_2}}}$
C.$\dfrac{{{K_2}{K_3}^3}}{{{K_1}}}$
D.$\dfrac{{{K_2}{K_3}}}{{{K_1}}}$

Answer 1

The equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium. The chemical equilibrium is a state approached by a dynamic chemical system after sufficient time has elapsed at which its composition has no measurable tendency to further change. In simple words, at chemical equilibrium, the rate of the forward reaction is equal to the rate of backward reaction, and there is no net change in the concentration of reactants and products. The equilibrium constant expresses the relationship between products and reactants at equilibrium. It is numerically equal to the ratio of the product of the concentration of products to the product of reactants, each raised to some power equal to their stoichiometric coefficients.

Complete answer:
The equilibrium constant has some properties:
When a reaction is reversed, the new equilibrium constant is equal to the reciprocal of the original equilibrium constant.
When a chemical reaction is multiplied by a natural number the new equilibrium constant is equal to the original equilibrium raised to the power equal to the number the reaction was multiplied.
Taking the above points in mind, let us predict the equilibrium constant K,
$2N{H_3} + \dfrac{5}{2}{O_2}\xrightarrow{k}2NO + 3{H_2}O$
The above reaction is obtained by reversing the first equation and adding it to the second chemical reaction and three times the third reaction:
$2N{H_3} + {N_2} + {O_2} + 3{H_2} + \dfrac{3}{2}{O_2} \rightleftharpoons {N_2} + 3{H_2} + 2NO + 3{H_2}O$
This gives, $2N{H_3} + \dfrac{5}{2}{O_2}\xrightarrow{k}2NO + 3{H_2}O$
Therefore, we have, $K = \dfrac{{{K_2}{K_3}^3}}{{{K_1}}}$

Thus, the correct answer is option C.

Note:
The first chemical reaction for the formation of ammonia is known as Haber’s Process. Haber’s process is the industrial method for the production of ammonia. This is done in the presence of iron as a catalyst and molybdenum as an activator. The process is performed at a temperature ranging from 400-500 degrees Celsius and 200 atmospheres of pressure. The process is exothermic.