Question

\[2{\text{ }}moles\] of \[{N_2}\] are mixed with \[6{\text{ }}moles\] of \[{H_2}\] in closed vessel of one liter capacity. If \[50\% \] \[{N_2}\] is converted into \[N{H_3}\] at equilibrium, the value of \[{K_c}\] for the reaction,
\[{N_2}_{\left( g \right)} + 3{H_2}_{\left( g \right)} \rightleftharpoons 2N{H_3}_{\left( g \right)}\]
A.4/27
B.27/4
C.1/27
D.9

Answer 1

Hint: We have to calculate the equilibrium constant \[\left( {{K_c}} \right)\]of the balanced reaction. We can use the law of mass action to solve this question. Before that, we must know the equilibrium constant is the value of its reaction quotient at chemical equilibrium.
The equilibrium constant for a given reaction depends on temperature only. For endothermic reactions equilibrium constant will increase on raising the temperature while for exothermic reactions equilibrium constant will decrease on raising the temperature.

Complete step by step answer:
In the question, they have given
\[2{\text{ }}moles\]of \[{N_2}\]and \[6{\text{ }}moles\]of \[{H_2}\]initially
Therefore, \[50\% {\text{ }}of{\text{ }}2{\text{ }}mol{\text{ }}{N_2} = {\text{ }}2 \times \dfrac{{50}}{{100}} = 1mol\]
So, \[1{\text{ }}mol{\text{ }}of{\text{ }}{N_2}\]has been reacted up to equilibrium establishment and \[1{\text{ }}mol{\text{ }}of{\text{ }}{N_2}\] left at equilibrium.
For reaction of \[1{\text{ }}mol{\text{ }}of{\text{ }}{N_2}\] , we need \[3{\text{ }}mol{\text{ }}of{\text{ }}{H_2}\]according to stoichiometry of reaction.
\[1{\text{ }}mol{\text{ }}of{\text{ }}{N_2}\] will form\[2{\text{ }}mol{\text{ }}of{\text{ }}N{H_3}\].
Balanced chemical reaction:
                                                   \[{N_2}_{\left( g \right)} + 3{H_2}_{\left( g \right)} \rightleftharpoons 2N{H_3}_{\left( g \right)}\]

Given	2mol	6mol	0
At equilibrium moles	2-1 = 1	6-3=3	2
Equilibrium conc.	1/1	3/1	2/1
	1M	3M	2M

We can write the equilibrium constant by use of law of mass value as
\[{K_c}{\text{ }} = {\text{ }}\dfrac{{{{\left[ {N{H_3}} \right]}^2}}}{{\left[ {{N_2}} \right].{{\left[ {{H_2}} \right]}^3}}}\] ;
 Here \[\left[ {N{H_3}} \right]\] = active mass or molar concentration of ammonia at equilibrium state.
\[\left[ {{N_2}} \right]\]= Active mass or molar concentration of Nitrogen gas at equilibrium state.
\[\left[ {{H_2}} \right]\]= Active mass or molar concentration of Hydrogen gas at equilibrium state.
We know the molar concentration of ammonia, nitrogen and hydrogen. Therefore, we can substitute the value in this equation
\[{K_c} = \dfrac{{{{(2)}^2}}}{{1x{{(3)}^3}}} = \dfrac{4}{{27}}\]
\[{K_c} = {\text{ }}4/27\]
Hence, the equilibrium constant for balanced reaction is \[{K_c} = {\text{ }}4/27\]

Note:
We are always using equilibrium molar concentrations in the expression of equilibrium constant\[\left( {{K_c}} \right)\]. Concentrations of reactants and products remain constant at equilibrium state. The value of the equilibrium constant does not depend on the volume, moles, concentration, pressure, etc. The equilibrium constant value for a reaction is the measurement of the yield of the reaction. The high value of the equilibrium constant shows the high yield of reaction and the low value of the equilibrium constant shows low yield of the reaction.