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Question: Given, 1 "mole of" \[{{N}_{2}}\]"and" 3 "moles of \({{H}_{2}}\) are placed in 1L vessels. Find the c...

Given, 1 "mole of" N2{{N}_{2}}"and" 3 "moles of H2{{H}_{2}} are placed in 1L vessels. Find the concentration of NH3N{{H}_{3}} at equilibrium, if the equilibrium constant (Kc)\left( {{K}_{c}} \right) at 400K "is" 427\dfrac{4}{27}

Explanation

Solution

A chemical reaction's equilibrium constant is the value of its reaction quotient at chemical equilibrium, a condition reached by a dynamic chemical system after a period of time has passed in which its composition shows no discernible tendency to change. The equilibrium constant is independent of the initial analytical concentrations of the reactant and product species in the mixture for a particular set of reaction circumstances.

Complete answer:
As a result, established equilibrium constant values may be used to calculate the composition of a system at equilibrium, given its beginning composition. The value of the equilibrium constant can, however, be influenced by reaction factors such as temperature, solvent, and ionic strength. Many chemical systems, as well as physiological processes like oxygen transport by haemoglobin in blood and acid–base balance in the human body, need an understanding of equilibrium constants.
For our question
N2+3H2→2NH3ΔH∘=−91.8 kJ/mol{{N}_{2}}+3{{H}_{2}}\to 2N{{H}_{3}}\quad \Delta {{H}^{{}^\circ }}=-91.8~\text{kJ/mol}
The equilibrium constant is written as
K=[ApBq][A]p[B]qK=\dfrac{[{{A}_{p}}{{B}_{q}}]}{{{[A]}^{p}}{{[B]}^{q}}}
Volume = 1 L
Temperature = 400 K
Equilibrium constant, Kc=427{{K}_{c}}=\dfrac{4}{27}
The net balanced chemical reactions
N2+3H2→2NH3{{N}_{2}}+3{{H}_{2}}\to 2N{{H}_{3}}\quad

Initial Concentration120
At equilibrium1-x3-3x2x

The equilibrium constant Kc=[NH3]2[N2]1[H2]3{{K}_{c}}=\dfrac{{{[N{{H}_{3}}]}^{2}}}{{{[{{N}_{2}}]}^{1}}{{[{{H}_{2}}]}^{3}}}
Now from the question
Kc=[NH3]2[N2]1[H2]3=427=(2x)2(1−x)(3−3x)2{{K}_{c}}=\dfrac{{{[N{{H}_{3}}]}^{2}}}{{{[{{N}_{2}}]}^{1}}{{[{{H}_{2}}]}^{3}}}=\dfrac{4}{27}=\dfrac{{{(2x)}^{2}}}{(1-x){{(3-3x)}^{2}}}
Find the value of x
427=(2x)2(1−x)(3−3x)2\dfrac{4}{27}=\dfrac{{{(2x)}^{2}}}{(1-x){{(3-3x)}^{2}}}
x=0.381Mx=0.381M
Concentration of NH3N{{H}_{3}}=2x=2(0.381)M2x=2(0.381)M
Concentration of NH3N{{H}_{3}}==0.762M=0.762M

Note:
The determination of stability constant values beyond the usual range for a specific procedure is one application of a stepwise constant. Many metal EDTA complexes, for example, are outside the potentiometric method's range. Competition with a weaker ligand dictated the stability constants for such complexes.