Question

A mixture of \[{{\text{H}}_{\text{2}}}{\text{, }}{{\text{N}}_{\text{2}}}\] and \[{\text{N}}{{\text{H}}_3}\] with molar concentration \[{\text{5 }} \times {\text{ 1}}{{\text{0}}^{ - {\text{3}}}}{\text{mol/L, 4 }} \times {\text{ 1}}{{\text{0}}^{ - {\text{3}}}}{\text{mol/L}}\] and \[{\text{2 }} \times {\text{ 1}}{{\text{0}}^{ - {\text{3}}}}{\text{mol/L}}\] respectively was prepared and heated to \[{\text{500 K}}\] .The value of K, for the reaction \[{{\text{N}}_{\text{2}}}\left( g \right){\text{ + 3 }}{{\text{H}}_{\text{2}}}\left( g \right){\text{ }} \rightleftharpoons {\text{ 2 N}}{{\text{H}}_{\text{3}}}\left( g \right)\] at this given temperature is 60.Predict whether ammonia tends to form or decompose at this stage of concentration

Answer 1

Hint:From the given values of the concentrations of the reactants and products, calculate the value of the reaction quotient. Then compare the value of the reaction quotient with the value of the equilibrium constant and decide if ammonia is formed or decomposed at present concentration.

Complete answer:
Nitrogen gas reacts with hydrogen gas to form ammonia gas. Write the balanced chemical equation for this reaction:
\[{{\text{N}}_{\text{2}}}\left( g \right){\text{ + 3 }}{{\text{H}}_{\text{2}}}\left( g \right){\text{ }} \rightleftharpoons {\text{ 2 N}}{{\text{H}}_{\text{3}}}\left( g \right)\]
Write the expression for the equilibrium constant for the above reaction
\[\Rightarrow {{\text{K}}_{\text{c}}}{\text{ = }}\dfrac{{{{\left[ {{\text{N}}{{\text{H}}_{\text{3}}}} \right]}^2}}}{{\left[ {{{\text{N}}_{\text{2}}}} \right] \times {{\left[ {{{\text{H}}_{\text{2}}}} \right]}^3}}}\]
\[\Rightarrow {{\text{K}}_{\text{c}}}{\text{ = 60}}\]
Write the expression for the reaction quotient and substitute values of various concentrations to calculate the reaction quotient
\[\Rightarrow {{\text{Q}}_{\text{c}}}{\text{ = }}\dfrac{{{{\left[ {{\text{N}}{{\text{H}}_{\text{3}}}} \right]}^2}}}{{\left[ {{{\text{N}}_{\text{2}}}} \right] \times {{\left[ {{{\text{H}}_{\text{2}}}} \right]}^3}}}\]
\[\Rightarrow {{\text{Q}}_{\text{c}}}{\text{ = }}\dfrac{{{{\left( {{\text{2 }}\times {{\text{10}}^{{\text{ - 3}}}}} \right)}^2}}}{{\left( {{\text{5 }}\times {{\text{10}}^{{\text{ - 3}}}}} \right) \times {{\left( {4{\text{ }}\times {{\text{10}}^{{\text{ - 3}}}}} \right)}^3}}}\]
\[\Rightarrow {{\text{Q}}_{\text{c}}}{\text{ = }}100\]
But \[{{\text{K}}_{\text{c}}}{\text{ = 60}}\]
Hence, \[{{\text{Q}}_{\text{c}}} > {{\text{K}}_{\text{c}}}\]
The value of the reaction quotient is greater than the value of the equilibrium constant. The concentration of ammonia is greater than its equilibrium value. The concentrations of nitrogen and hydrogen are smaller than their equilibrium values.
The equilibrium will shift towards reactants. More and more ammonia will decompose to form more nitrogen and more hydrogen.


Note:
A reaction is at equilibrium when the value of the reaction quotient is equal to the value of the equilibrium constant. At this point, the instantaneous concentrations are equal to the equilibrium concentrations.