Question

How to calculate\[{{\text{K}}_c}\]?\[{{\text{N}}_2}\left( g \right){\text{ + }}{{\text{O}}_2}\left( g \right) \rightleftarrows \,\,\,2{\text{NO}}\left( g \right)\]
The equilibrium concentrations of the gases at \[{\text{1500}}\,{\text{K}}\]are
\[{\text{ }}{{\text{O}}_2} = 1.7 \times {10^{ - 3}}\,{\text{M}}\]
\[{\text{ }}{{\text{N}}_2} = 6.4 \times {10^{ - 3}}\,{\text{M}}\]
\[{\text{ NO}} = 1.1 \times {10^{ - 5}}\,{\text{M}}\]
Calculate the value of\[{{\text{K}}_c}\] at \[{\text{1500}}\,{\text{K}}\]from this date.

Answer 1

Hint:Here, \[{{\text{K}}_c}\]is the equilibrium constant which is calculated by taking the ratio of the concentration of products to the concentration of the reactants at equilibrium condition.
It is a measure of the equilibrium position of the reaction. If the value is less than 1 indicates the concentration of reactant is higher than the concentration of product and hence, favours the formation of the products.If the value is greater than 1 indicates the concentration of products is higher than the concentration of reactants and hence, favours the formation of the products.

Complete step-by-step answer:Here, the reaction given is \[{{\text{N}}_2}\left( g \right){\text{ + }}{{\text{O}}_2}\left( g \right) \rightleftarrows \,\,\,2{\text{NO}}\left( g \right)\].
The \[{{\text{K}}_c}\] for the reaction is calculated by taking the ratio of the concentration of \[{\text{NO}}\] raised to its power 2, to the product of the concentration of \[{{\text{N}}_2}\]and\[{{\text{O}}_2}\].
The equilibrium constant expression for the reaction given is as follows:
\[{{\text{K}}_c} = \dfrac{{{{\left[ {{\text{NO}}} \right]}^2}}}{{\left[ {{{\text{N}}_2}} \right]\left[ {{{\text{O}}_2}} \right]}}\]
Here, substitute \[1.7 \times {10^{ - 3}}\,{\text{M}}\] for \[\left[ {{{\text{O}}_2}} \right]\], \[6.4 \times {10^{ - 3}}\,{\text{M}}\] for \[\left[ {{{\text{N}}_2}} \right]\], and \[1.1 \times {10^{ - 5}}\,{\text{M}}\] for \[\left[ {{\text{NO}}} \right]\].
\[\Rightarrow {{\text{K}}_c} = \dfrac{{{{\left( {1.1 \times {{10}^{ - 5}}\,{\text{M}}} \right)}^2}}}{{\left( {6.4 \times {{10}^{ - 3}}\,{\text{M}}} \right)\left( {1.7 \times {{10}^{ - 3}}\,{\text{M}}} \right)}}\]
\[\Rightarrow {{\text{K}}_c} = \dfrac{{1.21 \times {{10}^{ - 10}}\,{{\text{M}}^2}}}{{1.088 \times {{10}^{ - 5}}\,{\text{M}}}}\]
\[\therefore {{\text{K}}_c} = 1.112 \times {10^{ - 5}}\,{\text{M}}\]
Thus, the value equilibrium constant for reaction at \[{\text{1500}}\,{\text{K}}\] is \[1.1 \times {10^{ - 5}}\,{\text{M}}\].
Here, we can see that the value of the equilibrium constant is less than one indicates reactants concentration is higher than the concentration of products, and hence, the reaction proceeds in direction of the product formation.

Note:In equilibrium constant the concentration is equilibrium concentration while in reaction quotients the concentration is other than equilibrium concentration.The values of the reaction quotient and equilibrium constants are used to decide whether the formation reactant or product is favoured. That is the direction of the reaction is decided using the values of the reaction quotient and equilibrium constant.