Question

At equilibrium, the concentration of ${{{N}}_2},\left[ {{{{N}}_2}} \right] = 3 \times {10^{ - 3}}{{M}}$, ${{{O}}_2},\left[ {{{{O}}_2}} \right] = 4.2 \times {10^{ - 3}}{{M}}$ and ${{NO,}}\left[ {{{NO}}} \right]{{ = 2}}{{.8}} \times {{1}}{{{0}}^{ - 3}}{{M}}$ in a diluted amount at $800{{K}}$. What will be the ${{{K}}_{{c}}}$ for the reaction?
${{{N}}_{2\left( {{g}} \right)}}{{ + }}{{{O}}_{2\left( {{g}} \right)}} \rightleftharpoons 2{{N}}{{{O}}_{\left( {{g}} \right)}}$

Answer 1

Hint:The forward and reverse reactions proceed at the same rate at equilibrium. Also, when equilibrium is achieved, both the reactants and products remain constant. Equilibrium constant expression is applicable only when the reactant concentrations and product concentrations are constant at equilibrium.

Complete step by step solution:
Consider an equilibrium system,
${{xA + yB}} \rightleftharpoons {{zC}}$
The partial pressures of the gases will be equal at equilibrium. When the time moves on, products B and C get accumulated and the reactant A gets depleted. Thus the rate of forward reaction decreases and the rate of reverse reaction increases. Thus the system will be in equilibrium.
Equilibrium constant can be calculated using the concentrations of both reactants and products.
At a certain temperature, the product of the concentrations of products raised to the corresponding stoichiometric coefficient divided by the product of concentrations of reactants raised to corresponding stoichiometric coefficients.
The equilibrium constant, ${{{K}}_{{c}}} = \dfrac{{{{\left[ {{C}} \right]}^{{z}}}}}{{{{\left[ {{A}} \right]}^{{x}}}{{\left[ {{B}} \right]}^{{y}}}}}$
The given reaction is ${{{N}}_{2\left( {{g}} \right)}}{{ + }}{{{O}}_{2\left( {{g}} \right)}} \rightleftharpoons 2{{N}}{{{O}}_{\left( {{g}} \right)}}$
It is given that the the concentration of ${{{N}}_2},\left[ {{{{N}}_2}} \right] = 3 \times {10^{ - 3}}{{M}}$
$\left[ {{{{O}}_2}} \right] = 4.2 \times {10^{ - 3}}{{M}}$ , $\left[ {{{NO}}} \right]{{ = 2}}{{.8}} \times {{1}}{{{0}}^{ - 3}}{{M}}$
The stoichiometric coefficients, ${{x = 1}}$, ${{y = 1}}$, ${{z = 2}}$.
Substituting these values, we get
\[{{{K}}_{{c}}} = \dfrac{{{{\left( {{{2}}{{.8}} \times {{1}}{{{0}}^{ - 3}}} \right)}^2}}}{{\left( {3 \times {{10}^{ - 3}}} \right)\left( {4.2 \times {{10}^{ - 3}}} \right)}}\]
On simplification, we get
\[{{{K}}_{{c}}} = \dfrac{{7.84 \times {{10}^{ - 6}}}}{{1.26 \times {{10}^{ - 5}}}} \rightleftharpoons {{{K}}_{{c}}} = 0.622\]
Thus the equilibrium constant, \[{{{K}}_{{c}}} = 0.622\]

Additional information:
Equilibrium constant does not depend on the initial concentrations of reactants and products. It is also temperature dependent. If the equilibrium constant has a very large value, then the reaction is approaching completion.

Note:
Equilibrium constant for the reverse reaction is the inverse of the equilibrium constant in the forward reaction. If it is greater than one, then it is product favored. If it is less than one, then it is reactants favored. Here, it is less than one. So it is reactant favored.