Question

Equilibrium Constant ${{\text{K}}_{\text{1}}}$and ${{\text{K}}_{2}}$for the following equilibria are related as:
a.\[\text{N}{{\text{O}}_{\left( \text{g} \right)}}\text{ + }\dfrac{\text{1}}{\text{2}}{{\text{O}}_{\text{2}}}_{\left( \text{g} \right)}\rightleftharpoons \text{N}{{\text{O}}_{\text{2}}}_{\left( \text{g} \right)}\]
b.$\text{2N}{{\text{O}}_{\text{2}}}_{\left( \text{g} \right)}\rightleftharpoons \text{2N}{{\text{O}}_{\left( \text{g} \right)}}\text{+ }{{\text{O}}_{\text{2}}}_{\left( \text{g} \right)}$
A.${{\text{K}}_{\text{1}}}\text{=}\sqrt{{{\text{K}}_{\text{2}}}}$
B.${{\text{K}}_{2}}\text{=}\frac{\text{1}}{{{\text{K}}_{1}}}$
C.${{\text{K}}_{\text{1}}}\text{= 2}{{\text{K}}_{\text{2}}}$
D.${{\text{K}}_{2}}\text{=}\frac{\text{1}}{\text{K}_{1}^{2}}$

Answer 1

Hint:When the rates of the forward reaction and the backward reaction become equal then there exists a dynamic equilibrium in the reaction medium. The equilibrium is guided by different parameters such as the temperature, pressure, concentrations, etc.

Complete step by step answer:
The equilibrium constant is equal to the ratio of the concentration of the products to the concentration of the reactants raised to their stoichiometric coefficient in the balanced rate equation.
As per the given equations, the equilibrium constant for the first equation,
\[\text{N}{{\text{O}}_{\left( \text{g} \right)}}\text{ + }\dfrac{\text{1}}{\text{2}}{{\text{O}}_{\text{2}}}_{\left( \text{g} \right)}\rightleftharpoons \text{N}{{\text{O}}_{\text{2}}}_{\left( \text{g} \right)}\] is:
 ${{\text{K}}_{\text{1}}}\text{=}\dfrac{\left[ \text{N}{{\text{O}}_{\text{2}}} \right]}{\left[ \text{NO} \right]{{\left[ {{\text{O}}_{\text{2}}} \right]}^{\dfrac{\text{1}}{\text{2}}}}}$$\Rightarrow \left[ \text{N}{{\text{O}}_{\text{2}}} \right]\text{=}{{\text{K}}_{\text{1}}}\left[ \text{NO} \right]{{\left[ {{\text{O}}_{\text{2}}} \right]}^{\dfrac{\text{1}}{\text{2}}}}$…… 1
For the second reaction $\text{2N}{{\text{O}}_{\text{2}}}_{\left( \text{g} \right)}\rightleftharpoons \text{2N}{{\text{O}}_{\left( \text{g} \right)}}\text{+ }{{\text{O}}_{\text{2}}}_{\left( \text{g} \right)}$,
\[{{\text{K}}_{2}}\text{=}\dfrac{{{\left[ \text{NO} \right]}^{2}}\left[ {{\text{O}}_{\text{2}}} \right]}{{{\left[ \text{N}{{\text{O}}_{\text{2}}} \right]}^{2}}}\]\[\Rightarrow \left[ \text{N}{{\text{O}}_{\text{2}}} \right]\text{=}\sqrt{\dfrac{{{\left[ \text{NO} \right]}^{2}}\left[ {{\text{O}}_{\text{2}}} \right]}{{{\text{K}}_{2}}}}\]……2
Equating, equation 1 and 2, we get,
${{\text{K}}_{\text{1}}}\left[ \text{NO} \right]{{\left[ {{\text{O}}_{\text{2}}} \right]}^{\dfrac{\text{1}}{\text{2}}}}=\sqrt{\dfrac{{{\left[ \text{NO} \right]}^{2}}\left[ {{\text{O}}_{\text{2}}} \right]}{{{\text{K}}_{2}}}}$
$\Rightarrow {{\text{K}}_{\text{1}}}\left[ \text{NO} \right]{{\left[ {{\text{O}}_{\text{2}}} \right]}^{\dfrac{\text{1}}{\text{2}}}}=\left[ \text{NO} \right]{{\left[ {{\text{O}}_{\text{2}}} \right]}^{\dfrac{\text{1}}{\text{2}}}}\sqrt{\dfrac{1}{{{\text{K}}_{2}}}}$
Solving this:
$\Rightarrow {{\text{K}}_{\text{1}}}=\sqrt{\dfrac{1}{{{\text{K}}_{2}}}}$
$\Rightarrow {{\text{K}}_{2}}\text{=}\dfrac{\text{1}}{\text{K}_{1}^{2}}$

Therefore, the correct answer is option D.

Note:
There are different types of equilibrium constants depending upon the type of the reactants involved. Stepwise formation constants are those types of equilibrium constants in which the equilibrium concentration of the reactants and the products involved in each step is written.
The association and the dissociation constants are those that show the concentrations of the dissociations and associations. In case of complex reactions, the equilibrium constant involves the concentration of the metal complex formed after the reaction and the concentration of the metal ions and the ligands present in the medium before the reaction and are called “the stability constants.”
Hydrolysis constants denote the concentrations of the hydroxide ion and the hydronium ion that is formed in the medium due to the dissociation in the aqueous solutions.