Question

Equilibrium constants $\mathop K\nolimits_1 $ and $\mathop K\nolimits_2 $ for the following equilibria are related as:
A.\[NO\left( g \right) + \dfrac{1}{2}\mathop O\nolimits_2 \overset {{K_1}} \leftrightarrows \mathop {NO}\nolimits_2 \left( g \right)\]
B.\[2NO2\left( g \right)\overset {{K_2}} \leftrightarrows 2NO\left( g \right) + O2\left( g \right)\] (A) $\mathop K\nolimits_2 = \dfrac{1}{{\mathop K\nolimits_1 }}$(B) $\mathop K\nolimits_2 = \mathop K\nolimits_1^2 $

(C)$\mathop K\nolimits_2 = \dfrac{{\mathop K\nolimits_1 }}{2}$
D.\[\mathop K\nolimits_2 = \dfrac{1}{{\mathop K\nolimits_1^2 }}\]

Answer 1

Hint: In a reversible reaction, at equilibrium the product of the molar concentration of products, each raised to the power equal to its coefficient, divided by the product of the molar concentration of the reactant each raised to the power equal to its coefficient, is constant at a constant temperature and is called equilibrium constant.

Complete step by step answer:
Firstly, we will find the equilibrium constant for both the reactions
Where $\mathop K\nolimits_1 $ is equilibrium constant of first reaction which is product of the molar concentration of the reactant each raised to the power equal to its coefficient
$\mathop K\nolimits_1 = \dfrac{{\mathop {[\mathop {NO}\nolimits_2 ]}\nolimits^{} }}{{[\mathop {NO}\limits^{} ][\sqrt {\mathop O\nolimits_2 } ]}}............eqn1$
Where $\mathop K\nolimits_2 $ is equilibrium constant of second reaction that is product of the molar concentration of the reactant each raised to the power equal to its coefficient

$\mathop K\nolimits_2 = \dfrac{{[\mathop O\nolimits_2 ]\mathop {[NO]}\nolimits^2 }}{{\mathop {[\mathop {NO}\nolimits_2 ]}\nolimits^2 }}........eqn2$

To remove root or power $\dfrac{1}{2}$ from equation 1 we square both sides so equation becomes
$\mathop K\nolimits_1^2 = \dfrac{{\mathop {[\mathop {NO}\nolimits_2 ]}\nolimits^2 }}{{\mathop {[NO]}\nolimits^2 \mathop {[\sqrt {\mathop O\nolimits_2 } ]}\nolimits^2 }}............eqn3$

It becomes
$\mathop K\nolimits_1^2 = \dfrac{{\mathop {[\mathop {NO}\nolimits_2 ]}\nolimits^2 }}{{\mathop {[NO]}\nolimits^2 [\mathop O\nolimits_2 ]}}............eqn3$

Now if we reverse this equation it becomes equal to rate of reaction K2
$\dfrac{1}{{\mathop K\nolimits_1^2 }} = \dfrac{{[\mathop O\nolimits_2 ]\mathop {[\mathop {NO}\nolimits_2 ]}\nolimits^2 }}{{\mathop {[NO]}\nolimits^2 }}............eqn4$

So, now \[\mathop {\mathop k\nolimits_2 = \dfrac{1}{{\mathop K\nolimits_1^2 }}}\nolimits_{}^{} \] and hence option D is correct.


Note:
Law of chemical equilibrium and equilibrium constant as the law of mass action. It states that the rate at which a substance reacts is proportional to its active mass. Active mass means molar concentration i.e., the number of gram moles of the solute per unit mass and the rate of a chemical reaction is proportional to the product of the active masses of the reacting volume.