Question

For a reaction equilibrium,\[{{N}_{2}}{{O}_{4}}\left( g \right)\to 2N{{O}_{2}}\left( g \right)\], the concentration of \[{{N}_{2}}{{O}_{4}}\left( g \right)\] and \[N{{O}_{2}}\] ​ at equilibrium are \[4.8\times {{10}^{-2}}\] and \[1.2\times {{10}^{-2}}\]mol/L respectively.
The value of \[K\] for the reaction is:
(A) \[3\times {{10}^{-3}}\]
(B) \[3.3\times {{10}^{-3}}\]
(C) \[3\times {{10}^{-1}}\]
(D) \[3.3\times {{10}^{-1}}\]

Answer 1

Hint: The equilibrium constant (\[{{K}_{eq}}\]) is a formula that illustrates the relationship between the concentrations of the reactants and the products.

Complete Step by Step Solution:
The equilibrium constant, which is frequently denoted by the letter [K] when a chemical reaction reaches equilibrium, provides information about how reactants and products interact. One method for determining the equilibrium constant of concentration (denoted by [K_c]) of a chemical reaction at equilibrium is the ratio of the concentration of the products to the concentration of the reactants, both elevated to their respective stoichiometric coefficients. It is important to keep in mind that different types of equilibrium constants are used to define interactions between reactants and products of equilibrium reactions in terms of different units. The ratio between the amounts of reactant and product for a chemical reaction can be described as the equilibrium constant.

At equilibrium, the rate of the forward reaction = the rate of the backward reaction
That is, \[{{R}_{f}}={{R}_{b}}\] or, \[{{K}_{f}}\times \alpha \times {{\left[ A \right]}^{a}}{{\left[ B \right]}^{b}}={{K}_{b}}\times \alpha \times {{\left[ C \right]}^{c}}{{\left[ D \right]}^{d}}\]
The rate constants are constant at a certain temperature. An equilibrium constant is the ratio of the forward reaction rate constant to the backward reaction rate constant, which must be a constant (\[{{K}_{eq}}\]).
\[{{K}_{eq}}={{K}_{f}}/{{K}_{b}}={{\left[ C \right]}^{c}}{{\left[ D \right]}^{d}}/{{\left[ A \right]}^{a}}{{\left[ B \right]}^{b}}={{K}_{c}}\]
Where\[{{K}_{c}}\] indicates the equilibrium constant measured in moles per litre.

According to our question,
The equilibrium reaction is \[{{N}_{2}}{{O}_{4}}\left( g \right)\to 2N{{O}_{2}}\left( g \right)\]
Using formula \[{{K}_{c}}={{\left[ C \right]}^{c}}{{\left[ D \right]}^{d}}/{{\left[ A \right]}^{a}}{{\left[ B \right]}^{b}}\]
Given, \[{{N}_{2}}{{O}_{4}}\left( g \right)\]= \[4.8\times {{10}^{-2}}\] mol/L
\[N{{O}_{2}}\]= \[1.2\times {{10}^{-2}}\] mol/L
Substituting the values, \[{{K}_{c}}={{\left[ N{{O}_{2}} \right]}^{2}}/\left[ {{N}_{2}}{{O}_{4}} \right]\]
\[{{K}_{c}}={{\left( 1.2\times {{10}^{-2}} \right)}^{2}}/4.8\times {{10}^{-2}}\]
\[{{K}_{c}}=3\times {{10}^{-3}}\] mol/L
The correct option is A.

Note: There is a separate formula for calculating the values for liquids and gases, respectively. When conditions are ideal, temperature and concentration have little effect on the calculation, but when conditions are not ideal, they have a substantial impact.