Question

For the reaction \(2 \mathrm{NO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{5}(\dot{g})\), if the equilibrium constant is \(K_{p}\), then the equilibrium constant for the reaction \(2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\) would be : (a) \(K_{p}^{2}\) (b) \(\frac{2}{K_{p}}\) (c) \(\frac{1}{K_{p}^{2}}\) (d) \(\frac{1}{\sqrt{K_{p}}}\)

Short answer

\(\frac{1}{K_p^2}\)

Step-by-step solution

Identify the Original Reaction and its Equilibrium Constant

The original reaction is given as: \(2 \mathrm{NO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{5}(g)\) with the equilibrium constant \(K_p\).

Write the Reverse Reaction

The reverse reaction would be: \(\mathrm{N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)\). The equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction, or \(1/K_p\).

Double the Reverse Reaction

If we double the coefficients of the reverse reaction, we get: \(2 \mathrm{N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons 4 \mathrm{NO}_{2}(g)+ \mathrm{O}_{2}(g)\). The equilibrium constant \(K_{p-new}\) for the reaction when it is doubled is \(K_{p-new} = \left(1/K_p\right)^2\).

Choose the Correct Option

The correct option reflecting the equilibrium constant for the doubled reverse reaction is \(\frac{1}{\sqrt{K_p}}\). This simplifies to \(\frac{1}{K_p^2}\) as the equilibrium constant for the reaction in this case.

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as K, is a number that defines the ratio of the concentration of the products to the concentration of the reactants for a reversible chemical reaction at equilibrium. The equilibrium constant is crucial for predicting the extent of a reaction and understanding how different conditions affect the chemical system.

For the reaction given in the exercise, the equilibrium constant expressed in terms of partial pressures (Kp) represents the state of equilibrium for the forward reaction. When the direction of the reaction is reversed, as in the solution steps, the new equilibrium constant becomes the reciprocal of the original one. Furthermore, when reaction coefficients are multiplied (as in doubling the reaction), the equilibrium constant is raised to the power corresponding to the multiplier (squared in the exercise's case). Understanding how to manipulate the equilibrium constant formula depending on the changes in the chemical equation is a valuable skill in physical chemistry.

Le Chatelier's Principle

Le Chatelier's principle provides a qualitative means of predicting the shift in direction a chemical reaction will take when it is subjected to changes in concentration, temperature, or pressure. In essence, this principle suggests that if an external stress is applied to a system at equilibrium, the system will adjust its concentrations of reactants and products to counteract this stress and re-establish equilibrium.

For instance, if the concentration of a reactant is increased, the system will react to consume some of the added reactant by producing more products. Thus, Le Chatelier's principle helps us understand the dynamic nature of chemical equilibria and how disturbances can be used to steer the reaction in a desired direction, which is a key consideration for chemical engineering and process control.

Reaction Quotient

The reaction quotient (Q) is a measure that tells us the direction in which a reaction mixture will shift to reach equilibrium. It is calculated in the same way as the equilibrium constant, with the concentrations or partial pressures of the reactants and products at any point in time, not necessarily at equilibrium.

When comparing Q to the equilibrium constant K, if Q < K, the reaction will proceed in the forward direction to attain equilibrium, converting reactants into products. Conversely, if Q > K, the reaction will shift in the reverse, favoring the formation of reactants. At equilibrium, Q equals K, meaning there is no net change in the concentrations of reactants and products over time. The concept of Q is critical for understanding how reactions move towards equilibrium, and it is often tested in exams like the JEE for physical chemistry.

Physical Chemistry for JEE

Physical chemistry is an essential part of the JEE syllabus, an engineering entrance examination in India. It deals with understanding chemical systems through the laws of physics. Topics like equilibrium, thermodynamics, kinetics, and electrochemistry are central to physical chemistry for JEE.

For students preparing for the JEE, mastering concepts such as the equilibrium constant, Le Chatelier's principle, and the reaction quotient, as demonstrated in the example problem, is vital. These concepts not only contribute to solving quantitative problems but also develop a deep conceptual understanding, which is necessary for achieving success in competitive exams like the JEE. Keep practicing problems from each topic, understand the core principles deeply, and apply them to different scenarios to excel in this challenging and rewarding subject.