Problem 203 - A Complete Resource Book in Chemistry for JEE Main

Question

The equilibrium constant \((\mathrm{K})\) for the reaction \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}(\mathrm{g}) \quad\) at temperature \(\mathrm{T}\) is \(4 \times 10^{-4} .\) The value of \(K_{c}\) for the reaction, \(\mathrm{NO}(\mathrm{g}) \rightarrow \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) at the same temperature is (a) \(4 \times 10^{-4}\) (b) \(0.02\) (c) \(2.5 \times 10^{2}\) (d) \(50.0\)

Step-by-Step Solution

Answer

The value of \(K_c\) is 50.0 (option d).

Step 1: Understand the given equilibrium constant

The equilibrium constant \(K\) for the reaction \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}(\mathrm{g})\) at temperature \(T\) is \(4 \times 10^{-4}\). This means that \(K = \frac{[\mathrm{NO}]^2}{[\mathrm{N}_2][\mathrm{O}_2]} = 4 \times 10^{-4}\).

Step 2: Relate original and reverse reactions

For the reverse reaction \(\mathrm{NO}(\mathrm{g}) \rightarrow \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\), the equilibrium constant \(K'\) is the reciprocal of the given reaction's \(K\). Thus, \(K' = \frac{1}{K} = \frac{1}{4 \times 10^{-4}}\).

Step 3: Adjust the reaction coefficients

The reverse reaction is half of \(2\mathrm{NO}(\mathrm{g}) \rightarrow \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\), so the equilibrium constant for the reaction is \(\sqrt{\frac{1}{4 \times 10^{-4}}}\) because it's a half reaction.

Step 4: Calculate the equilibrium constant value

First, calculate \(\frac{1}{4 \times 10^{-4}} = 2.5 \times 10^3\). Then, take the square root to get \(\sqrt{2.5 \times 10^3} = 50\).

Step 5: Identify the correct option

The equilibrium constant \(K_c\) for the reaction \(\mathrm{NO}(\mathrm{g}) \rightarrow \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) at the same temperature is \(50.0\). Thus, the correct answer is option (d).

Key Concepts

Reaction Equilibrium

In the realm of chemistry, understanding how components of a reaction reach a steady state is critical. This balanced point is referred to as reaction equilibrium. At equilibrium, the forward and backward reactions occur at equal rates, which means the concentrations of the reactants and products remain constant over time. However, it is important to note that this equilibrium is dynamic, not static; molecules continue to react, but no net change in concentrations is observed.

For a given reaction such as oindent\(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}(\mathrm{g})\), the reaction equilibrium exists when the rates of generating \(\mathrm{NO}\) from \(\mathrm{N}_2\) and \(\mathrm{O}_2\) and the reverse depletion of \(\mathrm{NO}\) back to \(\mathrm{N}_2\) and \(\mathrm{O}_2\) are equal. The equilibrium state is heavily influenced by conditions such as temperature, pressure, and concentration of involved gases.

Chemical Reactions

The ongoing transformations in which substances are converted into different species are known as chemical reactions. Each chemical reaction can be represented with a balanced chemical equation, showing how atoms and molecules rearrange to form different substances. For example, the reaction \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}(\mathrm{g})\) is a straightforward representation of nitrogen and oxygen gas combining to form nitric oxide.

The reverse of this process is given by the equation \(\mathrm{NO}(\mathrm{g}) \rightarrow \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\).

Reversibility is a key feature in some chemical reactions, implying the reactants can form products and products can convert back into reactants under the right conditions. This coexistence of forward and backward reactions is encapsulated through the concept of equilibrium, expressed quantitatively using the equilibrium constant \(K\). The equilibrium constants for the forward and reverse reactions are inverses of each other, as seen in our example.

Le Chatelier's Principle

Le Chatelier's Principle is an essential concept when dealing with reaction equilibrium. It states that when a system at equilibrium experiences a change in concentration, temperature, or pressure, the system will adjust to minimize this change and restore a new equilibrium state. In simple terms, if a stress is applied to a system, the system shifts in the direction that will counteract that stress.

For example, consider an increase in concentration of reactants in the equation \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}(\mathrm{g})\). According to Le Chatelier's Principle, the system will respond by forming more \(\mathrm{NO}\) to counterbalance the added reactants. Similarly, if the temperature is increased for an endothermic reaction, the equilibrium will shift toward the products to absorb the extra heat.

Understanding how to manipulate conditions using Le Chatelier's Principle is crucial for optimizing the yields in chemical industry processes and is a powerful tool in the chemist’s toolkit.