Problem 72 - A Complete Resource Book in Chemistry for JEE Main

Question

If equilibrium constants of reaction, \(\mathrm{N}_{1}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}\) is \(\mathrm{K}_{1}\), and \(1 / 2 \mathrm{~N}_{2}+1 / 2 \mathrm{O}_{2} \rightleftharpoons \mathrm{NO}\) is \(\mathrm{K}_{2}\) then (a) \(\mathrm{K}_{1}=\mathrm{K}_{2}\) (b) \(\mathrm{K}_{1}=2 \mathrm{~K}_{2}\) (c) \(\mathrm{K}_{2}=\sqrt{\mathrm{K}}_{1}\) (d) \(\mathrm{K}_{1}=1 / 2 \mathrm{~K}_{2}\)

Step-by-Step Solution

Answer

(c) \( \mathrm{K}_{2} = \sqrt{\mathrm{K}_{1}} \)

Step 1: Understand the Given Reactions

The balanced chemical equation for the first reaction is \( \mathrm{N}_{2} + \mathrm{O}_{2} \rightleftharpoons 2\mathrm{NO} \) with equilibrium constant \( \mathrm{K}_{1} \). The second reaction given is \( \frac{1}{2} \mathrm{N}_{2} + \frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{NO} \) with equilibrium constant \( \mathrm{K}_{2} \).

Step 2: Relate the Reactions

The second reaction is essentially half of the first reaction. Each of the reactants in the first reaction is divided by 2, producing only one mole of \( \mathrm{NO} \) instead of two as in the first reaction.

Step 3: Determine Relationship Between K1 and K2

The equilibrium constant for a reaction that is halved can be found by taking the square root of the original equilibrium constant. Therefore, \( \mathrm{K}_{2} = \sqrt{\mathrm{K}_{1}} \). This is due to the fact that if the reaction coefficients are halved, so is the power to which the concentration terms are raised in the expression for the equilibrium constant.

Key Concepts

Equilibrium Constant

The equilibrium constant, often denoted as \( K \), is a crucial aspect of understanding chemical equilibrium. It provides insight into the extent a chemical reaction proceeds under a given set of conditions. Specifically for the reaction \( \text{N}_{2} + \text{O}_{2} \rightleftharpoons 2\text{NO} \), the equilibrium constant \( K_1 \) is a reflection of the concentration of the products and reactants at equilibrium.
The value of the equilibrium constant isn't arbitrary; it is grounded on the specifics of the reaction equation, temperature, and the nature of the chemical substances involved.
A larger equilibrium constant suggests a reaction that favors products at equilibrium, while a smaller one indicates a reaction favoring reactants.

• At \( K_1 \), the reaction allows for the formation of \( 2\text{NO} \) molecules.
• The equilibrium constant changes with modification of reaction steps, such as dividing the reactants and products by a factor.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. They can proceed to various extents, which is often determined by factors like temperature, pressure, and concentration of reactants. For example, the reaction \( \text{N}_{2} + \text{O}_{2} \rightleftharpoons 2\text{NO} \) involves nitrogen and oxygen forming nitric oxide at equilibrium.
In this reaction, the stoichiometry is balanced, indicating the precise ratios in which reactants combine and products form.
During the reverse reaction, products convert back into reactants, making equilibrium a dynamic state. The interplay between forward and reverse processes reaches a point where the concentrations of both reactants and products remain consistent over time, establishing what is known as chemical equilibrium.

Reaction Stoichiometry

Reaction stoichiometry specifies the mole ratios of reactants and products in a balanced chemical equation. For the equation \( \text{N}_{2} + \text{O}_{2} \rightleftharpoons 2\text{NO} \), stoichiometry ensures that one mole of nitrogen reacts with one mole of oxygen to produce two moles of nitric oxide.
In the alternative equation \( \frac{1}{2}\text{N}_{2} + \frac{1}{2}\text{O}_{2} \rightleftharpoons \text{NO} \), stoichiometry is modified such that half a mole of each reactant is required to produce one mole of the product.
The stoichiometric changes affect the calculation of the equilibrium constant. It reflects how each mole count in the equation influences the concentration terms in the equilibrium expression.

• The full reaction has stoichiometric coefficients equating to: \( \frac{1}{1} \) for each reactant and \( \frac{2}{1} \) for the product \( \text{NO} \).
• The altered reaction has coefficients of \( \frac{1}{2} \) for both \( \text{N}_{2} \) and \( \text{O}_{2} \), and \( 1 \) for \( \text{NO} \).

Equilibrium Expressions

Equilibrium expressions are mathematical representations of how reactant and product concentrations relate at equilibrium, giving us the equilibrium constant \( K \). For the reaction \( \text{N}_{2} + \text{O}_{2} \rightleftharpoons 2\text{NO} \), the expression is structured as follows:
\[K_1 = \frac{[\text{NO}]^2}{[\text{N}_2][\text{O}_2]}\]A different form applies for half the reaction:\[K_2 = \frac{[\text{NO}]}{[\text{N}_2]^{1/2}[\text{O}_2]^{1/2}}\]These equations show how the stoichiometry impacts the exponents of concentration terms in the equilibrium expression.

• Reducing the stoichiometry by half affects the exponents, consequently affecting \( K \).
• The equilibrium constant \( K_2 \) simplifies to \( \sqrt{K_1} \) due to these changes in reaction partitioning.