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

Question

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

Step-by-Step Solution

Answer

Option (c) is correct: \( K_2 = \sqrt{K_1} \).

Step 1: Understand the reactions and equilibrium constants

The given reactions are: \( \mathrm{N}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO} \) with equilibrium constant \( K_1 \) and \( \frac{1}{2} \mathrm{N}_{2} + \frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{NO} \) with equilibrium constant \( K_2 \). Notice that the second reaction is exactly half of the first reaction.

Step 2: Relationship between reactions and constants

If you multiply a chemical reaction by a factor, the equilibrium constant for the new reaction is that factor raised to the power of the equilibrium constant of the original reaction. In this case, the second reaction is one-half of the first reaction.

Step 3: Determine equilibrium constant relationships

Given that the second reaction is half the stoichiometric coefficients of the first, the relationship between their constants will be:\[ K_2 = (K_1)^{1/2} \]. This arises from the fact that halving the coefficients corresponds to taking the square root of the equilibrium constant.

Step 4: Identify the correct option

From Step 3, we derived that \( K_2 = \sqrt{K_1} \), which matches option (c). Therefore, option (c) \( K_2 = \sqrt{K_1} \) is the correct answer.

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the concentrations of reactants and products remain constant over time. This does not mean the reactants and products are equal in concentration but that their ratio remains unchanged because the forward and reverse reactions are occurring at the same rate.
In the context of the given reactions, the equilibrium constants, \( K_1 \) and \( K_2 \), provide a measure of the balance between the nitrogen, oxygen, and nitrogen monoxide (NO) molecules in their respective equilibrium states. An equilibrium constant expresses the ratio of the concentrations of products to reactants at equilibrium. It can be denoted as:

• \( K = \frac{[\text{products}]}{[\text{reactants}]} \)

This expression is crucial because it tells us how far a reaction will proceed before reaching equilibrium.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. These processes can be represented by chemical equations, showcasing how substances interact to form new products. For instance, the provided reactions with nitrogen, oxygen, and NO illustrate different molecular combinations.
Each side of a chemical equation must be balanced, meaning the number of atoms for each element is the same on both sides. The reaction \( \mathrm{N}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO} \) demonstrates a direct combination reaction.
Such equations can be manipulated by altering coefficients, which directly affects the equilibrium constant. If a reaction equation is altered, such as being halved as indicated in the exercise, the nature of the equilibrium constant changes too. This manipulation impacts calculations of equilibrium constants and predictions of reaction behavior.

Stoichiometry

Stoichiometry is the aspect of chemistry that pertains to the quantitative relationships or ratios between reactants and products in a chemical reaction. It helps determine the amount of each substance needed or produced in a reaction. This concept underpins the idea of scaling chemical reactions, as seen in the exercise transformed from \( \mathrm{N}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO} \) to \( \frac{1}{2} \mathrm{N}_{2} + \frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{NO} \).
When a chemical equation is multiplied or divided by a certain factor, stoichiometry defines how the equilibrium constant is adjusted. For example, halving the stoichiometric coefficients of the original reaction requires taking the square root of the equilibrium constant, as the reaction example demonstrates.

• This ensures the stoichiometric balance and reflects proportional changes in reaction dynamics.

Understanding stoichiometry and these relationships is essential for predicting the efficiencies and equilibriums of chemical reactions.