Problem 6 - A Complete Resource Book in Chemistry for JEE Main

Question

If equilibrium constant for the reaction \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) is \(\mathrm{K}_{\mathrm{c}}\), then the equilibrium con- stant for the reaction \(\mathrm{NH}_{3}=1 / 2 \mathrm{~N}_{2}+\frac{3}{2} \mathrm{H}_{2}\) will be (a) \(\frac{1}{\mathrm{~K}_{\mathrm{c}}}\) (b) \(\frac{1}{\mathrm{~K}^{2}}\) (c) \(\sqrt{K}\) (d) \(\frac{1}{i \mathrm{~K}_{c}}\)

Step-by-Step Solution

Answer

The correct answer is (a) \(\frac{1}{K_c}\).

Step 1: Understand Equilibrium Constant Definition

The equilibrium constant, denoted as \( K_c \), for a reaction relates to the concentration of products over reactants at equilibrium raised to the power of their stoichiometric coefficients.

Step 2: Original Reaction Equilibrium Constant

Given the original reaction \( \mathrm{N}_2 + 3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3 \),the equilibrium constant \( K_c \) is expressed as:\[ K_c = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3} \]

Step 3: Reverse the Reaction

The new reaction is the reverse of the original: \( \mathrm{NH}_3 \rightleftharpoons \frac{1}{2} \mathrm{N}_2 + \frac{3}{2} \mathrm{H}_2 \).Reversing the reaction inverses the equilibrium constant: \( K' = \frac{1}{K_c} \).

Step 4: Adjust for Stoichiometry

The stoichiometry of the reversed reaction equation is halved as compared to the original i.e., \( 2 \mathrm{NH}_3 \) in the original reaction becomes \( \mathrm{NH}_3 \) in the reversed reaction.This implies:\[ K' = \frac{1}{K_c^{1/2}} = \frac{1}{\sqrt{K_c}} \] Hence, the answer will be \( \frac{1}{K_c^{-0.5}} \) equivalent to \( K_c^{-1} \).

Step 5: Choose the Correct Option

From the calculations, the inverse of the equilibrium constant with the adjustmentfor the half reaction is \( \frac{1}{K_c} \). Analyze this with the options given. The correct answer corresponds to option (a), \( \frac{1}{K_c} \).

Key Concepts

Reversible Reactions

In the realm of chemistry, many reactions are not just a one-way street. They can proceed in both directions. Such reactions, where products can revert to reactants under the same conditions, are known as reversible reactions. An example of this is the synthesis of ammonia, which is represented as \[ \mathrm{N}_2 + 3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3 \]. The double arrow signifies that the reaction can go both ways. This means that when the reaction reaches a state of balance, the reactants and products are constantly forming and decomposing. However, their overall concentrations remain unchanged.

Reversible reactions are vital in chemical equilibrium. They help in maintaining a balance, as they adjust itself when changes occur in concentration, temperature, or pressure. In such reactions, it’s common to see a dynamic but stable condition where no net change is observed. Understanding this concept is crucial for calculating equilibrium constants, which provide insight into how far a reaction has proceeded in either direction.

Stoichiometry Adjustments

Stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. When dealing with equilibrium constants, understanding stoichiometry is essential because these constants are derived from the balanced chemical equation. In our example, the balanced reaction \(\mathrm{N}_2 + 3\mathrm{H}_2 \rightleftharpoons 2\mathrm{NH}_3\) gives rise to an equilibrium constant \(K_c\) based on this stoichiometry.

In reversible reactions, when the reaction is flipped or altered, such as changing \(2 \mathrm{NH}_3\) to \(\mathrm{NH}_3 \rightleftharpoons \frac{1}{2} \mathrm{N}_2 + \frac{3}{2} \mathrm{H}_2\), stoichiometry adjustments are necessary. Here, the stoichiometric coefficients in the balanced equation play a role in changing the expression for the equilibrium constant. For example, the coefficient change by a factor of 2 requires you to adjust the exponent in the equilibrium constant. It ensures the proportions between products and reactants reflect accurately. This results in the equilibrium constant \( K'\) being adjusted appropriately as \( \frac{1}{\sqrt{K_c}} \) to reflect these changes.

Chemical Equilibrium

Chemical equilibrium represents a state where the forward and reverse reactions occur at equal rates, resulting in no net change in the concentration of reactants and products. At this stage, we can use the equilibrium constant \(K_c\) to describe the ratio of the concentrations of the products to the reactants, each raised to the power of their respective stoichiometric coefficients. For the reaction \(\mathrm{N}_2 + 3\mathrm{H}_2 \rightleftharpoons 2\mathrm{NH}_3\), \(K_c\) equates to \(\frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3}\).

At equilibrium, this ratio remains constant unless subjected to external changes, such as in temperature. Understanding chemical equilibrium allows chemists to predict how a reaction responds to changes and to ascertain the extent of a reaction. This can be critical in industries that rely on chemical processes, as achieving a desired balance can optimize production. Moreover, knowing how to calculate and adjust equilibrium constants provides insights into reaction dynamics, supporting effective control and manipulation of chemical processes.