Problem 117 - A Complete Resource Book in Chemistry for JEE Main

Question

The equilibrium constant value for the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) is \(1.48 \times 10^{-5}\), the value for the reaction \(1 / 2 \mathrm{~N}_{2}(\mathrm{~g})+3 / 2 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})\) is \(\mathrm{n} \times 10^{-3}\) where \(\mathrm{n}\) is (a) \(1.85\) (b) \(3.85\) (c) \(4.85\) (d) 10

Step-by-Step Solution

Answer

The value of \( n \) is (b) 3.85.

Step 1: Understand Formula for Equilibrium Constant

The equilibrium constant for a chemical reaction is the ratio of the concentrations of the products to the reactants, each raised to the power of their coefficients in the balanced equation. For example, for the reaction \(aA + bB \rightleftharpoons cC + dD\), the equilibrium constant \(K\) can be expressed as:\[ K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]

Step 2: Identify the Given Reaction and its Equilibrium Constant

The problem provides the equilibrium constant for the reaction \(\mathrm{N}_2(\mathrm{~g}) + 3\mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2\mathrm{NH}_3(\mathrm{~g})\), which is \(1.48 \times 10^{-5}\). This is the reaction we start with.

Step 3: Determine the Relationship Between Reactions

The given problem asks to find the equilibrium constant for the reaction \(1 / 2 \mathrm{~N}_2(\mathrm{~g}) + 3 / 2 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_3(\mathrm{~g})\), which is half of the original equation provided. This implies we are dealing with the square root relationship of the equilibrium constants.

Step 4: Calculate the Equilibrium Constant for the New Reaction

When we multiply or divide the entire chemical equation by a number, the equilibrium constant is raised to the power of that number. Since we are reducing the first reaction by a factor of 2, the equilibrium constant for the new reaction is the square root of the original equilibrium constant:\[ K_{new} = (K_{original})^{1/2} = (1.48 \times 10^{-5})^{1/2} \]

Step 5: Compute and Find the Value of n

Calculate the value of \((1.48 \times 10^{-5})^{1/2}\):\[ (1.48 \times 10^{-5})^{1/2} = \sqrt{1.48 \times 10^{-5}} \approx 3.85 \times 10^{-3} \]Thus, \( n \) in \( n \times 10^{-3} \) is approximately \(3.85\).

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K \), is a fundamental concept in chemical reactions signifying the ratio of the concentrations of products to reactants at equilibrium. It reflects the point at which a chemical system is balanced. For the generic reaction \(aA + bB \rightleftharpoons cC + dD\), the equilibrium constant is mathematically expressed as:

• \[ K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \] where [A], [B], [C], and [D] are the molar concentrations.

The exponent for each concentration corresponds to the stoichiometric coefficient from the balanced equation. These coefficients emphasize the impact each substance has on the equilibrium. A large \( K \) value indicates a reaction favoring products, while a small \( K \) signifies an equilibrium favoring reactants. Understanding \( K \) helps predict the position of equilibrium and thus the outcome of reactions. Knowing this allows chemists to control reactions, optimizing yields or minimizing undesired side products.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. They can occur through various pathways and can be reversible or irreversible. Reversible reactions, such as \( \mathrm{N}_2(g) + 3\mathrm{H}_2(g) \rightleftharpoons 2\mathrm{NH}_3(g) \), reach a state of chemical equilibrium where the rate of the forward reaction equals the rate of the reverse reaction.

• In equilibrium, concentrations of reactants and products are constant, though the chemical processes continue at a molecular level.
• The concept of dynamic equilibrium is crucial as it shows that even at equilibrium, reactions are ongoing though concentrations remain stable.

Comprehensively understanding chemical reactions includes recognizing these types of equilibrium states. It highlights the balance of chemical changes, helping us manipulate conditions to shift equilibrium as needed.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships of the substances involved in chemical reactions. It hinges on the balanced chemical equation, which provides the framework for making calculations about how much reactant is required or product yielded.

• For example, in the reaction \( \mathrm{N}_2(g) + 3\mathrm{H}_2(g) \rightleftharpoons 2\mathrm{NH}_3(g) \), the coefficients 1, 3, and 2 indicate that one mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia.

Stoichiometry allows us to calculate the equilibrium constant of a modulated reaction, such as when the balanced equation is halved. By applying stoichiometric principles, we understand that the equilibrium constant will adjust based on these coefficient changes. When the reaction is divided by two, the equilibrium constant is recalculated by taking the square root of the original \( K \), emphasizing the power in manipulating stoichiometry for desired chemical outcomes.