Problem 105 - A Complete Resource Book in Chemistry for JEE Main

Question

What will be the value of equilibrium constant \(\left(\mathrm{K}_{1}\right)\) for the reaction, \(\mathrm{HI}(\mathrm{g}) \rightleftharpoons 1 / 2 \mathrm{H}_{2}(\mathrm{~g})+1 / 21_{2}(\mathrm{~g})\), if its value for the reaction \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}\) is \(64 ?\) (a) \(1 / 64\) (b) \(1 / 8\) (c) 64 (d) 8

Step-by-Step Solution

Answer

The value of the equilibrium constant \(K_1\) is \(\frac{1}{8}\); the answer is (b).

Step 1: Identify the Given Reaction

We are given the equilibrium constant \(K_2 = 64\) for the reaction \(\mathrm{H}_{2}(\mathrm{g}) + \mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2\mathrm{HI}(\mathrm{g})\). We need to find the equilibrium constant \(K_1\) for the reaction \(\mathrm{HI}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{H}_{2}(\mathrm{g}) + \frac{1}{2} \mathrm{I}_{2}(\mathrm{g})\).

Step 2: Express the Reaction in Terms of the Given Reaction

The reaction \(\mathrm{HI}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{H}_{2}(\mathrm{g}) + \frac{1}{2} \mathrm{I}_{2}(\mathrm{g})\) is the reverse of the reaction \(2\mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{g}) + \mathrm{I}_{2}(\mathrm{g})\). This reaction is obtained by both reversing and halving the given reaction.

Step 3: Determine the Relationship Between Constants

Reversing a reaction involves taking the reciprocal of its equilibrium constant. Halving the coefficients in a balanced reaction affects the equilibrium constant by taking the square root.So, \(K_1 = \frac{1}{\sqrt{K_2}}\).

Step 4: Calculate the Equilibrium Constant K1

Substitute the known value of \(K_2\), which is 64, into the relationship formula:\[ K_1 = \frac{1}{\sqrt{64}} = \frac{1}{8} \]

Step 5: Select the Correct Answer

After calculating, we find \(K_1 = \frac{1}{8}\). Therefore, the correct answer is option (b) \(\frac{1}{8}\).

Key Concepts

Understanding Equilibrium Constants

In chemical reactions, the equilibrium constant, denoted as \( K \), tells us about the balance between the products and reactants when the reaction has reached a state of equilibrium. This constant is significant because it helps in predicting the direction of the reaction and the concentration of compounds at equilibrium.

For a general reaction, \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant \( K \) is expressed as:

• \( K = \frac{[C]^c [D]^d}{[A]^a [B]^b} \)

The concentrations of the reactants and products are each raised to the power of their coefficients in the balanced chemical equation.

The value of \( K \) can show whether the equilibrium favors products over reactants. A large \( K \) value means more products than reactants at equilibrium, while a smaller \( K \) value indicates the opposite. It's important to note that \( K \) only changes with temperature, making it a reliable indicator of reaction conditions at a constant temperature.

The Nature of Reversible Reactions

Reversible reactions are key in chemistry since they can proceed in both directions, from reactants to products and vice versa. This duality allows the reactions to reach a state of equilibrium, where the rate of the forward reaction equals the rate of the reverse reaction.

In the context of reversible reactions, we consider reactions like:

• \( ext{A + B} \rightleftharpoons ext{C + D} \)

As the reaction progresses, the concentration of reactants decreases while the concentration of products increases, until they balance out. This dynamic balance is what we call chemical equilibrium.

Reversible reactions illustrate the dynamic, ever-changing nature of chemical processes, emphasizing that reactions don't necessarily go to completion. Instead, they reach a balanced state where both reactants and products coexist.

Thermodynamics in Chemistry

Thermodynamics in chemistry deals with the study of energy, heat, and work in chemical processes. It helps us understand how and why reactions occur, providing insights into reaction feasibility and the amount of energy released or absorbed.

There are three main laws of thermodynamics which govern chemical reactions:

• The First Law, which states that energy is conserved in an isolated system.
• The Second Law, which explains that the total entropy (disorder) of an isolated system always increases over time.
• The Third Law, which claims that the entropy of a perfect crystal approaches zero as the temperature approaches absolute zero.

These laws guide us in predicting reaction directions, determining the spontaneity of processes, and calculating changes in energy. They form the backbone for understanding how reactions reach equilibrium, as energy distribution is a driving factor for achieving chemical stability.