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 equilibrium constant \( \mathrm{K}_1 \) is \( \frac{1}{8} \).

Step 1: Understand the Given Reactions

We are given two chemical reactions. The first reaction is \(\mathrm{HI}( ext{g}) \rightleftharpoons \frac{1}{2} \mathrm{H}_{2}( ext{g}) + \frac{1}{2} \mathrm{I}_{2}( ext{g})\), for which we need to find the equilibrium constant \( \mathrm{K}_1 \). The second reaction is \(\mathrm{H}_{2}( ext{g}) + \mathrm{I}_{2}( ext{g}) \rightleftharpoons 2 \mathrm{HI} \), with an equilibrium constant given as \(64\).

Step 2: Establish Relationship Between Reactions

Note that the first reaction is the reverse of the second reaction, with coefficients halved. Thus, if the equilibrium constant for \(\mathrm{H}_{2} + \mathrm{I}_{2} \rightleftharpoons 2\mathrm{HI}\) is \(64\), then for the reverse reaction \(2\mathrm{HI} \rightleftharpoons \mathrm{H}_{2} + \mathrm{I}_{2}\), the equilibrium constant would be \(\frac{1}{64}\), due to the principle that the equilibrium constant of a reverse reaction is the reciprocal.

Step 3: Modify for Coefficient Change

The given first reaction is \(\text{HI} \rightleftharpoons \frac{1}{2} \mathrm{H}_{2} + \frac{1}{2} \mathrm{I}_{2}\). This reaction has coefficients that are half of the reversed second reaction. Adjusting the equilibrium constant for this change, the new constant \( \mathrm{K}_1 \) is given by taking the square root of \(\frac{1}{64}\), because adjusting coefficients affects the constant proportionally.

Step 4: Calculate the New Equilibrium Constant

Calculate the equilibrium constant for the adjusted reaction: \( \mathrm{K}_1 = \sqrt{\frac{1}{64}} = \frac{1}{8} \). This is because the root reflects the direct modification of mole ratio provided by coefficients.

Key Concepts

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction and its reverse proceed at the same rate. It means that the concentrations of reactants and products remain constant over time, although they are not necessarily equal. This state is dynamic because reactions continue to occur, but no net change is observed in the system's concentrations.

In a chemical reaction such as \(\text{A} + \text{B} \rightleftharpoons \text{C} + \text{D}\), equilibrium is reached when:

• The rate of the forward reaction (forming products) equals the rate of the reverse reaction (forming reactants).
• The concentrations of \(\text{A}, \text{B}, \text{C},\) and \(\text{D}\) no longer change significantly.

At equilibrium, the equilibrium constant \(K\) is calculated using the expression \(K = \frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]}\). This expression, derived from the law of mass action, highlights the ratio of product to reactant concentrations at equilibrium in a specific reaction. The larger the \(K\), the more products favored at equilibrium. Conversely, a smaller \(K\) suggests more reactants are present at equilibrium. The value of \(K\) is constant at a given temperature, though it can shift with temperature changes both in magnitude and in favorability towards reactants or products.

Reversible Reactions

Reversible reactions refer to reactions where the conversion of reactants to products can occur in both directions: the forward and reverse processes. An example is the decomposition and formation of hydrogen iodide:

\( \text{H}_2(\text{g}) + \text{I}_2(\text{g}) \rightleftharpoons 2 \text{HI}(\text{g}) \).

In this reaction, hydrogen and iodine gas react to form hydrogen iodide, while hydrogen iodide can break down back into hydrogen and iodine gases under certain conditions. The ability of reactions to proceed in both directions is essential for establishing chemical equilibrium, where products and reactants co-exist in balanced concentrations.

Reversible reactions are fundamental to calculating the equilibrium constant since any reversible reaction could theoretically reach equilibrium. It is important to understand that when dealing with reversible reactions, the conditions (such as temperature, pressure, and concentration) determine the extent to which the reaction favors the forward or reverse process, thus affecting the equilibrium constant \(K\).

This dual pathway allows chemists to manipulate the conditions to favor the formation of products or the conservation of reactants, making reversible reactions a cornerstone of chemical manufacturing and optimization.

Reaction Coefficients

Reaction coefficients are numerical values that indicate the relative amounts of each substance involved in a chemical reaction, described in a balanced chemical equation. These coefficients are crucial because they determine the stoichiometric balance of a reaction.

For the chemical equation \(2\text{A} + 3\text{B} \rightarrow 4\text{C}\), the coefficients are 2, 3, and 4, representing the stoichiometry of reactants A and B and product C. When determining the equilibrium expression, these coefficients become exponents in the equilibrium constant expression:

\(K = \frac{[\text{C}]^4}{[\text{A}]^2[\text{B}]^3}\).

Changes in these coefficients alter the equilibrium constant expression. If you halve the coefficients for reversible reactions, you must adjust the equilibrium constant accordingly - usually by taking the appropriate root. For instance, in the scenario from the exercise, halving all coefficients resulted in using the square root of the original reaction's constant.

This mathematical relationship highlights why understanding reaction coefficients is vital. They affect not just the quantities of substances involved but also influence the calculated equilibrium constant \(K\), revealing the extent of a reaction's favorability.