Problem 159 - A Complete Resource Book in Chemistry for JEE Main

Question

For the reaction \(2 \mathrm{HI}(\mathrm{g})=\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g})\) if the degree of dissociation is \(\alpha\) and \(\mathrm{K}_{\mathrm{p}}\) is the equilibrium constant, then (a) \(\alpha=\frac{\sqrt{2 \mathrm{~K}_{\mathrm{p}}}}{1+\sqrt{2 \mathrm{k}_{\mathrm{p}}}}\) (b) \(\alpha=\frac{2 \sqrt{K_{\mathrm{P}}}}{1+2 \sqrt{\mathrm{K}_{\mathrm{p}}}}\) (c) \(\sqrt{2 \mathrm{~K}_{\mathrm{p}}}=\left(\frac{\alpha}{1-\alpha}\right)\) (d) \(2 \sqrt{K_{\mathrm{p}}}=\left(\frac{\alpha}{1-\alpha}\right)\)

Step-by-Step Solution

Answer

The correct answer is (c): \(\sqrt{2K_p} = \frac{\alpha}{1-\alpha}\).

Step 1: Understand the Reaction and Initial Conditions

The given reaction is \(2\mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{g}) + \mathrm{I}_2(\mathrm{g})\). Initially, let's assume the moles of \(\mathrm{HI}\) is \(2n\). As dissociation occurs, \(\alpha\) represents the degree of dissociation, so \(n\alpha\) moles of \(\mathrm{HI}\) dissociate into \(n\alpha\) moles of \(\mathrm{H}_2\) and \(n\alpha\) moles of \(\mathrm{I}_2\).

Step 2: Calculate the Equilibrium Concentrations

At equilibrium, the moles of \(\mathrm{HI}\), \(\mathrm{H}_2\), and \(\mathrm{I}_2\) are \((2n - 2n\alpha) = 2n(1-\alpha)\), \(n\alpha\), and \(n\alpha\), respectively. Total moles at equilibrium are \(2n(1-\alpha) + n\alpha + n\alpha = 2n\).

Step 3: Express Partial Pressures

Since the total number of moles at equilibrium remains constant, the partial pressures can be expressed as fractions of total pressure \(P\). Therefore, partial pressures are \(P_{\mathrm{HI}} = P \frac{2(1-\alpha)}{2} = P(1-\alpha)\), \(P_{\mathrm{H}_2} = P \frac{\alpha}{2}\), and \(P_{\mathrm{I}_2} = P \frac{\alpha}{2}\).

Step 4: Write the Expression for \(K_p\)

The equilibrium constant \(K_p\) for the reaction in terms of partial pressures is given by: \(K_p = \frac{(P_{\mathrm{H}_2})(P_{\mathrm{I}_2})}{(P_{\mathrm{HI}})^2} = \frac{(\frac{P\alpha}{2})(\frac{P\alpha}{2})}{(P(1-\alpha))^2} = \frac{P^2\alpha^2}{4P^2(1-\alpha)^2}\). Simplifying, we get \(K_p = \frac{\alpha^2}{4(1-\alpha)^2}\).

Step 5: Solve for \(\alpha\)

Rearranging the equation from the previous step, we get \(\alpha^2 = 4K_p(1-\alpha)^2\). Solving for \(\alpha\), we find \(\sqrt{2K_p} = \frac{\alpha}{1-\alpha}\). This matches with answer (c) from the options given.

Key Concepts

Degree of Dissociation

The degree of dissociation, represented by \( \alpha \), is a measure of how much a compound breaks down into its components during a chemical reaction. In the reaction \( 2\mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{g}) + \mathrm{I}_2(\mathrm{g}) \), \( \alpha \) indicates the fraction of hydrogen iodide (HI) that dissociates into hydrogen (\( \mathrm{H}_2 \)) and iodine (\( \mathrm{I}_2 \)) gases. Initially, if we assume 2 moles of HI, the moles of HI that dissociate are \( 2n\alpha \). Thus, the degree of dissociation effectively helps in calculating the number of product moles formed as a result of partial or complete dissociation.

Understanding this concept is crucial because it directly impacts the calculation of equilibrium concentrations and partial pressures within the reaction vessel. It is a fundamental parameter that describes how far chemical equilibrium has been achieved.

Partial Pressures

Partial pressure is the pressure exerted by a single type of gas in a mixture of gases in a closed container. At equilibrium, each component gas in the reaction \( 2\mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{g}) + \mathrm{I}_2(\mathrm{g}) \) has its own partial pressure that contributes to the total pressure \( P \) of the system.

For HI, the partial pressure \( P_{\mathrm{HI}} \) is \( P(1-\alpha) \), whereas for both \( \mathrm{H}_2 \) and \( \mathrm{I}_2 \), it is \( \frac{P\alpha}{2} \). These partial pressures are essential when expressing the equilibrium constant \( K_p \) in terms of pressure, which subsequently helps in understanding how pressure changes affect the equilibrium state.

By analyzing partial pressures, we can predict how the system behaves under different conditions, such as changes in volume or temperature.

Chemical Equilibrium

Chemical equilibrium refers to a dynamic state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the backward reaction. This equilibrium is reached when the concentrations of the reactants and products remain constant over time. In the reaction \( 2\mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{g}) + \mathrm{I}_2(\mathrm{g}) \), equilibrium is achieved when the dissociation of HI into \( \mathrm{H}_2 \) and \( \mathrm{I}_2 \) and their recombination back into HI occurs at the same rate.

Important to this concept is knowing that, despite the dynamic nature of equilibrium, there is no net change in the concentration of reactants and products. This is crucial when determining equilibrium concentrations and calculating the equilibrium constant \( K_p \), a value that provides insight into the position of equilibrium.

• A larger \( K_p \) suggests a position favoring products.
• A smaller \( K_p \) indicates a position favoring reactants.

Understanding these aspects of chemical equilibrium allows chemists to optimize conditions for desired yields in chemical processes.

Equilibrium Concentrations

Equilibrium concentrations are the concentrations of the reactants and products in a chemical reaction once equilibrium has been reached. For the reaction \( 2\mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{g}) + \mathrm{I}_2(\mathrm{g}) \), equilibrium concentrations are determined by the initial moles of reactants and the degree of dissociation \( \alpha \).

At equilibrium:

• The concentration of HI is \( 2n(1-\alpha) \).
• The concentrations of \( \mathrm{H}_2 \) and \( \mathrm{I}_2 \) are both \( n\alpha \).

These concentrations are pivotal in calculations involving \( K_p \) because they allow you to formulate accurate expressions of the equilibrium constant in terms of measurable pressures or concentrations.

Understanding equilibrium concentrations directly informs decisions in industrial and laboratory settings by predicting the outcomes of altering conditions such as pressure or temperature on the reaction's equilibrium state.