Problem 156 - A Complete Resource Book in Chemistry for JEE Main

Question

The ionization constant of \(\left[\mathrm{NH}_{4}^{+}\right]\)in water is \(5.6 \times 10^{-10}\) at \(25^{\circ} \mathrm{C}\). The rate constant for the reaction of \(\left[\mathrm{NH}_{4}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\)to form \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\) is \(3.4 \times 10^{10}\) litmol \(^{-1}\) \(\sec ^{-1}\) at \(25^{\circ} \mathrm{C}\). The rate constant for the proton transfer form water to \(\mathrm{NH}_{3}\) in lit \(\mathrm{mol}^{-1} \mathrm{sec}^{-1}\) is (a) \(6.07 \times 10^{5}\) (b) \(6.07 \times 10^{-5}\) (c) \(6.07 \times 10^{-3}\) (d) \(6.07 \times 10^{-4}\)

Step-by-Step Solution

Answer

The correct answer is (a) \(6.07 \times 10^5\).

Step 1: Write the Equilibrium Expression

The ionization of \([\mathrm{NH}_{4}^{+}]\) in water can be written as an equilibrium:\[\mathrm{NH}_{4}^{+} \rightleftharpoons \mathrm{NH}_{3} + \mathrm{H}^{+}\]The ionization constant \(K_i\) for this reaction is given as \(5.6 \times 10^{-10}\).

Step 2: Write the Rate Equation for the Forward Reaction

The forward reaction is the conversion of \([\mathrm{NH}_{4}^{+}]\) and \([\mathrm{OH}^{-}]\) to \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\), with a rate constant \(k_f = 3.4 \times 10^{10}\) litmol \(^{-1}\) \(\sec ^{-1}\). This can be expressed as:\[\text{Rate}_{\text{forward}} = k_f \times [\mathrm{NH}_4^+] \times [\mathrm{OH}^-]\]

Step 3: Calculate the Rate Constant of the Reverse Reaction

The reverse reaction involves the formation of \([\mathrm{NH}_{4}^{+}]\) from \(\mathrm{NH}_{3}\) accepting a proton from water. The rate constant \(k_r\) for the reverse reaction can be calculated using the relation:\[ K_i = \frac{k_f}{k_r} \quad \Rightarrow \quad k_r = \frac{k_f}{K_i} \]Substitute the given values:\[k_r = \frac{3.4 \times 10^{10}}{5.6 \times 10^{-10}}\]

Step 4: Solve for the Rate Constant of the Reverse Reaction

Calculate \(k_r\) using the given values:\[k_r = \frac{3.4 \times 10^{10}}{5.6 \times 10^{-10}} = 6.07 \times 10^5 \text{ litmol}^{-1} \text{s}^{-1}\]

Step 5: Choose the Correct Answer

The rate constant for the proton transfer from water to \(\mathrm{NH}_3\) at \(25^{\circ}\text{C}\) is \(6.07 \times 10^5 \text{ litmol}^{-1} \text{s}^{-1}\), as calculated. So, the correct answer is option (a).

Key Concepts

Understanding Proton Transfer

In the context of chemical reactions, a proton transfer is a process where a proton, represented by \(\mathrm{H}^+\), moves from one molecule to another. This is a fundamental concept in understanding acid-base interactions. For instance, when ammonium ion \([\mathrm{NH}_4^+]\) is in water, it can donate a proton to water, forming ammonia \(\mathrm{NH}_3\) and a hydronium ion \(\mathrm{H}_2\mathrm{O}\).

Proton transfer reactions can significantly affect the properties of solutions, influencing pH and other chemical behaviors.

• The transfer leads to a rearrangement of charges, influencing how molecules interact.
• These reactions are typically fast and reversible, meaning they can reach equilibrium quickly.

It's essential to understand the fast nature of proton transfers as it explains the dynamic equilibrium in solutions, where most of the proton transfers occur continuously.

Concept of Reaction Rate

The reaction rate is a measure of how quickly a chemical reaction proceeds. It can be influenced by various factors, including temperature, concentration of reactants, and the presence of catalysts. For the ionization of \([\mathrm{NH}_4^+]\) and its reaction with \([\mathrm{OH}^-]\) to form ammonia \(\mathrm{NH}_3\) and water, the reaction rate is determined by the rate constant \(k_f = 3.4 \times 10^{10} \text{ litmol}^{-1} \text{s}^{-1}\).

Key points to understand include:

• The rate constant \(k_f\) signifies how quickly the forward reaction occurs under given conditions. Higher values of \(k_f\) suggest faster reactions.
• An increase in reactant concentration generally increases the rate, as more molecules are available to interact.

Understanding reaction rates helps predict how different conditions affect the speed of a reaction, which is crucial in both industrial applications and biological processes.

Equilibrium Constant Explained

In chemistry, the equilibrium constant \(K_i\) describes the ratio of the concentration of products to reactants at equilibrium. For the ionization of \([\mathrm{NH}_4^+]\) in water, \(K_i\) stands at \(5.6 \times 10^{-10}\). This small value indicates a reaction where only a small amount of \([\mathrm{NH}_4^+]\) will convert to products under normal conditions.

Consider these critical points:

• A small \(K_i\) value often signifies a weak equilibrium, with the reactants largely present.
• Large \(K_i\) values suggest a strong tendency for products to form, indicating a strongly favored equilibrium in that direction.

Knowing the equilibrium constant helps understand the favorability and extent of a reaction. It is integral in calculating the concentration of species in a solution at equilibrium and determining how changes affect the system's equilibrium.