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 rate constant is \( 6.07 \times 10^{-5} \), matching option (b).

Step 1: Understand the reaction mechanism

The given reaction is \( \left[\mathrm{NH}_{4}^{+}\right] + \left[\mathrm{OH}^{-}\right] \rightarrow \mathrm{NH}_{3} + \mathrm{H}_{2}\mathrm{O} \), which involves ammonium ions forming ammonia and water. The problem provides the ionization constant \( K_a \) for \( \left[\mathrm{NH}_{4}^{+}\right] \).

Step 2: Apply the relation between rate constants and equilibrium constant

The relation \( k = \frac{k_f}{K_a} \) is applicable here, where \( k \) is the rate constant for the reverse reaction of proton transfer, \( k_f \) is the forward rate constant (given as \( 3.4 \times 10^{10} \text{ lit mol}^{-1} \text{sec}^{-1} \)), and \( K_a \) is the ionization constant \( 5.6 \times 10^{-10} \). This is because the reverse reaction equilibrium is related to the forward reaction equilibrium by the ionization constant.

Step 3: Calculate the rate constant for proton transfer

Using the formula \( k = \frac{3.4 \times 10^{10}}{5.6 \times 10^{-10}} \), calculate the value to determine \( k \). Here, we are solving for the reverse reaction rate constant. Calculate the expression to find:\[k = \frac{3.4 \times 10^{10}}{5.6 \times 10^{-10}} = 6.07 \times 10^{-5}\]lit \( \text{mol}^{-1} \text{sec}^{-1} \).

Step 4: Select the correct answer

The calculated rate constant for proton transfer from water to \( \mathrm{NH}_{3} \) is \( 6.07 \times 10^{-5} \), which matches option (b) \( 6.07 \times 10^{-5} \).

Key Concepts

Ionization Constant

Ionization constants are crucial in understanding the strength of an acid or base in a solution. In chemical kinetics, the ionization constant (\(K_a\)) helps us determine how well an acid can donate a proton to the solution, or how well a base can accept a proton. The ionization constant is defined specifically for reactions involving proton transfer. For an acid-base reaction in water, for example, the formula is:

• For acids: \(HA \rightleftharpoons H^+ + A^-\)
• \(K_a = \frac{[H^+][A^-]}{[HA]}\)

In the context of our exercise, the ionization constant of \([NH_4^+]\) is given as \(5.6 \times 10^{-10}\), telling us about the equilibrium between ammonium ions \([NH_4^+]\) and their ability to donate protons in a solution.

This value is very small, which means that ammonium ions only ionize a little in water, indicating it's a weak acid. Understanding \(K_a\) provides a clearer insight into the acid's behavior, allowing chemists to predict reactions involving ammonium ions and calculate related kinetic parameters more accurately.

Rate Constant

The rate constant, often denoted as \(k\), is pivotal in chemical kinetics. It determines the speed at which a reaction proceeds under specified conditions. Specifically, the term appears in rate laws which are expressions that relate to the concentration of reactants and the speed of the reaction.

In our exercise, the forward reaction rate constant \(k_f\) for the interaction between \([NH_4^+]\) and \([OH^-]\) is \(3.4 \times 10^{10} \text{ lit mol}^{-1} \text{ sec}^{-1}\). The rate constant’s units reflect a second order reaction because it combines the concentrations of two reactants.

• Rate equation for a reaction: \(\text{Rate} = k \times [A] \times [B]\)
• Units: \(\text{lit mol}^{-1} \text{ sec}^{-1}\) indicate second-order kinetics

In the context of determining reaction metrics, understanding \(k\) offers predictions about how quickly reactants convert into products over time, allowing chemists to control and optimize processes. In this scenario, it helps find the reverse rate constant for a related proton transfer reaction, showcasing the dynamic relationship between forward and reverse reaction constants.

Proton Transfer Reaction

Proton transfer reactions are an essential part of acid-base chemistry, characterized by the movement of protons between molecules or molecular entities. These reactions are often in equilibrium, involving the transfer of a proton from an acid to a base.

In the exercise provided, we examine the proton transfer between water and ammonia, emphasizing the relation between equilibrium and kinetic parameters. Specifically, the key equation to determine the proton transfer rate constant \(k\) is derived from:

• Given \(k_f = 3.4 \times 10^{10} \text{ lit mol}^{-1} \text{sec}^{-1}\)
• Using the relationship \(k = \frac{k_f}{K_a}\)
• This results in: \[k = \frac{3.4 \times 10^{10}}{5.6 \times 10^{-10}} = 6.07 \times 10^{-5} \text{ lit mol}^{-1} \text{sec}^{-1}\]

Here, the in-depth focus on proton transfer highlights the capacity of molecules like \([NH_3]\) to accept protons from water, essentially acting as a base. By calculating \(k\), we determine how quickly this reverse process occurs, thereby expanding our comprehension of the intricacies involved in such kinetic reactions. Understanding these metrics equips students and chemists to accurately model reactions and predict system behavior under various conditions.