Problem 157 - A Complete Resource Book in Chemistry for JEE Main

Question

Equilibrium constant of \(\mathrm{NH}_{4}^{+}\)to \(\mathrm{NH}_{3}\) and \(\mathrm{H}^{+}\)is \(10^{-10}\). The rate constant for \(\mathrm{NH}_{4}^{+}+\mathrm{OH}^{-} \rightarrow \mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O}\) is \(10^{10} .\) The rate constant for \(\mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{NH}_{4}^{+}+\mathrm{OH}^{-}\)is (a) \(10^{5}\) (b) \(10^{6}\) (c) \(10^{8}\) (d) \(10^{9}\)

Step-by-Step Solution

Answer

The calculated rate constant \(k_b\) is \(10^{20}\), possibly a typo in choices.

Step 1: Understand the Concepts

We are dealing with two related reactions and their rate constants. We have equilibrium constant K for the reaction \(\mathrm{NH}_{4}^{+} \rightleftharpoons \mathrm{NH}_{3} + \mathrm{H}^{+}\) and two rate constants for the forward reaction \(\mathrm{NH}_{4}^{+} +\mathrm{OH}^{-} \rightarrow \mathrm{NH}_{3} + \mathrm{H}_{2}\) and the backward reaction \(\mathrm{NH}_{3} + \mathrm{H}_{2} \rightarrow \mathrm{NH}_{4}^{+} + \mathrm{OH}^{-}\).

Step 2: Determine the Relationship Between Constants

For any reversible reaction at equilibrium, the relationship between the equilibrium constant (\(K\)) and the forward (\(k_f\)) and backward rate constants (\(k_b\)) is given by:\[ K = \frac{k_f}{k_b} \]

Step 3: Plug in the Values to Calculate

We are given \(K = 10^{-10}\) and \(k_f = 10^{10}\). We need to find \(k_b\), which is the rate constant for the reverse reaction. Using the relationship:\[ 10^{-10} = \frac{10^{10}}{k_b} \]

Step 4: Solve for \(k_b\)

Rearrange the formula to solve for \(k_b\): \[ k_b = \frac{10^{10}}{10^{-10}} \]Calculate:\[ k_b = 10^{10} \times 10^{10} = 10^{20} \]

Step 5: Match with Given Options

The calculated \(k_b\) is \(10^{20}\), which does not match any of the given options. Re-check the process or information given in the question, ensuring accuracy in interpretation.

Key Concepts

Equilibrium Constant

The concept of the equilibrium constant, often represented as \( K \), is fundamental in understanding chemical reactions that can proceed in both forward and reverse directions. In layman's terms, the equilibrium constant is a number that helps us understand the balance between the products and reactants in a chemical reaction when the reaction has reached a state where no further net change occurs. This state is known as chemical equilibrium.

For a reversible reaction like \( \mathrm{NH_4^+} \rightleftharpoons \mathrm{NH_3} + \mathrm{H^+} \), the equilibrium constant \( K \) can be expressed using the concentrations of the reactants and products at equilibrium:

\[ K = \frac{[\mathrm{NH_3}][\mathrm{H^+}]}{[\mathrm{NH_4^+}]} \]

**Key Points to Remember:**

• The equilibrium constant is temperature-dependent; it only remains constant at a specific temperature.
• A small \( K \) value (less than 1) indicates that the equilibrium position favors the reactants, while a large \( K \) value (greater than 1) suggests that the products are favored.
• In our exercise, the given \( K \) value of \( 10^{-10} \) indicates that there are more reactants than products at equilibrium.

Reaction Rate Constants

Reaction rate constants are a measure of the speed at which a reaction progresses. In simpler terms, they tell us how quickly the reactants are converted into products in a given reaction. These constants are crucial for both the forward and backward reactions of a reversible reaction.

For our exercise, the forward reaction is depicted as \( \mathrm{NH_4^+} + \mathrm{OH^-} \rightarrow \mathrm{NH_3} + \mathrm{H_2O} \) and the backward reaction is \( \mathrm{NH_3} + \mathrm{H_2O} \rightarrow \mathrm{NH_4^+} + \mathrm{OH^-} \). The constants for these types of reactions are referred to as \( k_f \) for the forward reaction and \( k_b \) for the backward reaction.

**Understanding Rate Constants:**

• The rate constant \( k \) is specific to each chemical reaction and depends on many factors, including temperature and the nature of the reaction.
• For the exercise, the forward rate constant \( k_f \) is given as \( 10^{10} \).
• We use the relationship \( K = \frac{k_f}{k_b} \) to find the backward reaction rate constant \( k_b \).

Reversible Reactions

A reversible reaction is a chemical reaction that can proceed in both forward and backward directions. This dual action allows the system to reach a dynamic equilibrium, where the rate of the forward reaction equals the rate of the backward reaction, resulting in no net change in the concentration of reactants and products over time.

In reversible reactions like \( \mathrm{NH_4^+} \rightleftharpoons \mathrm{NH_3} + \mathrm{H^+} \), it’s essential to remember:

**Characteristics of Reversible Reactions:**

• Reversible reactions do not fully convert reactants into products; instead, they settle into an equilibrium mixture of both.
• The position of equilibrium can be influenced by changing conditions such as temperature, pressure, and concentration (as per Le Chatelier's principle).
• In the context of our exercise, although the forward rate is rapid with \( k_f = 10^{10} \), the small \( K \) value of \( 10^{-10} \) implies that at equilibrium, most of the substance exists as reactants.
• The careful balance of forward and backward reactions is what characterizes reversible reactions, making them dynamic and responsive to changes in external conditions.