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

There is likely an error in the given options, as the calculated rate constant is \(10^{20}\), not matching any option.

Step 1: Understand the Concepts

In this problem, we are given rate constants for forward and reverse reactions, along with the equilibrium constant. Our goal is to find the unknown rate constant for the reverse reaction based on this information.

Step 2: Write the Equilibrium Constant Expression

The equilibrium constant \( K \) for a reaction is defined as \( K = \frac{k_{ ext{forward}}}{k_{ ext{reverse}}} \). Here, \( K = 10^{-10} \), and \( k_{\text{forward}} = 10^{10} \).

Step 3: Solve for the Unknown Rate Constant

To find the reverse rate constant \( k_{\text{reverse}} \), rearrange the equation from Step 2: \( k_{\text{reverse}} = \frac{k_{\text{forward}}}{K} \). Substitute the given values to find \( k_{\text{reverse}} = \frac{10^{10}}{10^{-10}} = 10^{20} \).

Step 4: Compare with Given Options

Compare the value calculated for the reverse rate constant \( 10^{20} \) with the options provided. Clearly, none of the given options (\(10^{5}, 10^{6}, 10^{8}, 10^{9}\)) match our calculated result. Therefore, there might be an issue with the options provided or further context needed.

Key Concepts

Rate Constant

The rate constant is a fundamental component of chemical kinetics, providing the rate at which a reaction progresses. It is denoted typically by the letter "k" and helps quantify the speed of either the "forward" or "reverse" reaction depending on the direction. Unlike other components of reactions, the rate constant remains unchanged when the concentrations of reactants or products alter. However, it is sensitive to changes in temperature, meaning it can vary if the temperature is different.

• The rate constant is unique to each specific reaction.
• It is influenced by the presence of catalysts that can lower the activation energy, thereby increasing the rate.
• Mathematically, it is involved in the rate equation: for a general reaction \(aA + bB \rightarrow cC + dD\), the rate might be described by \(\text{Rate} = k[A]^x[B]^y\).

Understanding the rate constant is crucial to predict how quickly a reaction will occur under specific conditions. Without it, determining the dynamics of a reaction system would be challenging.

Forward Reaction

A forward reaction refers to the process where reactants convert to products. In an equilibrium scenario, this forward process has a significant role in establishing the conditions of equilibrium. The given forward reaction in the problem is the transformation of ammonium ion \(\mathrm{NH}_{4}^{+}\) into ammonia \(\mathrm{NH}_{3}\) and water, via the interaction with hydroxide ion \(\mathrm{OH}^{-}\).

• The forward rate constant \(k_{\text{forward}}\) represents the rate of this conversion and helps us understand how efficient the process is.
• The larger the forward rate constant, the quicker the reactants turn into products.

This specific forward reaction rate constant is given as \(10^{10}\), implying a rapid conversion under specified conditions. In equilibrium understanding, knowing the rate helps in calculating other parameters like the reverse reaction rate or the equilibrium constant itself.

Reverse Reaction

In chemistry, a reverse reaction is the counterpart to the forward reaction, where products revert into reactants. In equilibrium, the continuous process of both forward and reverse reactions achieves a state where the concentrations of reactants and products remain stable.

• The reverse reaction for the scenario provided involves the recombination of ammonia and water to form \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{OH}^{-}\).
• The reverse rate constant \(k_{\text{reverse}}\) indicates how fast this reversal occurs.
• Finding \(k_{\text{reverse}}\) can involve rearranging known values, such as the equilibrium constant and the forward rate constant.

In our problem, the reverse rate constant helps balance the entire equation, ensuring the reaction can maintain its equilibrium state efficiently. Mastery of reverse reactions is essential for comprehensive insights into chemical equilibria and reaction kinetics.

Equilibrium Expression

The equilibrium expression is a mathematical equation that denotes the ratio of the concentrations of products to reactants when a chemical reaction is in a state of dynamic balance. The equilibrium constant \(K\) emerges from this expression and is a critical value in understanding the chemical equilibrium. For our given reaction, the equilibrium constant signifies how far the forward or reverse reaction predominates under equilibrium conditions.

• The equilibrium expression is derived from the Law of Mass Action and is given by: \(K = \frac{[\text{products}]}{[\text{reactants}]}\).
• In our specific case, it is represented by \(K = \frac{k_{\text{forward}}}{k_{\text{reverse}}}\).
• The value of \(K\) remains constant at a given temperature, regardless of how concentrations of products or reactants might fluctuate.

Understanding equilibrium expression is pivotal because it allows chemists to predict the direction of a reaction under non-equilibrium conditions and helps control the synthesis and yield of products in industrial chemical processes.