Problem 141 - A Complete Resource Book in Chemistry for JEE Main

Question

The equilibrium constant (K) of the reaction, \(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}\) at \(298 \mathrm{~K}\) is 100 . If the rate con- stant of the forward reaction is \(4 \times 10^{5}\), the rate constant of the reverse reaction is (a) 4 (b) \(4 \times 10^{2}\) (c) \(4 \times 10^{3}\) (d) \(4 \times 10^{5}\)

Step-by-Step Solution

Answer

The rate constant of the reverse reaction is \(4 \times 10^{3}\), option (c).

Step 1: Understand Equilibrium Constant Formula

The equilibrium constant (K) for a reaction is given by the formula \( K = \frac{k_f}{k_r} \), where \( k_f \) is the rate constant of the forward reaction and \( k_r \) is the rate constant of the reverse reaction.

Step 2: Substitute Known Values

Substitute the given values into the equilibrium constant formula: \( K = 100 \) and \( k_f = 4 \times 10^{5} \). The equation becomes \( 100 = \frac{4 \times 10^{5}}{k_r} \).

Step 3: Solve for Reverse Rate Constant

To find the reverse rate constant \( k_r \), rearrange the equation to solve for \( k_r \): \( k_r = \frac{4 \times 10^{5}}{100} \).

Step 4: Calculate Reverse Rate Constant

Perform the division: \( k_r = \frac{4 \times 10^{5}}{100} = 4 \times 10^{3} \).

Key Concepts

Rate Constant

The rate constant is a vital concept in chemical kinetics. It helps us understand how fast a reaction proceeds. In a chemical reaction, the rate constant is a proportionality constant that connects the reaction rate to the concentrations of the reactants. For a given reaction, the rate constant is affected by factors such as temperature and the presence of a catalyst.

For example, if you have a reaction: \( \text{A} + \text{B} \rightarrow \text{C} + \text{D} \), and if the reaction rate can be expressed as \( r = k_f [A][B] \), then \( k_f \) is the rate constant for the forward reaction.

• A higher rate constant indicates a faster reaction.
• A lower rate constant means a slower reaction.

Understanding the rate constant enables chemists to predict how reaction conditions will affect how quickly products form. It also helps in designing industrial processes for better efficiency.

Forward Reaction

In chemical reactions, the forward reaction refers to the process where reactants are converted into products as dictated by the reaction equation. For example, in the equation \( \text{A} + \text{B} \rightleftharpoons \text{C} + \text{D} \), moving from left to right, the reaction is progressing in the forward direction.

The rate of this forward reaction is determined by the rate constant of the forward reaction, denoted as \( k_f \). The forward reaction rate increases with higher concentrations of reactants and an increase in temperature.

• Greater concentrations of \( \text{A} \) and \( \text{B} \) lead to a higher chance of collisions and thus a faster forward reaction.
• Temperature increase can enhance molecular motion, leading to more effective collisions.

This understanding of forward reactions allows scientists to manipulate conditions to favor the formation of desired products in a chemical process.

Reverse Reaction

A reverse reaction is the opposite of the forward reaction. This occurs when products converted back into reactants. In the context of the reaction \( \text{A} + \text{B} \rightleftharpoons \text{C} + \text{D} \), the reverse reaction proceeds from right to left.

The rate constant for the reverse reaction, \( k_r \), determines how quickly this process occurs. It can be calculated using the equilibrium constant \( K \) and the forward rate constant \( k_f \) using the equation \( K = \frac{k_f}{k_r} \).

• If \( k_r \) is large, the reverse reaction is quick, implying that products are readily returned to reactants.
• A small \( k_r \) means the reverse process is slow, hence the equilibrium will favor the products.

Balancing the forward and reverse reactions is critical in a dynamic equilibrium. It helps achieve desired yields in chemical synthesis and optimization of reaction conditions.