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

Question

The equilibrium constant (K) of the reaction, \(\mathrm{A}+\mathrm{B}=\mathrm{C}+\mathrm{D}\) 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 \(4 \times 10^3\), so option (c) is correct.

Step 1: Understanding the relationship between rate constants and equilibrium constant

The equilibrium constant \( K \) is related to the rate constants of the forward \( k_f \) and reverse \( k_r \) reactions by the formula \( K = \frac{k_f}{k_r} \). \( K \) is the ratio of the forward rate constant to the reverse rate constant at equilibrium.

Step 2: Identifying given values

We know from the problem statement that \( K = 100 \) and the rate constant for the forward reaction \( k_f = 4 \times 10^5 \). We need to find the rate constant for the reverse reaction \( k_r \).

Step 3: Setting up the equation

Using the relationship \( K = \frac{k_f}{k_r} \), substitute the known values into the equation: \[ 100 = \frac{4 \times 10^5}{k_r} \].

Step 4: Solving for the reverse rate constant \( k_r \)

Rearrange the equation to solve for \( k_r \): \[ k_r = \frac{4 \times 10^5}{100} \].

Step 5: Simplifying the solution

Calculate \( k_r \): \[ k_r = 4 \times 10^3 \].

Key Concepts

Rate Constants

In the context of chemical reactions, rate constants are very important. They determine how fast a reaction takes place. For any given reaction, there are two key rate constants to consider:

• Forward Rate Constant ( \( k_f \) ): This is the rate at which reactants turn into products.
• Reverse Rate Constant ( \( k_r \) ): This is the rate at which products revert back to reactants.

Each reaction has its own specific rate constants, which depend on several factors like temperature and the nature of the reactants involved. At a given temperature, like the 298 K mentioned in our exercise, these constants remain the same unless conditions change.
Understanding and calculating these rates is fundamental to studying how reactions progress over time.

Forward and Reverse Reactions

In chemical reactions, processes can occur in both directions. This means that reactants convert to products (forward reaction) and products can also revert to reactants (reverse reaction).

The forward reaction can be represented as \( A + B \rightarrow C + D \), while the reverse reaction is \( C + D \rightarrow A + B \).

• Forward Reaction: This occurs when molecules combine, resulting in the formation of new products.
• Reverse Reaction: This happens when the formed products break back into the original reactants.

Forward and reverse reactions are constantly occurring at the molecular level. The balance between these two reactions determines the state of the reaction system, leading to the concept of chemical equilibrium, which allows us to use the equilibrium constant to compare them.

Chemical Equilibrium

When a reaction reaches chemical equilibrium, the rates of the forward and reverse reactions are equal. At this point, the concentration of reactants and products remains unchanged. This is not because reactions stop, but because the rate of making products equals the rate of forming reactants.

Equilibrium constants \( (K) \) are pivotal in defining this state. It is calculated as the ratio of the forward rate constant (\( k_f \) ) to the reverse rate constant (\( k_r \) ): \( K = \frac{k_f}{k_r} \).

• Equilibrium Constant (K): Value that describes the ratio of product concentrations to reactant concentrations at equilibrium. K can predict the direction in which a reaction is favored.
• Dynamic Nature: Despite no observable changes, reactions still proceed in both directions at the same rate.

In our exercise, a given equilibrium constant helped us find the reverse rate constant, emphasizing the connection between reaction rate and equilibrium.

Reaction Kinetics

Reaction kinetics is the study of rates at which chemical reactions occur. It examines how different conditions affect the speed and mechanisms of a reaction. By focusing on the kinetic aspects of a reaction, we can gain insights into the speed, steps, and conditions that favor or hinder a reaction.

• Temperature Influence: Generally, increasing temperature speeds up a reaction, as it gives molecules more energy to overcome activation barriers.
• Concentration Impact: Higher concentrations of reactants can increase the rate of reaction as there are more molecules available to collide and react.
• Catalysts: Substances that speed up reactions without being consumed. They work by lowering the activation energy required for a reaction.

Reaction kinetics not only helps in finding rate constants but also in optimizing industrial processes and understanding natural phenomena. With the rate constants and equilibrium constant, we earlier calculated in the exercise, we can better predict how quickly a reaction moves towards equilibrium.