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

Question

The equilibrium constant \((\mathrm{K})\) 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 \(4 \times 10^{3}\).

Step 1: Understanding the Relationship Between Rate Constants and Equilibrium Constant

The equilibrium constant \(K\) of a reaction is related to the rate constants of the forward \(k_{f}\) and reverse \(k_{r}\) reactions by the formula \(K = \frac{k_{f}}{k_{r}}\). Given that \(K=100\) and the forward reaction rate constant \(k_{f}=4 \times 10^5\), we can use these to find the reverse rate constant \(k_{r}\).

Step 2: Setting Up the Equation

Substitute the given values into the formula: \(100 = \frac{4 \times 10^{5}}{k_{r}}\). This equation reflects the relationship between the equilibrium constant and the rate constants of the forward and reverse reactions.

Step 3: Solving for the Reverse Rate Constant

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

Step 4: Calculate the Reverse Rate Constant

Perform the division to find \(k_{r}\): \(k_{r} = 4 \times 10^{3}\). This is the rate constant for the reverse reaction.

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K \), is a central concept in chemical equilibrium discussions. It represents the ratio of the concentration of products to reactants at equilibrium for a given reaction. For the reaction \( \mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D} \), the equilibrium constant can be expressed as:\[K = \frac{[\mathrm{C}][\mathrm{D}]}{[\mathrm{A}][\mathrm{B}]}\]where \([\mathrm{C}]\), \([\mathrm{D}]\), \([\mathrm{A}]\), and \([\mathrm{B}]\) are the molar concentrations of the respective species at equilibrium. A larger value of \( K \) (greater than 1) typically suggests that the products are favored at equilibrium, whereas a smaller \( K \) (less than 1) indicates the reactants are more prevalent.
Given in the exercise, \( K \) is 100, meaning products \( \mathrm{C} \) and \( \mathrm{D} \) are significantly more present at equilibrium than reactants \( \mathrm{A} \) and \( \mathrm{B} \). This balanced approach indicates a substantial forward progression of the reaction under the given conditions.

Rate Constants

Rate constants are vital in understanding reaction kinetics. They give us insight into the speed of both the forward and reverse reactions in an equilibrium scenario. In chemical reactions, each process of transitioning from reactants to products and products back to reactants has its own distinct rate constant.
**Forward Rate Constant \( k_f \):** This represents the speed at which the reactants \( \mathrm{A} \) and \( \mathrm{B} \) convert into products \( \mathrm{C} \) and \( \mathrm{D} \). In our context, \( k_f = 4 \times 10^5 \), indicating a rapid reaction proceeding to the right.
**Reverse Rate Constant \( k_r \):** Conversely, this constant accounts for the transformation of products back into reactants. This backward rate is crucial for maintaining equilibrium. By applying the formula:\[K = \frac{k_f}{k_r}\]we peer into the effectiveness and relative strengths of each reaction pathway. Analyzing these rate constants deepens our comprehension of how swiftly equilibrium is achieved.

Forward and Reverse Reactions

Chemical reactions are dynamic. Even at equilibrium, the reactions don't stop; rather, they balance. Forward and reverse reactions continuously occur, maintaining a state of dynamic equilibrium. Let's delve into these:

• Forward Reaction: In the reaction \( \mathrm{A} + \mathrm{B} \rightarrow \mathrm{C} + \mathrm{D} \), the forward reaction rate is dictated by the forward rate constant \( k_f \). A high \( k_f \) indicates rapid production of products.
• Reverse Reaction: Conversely, \( \mathrm{C} + \mathrm{D} \rightarrow \mathrm{A} + \mathrm{B} \) represents the reverse process, regulated by \( k_r \). The value of \( k_r \) dictates how quickly products can revert to reactants.

In equilibrium:

• The rate of the forward reaction equals the rate of the reverse reaction.
• Concentrations of reactants and products remain constant over time.

This continuous interchange guarantees that even without a visible change, a complex but perfectly balanced motion persists, much like a see-saw perfectly poised at its midpoint, reflecting the essence of chemical equilibrium.