Problem 68 - A Complete Resource Book in Chemistry for JEE Main

Question

For the reaction, \(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}\), the rate con- stants for the forward and backward reactions are found to be \(4.2 \times 10^{-2}\) and \(3.36 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) respectively. What is the equilibrium constant for the reaction? (a) \(11.5\) (b) \(12.5\) (c) \(8.0\) (d) \(6.0\)

Step-by-Step Solution

Answer

The equilibrium constant is 12.5, so the answer is (b).

Step 1: Understand the Relationship between Rate Constants and Equilibrium Constant

In a chemical reaction of the form \( \mathrm{A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D} \), the equilibrium constant \( K_{eq} \) can be calculated using the forward and backward rate constants. This relationship is given by the formula: \[ K_{eq} = \frac{k_{forward}}{k_{backward}} \] where \( k_{forward} \) is the rate constant for the forward reaction, and \( k_{backward} \) is the rate constant for the backward reaction.

Step 2: Plug in the Values

Substitute the provided values into the equilibrium constant formula. According to the problem, the forward rate constant \( k_{forward} = 4.2 \times 10^{-2} \mathrm{\ mol \ L}^{-1} \mathrm{\ s}^{-1} \) and the backward rate constant \( k_{backward} = 3.36 \times 10^{-3} \mathrm{\ mol \ L}^{-1} \mathrm{\ s}^{-1} \). Insert these values into the formula: \[ K_{eq} = \frac{4.2 \times 10^{-2}}{3.36 \times 10^{-3}} \]

Step 3: Calculate the Equilibrium Constant

Perform the division to calculate the equilibrium constant:\( K_{eq} = \frac{4.2 \times 10^{-2}}{3.36 \times 10^{-3}} \)Simplify the expression by dividing the two numbers:\( K_{eq} = 12.5 \)

Key Concepts

Rate Constant

When discussing chemical reactions, the term 'rate constant' frequently arises. A rate constant, denoted as \( k \), is a crucial factor that determines how quickly a reaction proceeds. Essentially, it is a numerical value that reflects the speed of a reaction under specific conditions.
For any given reaction, the rate constant differs based on whether it is for the forward or the backward reaction. * **Forward Rate Constant**: This constant indicates how fast the reactants are converted into products.
* **Backward Rate Constant**: This represents the speed at which products revert back to reactants.
Temperature and the presence of catalysts can affect the rate constant significantly. A general knowledge of rate constants helps students to predict the behaviour of reactions and determine reaction speeds.

Chemical Equilibrium

In the realm of chemistry, chemical equilibrium is a concept that describes a state in which the concentrations of all reactants and products remain constant over time. This is not because the reactions have stopped, but because the rates of the forward and reverse reactions are equal.
At this point, the reaction reaches a balance, or equilibrium. The equilibrium constant \( K_{eq} \) is a number that expresses the ratio of product concentrations to reactant concentrations at equilibrium. It is calculated using the formula: \[ K_{eq} = \frac{k_{forward}}{k_{backward}} \]This equation illustrates the relationship between rate constants and equilibrium. A larger \( K_{eq} \) value suggests a higher concentration of products at equilibrium, indicating the reaction favors the formation of products.

Reversible Reactions

Reversible reactions are fascinating because they can move in both directions: reactants to products and vice versa. This kind of reaction is represented by the double-headed arrow \( \rightleftharpoons \) in chemical equations.
Characteristics of reversible reactions include: * **Dynamic Balance**: Even at equilibrium, both the forward and backward reactions continue to occur.
* **Adjustments to Conditions**: Changes in temperature, pressure, or concentration can shift the position of equilibrium, affecting both the rate constants and the equilibrium constant. Understanding reversibility helps in grasping how chemical reactions can be manipulated to favor the production of certain substances. For instance, by altering conditions, one can push the reaction towards producing more products or more reactants, which is incredibly useful in industrial processes.