Problem 85 - A Complete Resource Book in Chemistry for JEE Main

Question

Equilibrium constant for the reaction \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g})\) is 81. If the velocity constant of the forward reaction is \(162 \mathrm{~L}\) \(\mathrm{mol}^{-1} \sec ^{-1}\), what is the velocity constant (in \(\mathrm{L} \mathrm{mol}^{-1}\) \(\sec ^{-1}\) ) for the backward reaction? (a) 13122 (b) 2 (c) 261 (d) 243

Step-by-Step Solution

Answer

The reverse rate constant \( k_r \) is 2 \( \mathrm{L} \mathrm{mol}^{-1} \sec^{-1} \); the answer is (b) 2.

Step 1: Understand the relationship between equilibrium and rate constants

The equilibrium constant \( K_{eq} \) for a reaction is the ratio of the rate constant of the forward reaction \( k_f \) to the rate constant of the reverse reaction \( k_r \). Mathematically, this is expressed as \( K_{eq} = \frac{k_f}{k_r} \).

Step 2: Plug in the known values

We are given \( K_{eq} = 81 \) and \( k_f = 162 \mathrm{~L} \mathrm{mol}^{-1} \sec^{-1} \). Substitute these values into the equation \( 81 = \frac{162}{k_r} \).

Step 3: Solve for the reverse rate constant \( k_r \)

To solve for \( k_r \), rearrange the equation from Step 2: \( k_r = \frac{162}{81}\). Calculate \( k_r \) to find \( k_r = 2 \mathrm{~L} \mathrm{mol}^{-1} \sec^{-1} \).

Key Concepts

Equilibrium Constant

The equilibrium constant, often symbolized as \( K_{eq} \), is a fundamental concept in chemical equilibrium. It represents the relationship between the concentrations of reactants and products at equilibrium in a chemical reaction. This constant is derived from the law of mass action, which states that at equilibrium at a constant temperature, the ratio of the product of concentrations of products to the product of concentrations of reactants is a constant for a given reaction.

In our given reaction, \[\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g})\rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g})\]the equilibrium constant \( K_{eq} \) is given as 81. This means that at equilibrium, the concentrations of hydrogen and carbon dioxide are much greater compared to the concentrations of water and carbon monoxide, suggesting the reaction favors the production of hydrogen and carbon dioxide under those conditions. Understanding \( K_{eq} \) helps predict the direction of the reaction and the extent to which reactants are converted to products at equilibrium.

Rate Constant

The rate constant, denoted as \( k \), plays a critical role in chemical kinetics. It is a parameter that quantifies the rate at which a reaction proceeds. The value of \( k \) is influenced by factors such as temperature and the presence of catalysts, but it is independent of the concentrations of reactants and products.

In the context of equilibrium, we distinguish between the rate constants of the forward and reverse reactions: \( k_f \) and \( k_r \) respectively. The equilibrium constant \( K_{eq} \) is determined by the ratio of these rate constants:\[K_{eq} = \frac{k_f}{k_r}\]
The exercise provides \( k_f = 162\mathrm{~L mol^{-1} sec^{-1}} \), and using the value of \( K_{eq} \), one can determine \( k_r \). Understanding the connection between \( K_{eq} \) and the rate constants is pivotal for analyzing how fast equilibrium is achieved.

Forward Reaction

The forward reaction refers to the process in which reactants are converted into products. In chemical notation, it is typically represented by the arrow pointing from reactants to products (e.g., \( A + B \rightarrow C + D \)).

For our specific reaction,\[\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g})\]the forward reaction involves the conversion of water vapor and carbon monoxide into hydrogen gas and carbon dioxide. The rate of this transformation is characterized by the forward rate constant \( k_f \), which is given as \( 162\mathrm{~L mol^{-1} sec^{-1}} \) in the exercise.

Understanding the forward reaction is important as it allows us to determine the speed at which products are formed from reactants before reaching equilibrium.

Reverse Reaction

The reverse reaction is the process by which products revert to form reactants. In our chemical equilibrium scenario, it is represented by the reaction going from products back to reactants, symbolized by the backward arrow (e.g., \( C + D \rightarrow A + B \)).

Taking our reaction into consideration:\[\mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g})\]
Here, hydrogen gas and carbon dioxide are converted back into water vapor and carbon monoxide. The reverse rate constant \( k_r \) quantifies how quickly this reaction occurs, which the solution calculates as \( 2\mathrm{~L mol^{-1} sec^{-1}} \). Understanding the reverse reaction and its rate constant helps in comprehending how and when equilibrium is reached, as both forward and reverse reactions occur simultaneously and at the same rate in dynamic equilibrium.