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 \(162 \mathrm{~L}\) \(\mathrm{mol}^{-1} \mathrm{sec}^{-1}\), 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

Short answer

The velocity constant for the backward reaction is 2 (option b).

Step-by-step solution

Understanding the Relationship between Reaction Rates and Equilibrium Constant

The equilibrium constant \( K_{eq} \) for a reversible reaction is given by the ratio of the forward reaction rate constant \( k_f \) to the backward reaction rate constant \( k_b \):\[ K_{eq} = \frac{k_f}{k_b}. \]In this scenario, \( K_{eq} = 81 \) and \( k_f = 162 \). We need to find \( k_b \).

Rearrange the Equation

Rearrange the expression for the equilibrium constant to solve for the backward reaction rate constant \( k_b \):\[ k_b = \frac{k_f}{K_{eq}}. \]

Substitute Given Values

Substitute the given values for \( k_f \) and \( K_{eq} \) into the rearranged equation:\[ k_b = \frac{162}{81}. \]

Perform the Calculation

Calculate \( k_b \) by dividing 162 by 81:\[ k_b = 2. \]

Select the Correct Answer

The calculated backward reaction rate constant \( k_b \) is 2. Compare this value with the provided options: (a) 13122, (b) 2, (c) 261, (d) 243. The correct answer is option (b) 2.

Key concepts

Reaction Rate Constants

In any chemical reaction, the reaction rate constant is a crucial factor that influences how quickly or slowly a reaction proceeds. It's essential to understand these constants when studying kinetics. The forward reaction rate constant, denoted as \( k_f \), and the backward reaction rate constant, denoted as \( k_b \), determine the speed of the forward and backward reactions, respectively. Each constant's unit depends on the order of the reaction and can be expressed in various forms, such as \( \text{L mol}^{-1} \text{sec}^{-1} \) for second-order reactions.

The relationship between these constants and the equilibrium constant \( K_{eq} \)—a measure of the reaction's balance at equilibrium—provides critical insights into the reaction's dynamics. Understanding this relationship requires recognizing that \( K_{eq} \) is defined as the ratio of \( k_f \) to \( k_b \). Knowing any two of these values allows you to determine the third. This formula effectively ties together reaction kinetics and thermodynamics, offering a comprehensive view of the reaction process.

Reversible Reactions

Reversible reactions are a fundamental concept in chemistry, where the reactants can transform into products and vice versa. Unlike irreversible reactions, where the reaction goes to completion, reversible reactions can reach an equilibrium state. At equilibrium, both the forward and backward reactions occur at the same rate, ensuring that the concentration of reactants and products remains constant over time.

This balance is depicted in the equation for the equilibrium constant \( K_{eq} \), which can be expressed as \( K_{eq} = \frac{k_f}{k_b} \), linking the reaction rate constants for the forward and backward reactions. Understanding reversible reactions helps chemists predict the direction of change in a system when external conditions, such as temperature or concentration, are altered. It also provides a basis for controlling reactions in industrial processes to maximize product yield efficiently.

Velocity Constants

Velocity constants, more commonly referred to as rate constants, play a pivotal role in reaction kinetics. They provide quantitative details about the speed at which reactants convert into products in a reaction. The forward velocity constant \( k_f \) can tell you how many reactant molecules transform to products per unit time under specific conditions, while the backward velocity constant \( k_b \) indicates the reverse process's rate.

In the context of equilibrium, these constants are not just simple numbers, but are directly related to the reaction's equilibrium position. A high forward velocity constant compared to the backward velocity constant suggests that the reaction favors product formation, and vice versa. Moreover, these constants allow you to calculate the equilibrium constant \( K_{eq} \) using the formula \( K_{eq} = \frac{k_f}{k_b} \). By understanding velocity constants, chemists can manipulate reaction conditions to favor either reaction direction, optimizing the system's efficiency in laboratory practices or large-scale production.