Problem 25 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

The reaction of methane with hydrogen atoms is an elementary process. \\[ \mathrm{CH}_{4}+\mathrm{H}^{*} \rightleftharpoons \mathrm{CH}_{3}^{*}+\mathrm{H}_{2} \\] The rate constant for the forward reaction at \(1000 \mathrm{K}\) is \(1.6 \times 10^{8} \mathrm{dm}^{3} \mathrm{mol}^{-1} \mathrm{s}^{-1}\). The equilibrium constant, \(K_{\mathrm{c}},\) at the same temperature is \(19.8 .\) Calculate the rate constant for the reverse reaction at \(1000 \mathrm{K} \text { . (Section } 9.6)\)

Step-by-Step Solution

Answer

The reverse rate constant is approximately \( 8.08 \times 10^6 \text{ dm}^3 \text{ mol}^{-1} \text{ s}^{-1} \).

Step 1: Understanding the Problem

We are given the rate constant for the forward reaction and the equilibrium constant. We need to calculate the rate constant for the reverse reaction using the provided data.

Step 2: Writing the Relationship between Rate Constants and Equilibrium Constant

For a reversible reaction, the equilibrium constant \( K_c \) is related to the rate constants of the forward \( k_{f} \) and reverse \( k_{r} \) reactions by the formula: \[ K_c = \frac{k_f}{k_r} \] In this problem, we know \( K_c = 19.8 \) and \( k_f = 1.6 \times 10^8 \text{ dm}^3 \text{ mol}^{-1} \text{ s}^{-1} \), and need to find \( k_r \).

Step 3: Rearranging the Formula

We can rearrange the formula to solve for the reverse rate constant \( k_{r} \): \[ k_r = \frac{k_f}{K_c} \]

Step 4: Substitute the Known Values

Substitute the known values of \( k_f \) and \( K_c \) into the equation: \[ k_r = \frac{1.6 \times 10^8 \text{ dm}^3 \text{ mol}^{-1} \text{ s}^{-1}}{19.8} \]

Step 5: Calculate the Reverse Rate Constant

Perform the calculation: \[ k_r = \frac{1.6 \times 10^8}{19.8} \approx 8.08 \times 10^6 \text{ dm}^3 \text{ mol}^{-1} \text{ s}^{-1} \] Thus, the rate constant for the reverse reaction is approximately \( 8.08 \times 10^6 \).

Key Concepts

Elementary Reaction

An elementary reaction is a chemical reaction where reactants convert to products in a single step. In such reactions, the mechanism involves basic molecular events, like bond breaking or forming, as opposed to complex multi-step processes. This simplicity means that the rate of an elementary reaction is directly proportional to the concentration of the reactants. For example, in the given exercise, the reaction between methane (\( \mathrm{CH}_4 \)) and hydrogen atoms (\( \mathrm{H}^* \)) directly yields methyl radicals (\( \mathrm{CH}_3^* \)) and hydrogen molecules (\( \mathrm{H}_2 \)).
What makes elementary reactions particularly interesting is that their stoichiometry directly tells us about the reaction orders. For instance, if two molecules of reactants are used in an elementary step, the reaction is second-order. This immediate insight is crucial for understanding kinetics without diving into the mechanism nuances as required for more complex reactions.

Equilibrium Constant

The equilibrium constant \( K_c \) is a crucial concept when dealing with reversible reactions. It provides a ratio that compares the concentration of products to reactants at equilibrium, a state where the forward and reverse reaction rates are equal. For our exercise, the equation \( \mathrm{CH}_4+\mathrm{H}^* \rightleftharpoons \mathrm{CH}_3^*+\mathrm{H}_2 \) has been given an equilibrium constant \( K_c = 19.8 \) at 1000 K. This numerical value indicates that at equilibrium, the concentration of products is significantly favored over the concentration of the reactants.
This ratio helps chemists understand how far the reaction proceeds before reaching its balanced state. A higher \( K_c \) value means that at equilibrium, the products dominate, while a lower \( K_c \) value suggests the reactants are more prevalent. Equilibrium constants are also fundamental in calculating reverse reactions' rate constants, as seen in the exercise, reinforcing the interconnected nature of kinetic and equilibrium concepts.

Rate Constant

Rate constants are proportionality factors in the rate equations that quantitatively describe reaction speed. In the exercise, the forward reaction rate constant \( k_f \) for the elementary reaction between methane and hydrogen atoms is given as \( 1.6 \times 10^8 \) dm\(^3\) mol\(^{-1}\) s\(^{-1}\).
This value tells us that the higher the concentration of \( \mathrm{CH}_4 \) and \( \mathrm{H}^* \), the faster the reaction will proceed in the forward direction. Rate constants vary with temperature and reaction specifics; thus, they're essential parameters for predicting how quickly a reaction reaches completion or equilibrium.

• A higher rate constant signifies a faster reaction.


• Rate constants are integral in comparing the dynamics of various reactions under different conditions.


• For reversible reactions, these constants are vital in calculating how equilibrium is approached, playing a role in both forward and reverse reactions, as seen with the relationship \( K_c = \frac{k_f}{k_r} \).

Reversible Reactions

Reversible reactions are processes where reactants convert to products and vice versa, indicating that the transformation can proceed in both directions. In chemical kinetics, particularly in our exercise, the reaction \( \mathrm{CH}_4+\mathrm{H}^* \rightleftharpoons \mathrm{CH}_3^*+\mathrm{H}_2 \) shows this duality.
The concept of reversibility is essential because it acknowledges the dynamic nature of chemical reactions - they're not one-way streets. Understanding this can help us grasp that at some point, both reactants and products exist in equilibrium.

• Equilibrium is the state where the rates of forward and backward reactions equalize.


• Reversible reactions are studied not just for their kinetics but also to understand equilibrium constants, shedding light on which side of the reaction is favored.

Reversible reactions emphasize the need for robust analysis, as these systems eventually aim to achieve a balance rather than transforming completely into products.