Problem 18 - Chemistry: An Atoms-Focused Approach

Question

At \(700 \mathrm{K}, K_{c}=8.7 \times 10^{6}\) for the gas-phase reaction between \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) forming \(\mathrm{NO}_{2}\). The rate constant for the reverse reaction at this temperature is \(0.54 M^{-1} \mathrm{s}^{-1}\) What is the value of the rate constant for the forward reaction at \(700 \mathrm{K} ?\)

Step-by-Step Solution

Answer

Answer: 4.698x10^6 M^-1s^-1

Step 1: Write down the given information

We are given the following information: Equilibrium constant, \(K_c = 8.7\times10^6\) Rate constant for the reverse reaction, \(k_{reverse} = 0.54\, M^{-1}s^{-1}\) Temperature, \(T = 700\, K\)

Step 2: Write down the relationship between equilibrium constant and rate constants for forward and reverse reactions

The relationship between the equilibrium constant, \(K_c\), and the rate constants for the forward and reverse reactions, \(k_{forward}\) and \(k_{reverse}\), can be written as follows: \(K_c = \frac{k_{forward}}{k_{reverse}}\)

Step 3: Calculate the rate constant for the forward reaction

Rearrange the formula from Step 2 to solve for the rate constant for the forward reaction: \(k_{forward} = K_c \times k_{reverse}\) Now, plug in the values given in the problem: \(k_{forward} = (8.7\times10^6) \times (0.54\, M^{-1}s^{-1})\) Compute the result: \(k_{forward} = 4.698\times10^6\, M^{-1}s^{-1}\)

Step 4: Report the result

The rate constant for the forward reaction at \(700\,K\) is \(4.698\times10^6\, M^{-1}s^{-1}\).

Key Concepts

Rate Constants

Rate constants are crucial in understanding how quickly chemical reactions proceed. They are denoted typically by the symbol "k" and can vary greatly between the forward and reverse processes of a reaction. At a given temperature, if you know the rate constant for one direction of a reaction and the equilibrium constant, you can easily find the other rate constant.

• In a chemical equation, the rate constant is affected by factors like temperature and the presence of a catalyst.
• The units of the rate constant vary depending on the order of the reaction, which describes how the rate depends on the concentration of the reactants.
• Higher rate constants indicate faster reactions.

In the provided exercise, we used the equilibrium position to derive the rate constant for the forward reaction, making these constants essential for calculating how long reactions take to reach equilibrium.

Reaction Kinetics

Reaction kinetics is the study of the speeds or rates at which chemical reactions occur. This branch of science helps explain how different variables affect these rates, giving insights into reaction mechanisms and pathways.

• Reaction kinetics helps in determining why certain reactions occur faster than others.
• Understanding kinetic information aids in predicting how a reaction will proceed and how external conditions like temperature or pressure changes will affect it.
• Kinetics often involves studying elementary steps to piece together a detailed reaction mechanism.

In the given exercise, kinetics comes into play as we relate the rate constants and equilibrium constants to understand the forward and reverse reactions' dynamics and speed.

Chemical Equilibrium

Chemical equilibrium occurs when a reaction reaches a state where the rate of the forward reaction equals the rate of the reverse reaction. In simple terms, the concentrations of reactants and products remain constant over time. This balance explains why certain concentrations remain stable even when reactions don't go to completion.

• The equilibrium constant, denoted as \(K_c\) for reactions involving concentration, quantifies the ratio of product concentration to reactant concentration at equilibrium.
• If \(K_c\) is very high, the reaction favors the products, while a low \(K_c\) indicates a reaction favoring the reactants.
• Le Chatelier's principle describes how a system at equilibrium responds to external changes, helping predict how changes in conditions will affect the equilibrium state.

In our exercise, we used the equilibrium constant to find the rate constant for the forward reaction, demonstrating the practical importance of understanding equilibrium in calculating how quickly systems approach balance.