The rate constant for the second-order reaction $$ 2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ is \(0.80 / M \cdot \mathrm{s}\) at \(10^{\circ} \mathrm{C}\). (a) Starting with a concentration of \(0.086 M,\) calculate the concentration of NOBr after 22 s. (b) Calculate the half-lives when \([\mathrm{NOBr}]_{0}=0.072 M\) and \([\mathrm{NOBr}]_{0}=0.054 M\)

Understanding the rate constant is crucial in the study of chemical reactions. The rate constant, denoted as 'k', is a critical factor in the kinetics of a chemical reaction. It provides the speed at which reactants turn into products. For a second-order reaction, such as the one between nitrogen oxide bromide (NOBr) and nitrogen oxide (NO) with bromine (Br₂), the rate constant has units of M^-1s^-1. This signifies that the reaction rate changes with the inverse of concentration over time.

The higher the value of the rate constant, the faster the reaction proceeds. It’s important to note that the rate constant is temperature-dependent and can be impacted by other factors such as catalysts. In the provided exercise, with a rate constant of 0.80 / M∙s at 10°C, we can expect the concentration of NOBr to decrease significantly over time as the reaction proceeds.

Calculating the concentration of reactants or products at any given time is a fundamental aspect of studying reaction kinetics. To calculate the concentration of NOBr after 22 seconds, we use the integrated rate law for a second-order reaction. This law relates the concentration of a reactant at time 't' to the initial concentration and the rate constant.

For the specific exercise, the formula \[ [A]_t = (\frac{1}{kt + [1/A_0]}) \] is used, where \( [A]_t \) represents the concentration at time \( t \) and \( [A]_0 \) is the initial concentration. By inputting the values \( [A]_0 = 0.086 M \) and \( k = 0.80 / M\cdot s \) along with \( t = 22s \) into this equation, students can calculate the remaining concentration of NOBr. This simplified approach allows for a direct solution without the need for complex calculus.

The concept of half-life in chemistry refers to the time it takes for half of the reactant to be converted into the product. For second-order reactions, the half-life is inversely proportional to both the rate constant and the initial concentration of the reactant. The formula for calculating the half-life of a second-order reaction is \( t_{1/2} = \frac{1}{k[A]_0} \).

This means that unlike first-order reactions, where the half-life is constant, the half-life in second-order reactions varies with the initial concentration. In the exercise, with initial NOBr concentrations of 0.072 M and 0.054 M, we would expect different half-lives for each scenario when applying the same rate constant. Hence, it illustrates the unique property of second-order reactions where the amount of substance you start with directly affects how quickly it will take to reduce to half its amount.

Chemical kinetics is the study of the rates of chemical processes. It examines how different variables such as concentration, temperature, and presence of a catalyst affect the rate of reaction. By understanding kinetics, we can predict the speed of a reaction under various conditions, optimize the conditions needed to produce desired products, and control the reaction rates for safety and efficacy.

In second-order kinetics, the reaction rate is proportional to the square of the concentration of one reactant or to the product of the concentrations of two reactants. As illustrated in the exercise, kinetic calculations require careful manipulation of the equations that define the reaction's behavior over time. Educators emphasizing the practical applications of kinetics help students understand the value of these calculations in real-world scenarios, such as the synthesis of chemicals, environmental modeling, and even in pharmacology where reaction rates can influence drug efficacy and safety.