Question

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 \(22 \mathrm{~s}\). (b) Calculate the half-lives when [NOBr] \(_{0}=0.072 M\) and \([\mathrm{NOBr}]_{0}=0.054 \mathrm{M}\)

Short answer

(a) [NOBr] ≈ 0.0342 M, (b) t_{1/2} ≈ 17.36 s (0.072 M) and t_{1/2} ≈ 23.15 s (0.054 M).

Step-by-step solution

Understanding the Second-Order Reaction

The given reaction is a second-order reaction, which means its rate depends on the square of the concentration of NOBr. The rate law can be expressed as \( r = k [NOBr]^2 \), where \( k = 0.80 / M \cdot s \). For second-order reactions, the integrated rate equation is \( \frac{1}{[A]} = \frac{1}{[A]_{0}} + kt \).

Calculating Concentration of NOBr after 22 seconds

Starting with an initial concentration \( [NOBr]_0 = 0.086 M \), we use the integrated rate law for second-order reactions: \[ \frac{1}{[NOBr]} = \frac{1}{0.086} + 0.80 \times 22 \]. Solving this, \( \frac{1}{[NOBr]} = 11.63 + 17.6 = 29.23 \). Therefore, \([NOBr] = \frac{1}{29.23} \approx 0.0342 \ M\).

Calculating Half-life for [NOBr]_{0} = 0.072 M

The half-life \( t_{1/2} \) for a second-order reaction is given by \( t_{1/2} = \frac{1}{k[NOBr]_{0}} \). Substitute \( k = 0.80 \) and \( [NOBr]_{0} = 0.072 M \): \[ t_{1/2} = \frac{1}{0.80 \times 0.072} \approx 17.36 \text{s} \].

Calculating Half-life for [NOBr]_{0} = 0.054 M

Similarly, for \([NOBr]_{0} = 0.054 M\), use the formula \( t_{1/2} = \frac{1}{k [NOBr]_{0}} \). Substitute \( k = 0.80 \) and \( [NOBr]_{0} = 0.054 M \): \[ t_{1/2} = \frac{1}{0.80 \times 0.054} \approx 23.15 \text{s} \].

Key concepts

Rate Constant

In chemical kinetics, the rate constant is a crucial factor determining how quickly a reaction occurs. It is denoted by the symbol \( k \) and varies with temperature. In our case of the second-order reaction involving NOBr, the rate constant is given as \( k = 0.80 \, / \, M \cdot s \).
This units \( / \, M \cdot s \) indicate it is a second-order reaction, as these units balance the equation in the rate law.

• Rate constant helps predict how fast reactions proceed.
• The value of \( k \) is specific to particular reactions and conditions like temperature and pressure.
• It influences the change in concentration of reactants/products over time.

Understanding and determining the rate constant simplifies analysis and predictions in chemical reactions.

Integrated Rate Law

The integrated rate law connects concentration, time, and the rate constant. For second-order reactions, the integrated rate law is expressed as:\[ \frac{1}{[A]} = \frac{1}{[A]_0} + kt \]
This formula allows us to determine the concentration of a reactant at any time \( t \).

• \( [A] \) is the concentration at time \( t \).
• \( [A]_0 \) is the initial concentration.
• \( k \) is the rate constant.
• \( t \) is the elapsed time.

Using the above law for NOBr after 22 seconds, you can substitute \( [NOBr]_0 \), \( k \), and \( t \) to find the final concentration.

Half-Life Calculations

In second-order reactions, the half-life, denoted as \( t_{1/2} \), varies with the initial concentration of the reactant. The formula for a second-order reaction is:\[ t_{1/2} = \frac{1}{k[A]_0} \]
This means that as the initial concentration \( [A]_0 \) decreases, the half-life increases.

• For \( [NOBr]_0 = 0.072 \, M \), the half-life is approximately \( 17.36 \, s \).
• For \( [NOBr]_0 = 0.054 \, M \), it increases to approximately \( 23.15 \, s \).
• The inverse relationship highlights how efficiently reactions can slow down as they proceed.

Understanding half-life calculations can provide insights into reaction dynamics and predict how long a reaction will proceed before a significant amount is converted.

Chemical Kinetics

Chemical kinetics is the study of reaction rates and the steps involved in a chemical reaction. It involves understanding not only the speed at which a reaction occurs but also the mechanism by which reactants turn into products.

• Kinetics helps identify factors affecting reaction rates, such as temperature, concentration, and catalysts.
• It enables chemists to design efficient processes by optimizing reaction conditions.
• Through various laws and mathematical models, such as rate laws and integrated rate laws, kinetics describes the quantitative aspects of chemical reactions.

By applying concepts like rate constants and half-life, chemical kinetics provides a comprehensive framework for interpreting and manipulating chemical reactions to achieve desired rates and yields.