Problem 54 - Chemistry An Atoms First Approach

Question

The rate law for the reaction $$2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ at some temperature is $$\text { Rate }=-\frac{\Delta[\mathrm{NOBr}]}{\Delta t}=k[\mathrm{NOBr}]^{2}$$ a. If the half-life for this reaction is 2.00 s when \([\mathrm{NOBr}]_{0}=\) \(0.900 \space M,\) calculate the value of \(k\) for this reaction. b. How much time is required for the concentration of NOBr to decrease to \(0.100 \space\mathrm{M} ?\)

Step-by-Step Solution

Answer

The rate constant \(k\) for the reaction is 0.5556 M^-1 s^-1, and the time required for the concentration of NOBr to decrease to 0.100 M is 13.33 s.

Step 1: Derive the formula for the second-order reaction

For a second-order reaction of the type \[2 NOBr(g) \longrightarrow 2 NO(g) + Br_{2}(g)\] The rate law can be written as: \[Rate = -\frac{\Delta[\text{NOBr}]}{\Delta t} = k[\text{NOBr}]^{2}\] where: \(\Delta [\text{NOBr}]\) = change in concentration of NOBr \(\Delta t\) = change in time k = rate constant Half-life, t (given) = 2.00 s

Step 2: Calculate the value of k for the reaction

First, we need to determine the initial concentration of NOBr: \[[\text{NOBr}]_{0} = 0.900\space M\] The half-life of a second-order reaction is given by the formula: \[t_{1/2} = \frac{1}{k[\text{NOBr}]_{0}}\] Now, substitute the given values in this formula and solve for k: \[2.00\space s =\frac{1}{k(0.900\space M)}\] \[k = \frac{1}{2.00\cdot 0.900}\] \[k = 0.5556\space M^{-1} s^{-1}\] The rate constant \(k\) for the reaction is 0.5556 M^-1 s^-1.

Step 3: Determine the time required for the concentration of NOBr to decrease to 0.100 M

The integrated rate law for the second-order reaction is: \[\frac{1}{[\text{NOBr}]_{t}} = kt + \frac{1}{[\text{NOBr}]_{0}}\] Now, we need to calculate the time it takes for the concentration of NOBr to decrease to 0.100 M: \[[\text{NOBr}]_{t} = 0.100\space M\] Substitute the values of \([\text{NOBr}]_{0}\), \([\text{NOBr}]_{t}\), and k in the integrated rate law: \[\frac{1}{0.100\space M} = (0.5556\space M^{-1} s^{-1})t + \frac{1}{0.900\space M}\] Solve the equation for t: \[t = \frac{\frac{1}{0.100\space M} - \frac{1}{0.900\space M}}{0.5556\space M^{-1} s^{-1}}\] \[t = 13.33\space s\] The time required for the concentration of NOBr to decrease to 0.100 M is 13.33 s.

Key Concepts

Rate Law

For a chemical reaction, the rate law is a mathematical expression that connects the rate of the reaction to the concentration of the reactants. In simpler terms, it tells us how fast the reaction takes place depending on how much of the starting materials are present. The rate is usually given in terms of how quickly the concentration of a substance changes over time, \(\frac{\Delta[reactant]}{\Delta t}\). In our reaction with \[2 \, \text{NOBr}(g) \longrightarrow 2 \, \text{NO}(g) + \text{Br}_{2}(g)\]\

• The rate law is expressed as \[\text{Rate} = -\frac{\Delta[\text{NOBr}]}{\Delta t} = k[\text{NOBr}]^{2}\].
• Here, \[k\] is the rate constant, a value that only changes with temperature.

The power to which the concentration is raised (which is 2 for NOBr in this case), shows us the order of the reaction with respect to NOBr. This helps us predict how changes in concentration affect the reaction speed. If you understand the rate law, you can determine how tweaking the concentration of reactants changes how fast things happen in the reaction.

Half-Life

Half-life is the time required for the concentration of a reactant to decrease to half of its initial amount. In chemical kinetics, this provides an understanding of how fast a reaction progresses over time. For our second-order reaction, half-life is different from first-order reactions as it depends on the initial concentration. The formula for the half-life \(t_{1/2}\) of a second-order reaction is:

• \[t_{1/2} = \frac{1}{k[\text{NOBr}]_{0}}\]

This implies that as the initial concentration \[\text{[NOBr]}_0\] increases, the half-life decreases. It's important to look at half-life in second-order reactions differently than in first-order ones due to this dependence on initial concentration. In the problem, knowing the initial concentration and that half-life is 2 seconds helps us solve for the rate constant. It's essential for predicting how long it takes for reactions to significantly deplete their reactants in various contexts.

Second-Order Reaction

A second-order reaction means that the rate depends on the concentration of one reactant raised to the second power. In simpler terms, if the concentration doubles, the reaction speed increases fourfold (since \(2^2=4\)). For second-order reactions, knowing the concentration and values in the rate law, you can calculate how things change over time.

• The general form of the rate law for second-order is \[\text{Rate} = k [A]^2\].
• This means the rate is proportional to the square of the concentration of the reactant.

Understanding these reactions helps us grasp how changes in concentration significantly influence reaction speed. For instance, if a large concentration of the reactant is present, the rate will be high, and the reaction will proceed more quickly. Appreciating the nature of a second-order reaction allows you to better predict and control chemical processes.

Rate Constant

The rate constant \(k\) is a crucial factor in the rate law, determining how quickly a reaction proceeds at a certain concentration. Each reaction has a unique \(k\) value depending on the conditions, specifically temperature. This constant provides a link between the concentration of reactants and the rate of the reaction:

• For the given reaction, \([NOBr]\), we've calculated \[k = 0.5556 \, M^{-1} \, s^{-1}\].

It stands as a unique fingerprint for a reaction's kinetics under specific conditions. If the temperature changes, \(k\) will, too, indicating a shift in how quickly molecules can form products. By understanding \(k\), we gain insights into the reaction's speed and how experimental modifications could affect outcomes, optimizing conditions for desired reaction speeds in real-world applications.