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 \(2.00 \mathrm{~s}\) when \([\mathrm{NOBr}]_{0}=\) \(0.900 \mathrm{M}\), calculate the value of \(k\) for this reaction. b. How much time is required for the concentration of NOBr to decrease to \(0.100 \mathrm{M}\) ?

Short answer

a. The value of the rate constant, \(k\), for this reaction is approximately \(0.556\:M^{-1}s^{-1}\). b. The time required for the concentration of NOBr to decrease to 0.100 M is approximately 14.2 seconds.

Step-by-step solution

Use Second Order Half-Life Formula

For a second order reaction, the half-life formula is: \(t_{1/2} = \cfrac{1}{k [\mathrm{NOBr}]_{0}}\), where \(t_{1/2}\) is the half-life, \(k\) is the rate constant, and \([\mathrm{NOBr}]_{0}\) is the initial concentration of NOBr. From the question, we have the half-life, \(t_{1/2} = 2.00\:s\), \([\mathrm{NOBr}]_{0} = 0.900\:M\). Next, we will find \(k\).

Solve for k

Rearrange the half-life formula to solve for \(k\): \(k = \cfrac{1}{t_{1/2} [\mathrm{NOBr}]_{0}}\). Substitute the given values: \(k = \cfrac{1}{(2.00\:s)(0.900\:M)}\). Now, calculate k: \(k \approx 0.556\:M^{-1}s^{-1}\). So, the value of the rate constant, \(k\), for this reaction is approximately \(0.556\:M^{-1}s^{-1}\). b. Time required for the concentration of NOBr to decrease to 0.100 M

Rearrange second order reaction formula

We start with the second-order reaction formula: \(\cfrac{1}{[\mathrm{NOBr}]_{t}} = kt + \cfrac{1}{[\mathrm{NOBr}]_{0}}\), where \([\mathrm{NOBr}]_{t}\) is the concentration of NOBr at time t. We want to find the time, t, for the concentration of NOBr to decrease to 0.100 M.

Substitute values and solve for t

We know the rate constant, \(k\), and the initial and final concentrations, \([\mathrm{NOBr}]_{0}\) and \([\mathrm{NOBr}]_{t}\). Insert these values into the formula: \(\cfrac{1}{0.100\:M} = (0.556\:M^{-1}s^{-1})(t) + \cfrac{1}{0.900\:M}\). Now, we will solve for t: \(t = \cfrac{\cfrac{1}{0.100\:M} - \cfrac{1}{0.900\:M}}{0.556\:M^{-1}s^{-1}}\). Calculate the value for t: \(t \approx 14.2\:s\). The time required for the concentration of NOBr to decrease to 0.100 M is approximately 14.2 seconds.

Key concepts

Rate Law

The rate law is an equation that connects the rate of a chemical reaction to the concentration of the reactants. In our exercise, the rate law for the reaction \[2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)\] is given by \[ \text{Rate} = -\frac{\Delta[\mathrm{NOBr}]}{\Delta t} = k[\mathrm{NOBr}]^{2} \]where \(k\) is the rate constant and \([\mathrm{NOBr}]\) is the concentration of nitrosyl bromide (NOBr). This particular form, where the rate depends on the concentration of NOBr squared, indicates that the reaction is second-order with respect to NOBr. Understanding the rate law is crucial for predicting how the concentration of reactants will change over time, which has practical applications in fields such as pharmaceuticals, environmental science, and material design.

Rate laws not only help in understanding the kinetics of a reaction but are also essential for determining the reaction mechanism. Therefore, being able to correctly write and interpret a rate law is a fundamental skill in the study of chemical kinetics.

Half-Life

Half-life, denoted as \(t_{1/2}\), is the time required for the concentration of a reactant to decrease to half of its initial value. For a second-order reaction, the half-life is inversely proportional to both the rate constant \(k\) and the initial concentration \([\mathrm{NOBr}]_{0}\). The formula for the half-life of a second-order reaction is \[t_{1/2} = \frac{1}{k [\mathrm{NOBr}]_{0}}\].

In simple terms, the half-life gives us an easy-to-understand measure of how quickly a reactant is consumed in a reaction. For example, in our exercise, the half-life when \([\mathrm{NOBr}]_{0} = 0.900 \mathrm{M}\) is 2.00 seconds. This information, combined with the rate constant, can help in predicting the duration required for the concentration to reach any other specified value.

Second-Order Reaction

A second-order reaction is one where the rate is proportional to the square of the concentration of one reactant, or to the product of the concentrations of two reactants. In mathematical terms, this can be represented as rate \(\propto [A]^2\) or rate \(\propto [A][B]\), where \([A]\) and \([B]\) are reactant concentrations. For our exercise, the reaction is second-order with respect to NOBr, which means that the rate of the reaction will quadruple if the concentration of NOBr is doubled.

Characteristics of Second-Order Reactions

• The units of the rate constants for second-order reactions are typically \(\text{M}^{-1}\text{s}^{-1}\) (concentration inverse time).
• Second-order reactions have a characteristic curved plot when graphing the reciprocal of the concentration of the reactant against time.
• To determine the time taken for the concentration to reduce to any given value, we often use an integrated form of the second-order rate law.

Understanding the specifics of second-order reactions helps in manipulating and controlling chemical processes.

Rate Constant

In chemical kinetics, the rate constant \(k\) is a proportionality factor that is part of the rate law equation. It relates the reaction rate to the concentrations of reactants. For a second-order reaction, the rate constant is determined using the formula \[k = \frac{1}{t_{1/2} [\mathrm{NOBr}]_{0}}\].

The value of \(k\) is affected by various factors such as temperature, ionic strength, and the presence of a catalyst. In the given exercise, the calculated rate constant for the decomposition of NOBr is approximately \(0.556 \mathrm{M}^{-1}\mathrm{s}^{-1}\).

Facts about Rate Constants:

• The rate constant is unique to each reaction and can be used to compare the rates of different reactions at the same temperature.
• It can provide insight into the reaction mechanism, although it alone does not determine the reaction order — that must be deduced from experimentally measured rates and concentrations.
• Rate constants are integral to the calculation of reaction rates and, therefore, for the design and optimization of industrial chemical processes.

Understanding the rate constant gives a glimpse into the dynamic world of chemical reactions, helping predict how fast a reaction will proceed under certain conditions.