Question

A certain reaction has the form \(\mathrm{aA} \longrightarrow\) Products At a particular temperature, concentration versus time data were collected. A plot of \(1 /[\mathrm{A}]\) versus time (in seconds) gave a straight line with a slope of \(6.90 \times 10^{-2}\). What is the differential rate law for this reaction? What is the integrated rate law for this reaction? What is the value of the rate constant for this reaction? If \([\mathrm{A}]_{0}\) for this reaction is \(0.100 M\), what is the first half-life (in seconds)? If the original concentration (at \(t=0\) ) is \(0.100 M\), what is the second half-life (in seconds)?

Short answer

The reaction is a second-order reaction with a differential rate law of \(\dfrac{d[\mathrm{A}]}{dt} = -k[\mathrm{A}]^2\). The integrated rate law is \(\dfrac{1}{[\mathrm{A}]} - \dfrac{1}{[\mathrm{A}]_0} = kt\). The rate constant, k, is \(6.90 \times 10^{-2}\,\text{s}^{-1}\). The first and second half-lives are 14.49 seconds and 28.99 seconds, respectively.

Step-by-step solution

Determine the order of the reaction

Since the plot of \(1/[\mathrm{A}]\) versus time gives a straight line, we can conclude that the reaction is a second-order reaction. Second-order reactions have the form: \(\dfrac{d[\mathrm{A}]}{dt} = -k[\mathrm{A}]^2\), where k is the rate constant.

Write the differential rate law

Now that we know the order of the reaction, we can write the differential rate law: \(\dfrac{d[\mathrm{A}]}{dt} = -k[\mathrm{A}]^2\)

Obtain the integrated rate law

We can find the integrated rate law by integrating the differential rate law: \(\int \dfrac{d[\mathrm{A}]}{[\mathrm{A}]^2} = -k \int dt\) This yields: \(\dfrac{1}{[\mathrm{A}]} - \dfrac{1}{[\mathrm{A}]_0} = kt\)

Calculate the value of the rate constant

We are given the slope of the plot as \(6.90 \times 10^{-2}\). Since the straight line represents the integrated rate law, the slope is equal to the rate constant k: \(k = 6.90 \times 10^{-2}\,\text{s}^{-1}\)

Find the first half-life

The first half-life can be found using the initial concentration and the rate constant. For a second-order reaction, the half-life is: \(t_{1/2} = \dfrac{1}{k[\mathrm{A}]_0}\) We're given the initial concentration as \([\mathrm{A}]_0 = 0.100\,\text{M}\), so: \(t_{1/2} = \dfrac{1}{(6.90 \times 10^{-2}\,\text{s}^{-1})(0.100\,\text{M})} = 14.49\,\text{s}\)

Find the second half-life

To find the second half-life, we need to find the new initial concentration for the next half-life period. After the first half-life, the remaining concentration would be half of the initial concentration: \([\mathrm{A}]_{1/2} = \dfrac{[\mathrm{A}]_0}{2} = \dfrac{0.100\,\text{M}}{2} = 0.050\,\text{M}\) Now, use the new initial concentration to find the second half-life: \(t_{1/2} = \dfrac{1}{k[\mathrm{A}]_{1/2}} = \dfrac{1}{(6.90 \times 10^{-2}\,\text{s}^{-1})(0.050\,\text{M})} = 28.99\,\text{s}\) So, the second half-life is 28.99 seconds.

Key concepts

Second-Order Reaction

Understanding the concept of a second-order reaction is an essential part of mastering reaction kinetics. A reaction is considered second-order if the rate of reaction depends on the concentration of one reactant squared or on the concentration of two different reactants. In simple terms, if the concentration of the reactant changes, the rate of reaction will change in a squared manner. This type of reaction can be represented by the equation:
\[ ext{A} + ext{A} ightarrow ext{products} \]
The hallmark of a second-order reaction is that a plot of the inverse of the concentration of the reactant, \(1/[ ext{A}]\), against time will give a straight line. This line indicates that the reaction follows second-order kinetics, and the slope of this line corresponds to the rate constant, \(k\). This graphical approach is a clear indicator of the order of the reaction and is a powerful tool in chemical kinetics.

Differential Rate Law

The differential rate law gives us valuable information about how the rate of a reaction changes with the concentration of reactants. For a second-order reaction specifically, the differential rate law is expressed as:
\[ \frac{d[ ext{A}]}{dt} = -k[ ext{A}]^2 \]
Here, \(\frac{d[ ext{A}]}{dt}\) represents the rate of change in concentration of \([\text{A}]\) over time. The negative sign indicates that the concentration of \(\text{A}\) decreases as the reaction progresses. The term \(k\) is the rate constant, which is unique to the reaction and its conditions, such as temperature.
Knowing the differential rate law allows chemists to predict how fast a reaction will proceed under given conditions, which is crucial for experiments and industrial processes. This equation is fundamental to understanding kinetics and requires a precise measurement of concentration and time.

Rate Constant

The rate constant, denoted \(k\), is a crucial element in the study of chemical kinetics as it quantifies the speed of a reaction. For any chemical reaction, the rate constant is a specific value that relates the speed of the reaction to the concentrations of the reactants. In the context of a second-order reaction, as we have seen:
\[ \frac{d[ ext{A}]}{dt} = -k[ ext{A}]^2 \]
Here, \(k\) serves as a proportionality factor. Determining \(k\) generally involves experimental observation, such as the slope of a plot of \(1/[ ext{A}]\) versus time. For the problem given, the slope is \(6.90 \times 10^{-2}\), thus setting \(k\) to this value.
This constant can vary with temperature and is unique to the specific reaction and conditions you're examining. It plays a vital role in comparing different reactions and their efficiencies, as well as determining half-life and understanding reaction mechanisms.

Half-Life Calculations

Half-life is a convenient measure of how long it takes for half of the reactant in a reaction to be consumed. For a second-order reaction, calculating the half-life involves both the initial concentration of the reactant and the rate constant. The formula used is:
\[ t_{1/2} = \frac{1}{k[ ext{A}]_0} \]
In this formula, \(t_{1/2}\) symbolizes the half-life, \(k\) is the rate constant, and \([ ext{A}]_0\) is the initial concentration of the reactant. From the given values, if the initial concentration \([ ext{A}]_0\) is \(0.100 \, ext{M}\), the first half-life is \(14.49 \, ext{s}\).
Given that each half-life changes as the concentration changes, the second half-life differs from the first because it starts with half the original concentration \(0.050 \, ext{M}\), leading to a longer time of \(28.99 \, ext{s}\). This dependency of half-life on concentration is a characteristic trait of second-order kinetics, as opposed to first-order reactions where half-life is constant regardless of concentration.