# 14.6: Second-Order Reactions

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The simplest kind of second-order reaction is one whose rate is proportional to the square of the concentration of one reactant. These generally have the form 2A → products. A second kind of second-order reaction has a reaction rate that is proportional to the product of the concentrations of two reactants. Such reactions generally have the form A + B → products. An example of the former is a dimerization reaction, in which two smaller molecules, each called a monomer, combine to form a larger molecule (a dimer).

The differential rate law for the simplest second-order reaction in which 2A → products is as follows:

$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{2\Delta t}=k[\textrm A]^2 \label{14.4.8}$

Consequently, doubling the concentration of A quadruples the reaction rate. For the units of the reaction rate to be moles per liter per second (M/s), the units of a second-order rate constant must be the inverse (M−1·s−1). Because the units of molarity are expressed as mol/L, the unit of the rate constant can also be written as L(mol·s).

For the reaction 2A → products, the following integrated rate law describes the concentration of the reactant at a given time:

$\dfrac{1}{[\textrm A]}=\dfrac{1}{[\textrm A]_0}+kt \label{14.4.9}$

Because Equation $$\ref{14.4.9}$$ has the form of an algebraic equation for a straight line, y = mx + b, with y = 1/[A] and b = 1/[A]0, a plot of 1/[A] versus t for a simple second-order reaction is a straight line with a slope of k and an intercept of 1/[A]0.

##### Note

Second-order reactions generally have the form 2A → products or A + B → products.

Simple second-order reactions are common. In addition to dimerization reactions, two other examples are the decomposition of NO2 to NO and O2 and the decomposition of HI to I2 and H2. Most examples involve simple inorganic molecules, but there are organic examples as well. We can follow the progress of the reaction described in the following paragraph by monitoring the decrease in the intensity of the red color of the reaction mixture.

Many cyclic organic compounds that contain two carbon–carbon double bonds undergo a dimerization reaction to give complex structures. One example is as follows:

For simplicity, we will refer to this reactant and product as “monomer” and “dimer,” respectively. The systematic name of the monomer is 2,5-dimethyl-3,4-diphenylcyclopentadienone. The systematic name of the dimer is the name of the monomer followed by “dimer.” Because the monomers are the same, the general equation for this reaction is 2A → product. This reaction represents an important class of organic reactions used in the pharmaceutical industry to prepare complex carbon skeletons for the synthesis of drugs. Like the first-order reactions studied previously, it can be analyzed using either the differential rate law (Equation $$\ref{14.4.8}$$) or the integrated rate law (Equation $$\ref{14.4.9}$$).

Table $$\PageIndex{1}$$: Rates of Reaction as a Function of Monomer Concentration for an Initial Monomer Concentration of 0.0054 M
Time (min) [Monomer] (M) Instantaneous Rate (M/min)
10 0.0044 8.0 × 10−5
26 0.0034 5.0 × 10−5
44 0.0027 3.1 × 10−5
70 0.0020 1.8 × 10−5
120 0.0014 8.0 × 10−6

To determine the differential rate law for the reaction, we need data on how the reaction rate varies as a function of monomer concentrations, which are provided in Table $$\PageIndex{1}$$. From the data, we see that the reaction rate is not independent of the monomer concentration, so this is not a zeroth-order reaction. We also see that the reaction rate is not proportional to the monomer concentration, so the reaction is not first order. Comparing the data in the second and fourth rows shows that the reaction rate decreases by a factor of 2.8 when the monomer concentration decreases by a factor of 1.7:

$\dfrac{5.0\times10^{-5}\textrm{ M/min}}{1.8\times10^{-5}\textrm{ M/min}}=2.8\hspace{5mm}\textrm{and}\hspace{5mm}\dfrac{3.4\times10^{-3}\textrm{ M}}{2.0\times10^{-3} \textrm{ M}}=1.7$

Because (1.7)2 = 2.9 ≈ 2.8, the reaction rate is approximately proportional to the square of the monomer concentration.

rate ∝ [monomer]2

This means that the reaction is second order in the monomer. Using Equation $$\ref{14.4.8}$$ and the data from any row in Table $$\PageIndex{1}$$, we can calculate the rate constant. Substituting values at time 10 min, for example, gives the following:

\begin{align}\textrm{rate}&=k[\textrm A]^2 \\8.0\times10^{-5}\textrm{ M/min}&=k(4.4\times10^{-3}\textrm{ M})^2 \\4.1 \textrm{ M}^{-1}\cdot \textrm{min}^{-1}&=k\end{align}

We can also determine the reaction order using the integrated rate law. To do so, we use the decrease in the concentration of the monomer as a function of time for a single reaction, plotted in part (a) in Figure $$\PageIndex{2}$$. The measurements show that the concentration of the monomer (initially 5.4 × 10−3 M) decreases with increasing time. This graph also shows that the reaction rate decreases smoothly with increasing time. According to the integrated rate law for a second-order reaction, a plot of 1/[monomer] versus t should be a straight line, as shown in part (b) in Figure $$\PageIndex{7}$$. Any pair of points on the line can be used to calculate the slope, which is the second-order rate constant. In this example, k = 4.1 M−1·min−1, which is consistent with the result obtained using the differential rate equation. Although in this example the stoichiometric coefficient is the same as the reaction order, this is not always the case. The reaction order must always be determined experimentally.

For two or more reactions of the same order, the reaction with the largest rate constant is the fastest. Because the units of the rate constants for zeroth-, first-, and second-order reactions are different, however, we cannot compare the magnitudes of rate constants for reactions that have different orders.

##### Example $$\PageIndex{1}$$

At high temperatures, nitrogen dioxide decomposes to nitric oxide and oxygen.

$\mathrm{2NO_2(g)}\xrightarrow{\Delta}\mathrm{2NO(g)}+\mathrm{O_2(g)}$

Experimental data for the reaction at 300°C and four initial concentrations of NO2 are listed in the following table:

 Experiment [NO2]0 (M) Initial Rate (M/s) 1 0.015 1.22 × 10−4 2 0.010 5.40 × 10−5 3 0.0080 3.46 × 10−5 4 0.0050 1.35 × 10−5

Determine the reaction order and the rate constant.

Given: balanced chemical equation, initial concentrations, and initial rates

Asked for: reaction order and rate constant

Strategy:

1. From the experiments, compare the changes in the initial reaction rates with the corresponding changes in the initial concentrations. Determine whether the changes are characteristic of zeroth-, first-, or second-order reactions.
2. Determine the appropriate rate law. Using this rate law and data from any experiment, solve for the rate constant (k).

Solution

A We can determine the reaction order with respect to nitrogen dioxide by comparing the changes in NO2 concentrations with the corresponding reaction rates. Comparing Experiments 2 and 4, for example, shows that doubling the concentration quadruples the reaction rate [(5.40 × 10−5) ÷ (1.35 × 10−5) = 4.0], which means that the reaction rate is proportional to [NO2]2. Similarly, comparing Experiments 1 and 4 shows that tripling the concentration increases the reaction rate by a factor of 9, again indicating that the reaction rate is proportional to [NO2]2. This behavior is characteristic of a second-order reaction.

B We have rate = k[NO2]2. We can calculate the rate constant (k) using data from any experiment in the table. Selecting Experiment 2, for example, gives the following:

\begin{align}\textrm{rate}&=k[\mathrm{NO_2}]^2 \\5.40\times10^{-5}\textrm{ M/s}&=k(\mathrm{\mathrm{0.010\;M}})^2 \\0.54\mathrm{\;M^{-1}\cdot s^{-1}}&=k\end{align}
##### Exercise $$\PageIndex{1}$$

When the highly reactive species HO2 forms in the atmosphere, one important reaction that then removes it from the atmosphere is as follows:

$2HO_{2(g)} \rightarrow H_2O_{2(g)} + O_{2(g)} \nonumber$

The kinetics of this reaction have been studied in the laboratory, and some initial rate data at 25°C are listed in the following table:

 Experiment [HO2]0 (M) Initial Rate (M/s) 1 1.1 × 10−8 1.7 × 10−7 2 2.5 × 10−8 8.8 × 10−7 3 3.4 × 10−8 1.6 × 10−6 4 5.0 × 10−8 3.5 × 10−6

Determine the reaction order and the rate constant.

Answer second order in HO2; k = 1.4 × 109 M−1·s−1

##### Note

If a plot of reactant concentration versus time is not linear, but a plot of 1/reaction concentration versus time is linear, then the reaction is second order.

##### Example $$\PageIndex{2}$$

If a flask that initially contains 0.056 M NO2 is heated at 300°C, what will be the concentration of NO2 after 1.0 h? How long will it take for the concentration of NO2 to decrease to 10% of the initial concentration? Use the integrated rate law for a second-order reaction (Equation $$\ref{14.4.9}$$) and the rate constant calculated above.

Given: balanced chemical equation, rate constant, time interval, and initial concentration

Asked for: final concentration and time required to reach specified concentration

Strategy:

1. Given k, t, and [A]0, use the integrated rate law for a second-order reaction to calculate [A].
2. Setting [A] equal to 1/10 of [A]0, use the same equation to solve for t.

Solution

A We know k and [NO2]0, and we are asked to determine [NO2] at t = 1 h (3600 s). Substituting the appropriate values into Equation 14.4.9,

\begin{align}\dfrac{1}{[\mathrm{NO_2}]_{3600}}&=\dfrac{1}{[\mathrm{NO_2}]_0}+kt=\dfrac{1}{0.056\textrm{ M}}+[(0.54 \mathrm{\;M^{-1}\cdot s^{-1}})(3600\textrm{ s})] \\&=2.0\times10^3\textrm{ M}^{-1}\end{align}

Thus [NO2]3600 = 5.1 × 10−4 M.

B In this case, we know k and [NO2]0, and we are asked to calculate at what time [NO2] = 0.1[NO2]0 = 0.1(0.056 M) = 0.0056 M. To do this, we solve Equation $$\ref{14.4.9}$$ for t, using the concentrations given.

$t=\dfrac{(1/[\mathrm{NO_2}])-(1/[\mathrm{NO_2}]_0)}{k}=\dfrac{(1/0.0056 \textrm{ M})-(1/0.056\textrm{ M})}{0.54 \;\mathrm{M^{-1}\cdot s^{-1}}}=3.0\times10^2\textrm{ s}=5.0\textrm{ min} \nonumber$

NO2 decomposes very rapidly; under these conditions, the reaction is 90% complete in only 5.0 min.

##### Exercise $$\PageIndex{2}$$

In the previous exercise, you calculated the rate constant for the decomposition of HO2 as k = 1.4 × 109 M−1·s−1. This high rate constant means that HO2 decomposes rapidly under the reaction conditions given in the problem. In fact, the HO2 molecule is so reactive that it is virtually impossible to obtain in high concentrations. Given a 0.0010 M sample of HO2, calculate the concentration of HO2 that remains after 1.0 h at 25°C. How long will it take for 90% of the HO2 to decompose? Use the integrated rate law for a second-order reaction (Equation $$\ref{14.4.9}$$) and the rate constant calculated in the exercise in Example $$\PageIndex{3}$$.

Answer 2.0 × 10−13 M; 6.4 × 10−6 s

In addition to the simple second-order reaction and rate law we have just described, another very common second-order reaction has the general form $$A + B \rightarrow products$$, in which the reaction is first order in $$A$$ and first order in $$B$$. The differential rate law for this reaction is as follows:

$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=-\dfrac{\Delta[\textrm B]}{\Delta t}=k[\textrm A][\textrm B] \label{14.4.10}$

Because the reaction is first order both in A and in B, it has an overall reaction order of 2. (The integrated rate law for this reaction is rather complex, so we will not describe it.) We can recognize second-order reactions of this sort because the reaction rate is proportional to the concentrations of each reactant.

Second-Order Integrated Rate Law Equation: https://youtu.be/hMSgk2Rm2xA

14.6: Second-Order Reactions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.