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The simplest sort of second-order reaction is one whose price is proportional to the square of the concentration of one reactant. These primarily have the create 2A → assets. A second type of second-order reaction has a reaction rate that is proportional to the product of the concentrations of 2 reactants. Such reactions mostly have the create A + B → assets. An example of the former is a dimerization reactivity, in which 2 smaller molecules, each referred to as a monomer, combine to create a bigger molecule (a dimer).

You are watching: Give the characteristic of a second order reaction having only one reactant.

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

< extrmrate=-dfracDelta< extrm A>2Delta t=k< extrm A>^2 label14.4.8>

Consequently, doubling the concentration of A quadruples the reactivity price. For the devices of the reactivity price to be moles per liter per second (M/s), the units of a second-order rate consistent should be the inverse (M−1·s−1). Due to the fact that the devices of molarity are expressed as mol/L, the unit of the rate continuous can additionally be created as L(mol·s).

For the reaction 2A → assets, the following integrated rate legislation defines the concentration of the reactant at a provided time:

=dfrac1< extrm A>_0+kt label14.4.9>

Due to the fact that Equation ( ef14.4.9) has actually the develop of an algebraic equation for a right line, y = mx + b, through y = 1/ and also b = 1/0, a plot of 1/ versus t for a basic second-order reactivity is a straight line through a slope of k and an intercept of 1/0.


Second-order reactions mainly have the create 2A → commodities or A + B → commodities.

Simple second-order reactions are prevalent. In addition to dimerization reactions, two various other examples are the decomposition of NO2 to NO and also O2 and the decomplace of HI to I2 and H2. Many examples involve straightforward inorganic molecules, yet tbelow are organic examples too. We have the right to follow the progression of the reaction described in the adhering to paragraph by surveillance the decrease in the intensity of the red color of the reactivity mixture.

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

Figure (PageIndex1)

For simplicity, we will certainly describe this reactant and also product as “monomer” and “dimer,” respectively. The methodical name of the monomer is 2,5-dimethyl-3,4-diphenylcyclopentadienone. The organized name of the dimer is the name of the monomer adhered to by “dimer.” Since the monomers are the very same, the basic equation for this reaction is 2A → product. This reactivity represents a vital course of organic reactions offered in the pharmaceutical industry to prepare facility carbon skeletons for the synthesis of drugs. Like the first-order reactions studied previously, it can be analyzed utilizing either the differential price legislation (Equation ( ef14.4.8)) or the integrated price law (Equation ( ef14.4.9)).

Table (PageIndex1): Rates of Reactivity as a Function of Monomer Concentration for an Initial Monomer Concentration of 0.0054 M Time (min) (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 identify the differential rate regulation for the reactivity, we require information on just how the reaction rate varies as a duty of monomer concentrations, which are offered in Table (PageIndex1). From the data, we view that the reaction price is not independent of the monomer concentration, so this is not a zeroth-order reactivity. We additionally watch that the reaction rate is not proportional to the monomer concentration, so the reactivity is not initially order. Comparing the information in the second and also fourth rows mirrors that the reactivity price decreases by a variable of 2.8 when the monomer concentration decreases by a element of 1.7:

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

price ∝ 2

This implies that the reactivity is second order in the monomer. Using Equation ( ef14.4.8) and the information from any type of row in Table (PageIndex1), we have the right to calculate the price continuous. Substituting values at time 10 min, for example, provides the following:

<eginalign extrmrate&=k< extrm A>^2 \8.0 imes10^-5 extrm M/min&=k(4.4 imes10^-3 extrm M)^2 \4.1 extrm M^-1cdot extrmmin^-1&=kendalign>

We deserve to additionally identify the reaction order making use of the integrated rate law. To perform so, we usage the decrease in the concentration of the monomer as a function of time for a solitary reaction, plotted in component (a) in Figure (PageIndex2). The dimensions display that the concentration of the monomer (initially 5.4 × 10−3 M) decreases via raising time. This graph additionally shows that the reaction rate decreases smoothly via enhancing time. According to the integrated rate law for a second-order reactivity, a plot of 1/ versus t have to be a right line, as displayed in part (b) in Figure (PageIndex7). Any pair of points on the line deserve to be provided to calculate the slope, which is the second-order price constant. In this example, k = 4.1 M−1·min−1, which is continual through the result derived making use of the differential rate equation. Although in this example the stoichiometric coreliable is the same as the reaction order, this is not always the situation. The reaction order have to constantly be identified experimentally.

Figure (PageIndex2): Dimerization of a Monomeric Compound, a Second-Order Reaction.​ These plots correspond to dimerization of the monomer in Figure (PageIndex6) as (a) the experimentally figured out concentration of monomer versus time and (b) 1/ versus time. The directly line in (b) is meant for an easy second-order reactivity.

For 2 or more reactions of the same order, the reactivity through the biggest price consistent is the fastest. Because the systems of the rate constants for zeroth-, first-, and second-order reactions are various, but, we cannot compare the magnitudes of rate constants for reactions that have actually various orders.

Exercise (PageIndex1)

When the extremely reenergetic species HO2 develops in the atmosphere, one important reactivity that then gets rid of it from the atmosphere is as follows:

<2HO_2(g) ightarrowhead H_2O_2(g) + O_2(g) onumber>

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

Experiment 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 continuous.

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


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

Example (PageIndex2)

If a flask that initially contains 0.056 M NO2 is heated at 300°C, what will certainly 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 included price legislation for a second-order reactivity (Equation ( ef14.4.9)) and also the price consistent calculated over.

Given: balanced slrfc.orgical equation, rate continuous, time interval, and initial concentration

Asked for: final concentration and also time forced to reach stated concentration


Given k, t, and
0, use the integrated price law for a second-order reaction to calculate . Setting equal to 1/10 of 0, use the exact same equation to fix for t.


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

Thus 3600 = 5.1 × 10−4 M.

B In this situation, we know k and also 0, and also we are asked to calculate at what time = 0.10 = 0.1(0.056 M) = 0.0056 M. To perform this, we solve Equation ( ef14.4.9) for t, using the concentrations provided.

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)-(1/_0)k=dfrac(1/0.0056 extrm M)-(1/0.056 extrm M)0.54 ;mathrmM^-1cdot s^-1=3.0 imes10^2 extrm s=5.0 extrm min onumber>

NO2 decomposes incredibly rapidly; under these problems, the reactivity is 90% complete in just 5.0 min.