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The reactions in which the sum of exponents in the corresponding rate law of the chemical reaction is equal to two are known as secondorder reactions. The rate of such reactions can be written either as
r = k [A] 2, or as r = k [A] [B].
From the above equation, we know that secondorder reactions are those chemical reactions that either depend upon the concentrations of two firstorder reactants or on the concentration of one secondorder reactant.
These two types of secondorder reactions can be described as follows
The sum of x and y (which corresponds to the order of the chemical reaction in question) equals two.
Examples of secondorder reactions include the following reactions
1. NH4CNO → H2NCONH2
Ammonium cyanate in water isomerized into urea.
2. H^{+} + OH^{} → H_{2}OC + O_{2} → CO +O
The two examples given above are the secondorder reactions depending on the concentration of two separate first order reactants.
3. CH3COOC2H5 + NaOH → CH3COONa + C2H5OH
Here, an example of the hydrolysis of an ester in the presence of a base, ethyl acetate in the presence of sodium hydroxide.
2NO2 → 2NO + O2
4. 2HI → I2 + H2
These reactions involve one second order reactant yielding the product.
5. 2 NOBr → 2 NO + Br2
In the gas phase, nitrosyl bromide decomposes into nitrogen oxide and bromine gas.
6. H+ + OH → H2O
Hydrogen ions and hydroxyl ions form water.
7. 2 NO2 → 2 NO + O2
Nitrogen dioxide decomposes into nitrogen monoxide and an oxygen molecule.
8. 2 HI → I2 + H2
Hydrogen Iodide decomposes into iodine gas and hydrogen gas.
9. + O3 → O2 + O2
During combustion, oxygen atoms and ozone can form oxygen molecules.
10. O2 + C → O + CO
Another combustion reaction, oxygen molecules react with carbon to form oxygen atoms and carbon monoxide.
11. O2 + CO → O + CO2
This reaction often follows the previous reaction. Oxygen molecules react with carbon monoxide to form carbon dioxide and oxygen atoms.
12. + H2O → 2 OH
One common product of combustion is water. This can react with all the loose oxygen atoms produced in the previous reactions to form hydroxides.
The differential rate law equation considering one second order reactant results a product in chemical reaction can be written as
In order to get the integrated rate equation, this differential form must be rearranged:
Integrating both the sides considering with the change in the concentration of reactant between time 0 and time t, we get the following equation

From the power rule of integration, we have:
Where C is the constant of Integration. Now, using this power rule in the previous equation, we get the following equation as
This is the required integrated rate expression of second order reactions.
We need to generalize [R]t as [R] and rearrange the integrated rate law equation of reactions of the second order to get the following reaction
Plotting a straight line (y=mx + c) corresponding to this equation (y = 1/[R] , x = t , m = k , c = 1/[R]0)
We can observe the slope of the straight line is equal to the value of the rate constant, k.
The time taken for half of the initial amount of reactant to undergo the reaction is known as the halflife of the secondorder reaction.
The following substitutions should be done to obtain the desired equation
[R] = [R]02
And, t = t1 / 2
Substituting these values in the integral form of the rate equation of second order reactions, we get:
Therefore, we get the required equation for the halflife of second order reactions
We can conclude from the above equation that the equation for the halflife implies that the halflife is inversely proportional to the concentration of the reactants.