Second Order Reaction

The sum of the power of concentration of reactants in the rate law expression is called the order of that chemical reaction. Reactions can be a first-order reaction, second-order reaction, pseudo-first-order reaction, etc. depending on the concentration of the reactants. In this article, we will discuss second-order reactions in detail. 

Suppose a reaction is \[ – aA + bB \rightarrow cC + dD \] 

Rate according to rate law expression \[ = kA^{x}B^{y}\]

Where x and y are concentrations of A and B, respectively. 

So, order of reaction will be = x + y

We can say x is the order of reaction with respect to A and y is the order of reaction with respect to B. 

Now if suppose x=1 and y = 1 then the reaction will be a 2nd order reaction. Reactions in which reactants are identical and form a product can also be second-order reactions. 

Many reactions such as decomposition of nitrogen dioxide, alkaline hydrolysis of ethyl acetate, decomposition of hydrogen iodide, formation of double-stranded DNA from two strands, etc. can be explained by second-order kinetics. 

What is Second Order Reaction? 

A reaction is called a second-order reaction when the overall order is two. Suppose if the reaction is as follows:

\[ A + A \rightarrow P \]

Or \[ 2A \rightarrow P \]

In these reactions, the rate is proportional to the square of the concentration of one reactant. The differential rate law for the above second-order reaction can be written as follows:

(Image to be added soon)

The rate of such reactions can also be written as \[ r = kA^{2} \]

Here k is the rate constant for the second-order reaction. Unit of reaction rate (r) is moles per liter per second (\[mol.L^{-1}.s^{-1}\]) and the unit of second-order rate constant is \[ M^{-1}.s^{-1} \] (M is molarity which can be expressed as mol/L).

If both the reactants are different in the reaction:

\[ A + B \rightarrow P\]

The rate for the above reaction can be written as follows:

\[ R = kA^{x}B^{y}\]

Where the sum of x and y is equal to two. 

Examples of Second-Order Reactions 

A few examples of second-order reaction are given below:

Nitrogen dioxide decomposes into nitrogen monoxide and oxygen. Reaction is given below:

\[ 2NO_{2} \rightarrow 2NO + O_{2} \]

Decomposition of hydrogen iodide – Hydrogen iodide breaks down into iodine and hydrogen. The reaction is given below :

\[ 2HI \rightarrow I_{2} + H_{2} \]

Decomposition of nitrosyl bromide:

\[2NOBr \rightarrow 2NO + Br_{2}\]

Hydrolysis of an ester in presence of a base –

\[ CH_{3}COOC_{2}H_{5} + NaOH \rightarrow  CH_{3}COONa + C_{2}H_{5}OH \]

Combustion Reaction –

\[ O_{2} + C \rightarrow O + CO \]

Integrated and differential Rate Equation for Second-Order Reactions 

We are considering here that equation where chemical reaction can be represented as follows –

\[ A + A \rightarrow P ....(1) \]

Generally, polymerization reactions follow the same as in the two monomer units combine and form a polymer. 

The differential rate law equation for the chemical equation (1) can be written as follows –

(Image to be added soon)_ _ _ _ _ (2) 

On rearranging the above equation (2), we get –

(Image to be added soon) _ _ _ _ _ (3)

On integrating the above equation (3) considering that concentration of the reactant changes between time 0 and time t, we get –

(Image to be added soon) _ _ _ _ _(4)

Applying the power rule of integration in equation (4), we get –

(Image to be added soon)

On simplifying equation (5), we get –

(Image to be added soon)

Equation (6) is the required integrated rate expression of second-order reactions. 

Second-Order Reaction Graph 

On rearranging the equation (6), we get –

(Image to be added soon)

On comparing equation (7) with straight-line equation or linear equation y = mx + c, we can write –

Y = 1/At (on y-axis)

X = t (on x-axis)

m = k (Slope)

c = 1/A0 (Intercept)

So, the graph can be drawn as follows –

(Image to be added soon)

It is clear from the graph that the slope is equal to the value of rate constant k. 

The Half-life of Second-Order Reactions 

The amount of time required by reactant/s in a reaction for undergoing decay by half is called the half-life of that reaction. In the same way, the amount of time required by reactant/s to undergo decay by half in the second-order reaction is called the half-life of the second-order reaction. So, while calculating the half-life of a reaction t becomes \[t_{\frac{1}{2}}\] and as \[t = t_{\frac{1}{2}}\] then At becomes \[ \frac{A_{0}}{2}\]. 

Now putting the values of t and A in equation (6), we get –

(Image to be added soon)

On solving equation (8), we get –

(Image to be added soon)

On simplifying equation (9), we get –

(Image to be added soon)

(Image to be added soon)

Equation (11) is the equation for the half-life of a second-order reaction. 

As we can see t1/2 is inversely proportional to the concentration of the reactant in second-order reactions. The graph is given below for the half-life of second-order reactions which is drawn between A and t. 

(Image to be added soon)

Although the graph looks very similar to first-order plots it decreases at a much faster rate as the graph shows above and the length of half-life increases while the concentration of the reactant decreases. This is the reason generally students find the concept of half-life for second-order reactions more difficult than first and zero-order reactions. The value of the rate constant of second-order reactions cannot be calculated directly from the half-life equation unless the initial concentration is known.  

Determination of the Half-life of reactions is largely used in the pharma field. For example, drug dosage interval is determined on the basis of the half-life period of the reaction of the drug. When chemical kinetics is used in pharma, it is called pharmacokinetics. It can also be defined as the branch of pharmacology concerned with the movement of drugs within the body. Another vital application of half-life in pharmacokinetics is that half-life for the drug reaction shows how tightly drugs bind to each ligand before it is undergoing decay. It is very important for drug design to know how tightly it binds with ligands. 

Conclusion

This was all about second-order reactions. You can get articles on related topics such as pseudo-first-order reactions, zero-order reactions, etc. as well on our website. If you want to get free PDFs of NCERT Solutions of Chemistry (for all classes), then register yourself on Vedantu or download the Vedantu learning app for Class 6-10, IITJEE and NEET.