Second Order Reaction - Examples, Graph, Equation, FAQs

Second Order Reaction - Examples, Graph, Equation, FAQs

Edited By Team Careers360 | Updated on Sep 24, 2024 09:46 PM IST

Chemical reactions control our daily lives, from the environment to the synthesis of key pharmaceuticals. In this respect, chemical kinetics explains phenomena limited to two-body events where the reacting molecules are together in solution and give a product or the rate of reaction can be controlled by two reactants interacting simultaneously with each other. These concepts, however, in an integrated capacity, act as almost a lifeline toward the betterment of perception of ordinary processes and further developments, especially in medicine, environmental science, and materials engineering.

This Story also Contains
  1. Detailed Explanation
  2. Some Solved Questions
  3. Conclusion

This paper attempts an in-depth discussion of second- and nth-order chemical kinetics reactions. We investigate some of the questions like how these kinds of reactions take place, how they are mathematically formulated, their various methods for establishing the order, practical applications, et cetera. At the end of this discussion, you should be able to see how exactly these principles bear on our comprehension of reaction mechanisms or how they are applied across the sciences.

Detailed Explanation

Second and nth-order reactions are cases where the rate of a chemical reaction is dependent on more than one power of concentration of reactants; unlike first-order reactions where they are dependent upon the concentration of one reactant raised to the power of one.

Also read -

1: Second-order reaction

The case in which the rate is proportional to the product of two concentrations of two reactants or to the square of the concentration of a single reactant is called a second-order reaction. Mathematically, it represents a reaction with two reactants A and B, where the rate might be defined by the formula:

$\frac{d[A]}{d t}=-k[A][B]$

For a reaction with a single reactant A:

$\frac{d[A]}{d t}=-k[A]^2$

One reason that second-order reactions are less common than first-order reactions is that often it requires two reactants to collide in a specific orientation for the reaction to occur. Thus, second-order reactions are less common than first-order reactions, but still very important in a great many chemical processes.

Related Topics,

JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%
Download EBook

2: nth Order Reaction

n-th order reactions The concept of second-order reactions can be generalized to any integer, with the index, once again, being the sum of exponents to which the concentrations appear in the rate law. For example, a third-order reaction with a single species A would be written:

$d t / d[A]=k[A] 3$

The nth-order reactions are variable and can occur in complex systems where more than one reactant or intermediate takes part to determine the overall rate of the reaction.

3: Methods of Determining Reaction Order

The experimental determination of the order of a chemical reaction is very important to learn about kinetics for designing appropriate conditions to get desired results. Following are some of the methods used:

- Initial Rate Method: The order of a reaction is often determined by measuring the initial rates of reaction varying the initial concentrations of reactants.

- Isolation Method: One may isolate some step or an intermediate of the sequence of reactions and study its influence on the overall rate.

Read more :

Integrated Rate Laws: These rate equations, derived directly from the experimental data, are integrated in order to learn about the order of a reaction and changes in concentration with time. a) Applications to Real Life and Academics:

Second and third-order reactions find applications in a wide range of scientific and industrial processes. In biological systems, enzymes often catalyze second or third-order reactions, influencing metabolic pathways and cellular functions. In industrial chemistry, understanding reaction order is essential for optimizing production and maintaining product quality. In academic research, driving the design of experiments and theoretical modeling improves our understanding of highly complex chemical systems.

Recommended topic video on (Second order reaction )


Some Solved Questions

1. Which of the following shows the expression of half-life period of second-order reaction?

$
\mathrm{A} \longrightarrow \mathrm{B}
$
(Where the initial concentration of $A$ is $[A]_0$ and the rate constant is $K$ )
1) $\frac{0.693}{\mathrm{~K}}$
2) $\frac{1}{\mathrm{~K}[\mathrm{~A}]_0}$
3) $\frac{[\mathrm{A}]_0}{2 \mathrm{~K}}$
4)none of above

Solution :
As we have learned,
Thus, $\mathbf{t}_{1 / 2}=\frac{1}{\mathrm{kA}_{\mathrm{o}}}$

Hence, the answer is the option (2)

2. Which of the following is the unit of rate constant for second-order reaction?

1) mol-1Ls-1

2)mol-2Ls-1

3)s-1

4)mol-1Ls-2

Solution

Unit for k dor second order = mol-1Ls-1

Hence, the answer is the option (1).


3. A second-order reaction requires 140 min to change the concentration of reactants from 0.16M to 0.02M The time required to become $0.08 \mathrm{M}=2 x$ min. What will be the value of x?

1)4 min

2)6 min

3)8 min

4) (correct)10 min

Solution

For second-order reaction

$\begin{aligned} & {[\mathrm{R}]_{\text {initial }}={ }^{\prime} \mathrm{a}^{\prime}=0.16 \mathrm{M}} \\ & {[\mathrm{R}]_{\text {final }}=0.02 \mathrm{M}}\end{aligned}$

$
\begin{aligned}
\mathrm{x} & =\mathrm{R}_{\text {initial }}-\mathrm{R}_{\text {final }} \\
& =0.16-0.02=0.14 \mathrm{M}
\end{aligned}
$


$
\begin{aligned}
& (a-x)=0.16-0.14=0.02 M \\
& K_2=\frac{1}{t} \times \frac{x}{a(a-x)}
\end{aligned}
$


$
\mathrm{k}_2=\frac{1}{140} \times \frac{0.14}{0.16 \times 0.02} \quad-(1)
$
Now, the time required to become a concentration

$\begin{aligned} & =0.08 \mathrm{M} \text { ie } \mathrm{x}=0.08 \mathrm{M} \\ & \mathrm{k}_2=\frac{1}{\mathrm{t}} \times \frac{0.08}{0.16 \times(0.16-0.08)}\end{aligned}$

Now put the value of \mathrm{ k_{2}} form (1) in (2), we get

$\begin{aligned} & \frac{1}{140} \times \frac{0.14}{0.16 \times 0.02}=\frac{1}{t} \times \frac{0.08}{0.16 \times 0.08} \\ & t=20 \mathrm{~min}=2 \mathrm{x} \mathrm{min} . \\ & x=10 \mathrm{~min} .\end{aligned}$
.
Hence, the answer is the option (4).

Conclusion

In general, the second and nth-order reactions of the chemical kinetics explain how the concentration of reactants influences the rate at which a reaction occurs. The current paper has explored the mathematical fundamentals of the aforementioned entities, methods for determining their order, and other uses in other science fields. With the understanding of the principles at hand, prediction, and control of the chemical reactions for a specific will be at the scientist's disposal.

Also check-

NCERT Chemistry Notes :

Frequently Asked Questions (FAQs)

1. What characterizes a second-order reaction from the rest of the orders of a reaction?

Second-order reactions involve rates dependent upon the product of two reactant concentrations or the square of a single reactant concentration

2. Can nth-order reactions have non-integer reaction orders?

No, the term 'nth order reactions' refers to reactions where the rate is proportional to the product of powers of reactant concentrations -these powers have to be integers.

3. How does reaction order impact the overall rate of reaction?

Reaction order is the way that the concentration of reactants affects the rate of reaction and kinetic and efficient ways of chemical procedures.

4. What are some practical examples of second-order reactions in everyday life?

Examples include the reaction of two gas molecules, some enzymatic reactions, and some chemical reactions in solution.

5. Why should the order of a reaction be determined experimentally?

The experimental determination of the order of a reaction provides very crucial information for the proper design of reaction conditions and prediction of reaction behavior, especially in understanding the mechanisms.

Articles

Get answers from students and experts
Back to top