First Order Reactions (Chemical Kinetics)

First Order Reactions (Chemical Kinetics) - Units, Example, FAQs

Edited By Team Careers360 | Updated on Sep 14, 2024 09:44 AM IST

The reaction rate is a significant concept in studying chemical kinetics, especially in first-order reactions. Imagine a case where your body mass is dissolving medication in the liver or where the pollutants are being degraded in the environment. It is exactly these principles that govern such processes as the decay of a reactant, that is first order reactions (in chemistry). The more one gets to grips with these fundamental concepts, the more one gets to see the science of the everyday.

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First Order Reactions
Graphical Representation of first-order reaction
Some Solved Examples
Conclusion

We will explore the amazing world of first-order reactions in chemical kinetics in this article. We will tackle how the reactions act, their synthetic representation, the practical implications, and more. You'll soon understand that first-order kinetics is a foundation for many disciplines such as pharmacology, environmental science, and materials engineering among others.

Detailed Explanation

First-order reactions describe a scenario where the rate of reaction is directly proportional to the concentration of only one reactant. Mathematically, this is expressed as

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

where [A] represents the concentration of the reactant and (k) is the rate constant. Throughout this article, we will explore various facts of first-order reactions, including their kinetics, rate laws, and practical implications.

First Order Reactions

The rate of the reaction is proportional to the first power of the concentration of the reactant

Let us consider a chemical reaction which occurs as follows:

$
\mathrm{A} \longrightarrow \mathrm{B}
$

We have,
$
\begin{aligned}
& \operatorname{rate}(\mathrm{r})=\mathrm{K}[\mathrm{A}]^1 \\
& \frac{-\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\mathrm{K}[\mathrm{A}] \\
& \Rightarrow \frac{\mathrm{d}[\mathrm{A}]}{[\mathrm{A}]}=-\mathrm{kdt}
\end{aligned}
$

Integrating both sides and putting the limits
$
\begin{aligned}
& \Rightarrow \int_{[\mathrm{A}]_0}^{[\mathrm{A}]_{\mathrm{t}}} \frac{\mathrm{d}[\mathrm{A}]}{[\mathrm{A}]}=-\mathrm{k} \int_0^{\mathrm{t}} \mathrm{dt} \\
& \Rightarrow \ln \left(\frac{[\mathrm{A}]_{\mathrm{t}}}{[\mathrm{A}]_0}\right)=-\mathrm{kt}
\end{aligned}
$

Simplifying the above expression we have,
$
\Rightarrow \ln \left(\frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}}\right)=\mathrm{kt}
$

In case we are dealing in terms of $\mathrm{a}$ and $\mathrm{x}$ (where $\mathrm{a}$ is the initial concentration of $\mathrm{A}$ and $\mathrm{x}$ is the amount of $\mathrm{A}$ dissociated at any time $\mathrm{t}$ )
$
\Rightarrow \mathrm{k}=\frac{1}{\mathrm{t}} \ln \left(\frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}}\right)=\frac{1}{\mathrm{t}} \ln \left(\frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\right)
$

Graphical Representation of first-order reaction

Units of k = time^-1

1: There is an issue in this area that is the first intermediary, the notation of the reaction being coated as that of a first-order-kinetic reaction mass called kinematics, making it an unknown number. This is the very phrase we use for our kinetic component contracts which are then the first instant impact factors.

Also, in this part, we explain the main characteristics of the first-order reactions and we introduce them thoroughly. Both the math that is employed to describe first-order kinetics and the recognition of the word constant (k) are given at the same time, and a clear understanding of the subject of half-life is explained.

2: More Variations of Rate Laws

Not only will the first-order reactions be the subject of our discussion but also other rate laws will be talked about in this part.

3: Half-Life of First-Order Reaction

Nothing is more on point in this discussion than the concept of half-life in the context of first-order reactions. We will explain half-life, and develop its formula for the first-order reactions, a mention of the importance of the concept in such areas as radioactive decay and drug metabolism will be made.

4: Graphs of First Order Kinetics

Diagrams provide a great way of grasping first-order reactions. We will show first concentration vs. time graphs which happen throughout the reaction in first-order reactions, on the other hand, the concentration will be plotted which changes linearly to time and ln(concentration) vs. time graphs which will represent graphically the behaviors of the reactions.

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This article will briefly describe the practical applications of first-order reactions in the real world. The first-order rate law, as we call it in chemical kinetics, has numerous practical uses. Some research initiatives include obtaining a better solution for decay constants through measures taken to ensure that the pollutants biodegrade fixes in the environment to discovering the optimal dosage of a drug that is administered to human beings, in medicine, the application of the first-order kinetics principles on the problem is essential. In the sphere of the academic universe, these issues are guiding the way for the implementation of experimental design and theoretical modeling in chemistry, environmental science, and their equivalents.

Some Solved Examples

Example 1.

1. For a first-order reaction, calculate the ratio between the time taken to complete 3/4th of the reaction and time taken to complete half of the reaction.

1)4/3

2)8/3

3) 2

4)1

Solution:

\begin{aligned} t_{1 / 2} & =\frac{0.69}{k}, \quad t_{3 / 4}=t_{75 \%} \\ t_{3 / 4} & =\frac{2.303}{k} \log \frac{a}{\left(a-\frac{3 a}{4}\right)} \\ & =\frac{2.303}{k} \log 4 \\ & =\frac{2.303}{k} \times 2 \times 0.3010=\frac{0.69 \times 2}{k} \\ \frac{t_{3 / 4}}{t_{1 / 2}} & =\frac{0.69 \times 2}{k} \times \frac{k}{0.69} \\ \frac{t_{3 / 4}}{t_{1 / 2}} & =2\end{aligned}

Hence, the answer is the option (1).

Example 2:

In a first order reaction, the concentration of the reactant, decreases from 0.8 M to 0.4 M in 15 minutes. The time taken (in minutes) for the concentration to change from 0.1 M to 0.025 M is

1) 30

2)15

3)7.5

4)60

Solution

The formula for the first-order reaction -

$
\ln \left[\frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\right]=\mathrm{kt}
$

The concentration of the reactant decreases from $0.8 \mathrm{M}$ to $0.4 \mathrm{M}$ in 15 minutes.
$
\begin{aligned}
& \text { So, } \\
& \mathrm{k}=\frac{2.303}{15} \log \frac{0.8}{0.4} \\
& \mathrm{k}=\frac{0.693}{15}
\end{aligned}
$

For next condition
$
\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{0.1}{0.025}
$

Comparing both equations -
$
\begin{aligned}
& \frac{0.693}{15}=\frac{2.303}{t} \log \frac{0.1}{0.025} \\
& \frac{\ln 2}{15}=\frac{1}{t} \ln 4
\end{aligned}
$

$\begin{aligned} & \frac{\ln 2}{15}=\frac{2}{\mathrm{t}} \ln 2 \\ & \mathrm{t}=30 \mathrm{~min}\end{aligned}$

Hence, the answer is the option (3).

Conclusion

Eventually, the first-order reactions are a fundamental phenomenon in the study of chemical kinetics where the rates of chemical changes are directly related to the concentrations of reactants. The paragraph has expounded the main points on the first-order kinetic principles by mathematically presenting them as well as illustrating the practical application, focusing on the small details and the big picture.

Also check-

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Frequently Asked Questions (FAQs)

1. Why is this reaction first-order?

It is that form of reaction in which the rate of reaction depends upon the concentration of a single reactant only. It is like a game. What controls the speed is how well just one player performs. If you double that reactant amount, the reaction rate doubles. And so on.

2. How do we find rate constant k for first-order reactions?

We determine how rates vary with time to allow one to calculate the value of the rate constant, k, for those reactions. Just like when one keeps a watch on how fast the juice in your glass runs out each time that you have been consuming it. We plot reactant concentration with time. Then look at the steepness of the line. The steeper the slope, the faster the reaction rate.

3. Why is half-life important in first-order reactions?

First-order reactions—in the half-life—amount to much. It shows that it is the time taken for the concentration of the reactant to become reduced to half from an initial amount. Something that remains constant over the whole lifetime of the reaction, which is key to understanding how that reaction runs.