Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 101

For a first-order reaction \(a \mathrm{~A} \longrightarrow\) products, where \(a \neq 1\), the rate is \(-\Delta[\mathrm{A}] / a \Delta t\), or in derivative notation, \(-\frac{1}{a} \frac{d[\mathrm{~A}]}{d t} .\) Derive the integrated rate law for the first-order decomposition of \(a\) moles of reactant.

Short Answer

Expert verified
Question: Derive the integrated rate law for a first-order reaction with "a" moles of reactant A given the rate law is -Δ[A] / aΔt or -1/a * d[A]/dt. Answer: The integrated rate law for this first-order reaction is [A] = [A]₀ * e^(-([A] / a)t).

Step by step solution

01

Write the given rate law equation

We are given the rate law: \(-\frac{1}{a}\frac{d[\mathrm{A}]}{dt}\)
02

Separate variables

Now we need to separate variables to prepare for integration: \(-\frac{1}{a}\frac{d[\mathrm{A}]}{dt} = -\frac{1}{a}\frac{d[\mathrm{A}]}{dt}\) Multiply both sides by dt and a: \(-d[\mathrm{A}] = a \frac{d[\mathrm{A}]}{[A]} dt\)
03

Integrate both sides

Now we have the equation set up for integration. We will integrate both sides with respect to time (t): \(\int -d[\mathrm{A}] = \int a\frac{d[\mathrm{A}]}{[A]} dt\)
04

Evaluate the integrals

Evaluate both integrals: \(-[\mathrm{A}] = a \ln{[\mathrm{A}]} + C\) Where C is the constant of integration.
05

Solve for C

To solve for C, we use the initial condition. At t=0, let [A] = [A]₀, which is the initial concentration of A: \(-[\mathrm{A}₀] = a\ln{[\mathrm{A}₀]} + C\) Therefore, C = -[A]₀ + a * ln([A]₀)
06

Substitute C back into the equation and solve for [A]

Now, we shall substitute the value of C back into the equation and solve for [A]: \(-[\mathrm{A}] = a \ln{[\mathrm{A}]} - [A]₀ + a \ln{[\mathrm{A}₀]}\) Now, isolate [A] on one side: \(\frac{[\mathrm{A}]}{a} = \ln{\frac{[\mathrm{A}]}{[\mathrm{A}₀}}\) Take the exponential on both sides: \(e^{\frac{[\mathrm{A}]}{a}} = \frac{[\mathrm{A}]}{[\mathrm{A}₀]}\) Solve for [A]: \([\mathrm{A}] = [\mathrm{A}₀] * e^{-\frac{[\mathrm{A}]}{a}}\) This is the integrated rate law for a first-order reaction with \(a\) moles of reactant A.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Kinetics
Chemical kinetics is the study of how fast chemical reactions occur and what factors affect their rates. It's fundamental in understanding reaction mechanisms and for the development of new chemicals and materials.

When we talk about the rate of a chemical reaction, we're referring to the speed at which the reactants are transformed into products. This depends on various factors such as temperature, pressure, concentration, and the presence of a catalyst. By analyzing the kinetics of a reaction, chemists can optimize conditions for production or prevent unwanted reactions, such as food spoilage or material corrosion.
Rate Law Equation
The rate law equation relates the rate of a reaction to the concentration of the reactants. It's an essential tool in chemical kinetics, allowing us to predict how the rate will change as conditions are altered. The general form of a rate law is: \(rate = k[A]^m[B]^n\), where \(k\) is the rate constant, \([A]\) and \([B]\) are the concentrations of reactants A and B, and \(m\) and \(n\) are the reaction orders with respect to A and B.

Understanding the rate law is crucial not only for chemists but also for industries such as pharmaceuticals, where reaction rates dictate production efficiency and product safety.
First-Order Reaction
A first-order reaction is one where the rate depends linearly on the concentration of one reactant. Mathematically, it's expressed as: \(rate = k[A]\).

In these reactions, the time it takes for the concentration of the reactant to decrease by half (the half-life) is constant and independent of its initial concentration. This is a hallmark of first-order kinetics and is highly relevant, for example, in pharmacology to understand how drugs are metabolized within the body.
Separation of Variables
Separation of variables is a mathematical method used to solve differential equations, which are equations involving derivatives. It involves rearranging the equation to isolate each variable (such as concentration and time in chemical kinetics) on different sides of the equation. This move sets the stage for integration, allowing us to find a relationship between the variables that describe the system's behavior over time.

This technique illuminates the pathway from rates to actual reactant concentration changes over time, bridging the gap between theory and experimental data.
Integration in Chemistry
Integration is a mathematical process that sums up changes to find the total effect over time or space. In chemistry, we use integration to transition from the rate of a reaction to the overall progress of the reaction, i.e., from considering the immediate speed of molecule transformation to understanding how much has transformed after a certain period.

The process can seem abstract, but it's like adding up all the tiny bits of reactant conversion that happen every split second to get the big picture. It's an invaluable tool for predicting concentrations of reactants and products at any point in the reaction.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The decomposition of \(\mathrm{A}\) at \(85^{\circ} \mathrm{C}\) is a zero-order reaction. It takes 35 minutes to decompose \(37 \%\) of an initial mass of \(282 \mathrm{mg}\). (a) What is \(k\) at \(85^{\circ} \mathrm{C}\) ? (b) What is the half-life of \(282 \mathrm{mg}\) at \(85^{\circ} \mathrm{C}\) ? (c) What is the rate of decomposition for \(282 \mathrm{mg}\) at \(85^{\circ} \mathrm{C} ?\) (d) If one starts with \(464 \mathrm{mg}\), what is the rate of its decomposition at \(85^{\circ} \mathrm{C} ?\)

The uncoiling of deoxyribonucleic acid (DNA) is a first-order reaction. Its activation energy is \(420 \mathrm{~kJ}\). At \(37^{\circ} \mathrm{C}\), the rate constant is \(4.90 \times 10^{-4} \mathrm{~min}^{-1}\). (a) What is the half-life of the uncoiling at \(37^{\circ} \mathrm{C}\) (normal body temperature)? (b) What is the half-life of the uncoiling if the organism has a temperature of \(40^{\circ} \mathrm{C}\left(\approx 104^{\circ} \mathrm{F}\right)\) ? (c) By what factor does the rate of uncoiling increase (per \({ }^{\circ} \mathrm{C}\) ) over this temperature interval?

The decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to \(\mathrm{NO}_{2}\) and \(\mathrm{NO}_{3}\) is a first-order gas-phase reaction. At \(25^{\circ} \mathrm{C}\), the reaction has a half-life of \(2.81\) s. At \(45^{\circ} \mathrm{C}\), the reaction has a half-life of \(0.313 \mathrm{~s}\). What is the activation energy of the reaction?

Diethylhydrazine reacts with iodine according to the following equation: $$\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2}(\mathrm{NH})_{2}(l)+\mathrm{I}_{2}(a q) \longrightarrow\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{~N}_{2}(l)+2 \mathrm{HI}(a q)$$ The rate of the reaction is followed by monitoring the disappearance of the purple color due to iodine. The following data are obtained at a certain temperature. $$ \begin{array}{cccc} \hline \text { Expt. } & {\left[\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2}(\mathrm{NH})_{2}\right]} & {\left[\mathrm{I}_{2}\right]} & \text { Initial Rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{h}) \\ \hline 1 & 0.150 & 0.250 & 1.08 \times 10^{-4} \\ 2 & 0.150 & 0.3620 & 1.56 \times 10^{-4} \\ 3 & 0.200 & 0.400 & 2.30 \times 10^{-4} \\ 4 & 0.300 & 0.400 & 3.44 \times 10^{-4} \\ \hline\end{array}$$ (a) What is the order of the reaction with respect to diethylhydrazine, iodine, and overall? (b) Write the rate expression for the reaction. (c) Calculate \(k\) for the reaction. (d) What must \(\left[\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2}(\mathrm{NH})_{2}\right]\) be so that the rate of the reaction is \(5.00 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{h}\) when \(\left[\mathrm{I}_{2}\right]=0.500 ?\)

Derive the integrated rate law, \([\mathrm{A}]=[\mathrm{A}]_{0}-k t\), for a zero-order reaction. (Hint: Start with the relation \(-\Delta[\mathrm{A}]=k \Delta \mathrm{t}\).)
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free