/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 187 Assertion: For a first order rea... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 187

Assertion: For a first order reaction, A (g) \(\rightarrow\) Product the time required to reduce successively the concentration of reactant by a constant fraction is always same. Reason: At any instant, the rate of a first order reaction is given by \(\mathrm{k}[\mathrm{A}]\).

Short Answer

Expert verified
The assertion is true, and the reason correctly explains it. Both are correct.

Step by step solution

01

Understanding the Assertion

The assertion states that for a first-order reaction, reducing the concentration of the reactant A by a constant fraction requires the same amount of time at each step. This implies that the reaction follows an exponential decay, which means that the time taken for the concentration of A to reach a certain percentage of its initial value is constant.
02

Analyzing the Reason

The reason given is that the rate of a first-order reaction is directly proportional to the concentration of the reactant A, expressed as: \( \text{Rate} = k[A] \). This means the rate depends linearly on the current concentration of A, which follows the first-order kinetics definition.
03

Connecting the Assertion and the Reason

In first-order reactions, the integrated rate law is given by \( [A] = [A]_0 e^{-kt} \), where \([A]_0\) is the initial concentration. The relation shows that the concentration decreases exponentially over time, and the time it takes for \([A]\) to decrease to a fraction of \([A]_0\) is inversely proportional to the rate constant \(k\). Thus, the assertion naturally follows from the mathematical description of a first-order reaction.
04

Conclusion and Verification

Both the assertion and the reason are correct. The reason explains that the rate being dependent on \( [A] \) directly leads to the characteristic that the time required to reduce the concentration by a consistent fraction remains constant in a first-order reaction. Therefore, we can confirm the assertion due to the explanation provided by the reason.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Rate Constant
The rate constant, often denoted by the symbol \(k\), is a fundamental component in the study of reaction kinetics. It characterizes the speed at which a reaction proceeds. For a first-order reaction, the rate constant is directly related to the reaction rate, as shown in the equation \(\text{Rate} = k[A]\). Here, \([A]\) represents the concentration of the reactant at any given time.

A few crucial aspects to remember about the rate constant in the context of first-order reactions include:
  • It is independent of the concentration of reactants. This means no matter the amount of substance present, the rate constant \(k\) remains unchanged.
  • The rate constant is crucial in calculating how quickly a reaction occurs, with larger \(k\) values indicating faster reactions.
  • It has specific units of \(s^{-1}\) for a first-order reaction, which denotes per second, aligning with the time factor in these kinds of reactions.
The unchanging nature of \(k\) with respect to concentration reinforces that the rate constant is intrinsic to the specific reaction occurring under the given conditions, making it an essential constant in kinetic analysis.
Exponential Decay
In a first-order reaction, the term exponential decay describes how the concentration of the reactant decreases over time. Mathematically, this behavior is captured by the formula \([A] = [A]_0 e^{-kt}\), where \([A]\) is the concentration of the reactant at time \(t\), \([A]_0\) is the initial concentration, and \(k\) is the rate constant.

Regular features of exponential decay include:
  • The reaction proceeds at a rate proportional to the concentration of the reactant, gradually declining as the concentration decreases.
  • This type of decay implies a constant fraction of the reactant disappears over equal time periods, which is a hallmark of first-order reactions.
  • The graph of concentration versus time displays a characteristic downward curve, never truly reaching zero, illustrating the ongoing nature of the decay.
Understanding exponential decay provides insight into how reactions naturally diminish over time, highlighting the predictability and uniformity of this type of kinetic behavior.
Integrated Rate Law
The integrated rate law for first-order reactions provides a mathematical description that links reactant concentration over time. For first-order reactions, the law is expressed by the equation \([A] = [A]_0 e^{-kt}\). This vital equation allows us to determine how much of a reactant remains at a given time or to calculate the time necessary for a reactant to reach a certain concentration.

Some important points about the integrated rate law for first-order reactions include:
  • It emphasizes the exponential relationship between concentration and time, ensuring precise predictions about reaction progress.
  • The formula is rearrangeable to calculate different variables, such as finding the rate constant \(k\) if time and concentration data are given.
  • By employing natural logarithms, we can linearize data for graph plotting, which aids in visually interpreting reaction kinetics.
Exploring the integrated rate law is essential for chemists and students to accurately quantify how first-order reactions evolve, ensuring a solid quantitative foundation for reaction analysis.

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

In the first-order reaction, half of the reaction is com pleted in 100 seconds. The time for \(99 \%\) reaction to occur will be (a) \(664.64 \mathrm{~s}\) (b) \(646.6 \mathrm{~s}\) (c) \(660.9 \mathrm{~s}\) (d) \(654.5 \mathrm{~s}\)

Which one of the following equations is correct for the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) ?\) (a) \(3 \frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{dt}}=2 \frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{dt}}\) (b) \(2 \frac{\mathrm{d}\left[\mathrm{NH}_{2}\right]}{\mathrm{dt}}=-3 \frac{\mathrm{d}\left[\mathrm{H}_{3}\right]}{\mathrm{dt}}\) (c) \(2 \frac{\mathrm{d}\left[\mathrm{NH}_{3}\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left[\mathrm{H}_{2}\right]}{\mathrm{dt}}\) (d) \(3 \frac{\mathrm{d}\left[\mathrm{NH}_{2}\right]}{\mathrm{dt}}=-2 \frac{\mathrm{d}\left[\mathrm{H}_{3}\right]}{\mathrm{dt}}\)

The reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) follows first order kinetics. The time taken for \(0.8\) mole of \(\mathrm{A}\) to produce \(0.6\) mole of is 1 hour. What is the time taken for conversion of \(0.9\) mole of \(\mathrm{A}\) to produce \(0.675 \mathrm{~mole}\) of B? (a) 2 hour (b) 1 hour (c) \(0.5\) hour (d) \(0.25\) hour

Consider the following reaction $$ \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) $$ The rate of this reaction in terms of \(\mathrm{N}_{2}\) at \(\mathrm{T}\) is \(-\mathrm{d}\left[\mathrm{N}_{2}\right] /\) \(\mathrm{dt}=0.02 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) What is the value of \(-\mathrm{d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}\) (in units of \(\left.\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\right)\) at the same temperature? (a) \(0.02\) (b) 50 (c) \(0.06\) (d) \(0.04\)

For an endothermic reaction, where \(\Delta \mathrm{H}\) represents the enthalpy of the reaction in \(\mathrm{kJ} /\) mole, the minimum value for the energy of activation will be (a) less than \(\Delta \mathrm{H}\) (b) zero (c) more than \(\Delta \mathrm{H}\) (d) equal to \(\Delta \mathrm{H}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation

Inorganic 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.