/*! 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 25 Under the same reaction conditio... [FREE SOLUTION] | 91影视

91影视

Under the same reaction conditions, initial concentration of \(1.386 \mathrm{~mol}\) \(\mathrm{dm}^{-3}\) of a substance becomes half in 40 seconds and 20 seconds through first order and zero order kinetics, respectively. Ratio \(\left(k_{1} / k_{0}\right)\) of the rate constant for first order \(\left(k_{1}\right)\) and zero order \(\left(k_{0}\right)\) of the reaction is \(-\) [2008](a) \(0.5 \mathrm{~mol}^{1} \mathrm{dm}^{3}\) (b) \(1.0 \mathrm{~mol} \mathrm{dm}^{3}\) (c) \(1.5 \mathrm{~mol} \mathrm{dm}^{-3}\) (d) \(2.0 \mathrm{~mol}^{-1} \mathrm{dm}^{3}\)

Short Answer

Expert verified
Option (a) 0.5 mol鈦宦 dm鲁.

Step by step solution

01

Understand the Reaction Half-life for First Order

For first order reactions, the half-life \( t_{1/2} \) is independent of the initial concentration. We use the formula: \[ t_{1/2} = \frac{0.693}{k_1} \] where \( k_1 \) is the rate constant for the first order reaction.
02

Calculate Rate Constant for First Order Reaction

Given that the half-life \( t_{1/2} \) for the first order reaction is 40 seconds, we can rearrange the formula from Step 1 to find \( k_1 \): \[ k_1 = \frac{0.693}{40} \approx 0.017325 \text{s}^{-1} \]
03

Understand Zero Order Kinetics

For zero order reactions, the half-life depends on the initial concentration \( [A]_0 \). The formula for half-life is: \[ t_{1/2} = \frac{[A]_0}{2k_0} \] where \( k_0 \) is the rate constant for zero order kinetics.
04

Calculate Rate Constant for Zero Order Reaction

Given the initial concentration \( [A]_0 = 1.386 \text{ mol} \text{ dm}^{-3} \) and half-life \( t_{1/2} = 20 \text{ seconds} \), we rearrange the formula from Step 3 to find \( k_0 \): \[ k_0 = \frac{[A]_0}{2 \times 20} = \frac{1.386}{40} = 0.03465 \text{ mol dm}^{-3} \text{s}^{-1} \]
05

Calculate the Ratio of Rate Constants

To find the ratio \( \frac{k_1}{k_0} \), divide the first order rate constant by the zero order rate constant:\[ \frac{k_1}{k_0} = \frac{0.017325}{0.03465} = 0.5 \text{ dm}^{3} \text{ mol}^{-1} \]
06

Choose the Correct Answer

The correct answer for the ratio \( \frac{k_1}{k_0} \) is option (a) which is \(0.5 \text{ mol}^{-1} \text{ dm}^{3}\).

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.

First Order Kinetics
First order kinetics describes a reaction where the rate is directly proportional to the concentration of one reactant. This means that as the concentration of the reactant decreases, the rate of the reaction also decreases at the same proportion. One of the key characteristics of first order reactions is that the half-life, which is the time it takes for half of the reactant to be consumed, remains constant regardless of the initial concentration. This is significant because it simplifies calculations and predictions regarding the behavior of such reactions.

For first order reactions, the half-life can be calculated using the formula:
  • \( t_{1/2} = \frac{0.693}{k_1} \)
Where \( k_1 \) is the rate constant. This constant provides insight into the reaction speed; a higher value of \( k_1 \) indicates a faster reaction. Understanding first order kinetics allows chemists to determine how long a reactant will last or how quickly a product will form under specific conditions.
Zero Order Kinetics
Zero order kinetics is quite different from first order kinetics. In zero order reactions, the rate is independent of the concentration of the reactant. This means that the reaction proceeds at a constant rate until the reactant is exhausted. Such behavior is typically observed when there is a constant supply of energy or a saturating catalytic surface, as can occur in certain industrial processes.

For zero order reactions, the half-life is dependent on the initial concentration, and it can be determined by the formula:
  • \( t_{1/2} = \frac{[A]_0}{2k_0} \)
Where \([A]_0\) is the initial concentration and \(k_0\) is the rate constant for zero order kinetics. The unique nature of zero order reactions, with their constant rate, can be particularly useful in controlled industrial reactions where maintaining a steady state is crucial. The understanding of zero mechanics helps chemists design better processes and predict outcomes over time.
Reaction Half-life
Reaction half-life provides a quick way to understand how fast a reaction progresses, by measuring the time it takes for the concentration of a reactant to decrease by half. This concept is integral to kinetics, as it reveals the dynamics of both first and zero order reactions.

In the context of first order kinetics, the half-life is invariant, meaning it doesn鈥檛 change regardless of how much reactant you start with. This makes it straightforward to apply in many chemical reactions.

Conversely, in zero order reactions, the half-life depends on the initial concentration of the reactant, offering a different perspective on how the reaction progresses. It illustrates how greater initial concentrations result in longer half-lives. Understanding half-life allows chemists and researchers to plan and adjust reactions efficiently, ensuring the desired progress rate. The distinction between the two helps in selecting the ideal conditions for industrial and laboratory processes, solving practical problems in chemistry and beyond.

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

For the first order reaction $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \longrightarrow 4 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) $$ (a) the concentration of the reactant decreases exponentially with time (b) the half-life of the reaction decreases with increasing temperature (c) the half-life of the reaction depends on the initial concentration of the reactant (d) the reaction proceeds to \(99.6 \%\) completion in eight half-life duration

\({ }_{90}^{234}\) Th disintegrates to give \({ }_{82}^{206} \mathrm{~Pb}\) as the final product. How many alpha and beta particles are emitted during this process?

In a bimolecular reaction, the steric factor \(P\) was experimentally determined to be 4.5. The correct option(s) among the following is(are) (a) The activation energy of the reaction is unaffected by the value of the steric factor (b) Experimentally determined value of frequency factor is higher than that predicted by Arrhenius equation (c) Since \(P=4.5\), the reaction will not proceed unless an effective catalyst is used (d) The value of frequency factor predicted by Arrhenius equation is higher than that determined experimentally

Two reactions \(\mathrm{R}_{1}\) and \(\mathrm{R}_{2}\) have identical pre- exponential factors. Activation energy of \(\mathrm{R}_{1}\) exceeds that of \(\mathrm{R}_{2}\) by \(10 \mathrm{~kJ} \mathrm{~mol}^{-1}\). If \(k_{1}\) and \(k_{2}\) are rate constants for reactions \(\mathrm{R}_{1}\) and \(\mathrm{R}_{2}\) respectively at \(300 \mathrm{~K}\), then \(\ln \left(k_{2} / k_{1}\right)\) is equal to : \(\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)\) (a) 8 (b) 12 (c) 6 (d) 4

One of the hazards of nuclear explosion is the generation of \(90 \mathrm{Sr}\) and its subsequent incorporation in bones. This nuclide has a half-life of 28.1 years. Suppose one microgram was absorbed by a new-born child, how much \({ }^{90} \mathrm{Sr}\) will remain in his bones after 20 years?

See all solutions

Recommended explanations on Chemistry Textbooks

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.