Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The rate constant (k) for a first order reaction is equal to \(4.2 \times 10^{-4} \mathrm{~s}^{-1}\). What is the half life for the reaction? a. \(3.7 \times 10^{3} \mathrm{~s}\) b. \(7.1 \times 10^{3} \mathrm{~s}\) c. \(2.71 \times 10^{3} \mathrm{~s}\) d. \(1.7 \times 10^{3} \mathrm{~s}\)

Short Answer

Expert verified
The half-life is approximately \(1.7 \times 10^{3} \, \mathrm{s}\). Option d.

Step by step solution

01

Identify the Formula for Half-Life

Since this is a first-order reaction, the half-life (\( t_{1/2} \) ) can be calculated using the formula \( t_{1/2} = \frac{0.693}{k} \), where \( k \) is the rate constant.
02

Substitute the Known Values

Insert the given rate constant \( k = 4.2 \times 10^{-4} \, \mathrm{s}^{-1} \) into the half-life formula: \( t_{1/2} = \frac{0.693}{4.2 \times 10^{-4}} \).
03

Perform the Calculation

Calculate the half-life using the values: \( t_{1/2} = \frac{0.693}{4.2 \times 10^{-4}} = 1.65 \times 10^{3} \, \mathrm{s} \).
04

Round and Compare

Round \( t_{1/2} = 1.65 \times 10^{3} \, \mathrm{s} \) to match the options and verify it with choice d, which is rounded to \( 1.7 \times 10^{3} \, \mathrm{s} \).

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 Vaia!

Key Concepts

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

First-Order Reaction
In chemical reactions, understanding the order of the reaction is crucial for predicting how the reaction behaves over time. A **first-order reaction** is characterized by a reaction rate that is directly proportional to the concentration of one reactant. Put simply, as the concentration of the reactant decreases, the rate at which the reaction occurs also decreases at a proportional rate. This means that first-order reactions have a constant half-life, regardless of the initial concentration of the reactant.

First-order reactions are represented by the rate law equation:
  • Rate = k[A]
Here, *Rate* is the rate of the reaction, *k* is the rate constant, and *[A]* is the concentration of the reactant.

Understanding the behavior of first-order reactions helps in predicting how long it will take for a reactant to deplete or transform into products, making it a foundational concept in chemistry.
Rate Constant
The **rate constant** is a key component in chemical kinetics, as it quantifies the speed at which a reaction progresses. It is specific to each reaction and varies depending on factors such as temperature and the presence of catalysts.

In the context of a first-order reaction, the rate constant is denoted by "k" and has units of reciprocal time, such as \( \text{s}^{-1} \). This signifies that it defines the rate in terms of how quickly the concentration of a reactant decreases within a unit of time.
  • The magnitude of the rate constant provides insight into the reaction's speed;
  • Larger values of *k* indicate faster reactions, whereas smaller values suggest slower changes.
To utilize the rate constant effectively, one must understand its integration into various formulas, like the half-life formula, to predict reaction characteristics accurately.
Chemical Kinetics
**Chemical kinetics** is the branch of chemistry that deals with understanding the speed of chemical reactions and the factors that influence this speed. It delves into the transformation of reactants to products, focusing on:
  • The reaction rate;
  • The steps involved in the reaction mechanism;
  • The effect of different variables such as concentration, temperature, and pressure.
In chemical kinetics, it is vital to know if a reaction follows zero, first, or second order as this dictates how calculations and predictions are made.

The study of chemical kinetics allows chemists and scientists to design experiments, plan chemical production, and even control processes involved in industrial applications. Deciphering kinetic data aids in optimizing reactions for yield, safety, and efficiency.
Half-Life Formula
The **half-life** of a reaction refers to the time required for the concentration of a reactant to reduce to half its initial value. For a first-order reaction, the half-life is a constant and can be calculated using a specific formula.The formula used for first-order reactions is:
  • and is expressed as \( t_{1/2} = \frac{0.693}{k} \).
In this expression, \( t_{1/2} \) represents the half-life, and \( k \) is the rate constant. The constant 0.693 arises from the natural logarithm of 2.This formula highlights the unique nature of first-order processes where the half-life is independent of the starting concentration. Therefore, whether you begin with a lot or just a little of a substance, it invariably takes the same amount of time to reduce by half.

Understanding and applying this formula is critical for laboratory experiments, drug dosages, and environmental assessments, ensuring that calculations remain consistent for varying concentrations.

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 rate law for the reaction \(\mathrm{RCl}+\mathrm{NaOH} \rightarrow \mathrm{ROH}+\mathrm{NaCl}\) is given by Rate \(=\mathrm{k}(\mathrm{RCl})\). The rate of the reaction is a. Halved by reducing the concentration of \(\mathrm{RCl}\) by one half. b. Increased by increasing the temperature of the reaction. c. Remains same by change in temperature. d. Doubled by doubling the concentration of \(\mathrm{NaOH}\).

The basic theory of Arrhenius equation is that (1) Activation energy and pre exponential factors are always temperature independent (2) The number of effective collisions is proportional to the number of molecule above a certain threshold energy. (3) As the temperature increases, the number of molecules with energies exceeding the threshold energy increases. (4) The rate constant in a function of temperature a. 2,3 and 4 b. 1,2 and 3 c. 2 and 3 d. 1 and 3

In a first order reaction the concentration of reactant decreases from \(800 \mathrm{~mol} / \mathrm{dm}^{3}\) to \(50 \mathrm{~mol} / \mathrm{dm}^{3}\) in \(2 \times\) \(10^{4} \mathrm{sec}\). The rate constant of reaction in \(\mathrm{sec}^{-1}\) is a. \(2 \times 10^{4}\) b. \(3.45 \times 10^{-5}\) c. \(1.386 \times 10^{-4}\) d. \(2 \times 10^{-4}\)

At \(380^{\circ} \mathrm{C}\), half life period for the first order decomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is \(360 \mathrm{~min}\). The energy of activation of the reaction is \(200 \mathrm{~kJ} \mathrm{~mol}^{-1}\). Calculate the time required for \(75 \%\) decomposition at \(450^{\circ} \mathrm{C}\) if half life for decomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is \(10.17 \mathrm{~min}\) at \(450^{\circ} \mathrm{C}\). a. \(20.4 \mathrm{~min}\) b. \(408 \mathrm{~min}\) c. \(10.2 \mathrm{~min}\) d. none

When the concentration of \(\mathrm{A}\) is doubled, the rate for the reaction: \(2 \mathrm{~A}+\mathrm{B} \rightarrow 2 \mathrm{C}\) quadruples. When the concentration of \(\mathrm{B}\) is doubled the rate remains the same. Which mechanism below is consistent with the experimental observations? a. Step I: \(2 \mathrm{~A} \rightleftharpoons \mathrm{D}\) (fast equilibrium) Step II: \(\mathrm{B}+\mathrm{D} \rightarrow \mathrm{E}\) (slow) Step III: \(\mathrm{E} \rightarrow 2 \mathrm{C}\) (fast) b. Step I: \(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{D}\) (fast equilibrium) Step II: \(\mathrm{A}+\mathrm{D} \rightarrow 2 \mathrm{C}\) (slow) c. Step \(\mathrm{I}: \mathrm{A}+\mathrm{B} \rightarrow \mathrm{D}\) (slow) Step II: \(\mathrm{A}+\mathrm{D} \rightleftharpoons 2 \mathrm{C}\) (fast equilibrium) d. Step I: \(2 \mathrm{~A} \rightarrow \mathrm{D}\) (slow) Step II: \(\mathrm{B}+\mathrm{D} \rightarrow \mathrm{E}\) (fast) Step III: \(\mathrm{E} \rightarrow 2 \mathrm{C}\) (fast)

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.

Sign-up for free