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

Hydrogen peroxide, which decomposes in a first-order reaction, has a half-life of 10 hours in air. How long will it take for hydrogen peroxide to decompose to \(10 \%\) of its original concentration?

Short Answer

Expert verified
It will take approximately \( 10 \cdot \frac{\ln(10)}{\ln(2)} \) hours or roughly 33 hours for the hydrogen peroxide to decompose to 10% of its original concentration.

Step by step solution

01

Understanding First-Order Reaction

First-order reactions have a characteristic that their rate of reaction is directly proportional to the concentration of the reactant. The half-life of a first-order reaction is constant regardless of the initial concentration. The formula relating time, initial concentration, and remaining concentration for first-order reactions is given by: \( t = \frac{\ln(\frac{[A]_0}{[A]})}{k} \), where \( t \) is the time, \( [A]_0 \) is the initial concentration, \( [A] \) is the remaining concentration, and \( k \) is the rate constant.
02

Determine the Rate Constant (k)

Since the half-life (\( t_{1/2} \)) is given, and for first-order reactions \( t_{1/2} = \frac{\ln(2)}{k} \), we can rearrange to solve for the rate constant \( k \): \( k = \frac{\ln(2)}{t_{1/2}} \). Plugging in the given half-life of 10 hours, we get: \( k = \frac{\ln(2)}{10} \).
03

Calculate the Time to Reach 10% Remaining Concentration

We want to find the time \( t \) when the concentration \( [A] \) reaches \( 10\% \) of the original concentration \( [A]_0 \). Let's set \( \frac{[A]}{[A]_0} = 0.10 \). Using the equation from Step 1 and substituting in the value for \( k \), we get: \[ t = \frac{\ln(\frac{[A]_0}{[A]})}{k} = \frac{\ln(\frac{1}{0.10})}{\frac{\ln(2)}{10}} = \frac{\ln(10)}{\frac{\ln(2)}{10}} \].
04

Perform the Calculation

Now, we can calculate the time \( t \) by evaluating the expression: \( t = \frac{10 \cdot \ln(10)}{\ln(2)} \).

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.

Half-Life of a Reaction
The half-life of a reaction is an important concept in chemical kinetics, representing the time it takes for half of the reactant to decompose or be consumed. For first-order reactions, the half-life is particularly significant because it remains constant regardless of the starting concentration of the reactant. This constancy allows for the straightforward calculation of the reaction rate constant and the estimation of reaction progress over time. Understanding the half-life provides insights into the stability and longevity of substances and is crucial for applications spanning from pharmaceuticals to environmental science. The half-life serves not only as a benchmark for the speed of a reaction but also aids in the design of experiments and the analysis of reaction mechanisms.
Reaction Rate Constant
The reaction rate constant, represented by the symbol k, is a crucial parameter in the field of chemical kinetics. It quantifies the speed of a chemical reaction, correlating the rate of reaction to the concentrations of reactants. In a first-order reaction, the rate constant allows us to connect the reaction rate to the concentration through a simple proportionality relationship. The value of k can be derived from experimental data, such as the half-life, through rearranging the first-order reaction formula. With the rate constant known, predicting the amount of reactant that will be consumed over a given period becomes possible. This constant is essential for chemists and engineers to design chemical processes and understand how different conditions affect the speed of reactions.
First-Order Reaction Kinetics
First-order reaction kinetics describes reactions where the rate is directly proportional to the concentration of a single reactant. These kinetics are characterized by their linear relationship between the natural logarithm of reactant concentration and time, which provides a straightforward method for analyzing reaction rates and predicting reaction behavior. The rate law for a first-order reaction is rate = k[A], where [A] is the concentration of the reactant. This relationship is the foundation for deriving other useful expressions, such as the mathematical formula to calculate the time needed for a reactant to reach a certain concentration, making it indispensable for studies in reaction mechanisms and chemical process optimization.
Chemical Kinetics
Chemical kinetics is the study of reaction rates and the factors affecting them, encompassing the analysis of how quickly reactions occur and the steps involved in their pathways. The field involves examining variables such as temperature, pressure, concentration, and the presence of catalysts, to understand the intricacies of reaction mechanisms. It's a cornerstone of understanding chemical processes and is vital for the development and improvement of industrial reactions, environmental modeling, and much more. By exploring the principles of chemical kinetics, students build a deeper understanding of how and why chemical reactions take place, learning to manipulate conditions to achieve the desired reaction rates.
Concentration and Time Relationship
The relationship between concentration and time is fundamental to the understanding of reaction kinetics, particularly in first-order reactions. The concentration of a reactant decreases exponentially over time, as described by the kinetic equations. This relationship allows for the derivation of a formula that includes the rate constant and enables calculation of the time required for a reactant to reach a certain level of decomposition. It's through this lens that chemists can predict concentrations at various times and simulate reaction progression. Mastery of this concept is essential for fields that depend on precise reaction control, such as the pharmaceutical industry, where the timing of drug degradation can impact efficacy and safety.

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 following question is based on Chemistry Outside the Classroom 13.1. The reaction of hydrogen and bromine appears to follow the mechanism shown, $$\begin{aligned}\mathrm{Br}_{2} & \longrightarrow 2 \mathrm{Br}^{*} \\\\\mathrm{Br} \cdot+\mathrm{H}_{2} & \longrightarrow \mathrm{HBr}+\mathrm{H} \\\\\mathrm{H} \cdot+\mathrm{Br}_{2} & \longrightarrow \mathrm{HBr}+\mathrm{Br} \\\2 \mathrm{Br} \cdot &\longrightarrow\mathrm{Br}_{2}\end{aligned}$$ (a) Identify the initiation step in the mechanism. (b) Identify any propagation steps. (c) Identify the termination step. (d) The mechanism also contains the reaction $$\mathrm{H} \cdot+\mathrm{HBr} \longrightarrow \mathrm{H}_{2}+\mathrm{Br}$$ How does this reaction affect the rate of formation of \(\mathrm{HBr}\) ?

What two factors influence the effectiveness of molecular collisions in producing chemical change?

Biological reactions usually involve the interaction of an enzyme with a substrate, the substance that actually undergoes the chemical change. In many cases, the rate of reaction depends on the concentration of the enzyme but is independent of the substrate concentration. What is the order of the reaction with respect to the substrate in such instances?

Radon- 220 is radioactive, and decays into polonium- 216 by emitting an alpha particle. This is a first-order process with a rate constant of \(0.0125 \mathrm{~s}^{-1}\). When the concentration of radon- 220 is \(1.0 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1}\), what is the rate of the reaction?

At a certain moment in the reaction, $$2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}$$ \(\mathrm{N}_{2} \mathrm{O}_{5}\), is decomposing at a rate of \(2.5 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\). What are the rates of formation of \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\) ?

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