/*! 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 31 Formic acid decomposes at \(550^... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Formic acid decomposes at \(550^{\circ} \mathrm{C}\) according to the equation $$\mathrm{HCO}_{2} \mathrm{H}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})$$ The reaction follows first-order kinetics. In an experiment, it is determined that \(75 \%\) of a sample of \(\mathrm{HCO}_{2} \mathrm{H}\) has decomposed in 72 seconds. Determine \(t_{1 / 2}\) for this reaction.

Short Answer

Expert verified
The half-life \(t_{1/2}\) for this reaction is approximately 36.09 seconds.

Step by step solution

01

Understand the Kinetics

This reaction follows first-order kinetics, which means the rate of reaction depends linearly on the concentration of formic acid. For first-order reactions, the relationship between concentration and time is given by the equation:\[ ln([A]_0/[A]) = kt \]where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time \(t\), and \(k\) is the rate constant.
02

Determine the Rate Constant, k

We are told that 75% of HCOâ‚‚H has decomposed in 72 seconds. This means 25% remains. Using the relationship for a first-order reaction we have:\[ ln([A]_0/[A]) = kt \]Substitute \([A]_0/[A] = 1/0.25 = 4\) and \(t = 72 \, \text{seconds}\) to find \(k\):\[ ln(4) = k \times 72 \]\[ k = \frac{ln(4)}{72} \approx 0.0192 \, \text{seconds}^{-1} \]
03

Calculate the Half-life, t_{1/2}

For first-order reactions, the half-life \(t_{1/2}\) is given by the formula:\[ t_{1/2} = \frac{0.693}{k} \]Substitute \(k = 0.0192 \, \text{seconds}^{-1}\):\[ t_{1/2} = \frac{0.693}{0.0192} \approx 36.09 \, \text{seconds} \]

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 91Ó°ÊÓ!

Key Concepts

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

Half-life calculation
The half-life of a reaction, denoted as \( t_{1/2} \), is the time it takes for half of the reactant to be consumed in a chemical reaction. It is an essential concept in understanding how quickly a reaction proceeds. For first-order reactions, the half-life is independent of the initial concentration of the reactant. This characteristic reliably informs us about the dynamics of chemical interactions without altering due to concentration changes. First-order reactions simplify the half-life calculation with the formula: \[ t_{1/2} = \frac{0.693}{k} \] where \( k \) is the rate constant. This formula highlights how half-life is a constant value, unchanged by varying amounts of starting material, making it extremely useful for predicting reaction schedules and durations. By knowing \( k \), you can straightforwardly determine how long it will take for half the reactant to decompose, offering a clear picture of the reaction's pace.
Rate constant determination
Determining the rate constant, \( k \), is fundamental for analyzing reaction kinetics, as it gives insight into the speed of a chemical reaction. The rate constant is unique to each reaction and varies with conditions such as temperature. For first-order reactions, the equation used is: \[ ln([A]_0/[A]) = kt \] In this equation: - \([A]_0\) is the initial concentration of the reactant. - \([A]\) is the concentration after time \( t \). By rearranging and solving for \( k \), researchers can quantify how a reaction progresses over time. For example, in a scenario where \( 75\% \) of a substance has decomposed, only \( 25\% \) remains. This remaining concentration can substitute in the equation to find \( k \). Using this process helps to illustrate the reaction rate, providing valuable insight into the efficiency of the reaction.
Chemical decomposition
Chemical decomposition is a process where a single compound breaks down into two or more simpler substances. This occurs through chemical reactions and often involves energy to disrupt chemical bonds. Decomposition reactions are categorized by their mechanism, conditions, and the substances they generate. In the context of the reaction \( \text{HCO}_2\text{H} \rightarrow \text{CO}_2 + \text{H}_2 \), decomposition results in the formation of carbon dioxide and hydrogen gas. These reactions are essential in both biological and industrial processes, underlining the role of energy changes and the resultant new product formation. Understanding the decomposition provides insights into reaction mechanisms and helps clarify conditions required for such transformations. It allows scientists to predict product formations and understand the energy dynamics involved.
Reaction kinetics
Reaction kinetics explores the rates of chemical processes and the steps that control these rates. It examines how different conditions like concentration, temperature, and catalysts affect the speed of a reaction. First-order kinetics specifically examines cases where the rate depends on the concentration of a single reactant. The rate laws derived from reaction kinetics provide mathematical relationships that describe how reactants transform into products over time. For first-order reactions, this relationship is a simple logarithmic decay, expressed as: - \[ ln([A]_0/[A]) = kt \] This formula allows us to calculate the time required for a given amount of reactant to decompose. Therefore, understanding reaction kinetics is crucial for predicting how fast reactions occur, optimizing conditions for industrial chemical processes, and enhancing productivity by controlling and applying heat or pressure efficiently. It lays the foundation for further studies on complex reaction networks.

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

At temperatures below \(500 \mathrm{K},\) the reaction between carbon monoxide and nitrogen dioxide $$ \mathrm{NO}_{2}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{NO}(\mathrm{g}) $$ has the following rate equation: Rate \(=k\left[\mathrm{NO}_{2}\right]^{2} .\) Which of the three mechanisms suggested here best agrees with the experimentally observed rate equation? Mechanism 1 \(\quad\) single, elementary step $$\mathrm{NO}_{2}+\mathrm{CO} \longrightarrow \mathrm{CO}_{2}+\mathrm{NO}$$ Mechanism \(2 \quad\) Two steps $$\begin{aligned}&\text { Slow } \quad \mathrm{NO}_{2}+\mathrm{NO}_{2} \longrightarrow \mathrm{NO}_{3}+\mathrm{NO}\\\&\text { Fast } \quad \mathrm{NO}_{3}+\mathrm{CO} \longrightarrow \mathrm{NO}_{2}+\mathrm{CO}_{2}\end{aligned}$$ Mechanism 3 \(\quad\) Two steps $$\begin{aligned}&\text { Slow } \quad \mathrm{NO}_{2} \longrightarrow \mathrm{NO}+\mathrm{O}\\\&\text { Fast } \quad \mathrm{CO}+\mathrm{O} \longrightarrow \mathrm{CO}_{2}\end{aligned}$$

Nitrogen oxides, \(\mathrm{NO}_{x}\) (a mixture of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) collectively designated as \(\mathrm{NO}_{x}\) ), play an essential role in the production of pollutants found in photochemical smog. The \(\mathrm{NO}_{x}\) in the atmosphere is slowly broken down to \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) in a first-order reaction. The average half-life of NO in the smokestack emissions in a large city during daylight is \(3.9 \mathrm{h}\) (a) Starting with \(1.50 \mathrm{mg}\) in an experiment, what quantity of \(\mathrm{NO}_{x}\) remains after \(5.25 \mathrm{h} ?\) (b) How many hours of daylight must have elapsed to decrease \(1.50 \mathrm{mg}\) of \(\mathrm{NO}_{x}\) to \(2.50 \times 10^{-6} \mathrm{mg} ?\)

What is the rate law for each of the following elementary reactions? (a) \(\mathrm{Cl}(\mathrm{g})+\mathrm{ICl}(\mathrm{g}) \longrightarrow \mathrm{I}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\) (b) \(\mathbf{O}(\mathrm{g})+\mathbf{O}_{3}(\mathrm{g}) \longrightarrow 2 \mathrm{O}_{2}(\mathrm{g})\) (c) \(2 \mathrm{NO}_{2}(\mathrm{g}) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g})\)

Hydrogenation reactions, processes wherein \(\mathrm{H}_{2}\) is added to a molecule, are usually catalyzed. An excellent catalyst is a very finely divided metal suspended in the reaction solvent. Tell why finely divided rhodium, for example, is a much more efficient catalyst than a small block of the metal.

Hundreds of different reactions can occur in the stratosphere, among them reactions that destroy the earth's ozone layer. The table below lists several (second-order) reactions of Cl atoms with ozone and organic compounds; each is given with its rate constant. $$\begin{array}{ll}\hline & \text { Rate Constant } \\\\\text { Reaction } & \left(298 \mathrm{K}, \mathrm{cm}^{3} / \mathrm{molecule} \cdot \mathrm{s}\right) \\\\\hline \text { (a) } \mathrm{Cl}+0_{3} \longrightarrow \mathrm{Cl} 0+0_{2} & 1.2 \times 10^{-11} \\\\\text {(b) } \mathrm{Cl}+\mathrm{CH}_{4} \longrightarrow\mathrm{HCl}+\mathrm{CH}_{3} & 1.0 \times 10^{-13} \\\\\text {(c) } \mathrm{Cl}+\mathrm{C}_{3} \mathrm{H}_{8} \longrightarrow\mathrm{HCl}+\mathrm{C}_{3} \mathrm{H}_{7} & 1.4 \times 10^{-10} \\\\\text {(d) } \mathrm{Cl}+\mathrm{CH}_{2} \mathrm{FCl} \longrightarrow\mathrm{HCl}+\mathrm{CHFCl} & 3.0 \times 10^{-18} \\\\\hline\end{array}$$ For equal concentrations of Cl and the other reactant, which is the slowest reaction? Which is the fastest reaction?

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.