/*! 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 68 Atmospheric chemistry involves h... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 68

Atmospheric chemistry involves highly reactive, odd-electron molecules such as the hydroperoxyl radical \(\mathrm{HO}_{2},\) which decomposes into \(\mathrm{H}_{2} \mathrm{O}_{2}\) and \(\mathrm{O}_{2} .\) Determine the rate law for the reaction and the value of the rate constant at \(298 \mathrm{K}\) by using the following data obtained at \(298 \mathrm{K}\). $$\begin{array}{cc} \text { Time }(\mu \mathrm{s}) & {\left[\mathrm{HO}_{2}\right](\mu M)} \\\0.0 & 8.5 \\\\\hline 0.6 & 5.1 \\\\\hline 1.0 & 3.6 \\\\\hline 1.4 & 2.6 \\\\\hline 1.8 & 1.8 \\\\\hline 2.4 & 1.1\end{array}$$

Short Answer

Expert verified
Answer: The rate law for the decomposition of hydroperoxyl radical (HO鈧�) into H鈧侽鈧� and O鈧� is Rate = k[H0鈧俔, where the reaction is first-order with respect to the HO鈧� concentration. The rate constant (k) at 298 K is approximately -0.063 (渭s鈦宦�).

Step by step solution

01

Calculate reaction rates from given concentration data.

We are given a table with the concentrations of HO鈧� at different time intervals. To determine the rates, we can compare the change in concentration over the change in time. We will calculate these changes as we progress from one reading to another: $\begin{array}{|c|c|} \hline \text{Time interval } (\mu s) & \frac{\Delta [\mathrm{HO}_{2}]}{\Delta t} (\mu M / \mu s) \\\\ \hline \hline 0.6 - 0.0 & \frac{5.1 - 8.5}{0.6 - 0.0} \\\\ \hline 1.0 - 0.6 & \frac{3.6 - 5.1}{1.0 - 0.6} \\\\ \hline 1.4 - 1.0 & \frac{2.6 - 3.6}{1.4 - 1.0} \\\\ \hline 1.8 - 1.4 & \frac{1.8 - 2.6}{1.8 - 1.4} \\\\ \hline 2.4 - 1.8 & \frac{1.1 - 1.8}{2.4 - 1.8} \\\\ \hline \end{array}$
02

Calculate the relative rates of reaction.

To determine the order of the reaction, we need to find the relative rates between the concentration changes. Using the calculated data above, we can relate the first reading with the rest of the readings: $\begin{array}{|c|c|} \hline \text{Relative time interval} (\mu s) & \frac{\text{Relevant rate}}{\text{Initial rate}} \\\\ \hline \hline 0.6 & \frac{ \text{Rate at 0.6} }{ \text{Rate at 0.0} } \\\\ \hline 1.0 & \frac{ \text{Rate at 1.0} }{ \text{Rate at 0.0} } \\\\ \hline 1.4 & \frac{ \text{Rate at 1.4} }{ \text{Rate at 0.0} } \\\\ \hline 1.8 & \frac{ \text{Rate at 1.8} }{ \text{Rate at 0.0} } \\\\ \hline 2.4 & \frac{ \text{Rate at 2.4} }{ \text{Rate at 0.0} } \\\\ \hline \end{array}$
03

Analyze the data for reaction order.

To find the reaction order, we will analyze the data from the above table. Observe whether the reaction rates are directly proportional to the concentration, square of the concentration, or some other relation. After analyzing the data, we find that the reaction is first-order with respect to the HO鈧� concentration. Thus, the rate law for the reaction is: Rate = k[H0鈧俔
04

Calculate the rate constant (k) at 298 K.

Now that we know the reaction is first-order, we can calculate the rate constant (k) at 298 K. Using the first reading from the table in step 1, we will have: Rate = k[H0鈧俔 \(\frac{\Delta [\mathrm{HO}_{2}]}{\Delta t} = \text{k} \times [\mathrm{HO}_{2}]\) \(k = \frac{\Delta [\mathrm{HO}_{2}]}{\Delta t \times [\mathrm{HO}_{2}]}\) Use the data from the time interval 0 - 0.6 碌s: \(k = \frac{5.1 - 8.5}{(0.6 - 0)(8.5)} \approx -0.063 (\mu s ^{-1})\) Hence, the rate constant k for the given reaction at 298 K is approximately -0.063 (\(\mu s ^{-1}\)).

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 Law Determination
In chemical kinetics, determining the rate law is essential for understanding how a reaction progresses over time. The rate law relates the rate of a chemical reaction to the concentration of its reactants. Typically, it's expressed as Rate = k[A]^m[B]^n, where k is the rate constant, and A and B are reactants with their respective reaction orders m and n.
This exercise looks at the hydroperoxyl radical HO鈧� and its decomposition. To determine the rate law, we calculated the changes in concentration over time to observe how the rate of reaction correlates with the concentration of HO鈧�. By scrutinizing these rates, we learned they were directly proportional to the concentration of HO鈧� itself.
This direct proportionality is characteristic of a first-order reaction. Thus, the rate law for this reaction simplifies to Rate = k[HO鈧俔. Understanding rate laws helps in predicting reaction behavior in different conditions, making it a crucial concept in reaction kinetics.
First-Order Reactions
First-order reactions are critical in chemistry as they describe processes where the rate is directly proportional to the concentration of a single reactant. If a reaction is first-order, a plot of the natural logarithm of the concentration of the reactant versus time will yield a straight line.
In our example, the hydroperoxyl radical HO鈧� decomposes in a first-order reaction. This means as the concentration of HO鈧� decreases, the rate at which it decomposes follows a predictable pattern. In first-order kinetics, the reactions exhibit an exponential decline of concentration.
When examining the provided data, observing the rate of reaction supports our conclusion of it being a first-order process. This revelation is instrumental in predicting how the concentration of HO鈧� will change over time, allowing for targeted interventions in atmospheric chemistry.
Rate Constant Calculation
Calculating the rate constant, k, for a reaction provides insight into how quickly the reaction proceeds under certain conditions. The rate constant is specific to a particular reaction at a given temperature.
Using the data provided at 298 K for the decomposition of HO鈧�, we determined that the reaction is first-order. We used the formula for rate constant k for first-order reactions: \(k = \frac{\Delta [\text{HO}_2]}{\Delta t \times [\text{HO}_2]}\).
Plugging in the values from the initial time interval of 0 - 0.6 碌s, we calculated k to be approximately -0.063 (碌s鈦宦�). Although negative in this context (due to concentration decrease), it reflects the rate at which HO鈧� decomposes. Knowing k helps in predicting the speed of reactions and comparing different reactions under various conditions.

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

Tropospheric Ozone Tropospheric (lower atmosphere) ozone is rapidly consumed in many reactions, including $$\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$$ Use the following data to calculate the instantaneous rate of the reaction at \(t=0.000 \mathrm{s}\) and \(t=0.052 \mathrm{s}\) $$\begin{array}{cc}\text { Time (s) } & {[\mathrm{NO}](\mathrm{M})} \\\0.000 & 2.0 \times 10^{-8} \\\\\hline 0.011 & 1.8 \times 10^{-8} \\\\\hline 0.027 & 1.6 \times 10^{-8} \\\\\hline 0.052 & 1.4 \times 10^{-8} \\\\\hline 0.102 & 1.2 \times 10^{-8} \\\\\hline\end{array}$$

Nitryl chloride, \(\mathrm{NO}_{2} \mathrm{Cl}\), is a reactive chlorine- containing species sometimes found in marine sediments in industrial areas. In the gas phase it decomposes to \(\mathrm{NO}_{2}\) and \(\mathrm{Cl}_{2}\) : $$2 \mathrm{NO}_{2} \mathrm{Cl}(g) \rightarrow 2 \mathrm{NO}_{2}(g)+\mathrm{Cl}_{2}(g)$$ Under laboratory conditions, the rate of formation of \(\mathrm{NO}_{2}(g)\) is \(5.7 \times 10^{-6} M \cdot \mathrm{s}^{-1}\) a. What is the rate of formation of \(\mathrm{Cl}_{2}(g) ?\) b. What is the rate of consumption of \(\mathrm{NO}_{2} \mathrm{Cl}(g)\) ?

Hydroperoxyl radicals react rapidly with ozone to produce oxygen and OH radicals: $$\mathrm{HO}_{2}(g)+\mathrm{O}_{3}(g) \rightarrow \mathrm{OH}(g)+2 \mathrm{O}_{2}(g)$$ The rate of this reaction was studied in the presence of a large excess of ozone. Determine the pseudo-first-order rate constant and the second-order rate constant for the reaction from the following data: $$\begin{array}{cll} \text { Time (ms) } & {\left[\mathrm{HO}_{2}\right](\mathrm{M})} & {\left[\mathrm{O}_{3}\right](\mathrm{M})} \\ \hline 0 & 3.2 \times 10^{-5} & 1.0 \times 10^{-3} \\ \hline 10 & 2.9 \times 10^{-5} & 1.0 \times 10^{-3} \\ \hline 20 & 2.6 \times 10^{-6} & 1.0 \times 10^{-3} \\ \hline 30 & 2.4 \times 10^{-6} & 1.0 \times 10^{-2} \\ \hline 80 & 1.4 \times 10^{-6} & 1.0 \times 10^{-3} \\ \hline \end{array}$$

The metabolism of table sugar (sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) ) begins with the hydrolysis of the disaccharide to glucose and fructose (both \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) ): $$\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 2 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)$$ The kinetics of the reaction were studied at \(24^{\circ} \mathrm{C}\) in a reaction system with a large excess of water, so the reaction was pseudo first order in sucrose. Determine the rate law and the pseudo-first-order rate constant for the reaction from the following data: $$\begin{array}{cc} \text { Time (s) } & {\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right](\mathrm{M})} \\ 0 & 0.562 \\ \hline 612 & 0.541 \\ \hline 1600 & 0.509 \\ \hline 2420 & 0.484 \\ \hline 3160 & 0.462 \\ \hline 4800 & 0.442 \\ \hline \end{array}$$

A proposed mechanism for the gas phase decomposition of hydrogen peroxide at an elevated temperature consists of three elementary steps: $$\begin{aligned} \mathrm{H}_{2} \mathrm{O}_{2}(g) & \rightarrow 2 \mathrm{OH}(g) \\ \mathrm{H}_{2} \mathrm{O}_{2}(g)+\mathrm{OH}(g) & \rightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{HO}_{2}(g) \\ \mathrm{HO}_{2}(g)+\mathrm{OH}(g) & \rightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{O}_{2}(g) \end{aligned}$$ If the rate law for the reaction is first order in \(\mathrm{H}_{2} \mathrm{O}_{2},\) which step in the mechanism is the rate-determining step?
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Analysis

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.