/*! 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 8 One loss mechanism for ozone in ... [FREE SOLUTION] | 91影视

91影视

One loss mechanism for ozone in the atmosphere is the reaction with the \(\mathrm{HO}_{2} \cdot\) radical: Using the following information, determine the rate law expression for this reaction: $$\begin{array}{ccc}\text { Rate }\left(\mathrm{cm}^{-3} \mathrm{s}^{-1}\right) & {\left[\mathrm{HO}_{2} \cdot\right]\left(\mathrm{cm}^{-3}\right)} & {\left[\mathrm{O}_{3}\right]\left(\mathrm{cm}^{-3}\right)} \\ \hline 1.9 \times 10^{8} & 1.0 \times 10^{11} & 1.0 \times 10^{12} \\\9.5 \times 10^{8} & 1.0 \times 10^{11} & 5.0 \times 10^{12} \\\5.7 \times 10^{8} & 3.0 \times 10^{11} & 1.0 \times 10^{12}\end{array}$$

Short Answer

Expert verified
The rate law expression for the reaction between ozone and the HO鈧偮 radical in the atmosphere is: Rate = k [HO鈧偮穄[O鈧僝

Step by step solution

01

Analyzing the concentration effect on the rate for HO鈧偮 radical

We will first inspect the effect of the concentration of HO鈧偮 on the reaction rate. To do this, we will compare experiments 1 and 3, in which we have the same [O鈧僝 but different [HO鈧偮穄 concentrations. From the table, we have: For experiment 1: \(\) Rate = 1.9 脳 10鈦 cm鲁/s, [HO鈧偮穄 = 1.0 脳 10鹿鹿 cm鈦宦 For experiment 3: \(\) Rate = 5.7 脳 10鈦 cm鲁/s, [HO鈧偮穄 = 3.0 脳 10鹿鹿 cm鈦宦 Now we will determine the relationship between the reaction rate and the concentration of HO鈧偮.
02

Calculating the order with respect to HO鈧偮 radical

To find the order of the reaction with respect to the HO鈧偮 radical, we will use the following relationship: \( \frac{\text{Rate}_3}{\text{Rate}_1} = \left(\frac{[\text{HO}_{2}\text{路}]_3}{[\text{HO}_{2}\text{路}]_1}\right)^{m} \) Plugging in the given values: \( \frac{5.7 \times 10^{8}}{1.9 \times 10^{8}} = \left(\frac{3.0 \times 10^{11}}{1.0 \times 10^{11}}\right)^{m} \) \( \Rightarrow 3 = 3^m \) Thus, m = 1, which implies the reaction is first order with respect to the concentration of the HO鈧偮 radical.
03

Analyzing the concentration effect on the rate for O鈧

Next, we will investigate the effect of the concentration of O鈧 on the reaction rate. To do this, we will compare experiments 1 and 2, in which we have the same [HO鈧偮穄 but different [O鈧僝 concentrations. From the table, we have: For experiment 1: \(\) Rate = 1.9 脳 10鈦 cm鲁/s, [O鈧僝 = 1.0 脳 10鹿虏 cm鈦宦 For experiment 2: \(\) Rate = 9.5 脳 10鈦 cm鲁/s, [O鈧僝 = 5.0 脳 10鹿虏 cm鈦宦 Now we will determine the relationship between the reaction rate and the concentration of O鈧.
04

Calculating the order with respect to O鈧

To find the order of the reaction with respect to O鈧, we will use the same relationship we used to derive the reaction order for the HO鈧偮 radical: \( \frac{\text{Rate}_2}{\text{Rate}_1} = \left(\frac{[\text{O}_{3}]_2}{[\text{O}_{3}]_1}\right)^{n} \) Plugging in the given values: \( \frac{9.5 \times 10^{8}}{1.9 \times 10^{8}} = \left(\frac{5.0 \times 10^{12}}{1.0 \times 10^{12}}\right)^{n} \) \( \Rightarrow 5 = 5^n \) Thus, n = 1, which implies the reaction is first order with respect to the concentration of O鈧.
05

Writing the rate law expression

Now that we know the reaction is first order with respect to both the HO鈧偮 radical and O鈧, we can write the rate law expression as follows: Rate = k [HO鈧偮穄^1 [O鈧僝^1 Or simply: Rate = k [HO鈧偮穄[O鈧僝 This is the rate law expression for the given reaction between ozone and the HO鈧偮 radical in the atmosphere.

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.

Chemical Kinetics and Reaction Rate Law
Chemical kinetics is the study of the rates at which chemical reactions proceed and the factors that influence these rates. The reaction rate law, also known as the rate equation, describes this behavior quantitatively, indicating how the reaction rate depends on the concentration of the reactants. By measuring how different concentrations affect the reaction speed, chemists can deduce the rate law for a specific reaction.

For instance, consider a reaction where the rate changes as the concentration of a reactant is altered. Suppose the rate doubles when the concentration of the reactant doubles; this suggests that the rate is directly proportional to the concentration of that reactant. This proportionality can be converted into a mathematical equation, which is the rate law. It captures the essence of kinetics by allowing predictions of how fast a reaction will occur under different conditions.
Determining Reaction Order
The reaction order is another key concept in chemical kinetics. It specifies the power to which the concentration of each reactant is raised in the rate law equation. For each reactant, the reaction order tells us how the rate of reaction will respond to a change in that reactant's concentration.

To determine the reaction order, scientists conduct a series of experiments altering reactant concentrations while observing the resulting change in reaction rates. Calculating the reaction order requires a mathematical analysis where experimental data is used to find a relationship between concentration changes and rate changes, as shown in the solution provided for the reaction between ozone and the HO鈧偮 radical. This is a fundamental step in developing a deep understanding of a reaction's mechanics.
The Phenomenon of Ozone Depletion
Understanding chemical kinetics is not just essential for laboratory reactions but also for environmental phenomena, such as ozone depletion. Ozone depletion refers to the thinning of the Earth's ozone layer, which is largely attributed to chemicals like chlorofluorocarbons (CFCs) and other ozone-depleting substances. The kinetics of ozone-depleting reactions involve various radicals, including the HO鈧偮 radical as illustrated in the exercise.

Scientists use the kinetics of these reactions to understand the rates at which ozone molecules are destroyed in the stratosphere. This understanding is crucial for developing strategies to manage and prevent further depletion of the ozone layer, which serves as a protective shield against the sun's harmful ultraviolet radiation.
Concentration Effect on Reaction Rates
The concentration effect is a fundamental principle illustrating how variations in the concentration of reactants influence the rate of a chemical reaction. When reactant concentrations are high, there are more particles available to collide and react, leading to a higher reaction rate. Conversely, a reduction in concentration typically results in a slower reaction.

The rate law for a reaction includes terms that mathematically describe the concentration effect. By analyzing how altering concentrations affects the reaction rate, we can deduce these terms, which are critical in predicting how a reaction will proceed under various conditions. This concept is pivotal in processes ranging from industrial synthesis to environmental chemistry.
Writing the Rate Law Expression
The rate law expression is a mathematical representation that relates the reaction rate to the concentration of each reactant, each raised to a power corresponding to their reaction order. For a given reaction, the rate law can only be determined experimentally and cannot be predicted solely by the reaction's stoichiometry.

The rate constant (k) in the expression is specific to the reaction and can also depend on temperature and the presence of a catalyst. Determining the rate law expression is a crucial step toward controlling a chemical process, whether for industrial application or understanding natural phenomena. The exercise showcased how to deduce the rate law for the reaction of ozone with the HO鈧偮 radical by analyzing experimental data, leading to the expression Rate = k [HO鈧偮穄[O鈧僝.

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

P35.32 The reaction of atomic chlorine with ozone is the first step in the catalytic decomposition of stratospheric ozone by \(\mathrm{Cl} \bullet\):$$\mathrm{Cl} \cdot(g)+\mathrm{O}_{3}(g) \rightarrow \mathrm{ClO} \cdot(g)+\mathrm{O}_{2}(g)$$ At \(298 \mathrm{K}\) the rate constant for this reaction is \(6.7 \times 10^{9} \mathrm{M}^{-1} \mathrm{s}^{-1}\) Experimentally, the Arrhenius pre-exponential factor was determined to be \(1.4 \times 10^{10} \mathrm{M}^{-1} \mathrm{s}^{-1} .\) Using this information determine the activation energy for this reaction.

In the stratosphere, the rate constant for the conversion of ozone to molecular oxygen by atomic chlorine is \(\mathrm{Cl} \cdot(g)+\mathrm{O}_{3}(g) \rightarrow \mathrm{ClO} \cdot(g)+\mathrm{O}_{2}(g)$$\left[\text { (half-life of } 5760 \text { years) } \mathrm{k}=\left(1.7 \times 10^{10} \mathrm{M}^{-1} \mathrm{s}^{-1}\right) e^{-260 K / T}\right]\) a. What is the rate of this reaction at \(20 \mathrm{km}\) where \\[\begin{array}{l}{[\mathrm{Cl}]=5 \times 10^{-17} \mathrm{M},\left[\mathrm{O}_{3}\right]=8 \times 10^{-9} \mathrm{M}, \text { and }} \\\T=220 \mathrm{K} ?\end{array}\\] b. The actual concentrations at \(45 \mathrm{km}\) are \([\mathrm{Cl}]=3 \times 10^{-15} \mathrm{M}\) and \(\left[\mathrm{O}_{3}\right]=8 \times 10^{-11} \mathrm{M} .\) What is the rate of the reaction at this altitude where \(T=270 \mathrm{K} ?\) c. (Optional) Given the concentrations in part (a), what would you expect the concentrations at \(20 . \mathrm{km}\) to be assuming that the gravity represents the operative force defining the potential energy?

Show that the ratio of the half-life to the one-quarter life, \(t_{1 / 2} / t_{1 / 4},\) for a reaction that is \(n\) th order \((n>1)\) in reac\(\operatorname{tant} A\) can be written as a function of \(n\) alone (that is, there is no concentration dependence in the ratio). (Note: The onequarter life is defined as the time at which the concentration is \(1 / 4\) of the initial concentration.

Consider the gas phase thermal decomposition of \(1.0 \operatorname{atm}\) of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COOC}\left(\mathrm{CH}_{3}\right)_{3}(g)\) to acetone \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}(g)\) and ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)(\mathrm{g}),\) which occurs with a rate constant of \(0.0019 \mathrm{s}^{-1} .\) After initiation of the reaction, at what time would you expect the pressure to be 1.8 atm?

Hydrogen abstraction from hydrocarbons by atomic chlorine is a mechanism for \(\mathrm{Cl} \cdot\) loss in the atmosphere. Consider the reaction of \(\mathrm{Cl} \cdot\) with ethane: $$\mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{Cl} \cdot(g) \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \cdot(g)+\mathrm{HCl}(g)$$ This reaction was studied in the laboratory, and the following data were obtained: $$\begin{array}{cc}\mathbf{T}(\boldsymbol{K}) & \mathbf{k}\left(\times \mathbf{1 0}^{-\mathbf{1 0}} \mathbf{M}^{-\mathbf{2}} \mathbf{s}^{-\mathbf{1}}\right) \\\\\hline 270 & 3.43 \\\370 & 3.77 \\ 470 & 3.99 \\\570 & 4.13 \\\670 & 4.23\end{array}$$ a. Determine the Arrhenius parameters for this reaction. b. At the tropopause (the boundary between the troposphere and stratosphere located approximately \(11 \mathrm{km}\) above the surface of Earth \(),\) the temperature is roughly \(220 \mathrm{K}\). What do you expect the rate constant to be at this temperature? c. Using the Arrhenius parameters obtained in part (a), determine the Eyring parameters \(\Delta H^{\dagger}\) and \(\Delta S^{\frac{1}{r}}\) for this reaction at \(220 \mathrm{K}\)

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.