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During a smog event, trace amounts of many highly reactive substances are present in the atmosphere. One of these is the hydroperoxyl radical, \(\mathrm{HO}_{2},\) which reacts with sulfur trioxide, \(\mathrm{SO}_{3}\). The rate constant for the reaction $$2 \mathrm{HO}_{2}(g)+\mathrm{SO}_{3}(g) \rightarrow \mathrm{H}_{2} \mathrm{SO}_{3}(g)+2 \mathrm{O}_{2}(g)$$ is \(2.6 \times 10^{11} M^{-1} s^{-1}\) at \(298 \mathrm{K} .\) The initial rate of the reaction doubles when the concentration of \(\mathrm{SO}_{3}\) or \(\mathrm{HO}_{2}\) is doubled. What is the rate law for the reaction?

Short Answer

Expert verified
Answer: The rate law for the reaction is Rate = 2.6 x 10鹿鹿 M鈦宦箂鈦宦 [HO鈧俔[SO鈧僝.

Step by step solution

01

Write the general form of the rate law

The general form of the rate law for this reaction can be written as: Rate = k[HO鈧俔鈦縖SO鈧僝岬 where k is the rate constant, [HO鈧俔 and [SO鈧僝 are the concentrations of hydroperoxyl radical and sulfur trioxide, respectively, and n and m are the orders of the reaction with respect to HO鈧 and SO鈧, respectively.
02

Find the order of the reaction with respect to HO鈧

Given that the initial rate doubles when the concentration of HO鈧 is doubled, we can set up the following equation: 2 (Rate1) = Rate2, where Rate1 = k[HO鈧俔鈦縖SO鈧僝岬 and Rate2 = k[2(HO鈧)]鈦縖SO鈧僝岬. Now, divide Rate2 by Rate1: 2 = [2(HO鈧)]鈦/[HO鈧俔鈦. Solve for n: 2 = 2鈦 => n = 1. So the order of the reaction with respect to HO鈧 is 1.
03

Find the order of the reaction with respect to SO鈧

Similarly, given that the initial rate doubles when the concentration of SO鈧 is doubled, we can set up the equation: 2 (Rate1) = Rate3, where Rate1 = k[HO鈧俔鈦縖SO鈧僝岬 and Rate3 = k[HO鈧俔鈦縖2(SO鈧)]岬. Divide Rate3 by Rate1: 2 = [2(SO鈧)]岬/[SO鈧僝岬. Solve for m: 2 = 2岬 => m = 1. So the order of the reaction with respect to SO鈧 is 1.
04

Write the rate law for the reaction

Now that we have the order of the reaction with respect to both reactants, we can write the rate law as: Rate = k[HO鈧俔鹿[SO鈧僝鹿 or Rate = 2.6 x 10鹿鹿 M鈦宦箂鈦宦 [HO鈧俔[SO鈧僝.

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Key Concepts

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

Rate Law
The rate law is a mathematical equation that relates the speed or rate of a chemical reaction to the concentration of the reactants involved. It is a fundamental concept in chemical kinetics, used to understand how different factors influence the speed of reactions.
The general form of a rate law is: Rate = k[A]鈦縖B]岬 Here, '[A]' and '[B]' represent the concentrations of the reactants, while 'n' and 'm' are the reaction orders for each reactant, which are experimentally determined values. They show how the rate is affected by the concentration of each reactant. The letter 'k' stands for the rate constant, which is unique for each reaction at a given temperature.
Understanding a reaction鈥檚 rate law provides insight into the reaction mechanism and how changes in conditions impact the overall process. In our example, the specific rate law was determined to be: Rate = k[HO鈧俔鹿[SO鈧僝鹿, meaning the rate is directly proportional to the concentration of both the hydroperoxyl radical and sulfur trioxide.
Reaction Order
The reaction order provides critical information about how the rate of a reaction is influenced by the concentration of the reactants. It tells us how many molecules of a particular reactant are involved in the rate-determining step.
In our smog event exercise, we determined the reaction orders with respect to the reactants.
  • For the hydroperoxyl radical ( HO鈧), when its concentration is doubled, the rate of reaction doubles, showing that it is first order with respect to HO鈧.
  • Similarly, the rate also doubles when the concentration of sulfur trioxide ( SO鈧) is doubled, indicating that the reaction is first order with respect to SO鈧 as well.
This means the reaction is overall second order since the sum of the individual orders is 1 (for HO鈧) + 1 (for SO鈧) = 2. Reaction orders are not necessarily related to the stoichiometric coefficients in a balanced chemical equation and are always determined empirically.
Rate Constant
The rate constant, represented by the symbol 'k', is a crucial parameter in the rate law equation. It provides a quantitative measure of how fast a reaction occurs and is unique to each chemical reaction. Importantly, 'k' is dependent on the temperature at which the reaction occurs.
For our specific reaction between the hydroperoxyl radical and sulfur trioxide, the rate constant is given as \( 2.6 \times 10^{11} \ M^{-1} s^{-1} \) at 298 K. This high value indicates that the reaction proceeds very quickly under these conditions. A large rate constant typically signifies a fast reaction, reflecting the ease with which reactant molecules interact to produce the products. Rate constants can be affected by changes in temperature, and they can provide insights into the energy profile of a reaction pathway.
Hydroperoxyl Radical
The hydroperoxyl radical, often symbolized as HO鈧, plays a significant role in atmospheric chemistry, particularly in processes related to air pollution such as smog formation. Radicals are molecules that have unpaired electrons, making them extremely reactive.
As seen in our reaction example, HO鈧 participates actively with sulfur trioxide to form hydrogen sulfite and oxygen. In atmospheric conditions, these radicals can be generated through photochemical reactions involving sunlight and pollutants, and they often contribute to complex chains of secondary reactions.
Given the high reactivity of radicals like HO鈧, understanding their reaction behavior is critical for modeling atmospheric chemical interactions, predicting pollution levels, and devising environmental protection strategies.
Sulfur Trioxide
Sulfur trioxide (SO鈧) is a chemical compound that plays a significant part in industrial and atmospheric chemistry. It is a colorless, corrosive gas that reacts vigorously with water, leading to the formation of sulfuric acid, a major component of acid rain.
In industrial settings, SO鈧 is often used in the production of sulfuric acid, which is essential for many manufacturing processes. In our example of a smog event, SO鈧 reacts with hydroperoxyl radicals, showing how this gas can contribute to harmful atmospheric reactions that impact air quality.
Understanding the dynamics of SO鈧 reactions is vital for developing effective strategies to mitigate environmental hazards, protect ecosystems, and improve public health by reducing exposure to harmful pollutants.

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Most popular questions from this chapter

The rate of a chemical reaction is too slow to measure at room temperature. We could either raise the temperature or add a catalyst. Which would be a better solution for making an accurate determination of the rate constant?

In the presence of water, NO and \(\mathrm{NO}_{2}\) react to form nitrous acid (HNO,) by the following reaction: $$\mathrm{NO}(g)+\mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 2 \mathrm{HNO}_{2}(a q)$$ When the concentration of NO or \(\mathrm{NO}_{2}\) is doubled, the initial rate of reaction doubles. If the rate of the reaction does not depend on \(\left[\mathrm{H}_{2} \mathrm{O}\right],\) what is the rate law for this reaction?

Compounds \(A\) and \(B\) react to give a single product, \(C .\) Write the rate law for each of the following cases and determine the units of the rate constant by using the units \(M\) for concentration and s for time: a. The reaction is first order in \(A\) and second order in \(B\). b. The reaction is first order in \(A\) and second order overall. c. The reaction is independent of the concentration of \(\mathrm{A}\) and second order overall. d. The reaction is second order in both \(\mathrm{A}\) and \(\mathrm{B}\).

On the basis of the frequency factors and activation energy values of the following two reactions, determine which one will have the larger rate constant at room temperature \((298 \mathrm{K})\). \(\mathrm{O}_{3}(g)+\mathrm{Cl}(g) \rightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g)\) \(A=2.9 \times 10^{-11} \mathrm{cm}^{3} /(\text { molecules } \cdot \mathrm{s}) \quad E_{2}=2.16 \mathrm{kJ} / \mathrm{mol}\) \(\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\) \(A=2.0 \times 10^{-12} \mathrm{cm}^{3} /(\text { molecules } \cdot \mathrm{s}) \quad E_{\mathrm{a}}=11.6 \mathrm{kJ} / \mathrm{mol}\)

In the reaction between nitrogen dioxide and ozone, $$2 \mathrm{NO}_{2}(g)+\mathrm{O}_{3}(g) \rightarrow \mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{O}_{2}(g)$$ how are the rates of change in the concentrations of the reactants and products related?

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