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Write the overall reaction that consists of the following elementary steps: (1) \(\quad \mathrm{N}_{2} \mathrm{O}_{5}(g) \rightarrow \mathrm{NO}_{3}(g)+\mathrm{NO}_{2}(g)\) (2) \(\quad \mathrm{NO}_{3}(g) \rightarrow \mathrm{NO}_{2}(g)+\mathrm{O}(g)\) (3) \(\quad 2 \mathrm{O}(g) \rightarrow \mathrm{O}_{2}(g)\)

Short Answer

Expert verified
Elementary Steps: 1. N2O5(g) → 2NO3(g) 2. NO3(g) → NO2(g) + O(g) 3. 2O(g) → O2(g) Answer: The overall reaction is N2O5(g) → 2NO2(g) + O2(g).

Step by step solution

01

Identify the intermediates

Look at each elementary step, and determine which species are formed in one step and consumed in a later step. These species are the intermediates that we will eliminate when combining the elementary steps. In this case, NO3(g) and O(g) are intermediates.
02

Combining Step 1 & Step 2

Add the two elementary steps together, making sure to eliminate any intermediates. If a species is formed in one step and consumed in the later step, eliminate it from both steps.In this case, NO3(g) is formed in Step 1 and consumed in Step 2, so we can eliminate it. The combined reaction will look like: N2O5(g) → 2NO2(g) + O(g).
03

Combining the combined reaction with Step 3

We can now add the combined reaction from step 2 with step 32021simple_formula(O(g) → O2(g)). Notice that there are two molecules of O(g) formed in the combined reaction, and two molecules consumed in Step 3, so they can be eliminated. Adding those reactions, we get: N2O5(g) → 2NO2(g) + O2(g). Therefore, the overall reaction for these elementary steps is: N2O5(g) → 2NO2(g) + O2(g).

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Key Concepts

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

Elementary Steps in Reaction Mechanisms
Understanding chemical reactions requires breaking down how reactant molecules change into products. In chemical kinetics, this breakdown is depicted through elementary steps, which are the individual stages that reactant molecules undergo during a complex reaction.

Each elementary step describes a single event in a reaction mechanism, such as the breaking of a bond or the formation of an intermediate. Mechanisms can consist of one or multiple elementary steps that collectively describe the sequential events leading to the final product. Importantly, these steps can often be classified into unimolecular or bimolecular processes, depending on whether one or two reactant molecules are involved, respectively.

For example, in the reaction where \( \text{N}_2\text{O}_5(g) \rightarrow \text{NO}_3(g) + \text{NO}_2(g) \), it is an elementary step in the mechanism where a single molecule of dinitrogen pentoxide decomposes into nitrogen dioxide and a nitrate radical. Understanding each elementary step is crucial, as they provide insights into the reaction rate and mechanism.
Reaction Intermediates
In the journey from reactants to products, a chemical reaction often proceeds through transitory species called reaction intermediates. These species are produced in one elementary step and consumed in a subsequent step within a reaction mechanism. They don't appear in the overall chemical equation for the reaction, as they are not present in the initial or final reaction mixture.

Characteristically unstable and fleeting, these intermediates can be radicals, ions, or neutral molecules. For instance, the nitrate radical (\(\text{NO}_3(g)\)) and atomic oxygen (\(\text{O}(g)\)) in our example are intermediates. They are formed and consumed within the reaction's pathway and thus provide a deeper understanding of the mechanism but do not feature in the final equation \(\text{N}_2\text{O}_5(g) \rightarrow 2\text{NO}_2(g) + \text{O}_2(g)\).

Identifying intermediates is an essential step because it allows chemists to understand the flow of the reaction and to potentially manipulate conditions to favor the production of desired products.
Combining Elementary Steps
The art of resolving complex chemical reactions into simpler parts comes full circle when we combine elementary steps to decipher the overall reaction equation. This involves adding up the elementary steps and canceling out the intermediates that appear on both sides of the resulting equations.

Let's take a closer look at our exercise. We started by identifying intermediates like \(\text{NO}_3(g)\) and \(\text{O}(g)\) in the initial steps. When we added the first two elementary steps, these intermediates were eliminated because they are not part of the products or reactants in the overall balanced equation. Lastly, we combined this with the third elementary step, where the atomic oxygen formed earlier reacted to produce \(\text{O}_2(g)\). Ultimately, combining these steps gave us the overall balanced chemical equation, which shows only the original reactants and the final products, providing a clear and concise depiction of the chemical transformation.

By understanding how to combine elementary steps and identify the fate of reaction intermediates, students can gain a comprehensive grasp of complex reaction mechanisms, simplifying their study of chemical kinetics.

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Most popular questions from this chapter

Any gas-phase reaction occurs more rapidly as the temperature of the gas increases. Why?

Ammonia reacts with nitrous acid to form an intermediate, ammonium nitrite (NH_/NO \(_{2}\) ), which decomposes to \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}:\) \(\mathrm{NH}_{3}(g)+\mathrm{HNO}_{2}(a q) \rightarrow \mathrm{NH}_{4} \mathrm{NO}_{2}(a q) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(\ell)\) a. The reaction is first order in ammonia and second order in nitrous acid. What is the rate law for the reaction? What are the units of the rate constant if concentrations are expressed in molarity and time in seconds? b. The rate law for the reaction has also been written as $$ \text { Rate }=k\left[\mathrm{NH}_{4}^{+}\right]\left[\mathrm{NO}_{2}^{-}\right]\left[\mathrm{HNO}_{2}\right] $$ Is this expression equivalent to the one you wrote in part \((a) ?\) c. With the data in Appendix 4, calculate the value of \(\Delta H_{\mathrm{rxn}}^{\circ}\) for the overall reaction \(\left(\Delta H_{\mathrm{f}}^{*}, \mathrm{HNO}_{2}=\right.\) \(-43.1 \mathrm{kJ} / \mathrm{mol})\) d. Draw an energy profile for the process with the assumption that \(E_{a}\) of the first step is lower than \(E_{\mathrm{a}}\) of the second step.

Dinitrogen pentoxide \(\left(\mathrm{N}_{2} \mathrm{O}_{5}\right)\) decomposes as follows to nitrogen dioxide and nitrogen trioxide: $$ \mathrm{N}_{2} \mathrm{O}_{5}(g) \rightarrow \mathrm{NO}_{2}(g)+\mathrm{NO}_{3}(g) $$ Calculate the average rate of this reaction between consecutive measurement times in the following table. $$\begin{array}{cc} \text { Time (s) } & {\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]} \\ & \text { (molecules/cm }^{3} \text { ) } \\ 0.00 & 1.500 \times 10^{12} \\ \hline 1.45 & 1.357 \times 10^{12} \\ \hline 2.90 & 1.228 \times 10^{12} \\ \hline 4.35 & 1.111 \times 10^{12} \\ \hline 5.80 & 1.005 \times 10^{12} \\ \hline \end{array}$$

The order of a reaction is independent of temperature, but the value of the rate constant varies with temperature. Why?

Does the concentration of a homogeneous catalyst appear in the rate law for the reaction it catalyzes?

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