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Write the overall reaction for the mechanism proposed below and identify any reaction intermediates. $$ \begin{aligned} &\text { Step } 1 \mathrm{AC}+\mathrm{B} \longrightarrow \mathrm{AB}+\mathrm{C} \\ &\text { Step } 2 \mathrm{AC}+\mathrm{AB} \longrightarrow \mathrm{A}_{2} \mathrm{~B}+\mathrm{C} \end{aligned} $$

Short Answer

Expert verified
The overall reaction is B \rightarrow A_2 B, and the reaction intermediates are AC and AB.

Step by step solution

01

Identify Reaction Intermediates

First, we need to identify any species that appear in the reaction mechanism but not in the overall reaction. These are called reaction intermediates. Here, both AC and AB appear as reactants and products within the steps but are not present in the final overall reaction.
02

Cancel Out the Intermediates and Spectator Species

Next, we add up the two steps of the mechanism and cancel the intermediates and any other species that appear on both sides of the reaction arrow. AB is created in Step 1 and consumed in Step 2, and AC is consumed in both steps. Additionally, C is formed in both steps.
03

Write the Overall Reaction

After cancellation, we combine the remaining species to give the overall reaction: Step 1 contributes B to the overall reaction, and Step 2 results in the product A2B. C does not appear since it cancels out.

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Key Concepts

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

Reaction Intermediates
In chemical reactions, some species are created and consumed within the process without appearing in the final overall equation. These species are known as reaction intermediates. Identifying these intermediates is key to understanding the reaction mechanism and simplifying complex reactions into more understandable steps.

Let's consider a simplified example where two steps are involved in a reaction mechanism. In the first step, compound AC reacts with B to form AB and C. In the second step, compound AC reacts with the newly formed AB to yield A2B and another molecule of C.

When we break down this mechanism, we see that both AC and AB are created and then consumed—they don't appear in the final products or reactants. Consequently, AC and AB are reaction intermediates. By recognizing these, we can focus on the substances that are truly altered from reactants to products, which is crucial for understanding the overall chemical change that has occurred.
Chemical Kinetics
The study of the rate at which chemical reactions occur and the factors that affect these rates is known as chemical kinetics. In kinetics, we are not only concerned with the end result of a reaction but also the 'path' taken to arrive there and how fast the reaction progresses.

In our example, the formation of AB from AC and B (Step 1), and then the reaction of AC with AB to form A2B (Step 2) will each have their own rate, determined by the reaction conditions such as temperature, pressure, and the presence of a catalyst. The intermediates play a vital role in kinetics since they can significantly influence the rate of the overall reaction. Understanding the role of intermediates can help chemists design reactions to proceed faster or slower, as needed for industrial or laboratory applications. Therefore, it is not just about whether the reactions occur, but how efficiently they do so, which is integral to the study and application of chemical kinetics.
Overall Reaction
The overall reaction is the simplified form of a reaction mechanism that shows only the starting reactants and the final products, excluding all the intermediates. In our given exercise, the overall reaction is found by adding together the steps of the mechanism and canceling out intermediates and any other species that appear on both sides.

By canceling out the intermediates AC and AB which are formed and used up within the reaction sequence, and acknowledging that C is a product in both steps and thus cancels itself out, the remaining reactants and products can be combined. The result is the overall reaction that provides a macroscopic perspective of the chemical change taking place.

This simplified reaction is what would be typically written in a chemical equation representing the process and is fundamental to understanding how a set of reactants is transformed into a set of products without needing to consider the more complex pathway of steps and intermediates.

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Most popular questions from this chapter

The first-order rate constant for the photodissociation of \(\mathrm{A}\) is \(6.85 \times 10^{-2} \mathrm{~min}^{-1}\). Calculate the time needed for the concentration of A to decrease to (a) \(\frac{1}{8}[\mathrm{~A}]_{0} ;\) (b) \(10 . \%\) of its initial concentration; (c) one-third of its initial concentration.

Determine the rate constant for each of the following firstorder reactions, in each case expressed for the rate of loss of \(A\) : (a) \(\mathrm{A} \longrightarrow \mathrm{B}\), given that the concentration of \(\mathrm{A}\) decreases to one-half its initial value in \(1000 . \mathrm{s} ;\) (b) \(\mathrm{A} \longrightarrow \mathrm{B}\), given that the concentration of A decreases from \(0.67 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) to \(0.53 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) in \(25 \mathrm{~s}\); (c) \(2 \mathrm{~A} \longrightarrow \mathrm{B}+\mathrm{C}\), given that \([\mathrm{A}]_{0}=0.153 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) and that after \(115 \mathrm{~s}\) the concentration of \(B\) rises to \(0.034 \mathrm{~mol} \cdot \mathrm{L}^{-1}\).

Because partial pressures are proportional to concentrations, rate laws for gas-phase reactions can also be expressed in terms of partial pressures, for instance, as Rate \(=\) \(k P_{\mathrm{X}}\) for a first-order reaction of a gas \(\mathrm{X}\). What are the units for the rate constants when partial pressures are expressed in torr and time is expressed in seconds for (a) zero-order reactions; (b) first-order reactions; (c) second-order reactions?

The Michaelis constant \(\left(K_{M}\right)\) is an index of the stability of an enzyme-substrate complex. Does a high Michaelis constant indicate a stable or an unstable enzyme-substrate complex? Explain your reasoning.

The rate law of the reaction \(2 \mathrm{NO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+\) \(2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) is Rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{H}_{2}\right]\), and the mechanism that has been proposed is Step \(1 \mathrm{NO}+\mathrm{NO} \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}\) Step \(2 \mathrm{~N}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}\) Step \(3 \mathrm{~N}_{2} \mathrm{O}+\mathrm{H}_{2} \longrightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}\) (a) Which step in the mechanism is likely to be rate determining? Explain your answer. (b) Sketch a reaction profile for the overall reaction, which is known to be exothermic. Label the activation energies of each step and the overall reaction enthalpy.

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