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According to collision theory, chemical reactions occur through molecular collisions. A unimolecular elementary process in a reaction mechanism involves dissociation of a single molecule. How can these two ideas be compatible? Explain.

Short Answer

Expert verified
Chemical reactions occur through molecular collisions according to the collision theory. Even a unimolecular elementary process, which involves the dissociation of a single molecule, can still be explained by this principle. The single molecule collides with other molecules or walls of a container and, provided it receives enough energy (activation energy) from the collision, it can dissociate. Thus, the two theories are compatible.

Step by step solution

01

Understanding Collision Theory

Collision theory postulates that for a reaction to occur, it is necessary for the reactive molecules to collide with one another. Not every collision causes a reaction; the molecules must possess a certain minimum amount of energy called the activation energy.
02

Understanding Unimolecular Elementary Process

Unimolecular elementary processes involve a single molecule. It could be that this molecule dissociates into two or more parts, changes structure, or reacts with light. It seems contrary to the collision theory as no collision appears to be involved in this process.
03

Relating Collision Theory and Unimolecular Processes

Considering a gas molecule in a container, it collides with other gas molecules as well as the walls of the container. While many of these collisions may not lead to a reaction, some can provide the molecule with sufficient energy (activation energy) to dissociate. This occurrence aligns with both the collision theory and the characteristic of a unimolecular process, making them compatible.

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Key Concepts

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

Unimolecular Process
A unimolecular process is an elementary chemical reaction involving only one molecule that changes in some way. This may involve dissociation, where the molecule splits into two or more smaller fragments. Alternatively, it might be a structural transformation or an interaction with light. Although it might initially seem that no collision is involved in unimolecular processes, this isn't entirely correct.
In the gas phase, molecules are constantly moving and frequently collide with each other or with the walls of their container. Each collision holds the potential to excite a molecule, providing it with enough energy to undergo a unimolecular reaction. Even if only one molecule appears in the process, the energy it requires to initiate change often results from prior interactions and collisions. Thus, unimolecular processes can still fit within the framework of collision theory.
Activation Energy
Activation energy is the minimum energy required for a chemical reaction to occur. Think of it as the initial push needed to start a process. In the context of collision theory, activation energy ensures that only collisions with sufficient energy will result in a reaction.
A molecule involved in a unimolecular process must obtain this activation energy to undergo transformation. This energy can come from previous collisions with other molecules or the walls of a container. For example, when a molecule in a gas collides, it can absorb enough energy to go over the energy barrier for reaction.
  • The activation energy threshold explains why not every molecular collision leads to a reaction.
  • It is a key factor in determining the rate of a chemical reaction.
  • Understanding activation energy is crucial for predicting how temperature changes can affect reaction rates.
It helps bridge the understanding between collision theory and reactions that initially seem unrelated to collisions, like unimolecular processes.
Reaction Mechanism
A reaction mechanism outlines the step-by-step sequence of elementary processes leading to a chemical reaction. Understanding these steps provides a clearer picture of how reactants ultimately become products.
Each step can involve bimolecular collisions or unimolecular changes. In a reaction mechanism, unimolecular processes emphasize the role of single-molecule transformations. Even in seemingly standalone steps, energy provided by earlier molecular collisions remains crucial.
For example, a reaction mechanism might start with a bimolecular collision providing activation energy, followed by a unimolecular step where that energy causes a change in a single molecule. This series of events shows how unimolecular processes and collision theory coexist in practice.
  • Reaction mechanisms break down complex processes into simpler units.
  • They help in predicting product formation and understanding reaction speed.
  • Studying mechanisms can assist in designing experiments or influencing conditions to desired outcomes.
Overall, reaction mechanisms are foundational to understanding how unimolecular processes and collision theory integrate into chemical reactions.

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Most popular questions from this chapter

The object is to study the kinetics of the reaction between peroxodisulfate and iodide ions. $$\begin{aligned} &\text { (a) } \mathrm{S}_{2} \mathrm{O}_{8}^{2-}(\mathrm{aq})+3 \mathrm{I}^{-}(\mathrm{aq}) \longrightarrow 2 \mathrm{SO}_{4}^{2-}(\mathrm{aq})+\mathrm{I}_{3}^{-}(\mathrm{aq}) \end{aligned}$$ The \(I_{3}^{-}\) formed in reaction (a) is actually a complex of iodine, \(\mathrm{I}_{2},\) and iodide ion, \(\mathrm{I}^{-}\). Thiosulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) also present in the reaction mixture, reacts with \(\mathrm{I}_{3}^{-}\) just as fast as it is formed. $$\text { (b) } 2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq})+\mathrm{I}_{3}^{-}(\mathrm{aq}) \longrightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}+3 \mathrm{I}^{-}(\mathrm{aq})$$ When all of the thiosulfate ion present initially has been consumed by reaction (b), a third reaction occurs between \(\mathrm{I}_{3}^{-}(\mathrm{aq})\) and starch, which is also present in the reaction mixture. $$\text { (c) } \mathrm{I}_{3}^{-}(\mathrm{aq})+\operatorname{starch} \longrightarrow \text { blue complex }$$ The rate of reaction (a) is inversely related to the time required for the blue color of the starch-iodine complex to appear. That is, the faster reaction (a) proceeds, the more quickly the thiosulfate ion is consumed in reaction (b), and the sooner the blue color appears in reaction (c). One of the photographs shows the initial colorless solution and an electronic timer set at \(t=0 ;\) the other photograph shows the very first appearance of the blue complex (after 49.89 s). Tables I and II list some actual student data obtained in this study. $$\begin{array}{l} \hline\text { TABLE I } \\ \text { Reaction conditions at } 24^{\circ} \mathrm{C}: 25.0 \mathrm{mL} \text { of the } \\ \left(\mathrm{NH}_{4}\right)_{2} \mathrm{S}_{2} \mathrm{O}_{8}(\text { aq) listed, } 25.0 \mathrm{mL} \text { of the } \mathrm{KI}(\mathrm{aq}) \\ \text { listed, } 10.0 \mathrm{mL} \text { of } 0.010 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq}), \text { and } 5.0 \mathrm{mL} \\ \text { starch solution are mixed. The time is that of the } \\ \text { first appearance of the starch-iodine complex. } \\ \hline & \text { Initial Concentrations, } \mathrm{M} \\ \hline \text { Experiment } & \left(\mathrm{NH}_{4}\right)_{2} \mathrm{S}_{2} \mathrm{O}_{8} & \mathrm{KI} & \text { Time, s } \\ \hline 1 & 0.20 & 0.20 & 21 \\ 2 & 0.10 & 0.20 & 42 \\ 3 & 0.050 & 0.20 & 81 \\ 4 & 0.20 & 0.10 & 42 \\ 5 & 0.20 & 0.050 & 79 \\ \hline \end{array}$$ $$\begin{array}{l} \hline \text { TABLE II } \\ \text { Reaction conditions: those listed in Table I for } \\ \text { Experiment } 4, \text { but at the temperatures listed. } \\ \hline \text { Experiment } & \text { Temperature, }^{\circ} \mathrm{C} & \text { Time, } \mathrm{s} \\ \hline 6 & 3 & 189 \\ 7 & 13 & 88 \\ 8 & 24 & 42 \\ 9 & 33 & 21 \\ \hline \end{array}$$ (a) Use the data in Table I to establish the order of reaction (a) with respect to \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) and to I \(^{-}\). What is the overall reaction order? [Hint: How are the times required for the blue complex to appear related to the actual rates of reaction? (b) Calculate the initial rate of reaction in Experiment 1 expressed in \(\mathrm{M} \mathrm{s}^{-1} .\) [Hint: You must take into account the dilution that occurs when the various solutions are mixed, as well as the reaction stoichiometry indicated by equations \((a),(b), \text { and }(c) .]\) (c) Calculate the value of the rate constant, \(k,\) based on experiments 1 and 2 (d) Calculate the rate constant, \(k\), for the four different temperatures in Table II. (e) Determine the activation energy, \(E_{\mathrm{a}}\), of the peroxodisulfate- iodide ion reaction. (f) The following mechanism has been proposed for reaction (a). The first step is slow, and the others are fast. $$\begin{array}{c} \mathrm{I}^{-}+\mathrm{S}_{2} \mathrm{O}_{8}^{2-} \longrightarrow \mathrm{IS}_{2} \mathrm{O}_{8}^{3-} \\ \mathrm{IS}_{2} \mathrm{O}_{8}^{3-} \longrightarrow 2 \mathrm{SO}_{4}^{2-}+\mathrm{I}^{+} \\ \mathrm{I}^{+}+\mathrm{I}^{-} \longrightarrow \mathrm{I}_{2} \\ \mathrm{I}_{2}+\mathrm{I}^{-} \longrightarrow \mathrm{I}_{3}^{-} \end{array}$$ Show that this mechanism is consistent with both the stoichiometry and the rate law of reaction (a). Explain why it is reasonable to expect the first step in the mechanism to be slower than the others.

The reaction \(2 \mathrm{NO}+\mathrm{Cl}_{2} \longrightarrow 2 \mathrm{NOCl}\) has the rate law: rate of reaction \(=k[\mathrm{NO}]^{2}\left[\mathrm{Cl}_{2}\right] .\) Propose a twostep mechanism for this reaction consisting of a fast reversible first step, followed by a slow step.

If the plot of the reactant concentration versus time is linear, then the order of the reaction is (a) zero order; (b) first order; (c) second order; (d) third order.

In the reaction \(2 \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+3 \mathrm{D},\) reactant \(\mathrm{A}\) is found to disappear at the rate of \(6.2 \times 10^{-4} \mathrm{M} \mathrm{s}^{-1}.\) (a) What is the rate of reaction at this point? (b) What is the rate of disappearance of \(\mathrm{B}\) ? (c) What is the rate of formation of D?

Three different sets of data of \([\mathrm{A}]\) versus time are giv the following table for the reaction \(A \longrightarrow\) prod [Hint: There are several ways of arriving at answer each of the following six questions. $$\begin{array}{cccccc} \hline \text { I } & & \text { II } & & \text { III } & \\ \hline \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } & \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } & \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } \\ \hline 0 & 1.00 & 0 & 1.00 & 0 & 1.00 \\ 25 & 0.78 & 25 & 0.75 & 25 & 0.80 \\ 50 & 0.61 & 50 & 0.50 & 50 & 0.67 \\ 75 & 0.47 & 75 & 0.25 & 75 & 0.57 \\ 100 & 0.37 & 100 & 0.00 & 100 & 0.50 \\ 150 & 0.22 & & & 150 & 0.40 \\ 200 & 0.14 & & & 200 & 0.33 \\ 250 & 0.08 & & & 250 & 0.29 \\ \hline \end{array}$$ What is the approximate half-life of the first-order reaction?

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