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Chapter 14: Problem 53

Reactions can be classified as unimolecular, bimolecular, and so on. Why are there no zero-molecular reactions? Explain why termolecular reactions are rare.

Short Answer

Expert verified
Zero-molecular reactions are impossible, as reactions involve at least one molecule. Termolecular reactions are rare due to the improbability of three molecules colliding favorably at once.

Step by step solution

01

Define Reaction Molecularity

The molecularity of a reaction is defined as the number of molecules that come together to react in an elementary reaction. It indicates the number of reacting species or molecules involved in a particular step of the reaction.
02

Analyze Zero-Molecular Reactions

A zero-molecular reaction would imply that no molecules are involved in the reaction process to convert reactants into products. However, reactions require at least one molecule to interact, therefore zero-molecular reactions are impossible physically or chemically.
03

Characterize Bimolecular Reactions

Bimolecular reactions involve two reacting molecules or species coming together to form products. These are common because two reactants colliding is practical due to molecular motion and concentration.
04

Understand Termolecular Reactions

Termolecular reactions involve three molecules colliding simultaneously to give products. Such events are rare because the probability of three reactant molecules simultaneously colliding with the proper orientation and sufficient energy is extremely low.
05

Summarize Reaction Probability

Reactions involving more than bimolecular collisions, such as termolecular reactions, have a low probability of occurrence. This is due to the complexity and specificity needed for three molecules to collide with correct orientation and energy.

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Key Concepts

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

Unimolecular Reactions
Unimolecular reactions are the simplest type of chemical reactions. They occur when a single molecule undergoes a transformation to produce one or more products. This type of reaction can happen when a molecule absorbs energy, often from heat or light, initiating a rearrangement or decomposition process.
An example is the decomposition of ozone (O_3) into oxygen (O_2) and an oxygen atom (O). These reactions typically have the reaction rate depending solely on the concentration of the single reactant.
This results in a first-order kinetic reaction where the rate law can be expressed as \( ext{Rate} = k[ ext{A}] \). Here, \( k \) is the rate constant, and \( [ ext{A}] \) is the concentration of the unimolecular reactant.
Bimolecular Reactions
Bimolecular reactions involved two reactant molecules coming together to form products. This is a common type of reaction because it relies on the collisions of two molecules. When these molecules collide with the right alignment and sufficient energy, a reaction can take place.
A good example is the reaction between nitrogen dioxide (NO_2) and carbon monoxide (CO) to produce nitrogen monoxide (NO) and carbon dioxide (CO_2).
The rate of a bimolecular reaction depends on the concentrations of both reactants, often following second-order kinetics. The rate expression may appear as \( ext{Rate} = k[ ext{A}][ ext{B}] \), where \( k \) is the rate constant, and \( [ ext{A}] \) and \( [ ext{B}] \) are the concentrations of the two reactants.
Termolecular Reactions
Termolecular reactions occur when three molecules collide simultaneously with the correct orientation and energy to form products.
Because these conditions are stringent, termolecular reactions happen less frequently compared to unimolecular and bimolecular reactions.
An example would be the reaction of two nitrogen dioxide (NO_2) molecules with an oxygen (O_2) molecule. Due to the low probability of simultaneous trios of collisions, actual termolecular reactions are rare, and many are believed to proceed through a series of bimolecular steps instead.
Termolecular reactions are typically described by third-order kinetics, with a rate law such as \( ext{Rate} = k[ ext{A}][ ext{B}][ ext{C}] \), where \( k \) is the rate constant, and \( [ ext{A}], [ ext{B}], [ ext{C}] \) are the concentrations of the three reactants.

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Most popular questions from this chapter

The rate law for the reaction: $$ 2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ is rate \(=k\left[\mathrm{H}_{2}\right][\mathrm{NO}]^{2}\). Which of the following mechanism can be ruled out on the basis of the observed rate expression? Mechanism I $$ \begin{array}{cc} \mathrm{H}_{2}+\mathrm{NO} \longrightarrow & \mathrm{H}_{2} \mathrm{O}+\mathrm{N} & (\text { slow }) \\ \mathrm{N}+\mathrm{NO} \longrightarrow & \mathrm{N}_{2}+\mathrm{O} & \text { (fast) } \\ \mathrm{O}+\mathrm{H}_{2} \longrightarrow & \mathrm{H}_{2} \mathrm{O} & \text { (fast) } \end{array} $$ Mechanism II $$ \begin{array}{ll} \mathrm{H}_{2}+2 \mathrm{NO} \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} & \text { (slow) } \\ \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \longrightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O} & \text { (fast) } \end{array} $$ Mechanism III $$ 2 \mathrm{NO} \rightleftarrows \mathrm{N}_{2} \mathrm{O}_{2} $$ (fast equilibrium) $$ \begin{array}{c} \mathrm{N}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} & \text { (slow) } \\ \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \longrightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O} & \text { (fast) } \end{array} $$

The following data were collected for the reaction between hydrogen and nitric oxide at \(700^{\circ} \mathrm{C}\) : $$ 2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g) $$ $$ \begin{array}{cccc} \text { Experiment } & {\left[\mathrm{H}_{2}\right](M)} & {[\mathrm{NO}](M)} & \text { Initial Rate }(M / \mathrm{s}) \\ \hline 1 & 0.010 & 0.025 & 2.4 \times 10^{-6} \\ 2 & 0.0050 & 0.025 & 1.2 \times 10^{-6} \\ 3 & 0.010 & 0.0125 & 0.60 \times 10^{-6} \end{array} $$ (a) Determine the order of the reaction. (b) Calculate the rate constant. (c) Suggest a plausible mechanism that is consistent with the rate law. (Hint: Assume that the oxygen atom is the intermediate.)

Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right),\) commonly called table sugar, undergoes hydrolysis (reaction with water) to produce fructose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) and glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right):\) $$ \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} $$ \(\begin{array}{ll}\text { fructose } & \text { glucose }\end{array}\) This reaction is of considerable importance in the candy industry. First, fructose is sweeter than sucrose. Second, a mixture of fructose and glucose, called invert sugar, does not crystallize, so the candy containing this sugar would be chewy rather than brittle as candy containing sucrose crystals would be. (a) From the following data determine the order of the reaction. (b) How long does it take to hydrolyze 95 percent of sucrose? (c) Explain why the rate law does not include \(\left[\mathrm{H}_{2} \mathrm{O}\right]\) even though water is a reactant. $$ \begin{array}{cc} \text { Time (min) } & {\left[\mathbf{C}_{12} \mathbf{H}_{22} \mathbf{O}_{11}\right](\boldsymbol{M})} \\ \hline 0 & 0.500 \\ 60.0 & 0.400 \\ 96.4 & 0.350 \\ 157.5 & 0.280 \end{array} $$

A protein molecule \(\mathrm{P}\) of molar mass \(\mathscr{M}\) dimerizes when it is allowed to stand in solution at room temperature. A plausible mechanism is that the protein molecule is first denatured (i.e., loses its activity due to a change in overall structure) before it dimerizes: \(\mathrm{P} \stackrel{k}{\longrightarrow} \mathrm{P}^{*}(\) denatured \() \quad\) (slow) $$ 2 \mathrm{P}^{*} \longrightarrow \mathrm{P}_{2} $$ (fast) where the asterisk denotes a denatured protein molecule. Derive an expression for the average molar mass (of \(\mathrm{P}\) and \(\left.\mathrm{P}_{2}\right), \bar{U},\) in terms of the initial protein concentration \([\mathrm{P}]_{0}\) and the concentration at time \(t,[\mathrm{P}]_{t},\) and \(\mathscr{M} .\) Describe how you would determine \(k\) from molar mass measurements.

Consider the following elementary steps for a consecutive reaction: $$ \mathrm{A} \stackrel{k_{1}}{\longrightarrow} \mathrm{B} \stackrel{k_{2}}{\longrightarrow} \mathrm{C} $$ (a) Write an expression for the rate of change of \(\mathrm{B}\). (b) Derive an expression for the concentration of B under "steady-state" conditions; that is, when \(\mathrm{B}\) is decomposing to \(\mathrm{C}\) at the same rate as it is formed from \(\mathrm{A}\).
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