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Determine the rate law for the reaction $$ \mathrm{I}(\mathrm{g})+\mathrm{I}(\mathrm{g})+\mathrm{M}(\mathrm{g}) \stackrel{k}{\Longrightarrow} \mathrm{I}_{2}(\mathrm{~g})+\mathrm{M}(\mathrm{g}) $$ where \(M\) is any molecule present in the reaction container. Give the units of \(k\). Determine the molecularity of this reaction. Is this reaction identical to $$ \mathrm{I}(\mathrm{g})+\mathrm{I}(\mathrm{g}) \stackrel{k}{\Longrightarrow} \mathrm{I}_{2}(\mathrm{~g}) $$ Explain.

Short Answer

Expert verified
The rate law is \( ext{Rate} = k[ ext{I}][ ext{I}][ ext{M}] \) with \( k \) units \( M^{-2}s^{-1} \). The reaction is termolecular, different from the second, which is bimolecular.

Step by step solution

01

Identify the reaction type

The given reaction is a bimolecular reaction involving two iodine atoms and one molecule \( M \). The third molecule \( M \) does not undergo any change, which suggests it may act as a catalyst or stabilizer. Therefore, the reaction is termolecular because it involves three molecules in the transition state.
02

Write the rate law

For a termolecular elementary reaction, the rate law is based on the reactants involved at the molecular level. The rate law expression can be written as: \[ ext{Rate} = k[ ext{I}][ ext{I}][ ext{M}] \] since the reaction involves two iodine molecules and one \( M \) molecule.
03

Determine the units of the rate constant \( k \)

The units of the rate constant \( k \) depend on the overall order of the reaction. This is a third-order reaction (two \( ext{I} \) atoms and one \( ext{M} \)). Thus, the rate has units of concentration per time (\( M/s \)), and the units of \( k \) can be determined from:\[ k = \frac{M/s}{M^3} = rac{1}{M^2s} \]Therefore, the units of \( k \) are \( M^{-2}s^{-1} \).
04

Determine the molecularity

The molecularity of a reaction is defined as the number of molecules that come together to react in an elementary step. Here, two iodine molecules and one \( M \) molecule are involved, so the molecularity of this reaction is three, meaning it is termolecular.
05

Comparison with the second reaction

The second reaction involves only two \( ext{I} \) atoms coming together to form \( ext{I}_2 \). This means that the second reaction is a bimolecular reaction, whereas the first reaction is termolecular. They are not identical because they involve different numbers and types of reacting species at the molecular level.

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Key Concepts

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

Termolecular Reaction
A termolecular reaction is a fascinating concept in chemistry, where three molecules come together simultaneously to form products. In the reaction given \[ \mathrm{I}(\mathrm{g}) + \mathrm{I}(\mathrm{g}) + \mathrm{M}(\mathrm{g}) \rightarrow \mathrm{I}_2(\mathrm{g}) + \mathrm{M}(\mathrm{g}) \],three molecules collide at the same time.Understanding termolecular reactions is crucial because they demonstrate the complexity and rarity of such interactions. These encounters are relatively uncommon due to the unlikelihood of three molecules colliding simultaneously with the proper orientation and energy.
  • Three-molecule collision: The simultaneous collision involves two iodine molecules \((\mathrm{I})\) and one other molecule \((\mathrm{M})\).
  • Complexity: These reactions are less frequent because of the strict conditions required for successful collisions.
Termolecular reactions are often studied to understand the dynamics and mechanisms of chemical reactions in gases.
Rate Constant Units
When analyzing chemical reactions, understanding the units of the rate constant \( k \) is vital. These units are derived from the rate law expression, which ties together the concentrations of reactants and the reaction rate. In the given problem, the reaction involves three reactants, making it a third-order reaction.For the rate law expression:\[\text{Rate} = k[\text{I}][\text{I}][\text{M}] \]where \( [\text{I}] \) and \( [\text{M}] \) are the concentrations of iodine and molecule \( M \), respectively.To find the units of \( k \):- Start with the units of rate: typically expressed as concentration per unit time, \( M/s \) (molarity per second).- For a third-order reaction, you divide by \( M^3 \) (since \([\text{I}]^2[\text{M}] = M^3\)), resulting in:\[k = \frac{M/s}{M^3} = \frac{1}{M^2s} \]Thus, the units of the rate constant \( k \) for this reaction are \( M^{-2}s^{-1} \). These units inform us about the specific relationship between the reaction rate and the reactant concentrations.
Molecularity
Molecularity is an essential concept in reaction kinetics, describing the number of molecules involved in an elementary reaction step. In the context of our exercise, recognizing the molecularity gives insight into the nature and complexity of the reaction.Molecularity is straightforward:
  • Termolecular: Three reactants, such as in the given reaction \([\text{I}] + [\text{I}] + [\text{M}]\), indicate a termolecular process with molecularity of three.
  • Bimolecular: When exactly two molecules react, similar to the second reaction \([\text{I}] + [\text{I}]\), the reaction is bimolecular.
It's important to remember that molecularity is only defined for elementary reactions, meaning it represents a single step in the reaction mechanism.Concluding from our problem, a termolecular reaction involves three molecules forming a reaction complex, which is quite distinct from a more common bimolecular interaction.
Bimolecular Reaction
Understanding bimolecular reactions allows us to explore the more common and simpler type of reaction mechanism. In these reactions, exactly two molecules collide and react to form products.Consider the expression:\[\text{I}(\text{g}) + \text{I}(\text{g}) \rightarrow \text{I}_2(\text{g})\]Here, two iodine gas molecules interact to form a single iodine molecule \((\text{I}_2)\).Important features of bimolecular reactions:
  • Two-molecule collision: It involves only two reacting species.
  • Simpler mechanism: Due to only requiring a two-molecule collision, the conditions needed for a successful reaction are less complex compared to termolecular reactions.
Bimolecular reactions are significantly more common and typically serve as fundamental steps in broader reaction mechanisms.

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Most popular questions from this chapter

A mechanism for ozone creation and destruction in the stratosphere is $$\begin{array}{c} \mathrm{O}_{2}(\mathrm{g})+h v \stackrel{j_{1}}{\longrightarrow} \mathrm{O}(\mathrm{g})+\mathrm{O}(\mathrm{g}) \\ \mathrm{O}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})+\mathrm{M}(\mathrm{g}) \stackrel{k_{2}}{\longrightarrow} \mathrm{O}_{3}(\mathrm{g})+\mathrm{M}(\mathrm{g}) \\ \mathrm{O}_{3}(\mathrm{g})+h v \stackrel{j_{3}}{\longrightarrow} \mathrm{O}_{2}(\mathrm{g})+\mathrm{O}(\mathrm{g}) \\ \mathrm{O}(\mathrm{g})+\mathrm{O}_{3}(\mathrm{g}) \stackrel{k_{4}}{\longrightarrow} \mathrm{O}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \end{array}$$ where we have used the symbol \(j\) to indicate that the rate constant is for a photochemical reaction. Determine the rate expressions for \(d[\mathrm{O}] / d t\) and \(d\left[\mathrm{O}_{3}\right] / d t .\) Assume that both intermediate species, \(\mathrm{O}(\mathrm{g})\) and \(\mathrm{O}_{3}(\mathrm{g}),\) can be treated by the steady-state approximation and thereby show that $$[\mathrm{O}]=\frac{2 j_{1}\left[\mathrm{O}_{2}\right]+j_{3}\left[\mathrm{O}_{3}\right]}{k_{2}\left[\mathrm{O}_{2}\right][\mathrm{M}]+k_{4}\left[\mathrm{O}_{3}\right]}$$ and $$\left[\mathrm{O}_{3}\right]=\frac{k_{2}[\mathrm{O}]\left[\mathrm{O}_{2}\right][\mathrm{M}]}{j_{3}+k_{4}[\mathrm{O}]}$$ Now substitute Equation 1 into Equation 2 and solve the resulting quadratic formula for \(\left[\mathrm{O}_{3}\right]\) to obtain $$\left[\mathrm{O}_{3}\right]=\left[\mathrm{O}_{2}\right] \frac{j_{1}}{2 j_{3}}\left\\{\left(1+4 \frac{j_{3}}{j_{1}} \frac{k_{2}}{k_{4}}[\mathrm{M}]\right)^{1 / 2}-1\right\\}$$ Typical values for these parameters at an altitude of \(30 \mathrm{km}\) are \(j_{1}=2.51 \times 10^{-12} \mathrm{s}^{-1}\) \(j_{3}=3.16 \times 10^{-4} \mathrm{s}^{-1}, k_{2}=1.99 \times 10^{-33} \mathrm{cm}^{6} \cdot\) molecule \(^{-2} \cdot \mathrm{s}^{-1}, k_{4}=1.26 \times 10^{-15} \mathrm{cm}^{3}\) molecule \(^{-1} \cdot \mathrm{s}^{-1}, \quad\left[\mathrm{O}_{2}\right]=3.16 \times 10^{17} \quad\) molecule \(\cdot \mathrm{cm}^{-3}, \quad\) and \(\quad[\mathrm{M}]=3.98 \times 10^{17}\) molecule \(\cdot \mathrm{cm}^{-3} .\) Find \(\left[\mathrm{O}_{3}\right]\) and \([\mathrm{O}]\) at an altitude of \(30 \mathrm{km}\) using Equations 1 and 2 Was the use of the steady-state assumption justified?

Antibiotic-resistant bacteria have an enzyme, penicillinase, that catalyzes the decomposition of the antibiotic. The molecular mass of penicillinase is \(30000 \mathrm{g} \cdot \mathrm{mol}^{-1}\). The turnover number of the enzyme at \(28^{\circ} \mathrm{C}\) is \(2000 \mathrm{s}^{-1}\). If \(6.4 \mu \mathrm{g}\) of penicillinase catalyzes the destruction of \(3.11 \mathrm{mg}\) of amoxicillin, an antibiotic with a molecular mass of \(364 \mathrm{g} \cdot \mathrm{mol}^{-1}\) in 20 seconds at \(28^{\circ} \mathrm{C},\) how many active sites does the enzyme have?

The turnover number for acetylcholinesterase, an enzyme with a single active site that metabolizes acetylcholine, is \(1.4 \times 10^{4} \mathrm{s}^{-1} .\) How many grams of acetylcholine can \(2.16 \times 10^{-6} \mathrm{g}\) of acetylcholinesterase metabolize in one hour? (Take the molecular mass of the enzyme to be \(4.2 \times 10^{4} \mathrm{g} \cdot \mathrm{mol}^{-1} ;\) acetylcholine has the molecular formula \(\mathrm{C}_{7} \mathrm{NO}_{2} \mathrm{H}_{16}^{+} .\) )

The standard Gibbs energy change of reaction for $$2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ is \(-457.2 \mathrm{kJ}\) at \(298 \mathrm{K}\). At room temperature, however, this reaction does not occur and mixtures of gaseous hydrogen and oxygen are stable. Explain why this is so. Is such a mixture indefinitely stable?

A proposed mechanism for the thermal decomposition of acetaldehyde $$\mathrm{CH}_{3} \mathrm{CHO}(\mathrm{g}) \stackrel{k_{\text {obs }}}{\longrightarrow} \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}(\mathrm{g})$$ is $$\begin{array}{c} \mathrm{CH}_{3} \mathrm{CHO}(\mathrm{g}) \stackrel{k_{1}}{\longrightarrow} \mathrm{CH}_{3}(\mathrm{g})+\mathrm{CHO}(\mathrm{g}) \\ \mathrm{CH}_{3}(\mathrm{g})+\mathrm{CH}_{3} \mathrm{CHO}(\mathrm{g}) \stackrel{k_{2}}{\longrightarrow} \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CH}_{3} \mathrm{CO}(\mathrm{g}) \\ \mathrm{CH}_{3} \mathrm{CO}(\mathrm{g}) \stackrel{k_{3}}{\longrightarrow} \mathrm{CH}_{3}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \\ 2 \mathrm{CH}_{3}(\mathrm{g}) \stackrel{k_{4}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6} \end{array}$$ Is this reaction a chain reaction? If so, identify the initiation, propagation, inhibition, and termination step(s). Determine the rate laws for \(\mathrm{CH}_{4}(\mathrm{g}), \mathrm{CH}_{3}(\mathrm{g}),\) and \(\mathrm{CH}_{3} \mathrm{CO}(\mathrm{g}) .\) Show that if you assume the steady-state approximation for the intermediate species, \(\mathrm{CH}_{3}(\mathrm{g})\) and \(\mathrm{CH}_{3} \mathrm{CO}(\mathrm{g}),\) the rate law for methane formation is given by $$\frac{d\left[\mathrm{CH}_{4}\right]}{d t}=\left(\frac{k_{1}}{k_{4}}\right)^{1 / 2} k_{2}\left[\mathrm{CH}_{3} \mathrm{CHO}\right]^{3 / 2}$$

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